Competitive Exam — Solutions & Explanations
M.Sc. Communication Engineering — Ninevah University / College of Electronics Engineering
Exams of 2019–2020 and 2020–2021 + Additional Question Banks (Microwave, Antennas, Transmission Lines, Modulation, Analysis)
All questions solved and explained
Before you start: This file covers two complete exams plus five additional question banks compiled from past Microwave, Antennas, Transmission-Line, Modulation, and Engineering-Analysis papers. A few answers in the original handwritten keys are wrong — these are flagged in an orange box with the physically correct answer. Questions repeated across papers are cross-referenced instead of being fully re-explained.
Exam 2019 – 2020 — Part 1 (MCQ)
Engineering Analysis
1
The conjugate of the complex number 1/(j−1) is:
A) −1/(j−1) B) 1/(j+1) C) −1/(j+1) D) 1/(j−1)
Answer: C — −1/(j+1)
The conjugate of the denominator (j−1) is (−j−1). So the conjugate of 1/(j−1) = 1/(−j−1) = −1/(j+1). (Only the sign of the imaginary part flips.)
2
Solve the simultaneous equations 2x+3y=8 and 3x+2y=7:
A) x=2, y=1 B) x=1, y=2 C) none of the above
Answer: B — x=1, y=2
Check by substitution: 2(1)+3(2)=8 ✔ and 3(1)+2(2)=7 ✔. (Elimination: multiply Eq.1 by 2 and Eq.2 by 3, then subtract.)
Signals & Power
3
V(t)=4sin(8πt)+cos(12πt) is applied across R=10Ω. The total average power is:
A) zero B) 0.1 W C) 1 W
Correct answer: P = 0.85 W
Power of each sinusoid = (A²/2)/R; total = their sum:
P = (4²/2)/10 + (1²/2)/10 = 0.8 + 0.05 = 0.85 W.
Note: The handwritten key marked "zero", which is wrong. The average power of a sinusoid is not zero. Use the RMS method: P = V²rms/R.
Electromagnetic Fields
4
The capacitance of a parallel-plate capacitor depends on:
A) positive charge on the plate B) negative charge on the plate C) positive & negative charge and the distance and εr D) plate area, distance between plates, and permittivity ε
Correct answer: D — plate area, distance between plates, and permittivity
C = εA/d. Capacitance depends only on the area A, the distance d, and the permittivity ε — never on the amount of charge (charge changes with voltage while C stays fixed).
Note: The handwritten key marked C (charges), which is physically wrong. The typed version marked D (area, distance, permittivity), which is correct.
Secure Communication
5
What is the advantage of RSA over DSS/DSA?
a) It can provide digital signature and encryption b) Fewer resources / faster because it is symmetric c) It is a block cipher d) It uses a one-time pad
Answer: a — it provides both digital signature and encryption
DSS provides a digital signature only, whereas RSA provides both a digital signature and data encryption — that is its key advantage.
6
The Data Encryption Standard (DES) is a:
a) bit cipher b) stream cipher c) block cipher d) substitution cipher
Answer: c — block cipher
DES encrypts data in 64-bit blocks at a time using a 56-bit key — it is a block cipher.
Optical Communication
7
Dispersion in an optical fiber is:
a) compression of the light pulse b) broadening of the transmitted light pulse along the channel c) overlapping of the pulse d) absorption
Answer: b — broadening of the light pulse
Dispersion causes broadening of the light pulse as it propagates, which limits the data rate because adjacent pulses overlap (ISI).
8
The optical fiber is designed such that:
a) core refractive index > cladding b) core < cladding c) core = cladding
Answer: a — core refractive index greater than cladding n_core > n_cladding
For total internal reflection to confine the light in the core, the core index must be higher than the cladding: n_core > n_cladding.
9
LEDs are not used as a light source for high speed because of:
a) higher cost b) poor reliability c) inability to provide power d) slow switching
Answer: d — slow switching
LEDs have slow switching speed and wide spectral width, so they cannot support high data rates — a laser diode is used instead for high speed.
Antennas & Propagation
10
In a line-of-sight (LOS) system, to maximize the received power one should:
a) increase the transmit antenna gain b) increase both transmit and receive antenna gains c) increase the receive antenna gain d) none
Answer: b — increase both transmit and receive antenna gains
From the Friis equation: Pr ∝ Gt × Gr. The received power is proportional to the product of both gains, so increasing both maximizes it.
Networks & Telephony
11
The broadcast channel in GSM is:
a) full duplex up-down b) half duplex uplink c) half duplex downlink
Answer: c — half duplex downlink only
The GSM broadcast channel (BCCH) is sent from the base station to the mobiles only (downlink) to broadcast synchronization and network info — so it is half duplex downlink.
12
In a telephony system, each 125 μs contains:
a) 30 time slots b) 32 voice channels c) 30 voice channels d) none
Answer: c — 30 voice channels
A PCM (E1) frame lasts 125 μs and contains 32 time slots: 30 voice channels plus 2 signaling slots (TS0 and TS16).
Radar & Microwave
a) pulse radar b) pulse-doppler radar c) CW radar d) any type
Answer: a — pulse radar (the handwritten key marked b — pulse-doppler)
Note: This one is disputed. A tracking radar is usually a pulse, pulse-doppler, or monopulse radar. The typed version marked "pulsed radar" and the handwritten one marked "pulse doppler". Confirm the exact definition used in your course.
14
The output S/N of an amplifier with NF=10dB and input S/N=25dB is:
a) 15 b) 35 c) 10 d) none
Answer: a — 15 dB
NF(dB) = (S/N)in(dB) − (S/N)out(dB)
10 = 25 − (S/N)out → (S/N)out = 15 dB.
Digital Communication
15
In digital communication, the matched filter in the receiver:
a) increases SNR b) decreases SNR c) does not affect SNR d) none
Answer: a — increases SNR
The matched filter is designed to give the maximum SNR at the sampling instant, minimizing the probability of detection error.
16
In data communication, transmitting a binary signal requires:
a) less bandwidth than analog b) more bandwidth than analog c) the same bandwidth d) none
Answer: b — more bandwidth compared to analog
A binary/digital signal has sharp edges (high-frequency components), so it needs a wider bandwidth than the equivalent analog signal.
17
A reflector (dish) antenna is used in satellite TV because it has:
a) low gain and narrow beamwidth b) high gain and narrow beamwidth c) low gain and narrow bandwidth d) low effective area
Answer: b — high gain and narrow beamwidth
A reflector dish focuses the radiation into a very narrow, high-gain beam, which is essential for long-distance satellite links.
Digital Signal Processing (DSP)
18
The similarity between the Fourier transform and the Z-transform:
a) both convert frequency to discrete time b) both convert time domain to frequency-spectrum domain c) both convert analog to digital d) both convert digital to analog
Answer: b — both convert the (discrete) time domain to the frequency-spectrum domain
The Z-transform operates on discrete signals and maps them from time to frequency; the DTFT is a special case of it at |z|=1.
Microwave
19
Considering its cutoff frequency, a waveguide behaves as a:
a) LPF b) HPF c) BPF
Answer: b — HPF (high-pass filter)
A waveguide passes only frequencies above the cutoff frequency fc and attenuates those below it → it behaves as a high-pass filter.
20
The device that lets a microwave travel in one direction with low loss and heavily attenuates the other direction is called:
a) isolator b) circulator c) attenuator d) tunnel diode
Answer: a — isolator
An isolator is a two-port ferrite device that passes the signal one way with low loss and strongly attenuates the reverse direction — used to protect the source from reflections.
21
A klystron operates on:
a) FM b) AM c) pulse modulation d) velocity modulation
Answer: d — velocity modulation
It varies the velocity of the electron beam in the buncher cavity so the electrons bunch together and deliver their energy to the catcher cavity — the principle of velocity modulation.
22
Spread-spectrum communication system:
Concept: spreading the signal over a very wide bandwidth
It distributes the signal energy over a much wider bandwidth than needed, providing: anti-jamming, security, and multiple access (CDMA).
23
A TEM wave can propagate inside a:
a) rectangular waveguide b) circular waveguide c) coaxial cable d) cavity
Answer: c — coaxial cable
The TEM mode needs two conductors, so it propagates in a coaxial cable and a two-wire line, but not in hollow waveguides (single conductor → only TE/TM modes).
Sampling & Probability
24
To reconstruct the analog signal from the sampled signal, the frequency must satisfy:
a) Fs = Fm b) Fs > Fm c) Fs ≥ 2Fm
Answer: c — Fs ≥ 2Fm (Nyquist theorem)
To avoid aliasing and fully reconstruct the signal, the sampling frequency must be at least twice the highest frequency in the signal.
25
The important parameter to determine the probability of error:
a) bit rate b) bandwidth c) Eb/No d) signal power
Correct answer: c — Eb/No (energy per bit to noise density)
The probability of error Pe ≈ ½·erfc(√(Eb/No)) — it depends directly on Eb/No.
Note: The handwritten key marked "bit rate", which is inaccurate. The governing parameter for error probability is Eb/No (the typed version marked it correctly).
Satellite
26
For a geosynchronous satellite, the free-space loss is highest in the band:
a) 6/4 GHz b) 30/20 GHz c) 14/12 GHz d) all have the same
Correct answer: b — 30/20 GHz (highest frequency = highest loss)
Free-space loss FSL = (4πdf/c)² is proportional to frequency squared. The distance to the satellite is fixed, so the highest-frequency band (Ka-band: 30/20 GHz) suffers the highest loss.
Note: The handwritten key marked "all have the same", which is wrong — the loss clearly depends on frequency, so it is not equal. Confirm with your instructor, because the key is physically incorrect here.
27
A limitation of the FDMA satellite system is:
a) downlink traffic is much higher than uplink b) less flexibility than TDMA in reassigning channels c) carrier frequency assignment is manual d) all of the above
Answer: d — all of the above
FDMA suffers from all of these: less flexibility, complex frequency assignment, and intermodulation problems — so all options are correct.
28
A satellite downlink at 12GHz has transmit power 4W and G=50dB. The EIRP in dBW is:
a) 54 b) 56 c) 62
Answer: b — 56 dBW
EIRP = 10·log(P) + G = 10·log(4) + 50 = 6 + 50 = 56 dBW.
Pulse Modulation & Networks
29
The modulation process using a train of fixed-width pulses whose polarity indicates whether the demodulator should rise or fall at each pulse:
a) PPM b) DM c) PAM d) PWM
Answer: b — DM (Delta Modulation)
DM sends one bit per sample: +Δ (rise) if the signal increased, −Δ (fall) if it decreased. "Polarity decides rise/fall" is an exact description of DM.
30
When a router runs out of buffer space, this is called:
a) source quench b) redirect c) information request d) low memory
Answer: a — source quench
An ICMP Source Quench message is sent by the router to the source during congestion to ask it to slow down its sending rate.
31
Beyond IP, UDP provides the services of:
a) routing & switching b) sending & receiving packets c) multiplexing & demultiplexing d) demultiplexing and error checking
Answer: d — demultiplexing and error checking
On top of IP, UDP adds two functions: multiplexing/demultiplexing via port numbers, and error checking via the checksum.
32
No collision in random access is provided by:
a) ALOHA b) CSMA/CD c) CSMA/CA d) Token
Answer: c — CSMA/CA (collision avoidance)
CSMA/CA = Collision Avoidance, preventing collisions before they happen (used in WiFi). Token is a controlled-access method, not random access.
33
An FIR filter can be described as:
Answer: a discrete-time / digital system with a finite-length impulse response
FIR = Finite Impulse Response. Its impulse response is finite (a limited number of values), it is always stable, and it can be made linear-phase — implemented in discrete time (digital).
34
A standing-wave pattern along a transmission line occurs:
Answer: when there is an impedance mismatch with the load
A mismatch causes reflection, so the incident wave interferes with the reflected wave, producing a fixed pattern of maxima and minima (standing wave). With a perfect match, no standing wave forms.
35
Fourier transform pair — the form of the equation:
DFT form: X(k) = Σ x(n)·e^(−j2πkn/N)
The written equation is the Discrete Fourier Transform (DFT): it converts N time samples to N frequency components, and the inverse (IDFT) does the reverse.
36
AM modulation — modulation index and bandwidth:
Answer: m = Am/Ac , BW = 2·fm
The AM modulation index = (message amplitude / carrier amplitude). The AM bandwidth = twice the highest message frequency (two sidebands).
Exam 2019 – 2020 — Part 2 (Essay & Calculation)
Q2
Write a brief description of the impulse response of a digital system.
The impulse response h(n) is the output of the system when the input is a unit impulse δ(n). It fully characterizes the system: for any other input, the output equals the convolution of the input with the impulse response: y(n) = x(n) ∗ h(n). For a linear time-invariant (LTI) system, h(n) alone completely represents the system (denoted h(n) or h(t)).
Q3
Compare an optical communication system with a satellite communication system in terms of: bandwidth, security, and other aspects.
Bandwidth: Optical is huge (THz level; practically many GHz per channel), while satellite is limited (transponder ~36–72 MHz).
Security: Optical is hard to tap without detection, while a satellite signal must be encrypted since it is broadcast over the air.
Other aspects: Optical has a long life and is upgradeable but is point-to-point and needs a local loop; satellite covers wide areas, is multipoint and mobile (sea/air) but has a short life (7–15 years) and is not upgradeable.
Q4
For two signals V1 (triangular) and V2 (sine), determine: average value, RMS value, and power dissipated in a 1Ω resistor.
For the sine wave (amplitude Vp): average Vavg = 0 (full cycle), Vrms = Vp/√2, power P = Vrms²/R. E.g. Vp=2: Vrms=1.41V, P=2W.
For the triangular wave (0 to Vp): average Vavg = Vp/2, Vrms = Vp/√3, power P = Vrms²/R.
Note: The exact numerical results depend on the peak values in the graph (Vp and the period). Use the formulas above and plug in the exact graph values on your paper.
Q5
For the shown AM waveform, find: type of modulation, carrier frequency, message frequency, and modulation index.
Answer: AM | fc=1/Tc | fm=1/Tm=1kHz | m=0.5
A) Type: amplitude modulation (the envelope follows the message).
B) Carrier frequency: fc = 1/Tc (reciprocal of the carrier period from the graph).
C) Message frequency: fm = 1/Tm = 1/(1ms) = 1000 Hz = 1 kHz.
D) Modulation index: m = (Vmax−Vmin)/(Vmax+Vmin) = (3−1)/(3+1) = 0.5.
Exam 2020 – 2021 — Part 1 (MCQ)
Most questions in this exam repeat or resemble the 2019-2020 exam. Here I focus on what is new or different, and give brief answers for the repeated ones with a reference to their earlier number.
1
V(t)=4sin(8πt)+cos(12πt), R=1Ω. The average power is:
a) 0.1 b) 0.3 c) zero d) none of these
Answer: d — none (the correct value is 8.5 W)
P = (4²/2 + 1²/2)/1 = 8.5 W — not among the options, so the answer is "none of these". (Note the difference from the 2019 version where R=10Ω.)
2-9
Questions on RSA, DES, dispersion, fiber design, LED, line-of-sight...
Repeated: see questions 5, 6, 7, 8, 9, 10 of the 2019-2020 exam (same answers)
Additional — "Light transmission in a material" → Answer: it always travels slower than the speed of light in vacuum, because n = c/v and n > 1 so v < c.
10
The Z-transform of the discrete signal {4, 1, 2̲, 7, 0, 3} (origin n=0 at the value 2):
a) 4z+z²+2z³+7z⁴+3z⁶ b) negative powers c) 4z²+z+2+7z⁻¹+3z⁻³
Answer: c — 4z² + z + 2 + 7z⁻¹ + 3z⁻³
X(z)=Σx[n]z⁻ⁿ. With the origin at the value 2: n=−2→4, n=−1→1, n=0→2, n=1→7, n=2→0, n=3→3:
= 4z² + z + 2 + 7z⁻¹ + 0 + 3z⁻³.
11
One property of the Fourier transform:
a) δ(t−a) ↔ e⁻ʲ²πᶠᵃ b) sin... c) cos...
Answer: a — δ(t−a) ↔ e⁻ʲ²πᶠᵃ (time-shifting property)
Shifting the impulse in time by a multiplies the spectrum by a phase factor e⁻ʲ²πᶠᵃ without changing its magnitude.
12
The convolution output of h(n)={1,1,1} and x(n)={1,2,3}:
a) {1,3,6,5,3} b) {1,1,0,0,1} c) {1,2,3,6,0}
Answer: a — {1, 3, 6, 5, 3}
y0=1, y1=3, y2=6, y3=5, y4=3. (Check: sum of output = 6×3 = 18 ✔.)
13
The broadcast channel in GSM:
a) full duplex up-down b) half duplex uplink c) half duplex downlink
Answer: c — half duplex downlink (repeat of Q11, 2019)
14
The digital subscriber line used to transmit data and voice:
a) SDSL b) ISDN c) GDMT
Answer: b — ISDN
ISDN (Integrated Services Digital Network) carries both voice and data over the same digital line.
15-18
Tracking radar, EIRP, S/N, mobile satellite service...
Quick answers:
• EIRP (power 10W, G=50dB): 10log(10)+50 = 60 dB.
• Output S/N (NF=10, S/Ni=25): 25−10 = 15 dB.
• Mobile Satellite Service: a two-way link.
• Tracking radar: see the note in Q13, 2019.
19
In an RF/microwave circuit, FM modulation is achieved using a:
a) PIN diode b) varactor diode c) phase shifter
Answer: b — varactor diode
The varactor's capacitance varies with voltage, so applying the message changes the resonant frequency → producing FM (used in a VCO).
20
The Fourier Series is used for:
a) a periodic signal b) a non-periodic signal c) a periodic signal expressed as exponentials/sinusoids
Answer: a periodic signal
The Fourier Series represents periodic signals as a sum of sinusoids/exponentials. The Fourier transform, in contrast, is for non-periodic signals.
21
A system is causal if and only if:
a) n is positive non-zero b) n is positive and zero c) n is negative non-zero d) the value is zero for negative n
Answer: d — its response is zero for negative n
A system is causal if its impulse response h(n)=0 for all n < 0 — i.e. the output does not depend on a future input.
22
The microwave device that passes a signal one way with low loss and attenuates the other:
a) circulator b) isolator c) directional coupler
Answer: b — isolator (repeat of Q20, 2019)
23
A fixed-amplitude pulse with varying width (depending on the modulator):
a) PPM b) PWM c) DM d) PCM
Answer: b — PWM (Pulse Width Modulation)
PWM: amplitude is fixed and the width changes with the message amplitude. (Note the difference from Q29-2019, where "fixed width, polarity decides rise/fall" = DM.)
24
In Go-Back-N ARQ, if the number of bits in the frame number = 2, then the window is:
a) sending=3, receiving=1 b) ... c) ... d) ...
Answer: a — sending window = 3, receiving window = 1
In Go-Back-N: sending window = 2ⁿ − 1 and receiving window = 1. For n=2: 2²−1 = 3, receiving = 1.
25
In binary, the parameter used to calculate the probability of error is:
a) bit rate b) bandwidth c) Eb/No d) signal power
Correct answer: c — Eb/No
Note: The handwritten key marked "bit rate", but the governing parameter for error probability is Eb/No (Pe ≈ ½erfc√(Eb/No)).
26
To solve the hidden-node problem we use:
a) CSMA/CD b) CSMA/CA c) ALOHA
Correct answer: b — CSMA/CA
The hidden-node problem is wireless and is solved by CSMA/CA with the RTS/CTS handshake. CSMA/CD is for wired networks and does not solve this problem.
Note: The handwritten key marked CSMA/CD — the correct choice for the hidden-node problem is CSMA/CA.
27
No collision in random access:
a) ALOHA b) CSMA/CD c) CSMA/CA d) Token
Answer: c — CSMA/CA
Useful rule: no collision in controlled access → Token; no collision in random access → CSMA/CA.
28
To reconstruct the signal, the sampling frequency must be:
a) fs ≥ fm b) fs > fm c) fs < fm d) fs ≥ 2fm
Answer: d — fs ≥ 2fm (Nyquist — repeat of Q24, 2019)
29
A geosynchronous satellite has the highest free-space loss in the band:
a) 6/4 b) 14/12 c) 30/20 d) all have the same
Correct answer: c — 30/20 GHz (highest frequency)
Note: See the detail in Q26, 2019. Loss increases with frequency, so the highest is the 30/20 GHz band — not "all the same" as in the handwritten key.
30
A standing wave on a transmission line occurs when:
a) two waves travel in the same direction b) two waves travel in opposite directions c) the line is matched
Answer: two waves traveling in opposite directions (incident + reflected) due to a mismatch
When the load is mismatched, part of the wave reflects, so the incident wave interferes with the reflected one (opposite directions), producing the standing wave. A matched line does not produce a standing wave.
31
To match a transmission line using the λ/4 method, Z₀ is found as (Zs = source impedance, ZL = load impedance):
a) Z₀=Zs b) Z₀=√(Zs·ZL) c) Z₀=2Zs
Answer: b — Z₀ = √(Zs · ZL)
The quarter-wave transformer matches two different impedances when its characteristic impedance is the geometric mean (square root of the product).
32
If r=f(θ) in polar coordinates, what is θ in Cartesian coordinates (x,y)?
a) tanθ = y/x b) 1+sinθ = y/x c) 1+cosθ = y/x
Answer: a — tanθ = y/x
From polar-to-Cartesian: x=r·cosθ, y=r·sinθ → tanθ = y/x, hence θ = tan⁻¹(y/x).
33
When the modulation index in FM is increased, the bandwidth:
a) increases b) decreases c) constant
Answer: a — increases
From Carson's rule: BW = 2(m+1)fm. Increasing the modulation index m directly increases the bandwidth.
34
Beamwidth is defined as:
a) angular separation between half-power points b) range of frequencies c) angular separation between two identical points on opposite sides of the pattern maximum
Answer: c — angular separation between two identical points on opposite sides of the pattern maximum
Beamwidth is the angular separation between two identical points on either side of the main lobe. When these are the half-power (−3dB) points, it is called HPBW. (Option b "range of frequencies" is wrong — that is bandwidth, not beamwidth.)
a) resists jamming b) transmits more than one signal at the same time on any frequency c) ... d) all of the above
Answer: b — transmits more than one signal at once (and if "all of the above" exists, it is the most complete)
Spread spectrum allows multiple simultaneous signals (CDMA), resists jamming, and provides security — so "all of the above" is a good answer when offered.
36
The spectrum in FM consists of:
a) one carrier and infinite sidebands b) one carrier and two sidebands c) many carriers
Answer: a — one carrier and infinite sidebands
FM theoretically produces an infinite number of sidebands (Bessel functions), but practically most of the power is contained within Carson's bandwidth.
37
In a 4-bit ADC, the LSB equals:
a) 6.25% b) 0.025% c) 0.12% d) 0.2%
Answer: a — 6.25%
Number of levels = 2⁴ = 16. The least-significant-bit value = 1/16 = 0.0625 = 6.25% of full scale.
38
In a reflex klystron, the modulation is:
a) velocity modulation b) frequency modulation c) amplitude modulation
Answer: a — velocity modulation
The reflex klystron uses velocity modulation of the electron beam in a single cavity (the electrons are returned by a repeller, causing bunching) — just like the ordinary klystron.
Exam 2020 – 2021 — Part 2 (Essay & Calculation)
AM ex.
AM wave: Vmax=100, Vmin=50, T=10ms. Find: m, Ac, fm, bandwidth, power efficiency, Am.
Full solution:
1) Modulation index: m = (Vmax−Vmin)/(Vmax+Vmin) = (100−50)/(100+50) = 50/150 = 0.33
2) Carrier amplitude: from Ac(1+m)=100 → Ac = 100/1.33 = 75.18
3) Message frequency: fm = 1/T = 1/(10ms) = 100 Hz
4) Bandwidth: BW = 2·fm = 2(100) = 200 Hz
5) Power efficiency: η = m²/(2+m²) = 0.109/2.109 ≈ 0.043 (4.3%)
6) Message amplitude: Am = m·Ac = 0.33 × 75.18 = 24.8
System
Determine the system properties: memory, causality, time-invariance, stability, linearity.
Memory: a system is "memoryless" if its output depends only on the present input (e.g. y(t)=3x(t)); it "has memory" if it depends on a past/future input.
Causality: the output does not depend on a future input → h(n)=0 for n<0.
Time-invariance: shifting the input in time shifts the output by the same amount. (A constant term like +2 does not break it, but a time-dependent coefficient like t·x(t) does → time-variant.)
Stability: a bounded input produces a bounded output (BIBO). E.g. x(t)=tan(t) is unstable.
Linearity: it satisfies superposition (additivity and scaling).
Plots
Plotting questions (PPM/PWM/PAM, LPF output, radiation pattern of sinθcos2φ).
• PPM/PWM/PAM: from a PPM signal, draw: PWM (pulses of varying width), PAM (pulses of varying amplitude following the signal), and the original analog signal (the sample envelope).
• LPF output with fc=120kHz: the filter passes frequency components below fc and removes those above it, smoothing the signal (removing the sharp edges).
• Radiation pattern for gain G=sinθ·cos2φ: drawn in three planes by fixing one angle and varying the other:
– xy plane: θ=90° → G=cos2φ
– xz plane: φ=0° → G=sinθ
– yz plane: φ=90° → G=−sinθ
Part 3 — Additional MCQ Bank (mixed subjects)
Source: a compiled MCQ set with no answer key. Every question below is solved from first principles. Subjects: Engineering Analysis, Signals/DSP, Analog & Digital Communication, Microwave, Antennas & Propagation, Optical, Satellite, and Networks.
Engineering Analysis & Signals
1
The minimum value of the function f(x)=x²+x+2 is:
a) 1/2 b) 3/4 c) 7/4 d) 1/4
Answer: c — 7/4
Vertex at x = −b/2a = −1/2. Then f(−½) = ¼ − ½ + 2 = 7/4. (Since a>0 the parabola opens up, so this is a minimum.)
2
The inverse of the matrix A=[[2,3],[4,1]] is:
a) [[0.3,0.4],[0.1,0.2]] b) [[0.1,−0.3],[−0.4,0.2]] c) [[−0.1,0.3],[0.4,−0.2]] d) [[−0.3,−0.4],[−0.1,0.2]]
Answer: c — [[−0.1, 0.3],[0.4, −0.2]]
det A = (2)(1) − (3)(4) = −10. A⁻¹ = (1/det)·[[d,−b],[−c,a]] = (1/−10)·[[1,−3],[−4,2]] = [[−0.1,0.3],[0.4,−0.2]].
3
Which is the linear differential equation?
a) y″ − sin(y″) = 5 b) y‴ + y·y″ = 0 c) y″ + 1 = cos(y′) d) None of the above
Answer: d — None of the above
A linear ODE has no nonlinear functions of y or its derivatives. (a) has sin(y″), (b) has the product y·y″, (c) has cos(y′) — all nonlinear. So none is linear.
4
v(t)=4cos(100πt)+sin(200πt) applied across R=10Ω. The total average power is:
a) zero b) 0.1 W c) 1 W d) None of these
Answer: d — None of these (correct value = 0.85 W)
P = (4²/2)/10 + (1²/2)/10 = 0.8 + 0.05 = 0.85 W — not listed, so "none of these".
5
A transmission line has a standing-wave pattern when:
a) the load equals the characteristic impedance b) two waves travel in the same direction c) the line is perfectly matched d) none of these
Answer: d — none of these
A standing wave forms only when there is a mismatch, producing an incident and a reflected wave travelling in opposite directions. Options (a) and (c) describe a matched line (no standing wave); (b) is wrong because the two waves move in opposite directions, not the same. Hence none of the given options.
6
The capacitance depends on:
a) charge on the plate b) potential difference c) the capacitor dimensions d) properties of the dielectric material e) all of the above
Answer: (c) and (d) — dimensions and dielectric material only
C = εA/d. Capacitance depends only on the geometry (area and separation → "dimensions") and the dielectric (ε). It does not depend on charge or voltage.
Note: "All of the above" (e) is not correct because (a) charge and (b) voltage do not affect C. If the exam forces one choice, pick the option that combines dimensions + dielectric; neither single option (c) nor (d) is complete on its own.
7
A linear-phase filter response is obtained when:
a) the phase varies linearly with frequency b) the phase varies inversely with frequency c) the phase does not vary linearly d) none
Answer: a — the phase varies linearly with frequency
Linear phase means ∠H(ω) = −ωτ (a straight line), giving a constant group delay and no phase distortion. FIR filters with symmetric coefficients achieve this.
8
FIR filters can be described as:
A) non-recursive B) no feedback C) recursive D) use feedback — a) A&B b) C&D c) A&D d) B&C
Answer: a — A and B (non-recursive, no feedback)
An FIR filter computes its output only from current and past inputs (no feedback of past outputs) → it is non-recursive. IIR filters are the recursive, feedback type.
9
The product of two odd signals is:
a) Odd b) Even c) both d) zero
Answer: b — Even
If f(−t)=−f(t) and g(−t)=−g(t), then f(−t)g(−t)=(−f)(−g)=fg → even.
10
Compared to x(t), the signal y(t)=x(0.5t−1) is:
a) compressed & shifted b) expanded & shifted c) shifted only d) amplitude-scaled by 2 and shifted
Answer: b — expanded (stretched) and shifted
A time factor 0.5 < 1 stretches the signal in time (expansion), and the −1 produces a time shift.
Analog & Digital Communication
11
AM-DSB-SC is preferred over AM-SSB because:
a) detection requires a less complex demodulation circuit b) the required bandwidth is smaller c) the consumed power is much smaller d) the total power is contained in the sidebands
Answer: a — its modulation/demodulation circuitry is simpler than SSB
SSB needs a sharp sideband filter or a Hilbert (phasing) network, which is hard to build. DSB-SC is simpler to generate. (b) and (c) are actually advantages of SSB, not DSB-SC, so they are wrong here.
12
In digital transmission, the modulation that requires the minimum bandwidth is:
a) PCM b) Differential PCM c) PAM d) Delta modulation
Answer: d — Delta modulation
Delta modulation sends only 1 bit per sample, giving the lowest bit rate of the listed schemes and therefore the smallest bandwidth.
13
Which of the following are not digital: PPM, PCM, PWM?
a) PPM & PCM b) PCM & PWM c) PPM & PWM d) all are digital
Answer: c — PPM and PWM are not digital
PPM and PWM are analog pulse modulations (position/width vary continuously). Only PCM produces a true digital (coded) signal.
14
In PCM, raising the number of quantization levels from 4 to 64 increases the bandwidth by about:
a) 3 times b) 4 times c) 8 times d) 16 times
Answer: a — 3 times
Bits per sample: 4=2²→2 bits, 64=2⁶→6 bits. PCM bandwidth ∝ bits, so the ratio = 6/2 = 3.
15
The number of bits in an A/D converter to obtain SNR ≥ 15 dB should be:
a) 1 b) 2 c) 3 d) 4
Answer: c — 3 bits
Quantization SNR ≈ 6.02n + 1.76 dB. For n=2 → ≈13.8 dB (<15); for n=3 → ≈19.8 dB (≥15). So 3 bits is the minimum.
Microwave
16
For any mode in a rectangular waveguide, propagation occurs:
a) above the cutoff frequency b) below cutoff c) only at cutoff d) at all frequencies
Answer: a — above the cutoff frequency
Below fc the mode is evanescent (attenuated). It only propagates for f > fc — the high-pass behaviour of a waveguide.
17
The device used to separate two microwave signals travelling in opposite directions is:
a) isolator b) PIN diode c) attenuator d) directional coupler
Answer: d — directional coupler
A directional coupler samples the forward and reverse travelling waves on separate ports, so it can separate (and measure) signals moving in opposite directions.
18
Which microwave device uses both a magnetic and an electric field?
a) TWT b) two-cavity klystron c) magnetron d) Gunn diode e) none
Answer: c — magnetron
The magnetron uses crossed fields: a radial electric field and an axial magnetic field that force the electrons into curved (cycloidal) paths.
19
A spread-spectrum communication system is:
a) resistant to jamming b) able to mask the signal from interceptors c) able to let many users share the same time/frequency allocation d) all of the above
Answer: d — all of the above
Spreading the energy over a wide band gives anti-jamming, low probability of intercept (masking), and multiple access (CDMA).
20
The main advantages of FIR filters compared to IIR filters are:
a) more accurate response & easy to implement b) linear phase & stable c) cheap & accurate frequency response
Answer: b — linear phase and (always) stable
FIR filters are inherently stable (no feedback poles) and can be designed with exactly linear phase. The price is a higher order than an equivalent IIR.
Antennas & Propagation
21
The radiation intensity U of a transmitting antenna varies with the distance R as:
a) 1/R b) 1/R² c) R d) None of these
Answer: d — it does not vary with R
Radiation intensity U = r²·W (power per unit solid angle). Since the power density W ∝ 1/R², the r² cancels it → U is independent of R in the far field.
22
In mobile systems the wave arrives at the receiver after:
a) ionospheric propagation b) tropo-scattering c) line of sight only d) direct, reflected and diffracted
Answer: d — direct, reflected, and diffracted (multipath)
In the mobile channel the signal reaches the receiver by several paths (direct + reflections + diffraction), causing multipath fading.
23
In LOS systems, to maximize received power one should:
a) increase the Tx gain only b) increase both antenna gains c) increase the Rx gain only d) none
Answer: b — increase both gains
By Friis, Pr ∝ Gt·Gr, so increasing both transmit and receive gains maximizes received power.
24
Interference between neighbouring base stations is avoided by:
a) assigning different groups of channels b) using different power levels c) using different antennas d) all of the above
Answer: a — assigning different channel groups (frequency reuse planning)
Adjacent cells use different frequency/channel groups so their signals do not interfere; the same group is reused only in cells far enough apart.
25
Fading is caused by: (1) multipath, (2) obstacles, (3) source-frequency variation, (4) amplitude/phase variation at the receiver.
a) (1)&(2) b) (1),(2)&(4) c) (2)&(3) d) all
Answer: b — (1), (2) and (4)
Fading results from multipath, obstacles/shadowing, and the resulting amplitude & phase fluctuations at the receiver. A frequency change at the source (3) is not a fading cause.
Optical Communication
26
Light is confined within the core of a simple optical fiber by:
a) total internal reflection b) diffraction c) refraction d) all of these
Answer: a — total internal reflection
Because n_core > n_cladding, rays beyond the critical angle undergo total internal reflection and stay trapped in the core.
27
In a p–n photodiode, the width of the depletion region:
a) does not vary with junction voltage, doping, profiles b) varies with junction voltage, doping and profiles c) varies with voltage only d) varies with voltage and doping only
Answer: b — varies with junction voltage, doping densities and profiles
The depletion width depends on the reverse-bias voltage and on the doping concentrations/profiles of the junction.
28
An optical receiver has BER = 10⁻⁹. At a data rate of 1 Tb/s, the number of errored bits each second is:
a) 100 b) 10 c) 1000 d) 1
Answer: c — 1000 bits
errors/s = BER × rate = 10⁻⁹ × 10¹² = 10³ = 1000.
29
An optical splice provides a connection between:
a) transmitter to fiber b) receiver to fiber c) fiber to fiber d) fiber to repeater
Answer: c — fiber to fiber
A splice permanently joins two fiber ends (fusion or mechanical) to extend a link.
Satellite & Networks
30
The limitations of FDMA satellite access are:
a) inefficient when downlink traffic ≫ uplink b) less flexible than TDMA in reassigning channels c) carrier-frequency assignment is hardware-controlled d) all of the above
Answer: d — all of the above
All three are recognized drawbacks of FDMA (plus intermodulation among carriers).
31
In geosynchronous satellites, free-space loss is highest in the band:
a) 6/4 GHz b) 14/12 GHz c) 30/20 GHz d) all the same
Answer: c — 30/20 GHz
FSL ∝ f²; the distance is fixed, so the highest frequency (Ka-band, 30/20 GHz) gives the highest loss.
32
The command that checks network connectivity is:
a) ipconfig b) tracert c) nslookup d) ping
Answer: d — ping
ping sends ICMP echo requests and reports whether the host replies (reachability) and the round-trip time.
33
An Ethernet switch forwards a received frame according to:
a) destination physical (MAC) address b) destination IP address c) source physical address d) source IP address
Answer: a — destination physical (MAC) address
A layer-2 switch reads the frame's destination MAC and forwards it out the matching port (it learns source MACs to build its table, but forwards by destination).
34
A computer network confined to one building belongs to:
a) LAN b) WAN c) Internet d) MAN
Answer: a — LAN (Local Area Network)
A network limited to a building/campus is a LAN; a MAN spans a city and a WAN spans large distances.
35
Why is IP-packet fragmentation sometimes required?
a) an IP packet can hold more data than a frame (MTU) b) applications send different amounts of data c) networks can be subnetted d) IP is connectionless
Answer: a — an IP packet can be larger than the link's MTU (frame size)
If a datagram exceeds the MTU of the next link, IP must fragment it into pieces that fit, to be reassembled at the destination.
36
Which is a passive attack?
a) masquerade b) replay c) modification of messages d) release of message contents
Answer: d — release of message contents
Passive attacks only eavesdrop (release of contents, traffic analysis). Masquerade, replay and modification all alter/inject traffic → active attacks.
37
EM waves are refracted when they:
a) pass into two media with the same refractive index b) are polarized at right angles c) meet a perfect conductor d) pass between two media of different refractive index
Answer: d — they cross a boundary between media of different refractive index
Refraction (bending) occurs when the wave speed changes — i.e. when n₁ ≠ n₂ across the interface (Snell's law).
38
The λ/4 transformer between a load and the source is used to:
a) reduce power dissipated in the load b) increase reflected power c) increase power delivered to the load (matching) d) none
Answer: c — to match the load so maximum power is delivered to it
With Z₀ = √(Zs·ZL), the quarter-wave section matches source to load, eliminating reflection and maximizing delivered power.
39
Frequency-hopping spread spectrum (FHSS) is classified into:
a) (a heading) b) slow frequency hopping c) fast frequency hopping
Answer: slow FH and fast FH (options b and c together)
In slow FHSS several bits are sent per hop; in fast FHSS the carrier hops several times per bit.
40
The digital telephony system uses the audio frequency range:
a) 300–2700 Hz b) 500–2500 Hz c) 300–3400 Hz
Answer: c — 300 to 3400 Hz
The standard telephone voice band is 300–3400 Hz, sampled at 8 kHz in digital (PCM) telephony.
Part 4 — Microwave Engineering Question Bank (Univ. of Mosul, 2009–2012)
These questions repeat heavily across the 2009/2010, 2010/2011 and 2011/2012 Microwave papers (Prof. K. Sayidmarie). They are consolidated here once by topic; answers match the circled keys and are verified by theory.
Tubes & Sources (Klystron, TWT, Magnetron, Gunn)
M1
In microwave oscillator circuits, the frequency can be varied by using:
a) point-contact diodes b) mechanical variation of a capacitor c) varactor diodes d) Gunn diodes
Answer: c — varactor diodes
A varactor's junction capacitance changes with the applied voltage, so it tunes the oscillator's resonant frequency.
M2
Conventional electronic tubes at RF/microwave frequencies suffer from:
a) inter-electrode capacitances and lead inductances b) bunching of electrons c) velocity modulation d) none
Answer: a — inter-electrode capacitances and lead inductances
At high frequency these parasitics plus electron transit time degrade the tube — which is why special tubes (klystron, TWT, magnetron) are used instead.
M3
TWT (travelling-wave tube) amplifiers are known as:
a) low-noise amplifiers b) high-power signal generators c) frequency multipliers d) none
Answer: a — low-noise (wideband) amplifiers
The TWT gives gain over a very wide bandwidth with relatively low noise, used as a microwave amplifier.
M4
Two-cavity klystrons are used as:
a) amplifiers at microwave frequencies b) high-power generators c) signal source d) none
Answer: a — amplifiers at microwave frequencies
The two-cavity klystron is an amplifier (buncher + catcher cavities). The single-cavity reflex klystron is the oscillator/signal source.
M5
The reflex klystron is used as:
a) high-power generator b) amplifier c) signal source in microwave systems d) none
Answer: c — signal source (low-power oscillator)
A repeller returns the beam to the single cavity, sustaining oscillation — used as a local oscillator / signal source.
M6
The common features between a TWT and a two-cavity klystron are:
a) velocity modulation and density modulation b) generation of microwave signals c) velocity modulation only d) none
Answer: a — velocity modulation and density (bunching) modulation
Both work by first velocity-modulating the electron beam, which then turns into density (current) modulation as electrons bunch.
M7
The role of the magnetic field in magnetrons is to:
a) accelerate electrons towards the anode b) force electrons to move in circular paths c) repel electrons towards the cathode d) none
Answer: b — force the electrons into circular (curved) paths
The axial magnetic field crossed with the radial electric field bends the electron trajectories, letting them give energy to the resonant cavities.
M8
An axial magnetic field and a radial electric field are used in:
a) reflex klystron b) magnetron c) TWT amplifier d) two-cavity klystron multiplier
Answer: b — magnetron
Crossed-field device: radial E plus axial B — the defining geometry of the magnetron.
M9
Frequency modulation of a microwave signal can be easily achieved in:
a) Gunn diodes b) two-cavity klystron amplifiers c) reflex klystrons d) TWT tubes
Answer: c — reflex klystrons
Varying the repeller voltage of a reflex klystron shifts its oscillation frequency → easy FM (electronic tuning).
M10
A condition required for a semiconductor to show the Gunn effect is that the energy difference between the upper and lower conduction-band valleys is:
a) smaller than 0.026 eV b) equal to the gap Eg=1.43 eV c) larger than the thermal energy (0.026 eV) at room temperature d) none
Answer: c — larger than the thermal energy (≈0.026 eV) at room temperature
The valley separation must exceed kT (≈0.026 eV) so electrons are not thermally excited into the upper, low-mobility valley until the field drives them there (transferred-electron mechanism).
M11
The modes in a reflex klystron:
a) give the same frequency but at different transit times b) result from excessive transit time c) are caused by spurious FM d) are only theoretical
Answer: a — they give the same (cavity) frequency but at different transit-time numbers (N+¾ cycles)
Each "mode" corresponds to a different number of cycles the returned electrons spend in the repeller space, all satisfying the same resonance.
Passive Components & Waveguides
M12
Waveguides can be considered as:
a) low-pass filters b) high-pass filters c) band-pass filters d) resonant circuits
Answer: b — high-pass filters
A hollow waveguide passes only frequencies above its cutoff fc and rejects lower ones → high-pass behaviour.
M13
The dominant mode in a waveguide is characterized by:
a) propagation constant equal to zero b) the lowest cutoff frequency c) cannot propagate without other modes d) phase constant that changes linearly with frequency
Answer: b — having the lowest cutoff frequency
The dominant mode (TE₁₀ in a rectangular guide) is the one with the lowest fc, so it propagates alone over the single-mode band.
M14
A wavemeter is used for:
a) matching a load to a source b) measuring RF and microwave frequencies c) measuring scattering parameters d) isolating incident from reflected waves
Answer: b — measurement of RF and microwave frequencies
A cavity wavemeter is tuned until it resonates (a dip in transmitted power), reading the unknown frequency off its calibrated scale.
M15
The isolator is used to:
a) reduce the power level b) produce a phase shift c) pass the signal in both directions at low loss d) pass the signal at low attenuation in one direction and not the reverse
Answer: d — pass low-loss in one direction and heavily attenuate the reverse
A ferrite isolator protects the source from reflections by absorbing the reverse wave.
M16
Ferrite materials used to make microwave devices are characterized by:
a) high conductivity & large permeability b) large permeability & small skin depth c) low loss & large relative permeability d) none
Answer: c — low loss and large relative permeability
Ferrites are low-loss magnetic insulators; their non-reciprocal behaviour in a bias field is exploited in isolators and circulators.
Scattering Parameters & Amplifiers
M17
The scattering matrix of a microwave device is symmetrical when:
a) the device has some amplification b) the device has no amplification c) the device is passive (reciprocal) d) true for all devices
Answer: c — the device is passive / reciprocal
For a reciprocal (passive, non-ferrite) network Sᵢⱼ = Sⱼᵢ, so the S-matrix is symmetric.
M18
The scattering parameters express the relation between:
a) voltage waves incident on the ports and those reflected/travelling away from the ports b) total voltages and total currents c) impedances and admittances d) none
Answer: a — incident voltage waves vs. reflected (outgoing) voltage waves
S-parameters relate the normalized outgoing waves b to the incoming waves a: b = S·a.
M19
A transistor amplifier is said to be unilateral when:
a) it is matched at input and output b) S₂₁ = 0 c) S₁₁ and S₂₂ = 0 d) S₁₂ = 0
Answer: d — when S₁₂ = 0
S₁₂ is the reverse transmission. If S₁₂=0 there is no feedback from output to input → the amplifier is unilateral.
Microwave — Calculation Problems
M-Q2
An air-filled rectangular waveguide (a=1.5 cm, b=0.8 cm) has H̄ = H₀[cos(πx/a)âz + j(βa/π)sin(πx/a)âx]·sin(3π×10¹⁰t − βz) A/m. Find: mode, cutoff frequency, propagation constant, E-field, transmitted power.
a) Mode: only the cos(πx/a) (m=1) variation in x and none in y → the TE₁₀ dominant mode.
Operating frequency: ω = 3π×10¹⁰ → f = ω/2π = 15 GHz.
b) Cutoff frequency (TE₁₀): fc = c/(2a) = 3×10⁸/(2×0.015) = 10 GHz.
c) Propagation constant: β = (2π/c)·√(f² − fc²) = (2π/3×10⁸)·√(15² − 10²)×10⁹ = 2π·(11.18×10⁹)/3×10⁸ ≈ 234 rad/m.
d) Electric field (TE₁₀): only Ey exists: Ey = −(jωμa/π)·H₀·sin(πx/a)·sin(ωt−βz).
e) Transmitted power: P = (|E₀|²·a·b)/(4·Z_TE), with Z_TE = η/√(1−(fc/f)²) = 377/√(1−(10/15)²) ≈ 506 Ω. Insert the peak E₀ from part (d) to get the numeric power.
Note: the final numeric power needs H₀ (not given as a number on the paper). The method and all formulas above are complete — plug in H₀ from your sheet.
M-Q3
A circular waveguide of 4 cm diameter. Find the frequency range for single-mode operation, identify that mode, and find the rectangular-waveguide dimensions with the same range.
Answer: single-mode band ≈ 4.395 – 5.74 GHz, mode = TE₁₁; equivalent rectangular guide a ≈ 3.41 cm, b ≈ 2.6 cm
Radius a = 2 cm = 0.02 m.
Lowest mode (TE₁₁): fc = 1.841·c/(2πa) = 1.841×3×10⁸/(2π×0.02) = 4.395 GHz.
Next mode (TM₀₁): fc = 2.405·c/(2πa) = 5.74 GHz.
So only the TE₁₁ mode propagates between 4.395 and 5.74 GHz.
Equivalent rectangular guide (TE₁₀): set its cutoff to 4.395 GHz: a = c/(2fc) = 3×10⁸/(2×4.395×10⁹) = 3.41 cm; set the upper limit (next mode) to 5.74 GHz: b = c/(2×5.74×10⁹) = 2.6 cm.
Part 5 — Antennas & Propagation Question Bank
Consolidated from the 4th-class Antennas papers and quizzes (Prof. K. Sayidmarie, 2017–2019). Repeated MCQ are merged; the recurring worked problems (horn design, microstrip patch, array factor, radiation-intensity) are solved at the end.
Propagation
A1
In radar systems, the received power from a target varies with the range R as:
a) 1/R b) 1/R² c) 1/R⁴ d) none
Answer: c — 1/R⁴
The wave spreads as 1/R² going to the target, and the echo spreads again as 1/R² coming back → overall Pr ∝ 1/R⁴ (the radar equation).
A2
The attenuation of microwaves in the atmosphere increases with:
a) increase in rain rate b) increase in rain rate and frequency c) increase in rain rate and decrease in frequency d) none
Answer: b — increase in both rain rate and frequency
Rain droplets scatter/absorb more at higher frequencies (shorter wavelengths), so attenuation rises with both rain rate and frequency.
A3
To maximize the power received by an antenna from a transmitter, the two antennas must be:
a) matched b) matched, polarization and angle aligned c) matched and polarization aligned d) none
Answer: b — impedance-matched, polarization-aligned, and beam (angle) aligned
Maximum transfer needs all three: conjugate impedance match, identical polarization, and the main beams pointed at each other.
A4
Ionospheric propagation is used between stations separated by:
a) less than 50 km b) 50–100 km c) larger than 200 km d) none
Answer: c — larger than 200 km
Sky-wave (ionospheric) propagation reflects HF signals off the ionosphere for long-distance links (hundreds–thousands of km), well beyond line-of-sight.
A5
In multipath propagation, the path-gain factor can vary between:
a) 0 and 1 b) 0 and 2 c) 1 and 2 d) none
Answer: b — 0 and 2
The direct and reflected waves add as |1 + Γe^{jφ}|; with |Γ|≈1 this ranges from 0 (full cancellation) to 2 (full reinforcement).
A6
In LOS propagation over spherical earth, the maximum distance between two antennas of heights 36 m and 49 m is:
a) 53.6 km b) 35.6 km c) 38 km d) none
Answer: a — 53.6 km
d(km) = 4.12(√h₁ + √h₂) = 4.12(√36 + √49) = 4.12(6+7) = 4.12×13 = 53.6 km (heights in metres).
A7
In multipath propagation, the reflected wave can:
a) only cancel the LOS wave b) may cancel or reinforce the LOS wave c) always reinforce the LOS wave
Answer: b — it may cancel or reinforce, depending on the phase
Depending on the extra path length (phase), the reflected wave adds constructively (reinforce) or destructively (cancel) with the direct wave.
A8
The refractive index of air decreases with height, therefore the effective earth radius looks ___ the actual one.
a) smaller than b) larger than c) the same as d) none
Answer: b — larger than the actual one
The downward bending of the ray (from the decreasing index) is equivalent to a flatter earth, modeled by an effective radius of 4/3 × R_earth — i.e. larger.
Basic Antenna Parameters
A9
The antenna gain relative to the isotropic radiator is expressed as:
a) dB b) dBi c) dBm d) none
Answer: b — dBi
"dBi" explicitly means decibels relative to an isotropic radiator (the reference for absolute gain).
A10
The angular separation between the half-power points of a radiation pattern is called the:
a) bandwidth b) beamwidth c) front-to-back ratio d) none
Answer: b — beamwidth (specifically HPBW)
The angle between the two −3 dB (half-power) points on either side of the main lobe is the half-power beamwidth.
A11
The input resistance at the center of a λ/2 dipole is approximately:
a) 36.5 Ω b) 50 Ω c) 73 Ω d) 300 Ω
Answer: c — ≈73 Ω
A half-wave dipole has a radiation resistance of about 73 Ω (a quarter-wave monopole is ≈36.5 Ω).
A12
As the length of a center-fed dipole is reduced below a quarter-wavelength, the radiation resistance:
a) decreases b) increases c) remains constant
Answer: a — decreases
Radiation resistance falls rapidly for short dipoles: Rr ∝ (L/λ)², so shortening the antenna lowers it.
A13
The folded dipole antenna has:
a) greater bandwidth than a half-wave dipole b) 300-Ω input impedance c) less bandwidth d) greater bandwidth AND 300-Ω input impedance
Answer: d — greater bandwidth and a 300-Ω input impedance
Folding raises the input impedance to ≈4×73 ≈ 300 Ω and widens the bandwidth — which is why it is used in TV/FM receivers.
A14
A vertical electric dipole antenna:
a) radiates uniformly in all directions b) radiates uniformly in all horizontal directions but stronger in the vertical c) radiates most strongly and uniformly in the horizontal directions d) does not radiate in the horizontal directions
Answer: c — strongest, uniform radiation in the horizontal plane (omnidirectional in azimuth)
Its pattern is a doughnut: a null along the dipole axis (vertical) and maximum, uniform radiation in the horizontal plane.
A15
The antenna with the smallest beamwidth among the following is the:
a) infinitesimal dipole b) λ/4 dipole c) half-wavelength dipole d) isotropic element
Answer: c — half-wavelength dipole
Of these, the λ/2 dipole is the most directive (HPBW ≈ 78°), so it has the smallest beamwidth. The isotropic source has the largest (radiates everywhere).
A16
The directivity and gain of an antenna have equal values when:
a) the antenna is lossless b) the antenna is well matched c) it has small radiation resistance d) always equal
Answer: a — when the antenna is lossless
G = e·D with efficiency e. If lossless, e=1 so G = D.
A17
The radiation intensity U(r,θ,φ) varies with the distance r as:
a) 1/r b) 1/r² c) r d) it does not vary with r
Answer: d — it does not vary with r
U = r²·W; since W ∝ 1/r² in the far field, the r² cancels and U is independent of r.
A18
An antenna is matched to a 50-Ω line, peak input current 1 A, efficiency 80%. The radiated power is:
a) 25 W b) 20 W c) 31.25 W d) none
Answer: b — 20 W
Input power Pin = ½·I²·Z₀ = ½·(1)²·50 = 25 W. Radiated = η·Pin = 0.8×25 = 20 W.
Arrays & Aperture Antennas
A19
Antenna arrays are used to achieve:
a) higher directivity and lower beamwidth b) higher directivity and larger beamwidth c) lower gain, higher efficiency d) higher gain and larger bandwidth
Answer: a — higher directivity and lower (narrower) beamwidth
Combining elements increases directivity, which narrows the main beam.
A20
For a given side-lobe level, the array with the lowest side lobes / smallest beamwidth between first nulls is:
a) Dolph-Tschebyscheff b) open-ended waveguide c) uniform d) end-fire
Answer: a — Dolph-Tschebyscheff array
The Dolph-Tschebyscheff array gives the optimum trade-off: for a given side-lobe level it has the narrowest beam (and vice-versa).
A21
To reduce the side lobes in a radiation pattern we:
a) increase the antenna size b) use larger element spacing c) taper the amplitude excitations of the array elements d) apply a linear phase shift
Answer: c — taper (reduce toward the edges) the amplitude excitations
Amplitude tapering lowers the side-lobe level (at the cost of a slightly wider main beam). A linear phase shift only steers the beam.
A22
To scan the main beam of an array antenna we apply:
a) amplitude taper b) change of element spacing c) a linear phase to the elements d) none
Answer: c — a progressive (linear) phase across the elements
A linear phase gradient steers the beam to θ₀ where β = −kd·cosθ₀ (the basis of phased arrays).
A23
The phase shift required to steer the main beam of a 4-element, λ/2-spaced array to 60° is:
a) 90° b) 45° c) 180° d) none
Answer: a — β = 90° (i.e. −90°)
With ψ = kd·cosθ + β = 0 at the beam: β = −kd·cos60° = −(2π/λ)(λ/2)(0.5) = −π/2 = −90°.
A24
The directivity of an end-fire array compared to a broadside array with the same number of elements is:
a) twice b) equal c) half d) none
Answer: a — twice
For the same length, an end-fire array has roughly double the directivity of a broadside array (it radiates one main lobe along the axis).
A25
The gain of a horn antenna can be increased by:
a) tapering the field at the aperture b) increasing the aperture area c) applying a linear phase across the aperture d) none
Answer: b — increasing the aperture area
G = ε·(4π/λ²)·A — a larger aperture area gives higher gain (up to the optimum flare beyond which phase error reduces it).
A26
A lossless aperture antenna has a gain of 10 dB and a physical area of 1.5λ². Its aperture efficiency is:
a) 53% b) 100% c) 35% d) none
Answer: a — 53%
G = 10 dB = 10 (linear). G = ε·(4π/λ²)·A → 10 = ε·(4π/λ²)(1.5λ²) = ε·18.85 → ε = 0.53 = 53%.
A27
The antenna with the smallest bandwidth among the following is the:
a) thin-wire monopole b) biconical dipole c) folded λ/2 dipole d) microstrip antenna
Answer: d — microstrip (patch) antenna
The resonant microstrip patch is high-Q and narrowband. The biconical dipole has the largest bandwidth.
A28
The antenna with the largest bandwidth is the:
a) thin-wire monopole b) microstrip antenna c) folded λ/2 dipole d) biconical dipole
Answer: d — biconical dipole
Its tapered (frequency-independent-like) geometry gives very wide bandwidth.
A29
The array which has no side lobes is the:
a) uniform b) binomial c) Dolph-Tschebyscheff d) end-fire
Answer: b — binomial array
A binomial array (excitations following Pascal's triangle, e.g. 1-2-1 or 1-3-3-1) for d ≤ λ/2 produces a pattern with no side lobes (at the cost of a wider main beam).
Antennas — Worked Problems
AP-1
Design an optimum-directivity pyramidal horn fed by an X-band guide at 10 GHz, with horn length from the apex ρ₁=ρ₂=10λ. Find the aperture dimensions (in λ), the directivity (dB), and the power to a matched load for incident density 10 µW/m².
λ = c/f = 3×10⁸/10×10⁹ = 3 cm.
a) Optimum aperture (in λ): a = √(3λρ₁) = √(3λ·10λ) = √30 λ ≈ 5.47λ (E-plane); b = √(2λρ₂) = √20 λ ≈ 4.47λ (H-plane). So aperture ≈ 5.47λ × 4.47λ.
b) Directivity: D = ½·(4π/λ²)·(a·b) = ½·(4π/λ²)·(5.47λ)(4.47λ) ≈ 153.6 ≈ 21.86 dB.
c) Power to matched load: Pr = W·Aem = W·(λ²/4π)·D = 10×10⁻⁶ ×(0.03²/4π)×153.6 ≈ 1.1 ×10⁻⁶ W ≈ 1.1 µW.
AP-2
Design a rectangular microstrip patch to resonate at 2 GHz on a substrate εr=4, h=1.6 mm. Find W, εreff, effective length, physical length L.
λ₀ = c/f = 15 cm.
a) Width: W = (c/2f)·√(2/(εr+1)) = (15/2)·√(2/5) = 4.74 cm.
b) Effective dielectric constant: εreff = (εr+1)/2 + (εr−1)/2·[1+12h/W]^{−1/2} = 2.5 + 1.5·[1+12(0.16/4.74)]^{−1/2} ≈ 3.76.
c) Length extension: ΔL = 0.412h·[(εreff+0.3)(W/h+0.264)]/[(εreff−0.258)(W/h+0.8)] ≈ 0.076 cm.
d) Physical length: L = c/(2f√εreff) − 2ΔL = (15/(2√3.76)) − 2(0.076) ≈ 3.87 − 0.15 ≈ 3.72 cm.
AP-3
A rotationally-symmetric antenna has U(θ)=K·cos²θ with total radiated power 10 W. Find HPBW, directivity, radiation resistance (I=1 A), max power density at 1 km, and gain (loss resistance 1 Ω).
HPBW: half power → cos²θ = 0.5 → θ = 45° → HPBW = 2×45° = 90°.
Find K: Prad = ∫∫U dΩ = K·2π·∫₀^{π/2}cos²θ·sinθ dθ = K·2π/3 = 10 → K = 4.77.
Directivity: D = 4π·Umax/Prad = 4π(4.77)/10 = 5.99 ≈ 7.78 dB.
Radiation resistance: Prad = ½·I²·Rr → 10 = ½·(1)²·Rr → Rr = 20 Ω.
Max power density at 1 km: Wmax = Umax/r² = 4.77/(10³)² = 4.77 µW/m².
Gain (Rloss=1 Ω): e = Rr/(Rr+Rloss) = 20/21 = 0.95 → G = e·D = 0.95×5.99 = 5.69 ≈ 7.55 dB.
AP-4
4-element uniform array along the z-axis, d=λ/2, β=0 (broadside). Give the normalized array factor, the HPBW and null-to-null beamwidth, and the phase to steer the beam end-fire (z-axis).
Array factor: (AF)ₙ = (1/N)·[sin(N·ψ/2)/sin(ψ/2)], with ψ = kd·cosθ + β, N=4, kd=π, β=0.
Broadside max at θ = 90°.
HPBW: ≈ 2[90° − cos⁻¹(1.391λ/(πNd))] = 2[90° − cos⁻¹(1.391/(π·4·0.5))] ≈ 52.6°.
Null-to-null: ≈ 2[90° − cos⁻¹(λ/(Nd))] = 2[90° − cos⁻¹(1/2)] = 60°.
To steer end-fire (θ=0): ψ=0 → β = −kd·cos0° = −π = −180°.
AP-5
3 half-wave dipoles along the y-axis, parallel to z, d=λ/2, with excitations 1 : 2 : 1 (binomial). Give the array factor in the principal planes and the no-sidelobe condition.
The 1-2-1 excitation is a 3-element binomial array → no side lobes for d=λ/2.
Array factor (3 elements): AF = 1 + 2cos(ψ) + ... → AF = 2cos²(ψ/2)·2, with ψ = kd·cosγ (γ measured from the array axis, y).
xy-plane: AF = 2cos²(π·sinφ) ; yz-plane: AF = 2cos²(π·sinθ) ; xz-plane: no variation (γ=90° → AF = constant).
The total pattern = element pattern (λ/2 dipole) × array factor.
Part 6 — Transmission Lines, Modulation & 2-Port Networks
Source has an answer key, which is mostly correct. Two answers are wrong and are flagged in orange with the corrected value.
Transmission Lines
T1
A transmission line can be represented as:
a) R & L in series and G & C in shunt b) R & G series, L & C shunt c) R & C series, G & L shunt d) none
Answer: a — R & L in series, G & C in shunt
The standard per-unit-length model: series resistance R and inductance L; shunt conductance G and capacitance C.
T2
The phase-shift constant of a lossless transmission line equals:
Answer: β = ω·√(LC)
For a lossless line (R=G=0): β = ω√(LC) and characteristic impedance Z₀ = √(L/C).
T3
Distortionless transmission line condition; find the inductance L.
Condition: L/R = C/G → L = RC/G
A line is distortionless when RC = LG, i.e. L = RC/G. Worked examples from the paper:
• R=10 Ω/km, C=9 nF/km, G=0.6 µS/km → L = (10×9×10⁻⁹)/(0.6×10⁻⁶) = 0.15 H/km.
• R=100 Ω/km, C=9 nF/km, G=0.6 µS/km → L = 1.5 H/km.
T4
A line of characteristic impedance 500 Ω is connected to a 500-Ω load. The reflection coefficient is:
a) 0 b) −1 c) +1 d) 2
Answer: a — 0
Γ = (ZL−Z₀)/(ZL+Z₀) = (500−500)/(500+500) = 0 → matched, no reflection.
T5
A quarter-wave line of Z₀=50 Ω is terminated by a 100-Ω resistor. The input impedance is:
a) 150 Ω b) 50 Ω c) 250 Ω d) 100 Ω
Correct answer: Zin = Z₀²/ZL = 50²/100 = 25 Ω
For a quarter-wave line: Zin = Z₀²/ZL = 2500/100 = 25 Ω.
Note: The key marked "50 Ω", which is wrong. The correct value is 25 Ω, which is not listed among the options — the question's options are flawed.
T6
The VSWR for a 50-Ω line terminated in (40+j30) Ω is:
a) 1.333 b) 0.8 c) 2 d) 1
Correct answer: c — VSWR = 2.0
Γ = (ZL−Z₀)/(ZL+Z₀) = (−10+j30)/(90+j30) = j0.333 → |Γ| = 0.333.
VSWR = (1+|Γ|)/(1−|Γ|) = 1.333/0.667 = 2.0.
Note: The key marked "1.333", which is only the numerator (1+|Γ|), not the full VSWR. The correct VSWR is 2.0 (option c).
T7
A quarter-wave transformer matching 100 Ω to 64 Ω has characteristic impedance:
Answer: Z₀ = √(100×64) = 80 Ω
The transformer impedance is the geometric mean of the two impedances it joins.
T8
Quick transmission-line facts (fill-ins, all verified):
• Characteristic impedance, C=100 pF/m, L=9 µH/m → Z₀ = √(L/C) = √(9×10⁻⁶/100×10⁻¹²) = 300 Ω.
• Lossy-line propagation constant: γ = √((R+jωL)(G+jωC)).
• Wavelength for β=12.56 rad/m: λ = 2π/β = 0.5 m.
• All energy is reflected if the load is open or short (Γ = ±1); a line repeats its impedance every nλ/2.
• VSWR of a matched line = 1; of an open circuit = ∞; reflection coefficient of a matched circuit = 0.
• SWR=2 → |Γ| = (2−1)/(2+1) = 1/3. |Γ|=0.7 → SWR = 1.7/0.3 ≈ 5.6.
• Vmax=20, Vmin=4 → SWR = 5. Vmin=20, Imax=0.25 → P = Vmin·Imax/2 = 2.5 W.
• Crank diagram: 15 V over 15 cm → 1 V/cm; 20 V over 10 cm → 2 V/cm.
Amplitude & Frequency Modulation
M1
Over-modulation occurs when the ratio Aa/Ab (message/carrier) is:
a) >1 b) <1 c) 1 d) 0
Answer: a — >1 (modulation index m > 1)
Over-modulation (m>1) distorts the envelope and causes carrier phase reversals → envelope detection fails.
M2
For a message m(t)=cos(2πfmt), the standard AM signal is:
a) cos·cos b) cos(2πfct) c) cos[2π(fc+fm)t] d) [1+cos(2πfmt)]·cos(2πfct)
Answer: d — Ac[1+m(t)]·cos(2πfct)
Standard (DSB full-carrier) AM has the form Ac[1+m(t)]cos(ωct) — carrier plus two sidebands.
M3
The output of an SSB-SC signal contains the frequency:
Answer: one sideband only, e.g. fc + fm (or fc − fm)
SSB transmits a single sideband — no carrier and no second sideband → bandwidth = fm.
M4
The message that can be retrieved completely by envelope detection for s(t)=Ac[1+m(θ)]cos(2πt/α) is:
a) 1.5cos b) 0.5cos c) 2sin d) 2cos
Answer: b — 0.5 cos(...) (|m| ≤ 1)
Envelope detection works only when |m(t)| ≤ 1. Only the 0.5-amplitude message avoids over-modulation.
M5
The bandwidth of NBFM is:
a) fm b) 2fm c) 2fm(β+1) d) fm(β+1)
Answer: b — 2fm
Narrowband FM behaves like AM in bandwidth: BW ≈ 2fm.
M6
The bandwidth of WBFM is:
a) fm b) 2fm c) 2fm(β+1) d) fm(β+1)
Answer: c — 2fm(β+1)
Carson's rule: BW = 2(Δf+fm) = 2fm(β+1).
M7
Which statement is true for FM?
a) carrier power rises with modulating level b) falls with level c) rises as level decreases d) carrier power is independent of the modulating level
Answer: d — carrier power is independent of the modulating level
FM has constant envelope, so the total transmitted power stays constant regardless of modulation — only the distribution among sidebands changes.
M8
The four elements in a PLL are: loop filter, amplifier, VCO, and:
a) up converter b) down converter c) phase detector d) frequency multiplier
Answer: c — phase detector
A PLL = phase detector + loop filter + amplifier + VCO, with the VCO output fed back to the phase detector.
M9
The de-emphasis network in a commercial FM receiver is a:
a) HPF b) LPF c) BPF d) none
Answer: b — LPF
De-emphasis (a low-pass filter) reverses the transmitter's pre-emphasis (high-pass boost), restoring a flat response and improving SNR.
M10
To generate FM using a PM modulator, the message m(t) should first be:
a) multiplied b) differentiated c) integrated d) unchanged
Answer: c — integrated
Since FM is PM of the integral of the message, passing m(t) through an integrator before a PM modulator yields FM.
M11
In mobile/TV transmission, the modulation used for video is:
a) PM b) VSB c) DSB-SC d) AM
Answer: b — VSB (Vestigial Sideband)
VSB saves bandwidth while preserving the low video frequencies — the standard for analog TV video.
M12
Numerical modulation results (verified):
• FM index, Δf=75 kHz, fm=100 Hz → β = Δf/fm = 750.
• FM index, Δf=5 kHz, fm=250 Hz → β = 20.
• 1 kV carrier, 30% AM → sideband amplitude = m·Ac/2 = 0.3×1000/2 = 150 V = 0.15 kV.
• AM-DSB efficiency at m=0.5: η = m²/(2+m²) = 0.25/2.25 ≈ 11%; maximum DSB-AM efficiency (m=1) = 33.3%.
• FM bandwidth, Δf=60 kHz, fm=5 kHz → BW = 2(60+5) = 130 kHz.
• SSB bandwidth for 6 kHz baseband = 6 kHz; for 9 kHz = 9 kHz.
• AM with Emax=60, Emin=40 → m = (60−40)/(60+40) = 0.2.
• DSB-AM carrier: BW=2fm, fmax=850 Hz, BW=50 Hz → fm=25, fc = 850−25 = 825 Hz.
• FM is the change of carrier frequency per the message; FM bandwidth > AM; increasing the modulating amplitude increases the number of sidebands; increasing modulating frequency increases AM bandwidth; recovering the information = demodulation; minimum bandwidth & power technique = SSB; in AM the carrier amplitude is varied.
Two-Port Networks
N1
The H-parameters hi and hf of a two-port network are obtained by:
a) shorting the output terminal b) opening the input c) shorting the input d) opening the output
Answer: a — by short-circuiting the output
hi = V₁/I₁ and hf = I₂/I₁ are both measured with the output shorted (V₂=0).
N2
How many variables does a two-port network consist of?
a) one b) two c) three d) four
Answer: d — four (V₁, I₁, V₂, I₂)
A two-port has two port voltages and two port currents; any 2×2 parameter set relates two of them to the other two.
N3
Which H-parameter is the open-circuit reverse voltage ratio?
a) hf (forward current gain) b) hi (input impedance) c) hr (reverse voltage ratio) d) ho (output admittance)
Answer: c — hr (the reverse voltage ratio)
hr = V₁/V₂ with the input open (I₁=0) — the open-circuit reverse voltage ratio.
Note: the printed option labels are garbled in the source; the correct standard symbol for the reverse voltage ratio is hr.
Part 7 — Engineering Analysis / Mathematics
Source has an answer key — all 15 answers verified below.
Vectors & Geometry
1
The unit vector of AB = −2i − 2j + k is:
a) (−2i+2j−k)/3 b) (−2i−2j+k)/3 c) (2i+2j−k)/3 d) (−2i+2j+k)/3
Answer: b — (−2i−2j+k)/3
|AB| = √((−2)²+(−2)²+1²) = √9 = 3. Unit vector = vector / magnitude = (−2i−2j+k)/3.
2
The angle between the planes x+y=1 and 2x+y−2z=2 is:
a) 45° b) 90° c) 60° d) 180°
Answer: a — 45°
Normals n₁=(1,1,0), n₂=(2,1,−2). cosθ = (n₁·n₂)/(|n₁||n₂|) = 3/(√2·3) = 1/√2 → θ = 45°.
3
For z = 1/(xy) over 1≤x≤2, 1≤y≤2, the volume bounded by R and the surface is:
a) 4 ln2 b) 2 ln2 c) (ln2)² d) 2(ln2)²
Answer: c — (ln2)²
∫₁²∫₁² (1/xy) dy dx = (∫₁² dx/x)(∫₁² dy/y) = (ln2)(ln2) = (ln2)² (separable integral).
Linear Algebra & Sequences
4
The eigenvalues of A = [[1,2],[2,1]] are:
a) −1, 3 b) −3, 1 c) 3, 1 d) −1, −3
Answer: a — −1 and 3
det(A−λI)=0 → (1−λ)²−4=0 → λ²−2λ−3=0 → λ = (2±√16)/2 = 3, −1.
5
The sequence {(−1)^{n+1}} is:
a) non-monotone & non-bounded b) monotone & bounded c) non-monotone & bounded d) monotone & non-bounded
Answer: c — non-monotone and bounded
The terms alternate 1, −1, 1, −1, … → not monotone, but bounded since |(−1)^{n+1}| ≤ 1.
6
The sequence {(3n²−1)/(10n+5n²)} is:
a) divergent b) convergent c) undefined
Answer: b — convergent
For large n the dominant terms give 3n²/5n² = 3/5, a finite limit → the sequence converges to 3/5.
7
The formula of the Maclaurin series is:
a) Σ cₙ(x−a)ⁿ b) Σ f⁽ⁿ⁾(0)/n!·xⁿ c) Σ cₙxⁿ
Answer: b — Σ f⁽ⁿ⁾(0)/n!·xⁿ
The Maclaurin series is the Taylor series centered at a=0.
8
The series Σ a·rⁿ⁻¹ (with r = aₙ₊₁/aₙ) is the general form of a:
a) power series b) geometric series c) Taylor series
Answer: b — geometric series
Σ a·rⁿ⁻¹ is the standard geometric series (constant ratio r between successive terms).
Statistics, Probability & ODE
9
The standard deviation of the data (8, 2, 4, 12, 1, 3) is:
a) 3.83 b) 5.22 c) 2.46 d) 4.31
Answer: a — 3.83
Mean μ = 30/6 = 5. Squared deviations: 9+9+1+49+16+4 = 88. Variance = 88/6 = 14.67; σ = √14.67 ≈ 3.83.
10
If P(E)=0.3, P(E∪F)=0.8, P(E∩F)=0.1, then P(F) is:
a) 1 b) 0.6 c) 0.5 d) 0.4
Answer: b — 0.6
Inclusion–exclusion: P(E∪F) = P(E)+P(F)−P(E∩F) → 0.8 = 0.3 + P(F) − 0.1 → P(F) = 0.6.
11
The differential equation (y·ln x − 1)·y dx = x dy is:
a) separable b) exact c) linear d) Bernoulli
Answer: a — separable
Rearranged: dy/[y(y·ln x −1)] = dx/x — variables separate into a function of y times dy equals a function of x times dx.
12
∫ f₂(x) dx is equal to: (options relate f₀, f₁, f₂)
Answer: none of the options (insufficient context)
No recurrence or definition is given relating f₀, f₁, f₂, so none of the listed options can be validated — the question is incomplete as printed.
13
The sum of observed values divided by the sample size is called:
a) variance b) range c) mode d) arithmetic mean
Answer: d — arithmetic mean
mean = (Σxᵢ)/n — exactly "sum of values over sample size".
14
The proportion of times that a value xᵢ appears in the sample is called:
a) cumulative relative frequency b) relative frequency c) cumulative absolute frequency d) absolute frequency
Answer: b — relative frequency
relative frequency = (number of times xᵢ appears)/(sample size) — a proportion.
15
The mode of {2, 6, 9, 2, 5, 4, 9, 7, 4, 10, 9, 4, 5, 9, 8} is:
a) 2 b) 8 c) 9 d) 4
Answer: c — 9
Counting: 9 appears 4 times (the most), 4 appears 3 times, others fewer. The mode is the most frequent value = 9.
Exam 2023 – 2024 — Communication Principles (Dr. Ali AlAbdullah & Dr. Firas Alsharbaty)
The MCQ part of this paper (AM-signal form, FM index β=750, 30% AM sideband = 0.15 kV, FM-from-PM by integration, WBFM = 2fm(β+1), video = VSB) repeats the Modulation bank — see M2, M5, M6, M10, M11, M12 in Part 6. Only the new worked problem and the balanced-modulator essay are added.
Q3
An AM wave is s(t)=10[1+0.2cos(4π×10⁴t)]cos(8π×10⁷t). Find: modulation depth, bandwidth, carrier & total sideband power, power efficiency, and sketch the spectrum.
Answers: m=0.2 | BW=40 kHz | Pc=50 W, Psb=1 W | η≈2%
Compare with s(t)=Ac[1+m·cos(2πfmt)]cos(2πfct): Ac=10, m=0.2, fm=2×10⁴=20 kHz, fc=4×10⁷=40 MHz.
1) Modulation depth: m = 0.2 (20%).
2) Bandwidth: BW = 2fm = 40 kHz.
3) Carrier power: Pc = Ac²/2 = 100/2 = 50 W (per 1 Ω). Total sideband power: Psb = Pc·m²/2 = 50×0.04/2 = 1 W (0.5 W each). Total Pt = Pc(1+m²/2) = 51 W.
4) Power efficiency: η = m²/(2+m²) = 0.04/2.04 ≈ 0.0196 ≈ 2%.
5) Spectrum: a carrier line at 40 MHz (amplitude 10, power 50 W) and two sidebands at 40 MHz ± 20 kHz = 39.98 & 40.02 MHz (amplitude mAc/2 = 1, power 0.5 W each).
Q2
DSB-SC balanced modulator: draw the block diagram, write vo(t), state the benefit of the filter, and explain why it is called a single-balanced modulator.
Block diagram: the carrier cos(ωct) drives two identical non-linear devices (diode/FET modulators). The message is applied with opposite polarity to the two branches — branch-1 sees +m(t), branch-2 sees −m(t). Their outputs are subtracted, then passed through a band-pass filter (BPF) centred at fc.
Output equation (square-law device vo=a·v+b·v²):
vo1 = a[cosωct+m] + b[cosωct+m]² , vo2 = a[cosωct−m] + b[cosωct−m]²
vo = vo1 − vo2 = 2a·m(t) + 4b·m(t)·cos(ωct).
Benefit of the filter: the BPF removes the unwanted baseband term 2a·m(t) (and harmonics), leaving only 4b·m(t)cos(ωct) — a clean DSB-SC signal centred at fc.
Why "single-balanced": the circuit cancels (balances out) the carrier only; it does not cancel the message feed-through. A double-balanced (ring) modulator cancels both the carrier and the message leakage.
Exam 2021 – 2022 — Communication Principles (Finals — Ahmed Abdulbari / Noor Saadallah)
Most MCQ/fill-ins in these papers repeat Part 6 (TL facts, FM index=20, NBFM=2fm, AM-DSB efficiency≈11%, SSB BW, Carson 130 kHz, crank diagram, VSWR, λ/4 transformer = 80 Ω, etc.) — see T1–T8, M1–M12. Only the new items are added below.
1
In AM, when the modulation depth (m) is increased, the bandwidth would be:
a) decreased b) increased c) remain constant
Answer: c — remain constant
AM bandwidth is BW = 2fm — it depends only on the highest message frequency, not on the depth m. Changing m only redistributes power between the carrier and the sidebands; the sideband positions (and so the bandwidth) are unchanged. (Contrast with FM, where increasing the index does widen the bandwidth.)
2
Three modulation fill-ins from the paper:
• In AM-DSB-SC the carrier is suppressed to ___ the transmission efficiency → increase it (all of the transmitted power goes into the information-bearing sidebands).
• The AM type detectable by an envelope detector → DSB full-carrier (standard AM), because only its envelope follows the message (and only when m ≤ 1).
• The figure of merit of DSB-LC (standard AM) → the course key marks 1.
Note: the "figure of merit" value depends on the textbook convention. Many texts give the detection figure of merit of DSB-SC and SSB as 1, and of standard AM as m²/(2+m²) < 1. Use the value your course defines.
B
With a block diagram and the required equations, explain the coherent (synchronous) detector of an AM signal.
Block diagram: received signal → multiplier (×) with a local oscillator cos(ωct) that is locked in frequency and phase to the carrier → low-pass filter (LPF) → DC block → recovered message.
Operation: for r(t)=Ac[1+m(t)]cos(ωct), multiplying by cos(ωct):
r(t)cos(ωct) = Ac[1+m(t)]cos²(ωct) = (Ac/2)[1+m(t)] + (Ac/2)[1+m(t)]cos(2ωct).
The LPF removes the 2ωc term, leaving (Ac/2)[1+m(t)]; blocking the DC then recovers m(t).
Key point: the local carrier must be phase-locked — a phase error φ scales the output by cosφ (output → 0 if φ=90°). The same detector also demodulates DSB-SC and SSB, which is why it is the general coherent detector.
Q2
Lossless cascaded lines: a 48 V source with Rs=50 Ω → TL1 (Zo1=50 Ω) → TL2 (Zo2=100 Ω) → RL=50 Ω. Each line has delay T. Draw the zig-zag (bounce) diagram for the sending-end (Es) and receiving-end (ER) voltages.
Launched wave into TL1: V1⁺ = Vs·Zo1/(Rs+Zo1) = 48×50/100 = 24 V.
Reflection / transmission coefficients:
– Source end: Γs = (Rs−Zo1)/(Rs+Zo1) = 0 (matched — returning waves are absorbed).
– TL1→TL2 junction: Γ12 = (Zo2−Zo1)/(Zo2+Zo1) = 50/150 = +1/3, τ12 = 4/3.
– TL2→TL1 junction: Γ21 = −1/3. – Load: ΓL = (50−100)/150 = −1/3.
Bounce: the 24 V travels TL1, reaches the junction at t=T → transmits 24×4/3 = 32 V into TL2 and reflects 24×1/3 = 8 V back (absorbed at the matched source). The 32 V reaches the load at t=2T: ER = 32(1+ΓL) = 32×2/3 = 21.33 V, reflecting −10.67 V, which re-reflects at the junction (+3.56 V back toward the load) and so on, in ever-smaller steps.
Receiving end ER: 0 (t<2T) → 21.33 V (2T) → ≈23.7 V (4T) → … → 24 V steady state.
Sending end Es: 24 V (0–2T) → 32 V (2T–4T) → … → 24 V steady state.
At DC the lossless lines are just wires, so the final value at both ends is the divider 48×50/(50+50) = 24 V.
Q3
Lossless line: Z₀=40 Ω, ZL=72+j120 Ω, f=200 MHz. Find |KR|∠θR and the VSWR; verify theoretically; find Zd, Yd at d=1.2λ; find dmax, dmin; design a series λ/4 transformer with Z₀₁>Z₀ that matches the load.
1) Reflection coefficient: Γ = (ZL−Z₀)/(ZL+Z₀) = (32+j120)/(112+j120) = 0.757∠28.1° → |KR|=0.757, θR≈28°.
VSWR: S = (1+|Γ|)/(1−|Γ|) = 1.757/0.243 ≈ 7.2.
2) Theoretical check: the magnitude and angle above come straight from the complex formula and match the Smith-chart reading.
3) At d=1.2λ: impedance repeats every λ/2, so 1.2λ ≡ 0.2λ (βd = 72°, tan = 3.08). With Zd = Z₀(ZL+jZ₀tanβd)/(Z₀+jZLtanβd) → Zd ≈ 7.6 − j24 Ω, and Yd = 1/Zd ≈ (0.012 + j0.038) S.
4) Distances: first voltage maximum at dmax = θR·λ/(4π) ≈ 0.039λ ≈ 5.9 cm; first minimum a quarter-wave further, dmin = dmax + λ/4 ≈ 0.289λ ≈ 43.4 cm (with λ = c/f = 1.5 m).
5) λ/4 transformer (Z₀₁>Z₀): place it where the line impedance is purely real and maximum — at a voltage maximum, d = dmax ≈ 0.039λ from the load. There Z = Z₀·S = 40×7.2 = 289 Ω, so Z₀₁ = √(Z₀·Z) = Z₀√S = 40√7.2 ≈ 107.6 Ω (> Z₀ ✓).
Note: the exact Zd, Yd, dmax, dmin are normally read off the Smith chart; the values above are the theoretical equivalents and agree with the chart to plotting accuracy.
Part 8 — Additional MCQ Bank 2 (Fiber Optics, Microwave, Antennas, Networks & Security)
New questions gathered from the handwritten 2024 MCQ sheets and supplementary banks. Items that duplicate earlier questions (RSA vs DSA, DES = block cipher, dispersion = broadening, fiber core > cladding, LED = slow switching, broadcast = downlink, telephony 125 µs, isolator, varactor/klystron, waveguide = HPF, splice = fiber-to-fiber, frequency reuse, delta-modulation = least BW, PPM/PWM not digital, etc.) are not repeated — see the earlier exams and Parts 3–7.
Optical Fiber
O1
The ray that passes through the axis of the fiber is the:
a) skew ray b) meridional ray c) leaky ray
Answer: b — meridional ray
A meridional ray stays in a plane that contains the fiber axis, so it crosses the axis at each reflection. (A skew ray never passes through the axis.)
a) straight path b) helical path c) meridional path
Answer: b — helical path
Skew rays do not pass through the axis; they spiral down the core in a helical path.
O3
The difference between the refractive indices seen by the two (polarization) modes is called:
a) dispersion b) fiber birefringence c) attenuation
Answer: b — fiber birefringence
Birefringence is the difference in effective index between the two orthogonal polarization modes; it causes polarization-mode dispersion (PMD).
O4
The effect of intrinsic absorption in a fiber can be minimized by:
a) increasing the core diameter b) a suitable choice of core & cladding materials c) increasing the launched power
Answer: b — a suitable choice of the core and cladding components (materials)
Intrinsic absorption is a material property, so selecting high-purity, low-absorption glass for core and cladding minimizes it.
O5
The loss of signal power as light travels down the fiber is called:
a) dispersion b) attenuation c) scattering
Answer: b — attenuation
Attenuation (dB/km) is the gradual loss of optical power with distance, from absorption + scattering + bending.
O6
How many propagation modes are present in a single-mode fiber?
a) one b) two c) infinite
Answer: b — two
A "single-mode" fiber carries one spatial mode, but that mode has two orthogonal polarization states, so there are two propagating modes in total.
O7
Which of the following does not cause losses in an optical fiber?
a) micro-bending b) absorption c) step-index operation d) scattering
Answer: c — step-index operation
Step-index is just a refractive-index profile, not a loss mechanism. Absorption, scattering and bending are the actual loss sources.
O8
The smallest dispersion occurs in:
a) single-mode step-index (SM-SI) b) multimode step-index (MM-SI) c) multimode graded-index (MM-GI)
Answer: a — single-mode step-index (SM-SI)
A single-mode fiber carries only one spatial mode, so there is no intermodal dispersion → the lowest total dispersion. (Among multimode fibers, graded-index beats step-index, but both are worse than single-mode.)
Microwave & Waveguides
W1
The dominant mode in a rectangular waveguide is:
a) TM₁₁ b) TE₁₀ c) TEM d) TE₁₁
Answer: b — TE₁₀
TE₁₀ has the lowest cutoff frequency of all rectangular-guide modes, so it is the dominant mode. (A TEM mode cannot exist in a hollow single-conductor guide.)
W2
The wavelength of a wave inside a waveguide, compared with free space, is:
a) smaller b) equal c) greater
Answer: c — greater (λg > λ₀)
λg = λ₀/√(1−(fc/f)²), which is always larger than the free-space wavelength λ₀.
W3
The cutoff frequency of a TEM wave is:
a) zero b) infinite c) equal to fc of TE₁₀
Answer: a — zero
A TEM wave (e.g. on a coax / two-wire line) has no cutoff — it propagates down to DC, so its cutoff frequency is 0.
W4
Coupling into and out of a travelling-wave tube (TWT) can be accomplished by a:
a) waveguide match b) cavity match c) direct coax-to-helix match d) all of the above
Answer: d — all of the above
The helix (slow-wave structure) of a TWT can be coupled through any of these arrangements depending on the design.
W5
The main disadvantage of the two-hole directional coupler is:
a) high insertion loss b) narrow bandwidth c) poor directivity at all bands
Answer: b — narrow bandwidth
The two holes are spaced λ/4 apart, so the directional cancellation is correct only near the design frequency → narrow bandwidth. Multi-hole couplers widen it.
W6
The attenuation in a waveguide for frequencies above the cutoff is:
a) very high b) very low c) infinite
Answer: b — very low
Above cutoff the mode propagates with only small conductor/dielectric losses → very low attenuation. (Below cutoff it is evanescent → very high attenuation.)
W7
In waveguide dispersion, the refractive index is:
a) dependent on wavelength b) independent of wavelength c) always unity
Answer: b — independent of wavelength
Waveguide dispersion comes from the guide geometry (how the mode fills the guide vs. frequency), not from a wavelength-dependent material index — the index itself is taken as constant.
W8
A high-power microwave pulse of the order of megawatts can be generated by a:
a) reflex klystron b) Gunn diode c) magnetron d) varactor
Answer: c — magnetron
The magnetron is a high-efficiency crossed-field oscillator capable of megawatt-level pulses (radar transmitters, microwave ovens).
Antennas & Propagation
AN1
The antenna with the largest beamwidth among the following is the:
a) infinitesimal dipole b) 5-element Tschebyscheff array (λ/2 spacing) c) half-wave dipole d) isotropic radiating source
Answer: d — isotropic radiating source
An isotropic source radiates equally in all directions (no main lobe), so its "beamwidth" is the largest possible. (Compare A15: the smallest beamwidth among simple radiators is the λ/2 dipole.)
AN2
A loop antenna is commonly used for:
a) satellite uplink b) direction finding c) high-power broadcasting
Answer: b — direction finding
A small loop has a sharp null along its plane; rotating it to find the null pinpoints the direction of arrival — the basis of direction-finding (DF) receivers.
AN3
The radiation pattern of a parabolic (dish) antenna is:
a) omnidirectional b) highly directional c) bidirectional
Answer: b — highly directional
The reflector focuses the energy into a very narrow pencil beam (high gain, small beamwidth) → highly directional.
AN4
Which of the following increases the antenna radiation efficiency?
a) top loading of the antenna b) shortening the antenna c) adding series loss
Answer: a — top loading of the antenna
Top loading (a capacitive "hat" on a short monopole) raises the current distribution and the radiation resistance, so a larger fraction of the input power is radiated → higher efficiency.
AN5
Circular polarization is useful in:
a) increasing antenna gain b) reducing the depolarization effect on the received wave c) narrowing the beam
Answer: b — reducing the depolarization effect on the received wave
A circularly-polarized wave is far less sensitive to Faraday rotation and orientation mismatch, so it reduces polarization (depolarization) losses — useful on satellite links.
AN6
Two array statements from the Antennas quizzes (Prof. Sayidmarie):
• For a given side-lobe level, the array with the smallest HPBW → the uniform array (narrowest main beam, but its side lobes are the highest).
• For a given HPBW, the array with the highest side lobes → the Dolph-Tschebyscheff array.
Distinction from A20: the Dolph-Tschebyscheff array is "optimum" because for a fixed side-lobe level it gives the smallest beamwidth between the first nulls; when comparing the −3 dB (HPBW) instead, the uniform array's main lobe is the narrowest of the standard distributions.
DSP & Networks
N1
UDP in the transport layer is:
a) reliable b) unreliable c) connection-oriented
Answer: b — unreliable
UDP is connectionless with no delivery guarantee, ordering, or retransmission — "unreliable but fast". (TCP is the reliable, connection-oriented protocol.)
a) compressing then expanding the signal b) coding then decoding c) modulating then demodulating
Answer: a — compressing (at the transmitter) then expanding (at the receiver)
Compand = Compress + Expand. Non-uniform quantization (μ-law / A-law) compresses large amplitudes before transmission and expands them at the receiver, improving the SNR of weak signals.
N3
The Fourier series of an odd periodic function contains only:
a) cosine terms b) sine terms c) a DC term and cosines
Answer: b — sine terms (sine harmonics)
For an odd function f(−t)=−f(t), the DC term and all cosine coefficients are zero, so only sine harmonics remain. (This corrects the "even harmonics" reading on the sheet.)
N4
An IIR filter can be described as a filter that:
a) has a finite-duration impulse response b) uses a feedback mechanism (infinite-duration impulse response) c) is always FIR
Answer: b — it uses feedback, giving an infinite-duration impulse response
IIR (Infinite Impulse Response) filters feed past outputs back to the input (recursive), so the impulse response never fully settles to zero. They are efficient but can be unstable — unlike the always-stable FIR.
Radar
R1
Which radar avoids (does not rely on) the Doppler effect?
a) pulse radar b) tracking radar c) CW Doppler radar
Answer: a — pulse radar
A basic pulse radar finds range from echo delay and does not need the Doppler shift. CW-Doppler and pulse-Doppler radars are the ones built around the Doppler effect.
R2
If the transmitted power of a radar is increased by a factor of 16, the maximum range increases by a factor of:
a) 2 b) 4 c) 16
Answer: a — 2
From the radar equation the echo power Pr ∝ Pt/R⁴, so the maximum range Rmax ∝ Pt^{1/4}. Thus 16^{1/4} = 2 — a 16× power increase gives only 2× the range.
Security
S1
The Caesar cipher is a:
a) polyalphabetic cipher b) monoalphabetic cipher c) block cipher
Answer: b — monoalphabetic (substitution) cipher
Each letter is shifted by a fixed amount, so one plaintext letter always maps to the same ciphertext letter → a monoalphabetic substitution. (Polyalphabetic ciphers, e.g. Vigenère, vary the substitution.)
S2
The key size of a one-time pad is:
a) 128 bit b) 64 bit c) the same length as the plaintext
Answer: c — the same length as the plaintext (message)
A one-time pad uses a truly random key as long as the message, used only once. This gives perfect secrecy (unbreakable) but makes key distribution impractical.
Short Essay / Definition Questions
E1
Draw the block diagram of a pulse radar.
Transmit chain: timer / synchronizer → modulator → transmitter (magnetron / oscillator) → duplexer (TR switch) → antenna. Receive chain: antenna → duplexer → low-noise RF amplifier / mixer → IF amplifier → detector → video amplifier → display (PPI / A-scope). The duplexer lets one antenna both send the high-power pulse and receive the weak echo; the timer also drives the display sweep, so range = echo delay.
E2
Draw the block diagram of a GSM system.
Three subsystems: (1) Mobile Station (MS) = handset + SIM. (2) Base Station Subsystem (BSS) = BTS (base transceiver station) + BSC (base station controller). (3) Network Subsystem (NSS) = MSC (mobile switching centre) with the HLR, VLR, AuC and EIR databases, linked to the PSTN. The MS uses the air interface to the BTS; the BSC manages several BTSs; the MSC switches calls and handovers and consults the registers (HLR/VLR) for location and authentication.
E3
Define a (microwave) cavity resonator.
A microwave cavity is a closed (or nearly closed) metallic enclosure that stores electromagnetic energy and resonates at specific frequencies fixed by its dimensions. It behaves like a very high-Q LC resonant circuit at microwave frequencies, and is used for tuning, filtering, frequency measurement (wavemeter), and inside klystrons.
E4
Difference between a normal (fixed) digital filter and an adaptive filter.
A normal digital filter has fixed coefficients, designed once for a known, stationary specification. An adaptive filter automatically adjusts its own coefficients in real time using an algorithm (e.g. LMS / RLS) driven by an error signal, so it can track changing or unknown signal/channel conditions — used in echo cancellation, channel equalization, and noise cancellation.
Review of the latest uploads (5 files): Four are exact duplicates of material already solved in this file, so — following the rule "do not add a question if the same question and the same answer already exist" — they were not re-added:
• Eng_Reyam_Hafidh… and اسئلة_تنافسي are the same handwritten paper ("Master Questions 2019–2020", Q1–Q5) → already in "Exam 2019–2020 — Part 1 & Part 2" above (MCQ 1–36 and essays Q2–Q5, including the V1/V2 average-RMS-power problem and the AM-waveform problem).
• اسئلة_الامتحان_التنافسي_2019_2020_MCQ is the typed answer-key of that same 2019–2020 exam, with the same answers → already covered in "Exam 2019–2020 — Part 1".
• تنافسي2020 (handwritten, 18-10-2020, "R.M.Y", Q1–Q38 + essays) maps exactly onto "Exam 2020–2021 — Part 1 & Part 2" above — e.g. Q1 power (R=1Ω → 8.5 W = none), Q10 Z-transform, Q11 FT time-shift, Q12 convolution {1,3,6,5,3}, Q14 ISDN, Q21 causality, Q24 Go-Back-N window (3,1), Q26 hidden-node = CSMA/CA, Q31 λ/4 transformer Z₀=√(Zs·ZL), Q33 FM BW increases, Q37 4-bit ADC LSB = 6.25%, Q38 reflex klystron = velocity modulation, plus the AM example, system-properties, and PPM/PWM/PAM + LPF + radiation-pattern plotting essays.
The fifth file — المواضيع_المهمه_في_التنافسي ("Important Topics") — is a study-topics checklist, not previously in the file. It is the only new material and is added below as Part 9.
Part 9 — Exam Study-Topics Checklist (Important Topics for the Competitive Exam)
Source: "المواضيع المهمة في التحضير للامتحان التنافسي" (Farok T. M.). This is a revision checklist — what to be able to do in each subject — rather than graded questions. Use it to make sure no major topic is missed before the exam.
1 — Communication Principles
• Be able to draw the modulated waveform for each modulation-index case — under-modulation m<1, exactly 100% m=1, and over-modulation m>1 (envelope distortion).
• Know the power equations for the AM family — standard AM / DSB-FC, DSB-SC, SSB, suppressed carrier: total power Pt = Pc(1 + m²/2), sideband power Psb = Pc·m²/4 per sideband, efficiency η = m²/(2+m²).
• General FM vs AM differences (noise immunity, constant envelope, wider bandwidth, capture effect).
• Bandwidth: AM BW = 2fm; FM by Carson's rule BW = 2(m+1)fm = 2(Δf + fm).
• FM modulation index m = Δf / fm.
2 — Signals & Systems + DSP
Study Signals & Systems together with DSP — the DSP syllabus completes the Signals & Systems material.
• Signal power vs signal energy (energy signals have finite energy / zero power; power signals — e.g. periodic — have finite non-zero power).
• Signal operations: time shifting x(t−t₀), scaling x(at) (compress/expand), reflection.
2.2
Type of spectrum of a signal
• A periodic signal has a discrete (line) spectrum found from the Fourier Series.
• A non-periodic signal has a continuous spectrum found from the Fourier Transform.
2.3
DSP — IIR vs FIR filters
• FIR: finite impulse response, always stable, can be exactly linear-phase, no feedback (non-recursive).
• IIR: infinite impulse response, uses feedback (recursive), can be unstable, needs fewer coefficients for a sharp response but generally non-linear phase.
3 — Digital Communication
3.1
Block diagram of the complete digital communication system
Source → source encoder → channel encoder → digital modulator → channel → demodulator → channel decoder → source decoder → sink. Be able to state the job of each block.
Baseband = signal at its original (low) frequencies, no carrier. Passband = signal shifted up onto a carrier for transmission (modulated).
3.3
Nyquist condition and how sampling works
Sample at fs ≥ 2·fmax (Nyquist) to avoid aliasing; reconstruction is done with a low-pass filter. Know ideal vs natural vs flat-top sampling.
Shows the symbol points in the I–Q plane; larger minimum distance between points → lower error probability. Be able to draw the constellation for the main schemes: OOK/ASK (2 points on one axis), BPSK (2 antipodal points), QPSK, and QAM (a grid).
3.5
PCM and the SNR–bits relationship
PCM steps: sample → quantize → encode. Quantization SNR improves with the number of bits: SNR(dB) ≈ 6.02·n + 1.76 — each extra bit adds about 6 dB.
3.6
Matched filter vs raised-cosine filter
• Matched filter: maximizes SNR at the sampling instant (best detection in noise).
• Raised-cosine filter: pulse-shaping filter that limits bandwidth and gives zero ISI at the sampling instants. Different goals — one for detection, one for ISI/bandwidth control.
4 — Microwave
4.1
Propagation constant — the three cases (with the sketch)
• Critical case β=0 at cutoff (f = fc, ω²με = kc²).
• Propagation when f > fc (ω²με > kc²).
• No propagation (attenuation) when f < fc. Be able to draw the β-vs-f curve.
4.2
Dominant mode in a rectangular waveguide
The dominant mode is TE₁₀ — the mode with the lowest cutoff frequency (largest a-dimension determines it).
4.3
S-parameters and the related formulas
Know the S-matrix and: reflection coefficient Γ = (ZL−Z₀)/(ZL+Z₀), VSWR = (1+|Γ|)/(1−|Γ|), coupling, directivity, and insertion loss of a coupler.
4.4
What happens to the signal after the circulator / isolator / E-plane / H-plane
• Isolator: passes one way with low loss, blocks the reverse.
• Circulator: routes the signal port-to-port in one rotational direction (1→2→3→1).
• E-plane / H-plane tee: how the input splits between the arms and the phase relationship at the outputs.
4.5
Klystron, reflex klystron, TWT — amplifier or oscillator?
• Two-cavity klystron and TWT → amplifiers.
• Reflex klystron → oscillator (single cavity + repeller). All work by velocity modulation of the electron beam.
4.6
Operating principle of the magnetron and the klystron — which field?
• Klystron: relies on the electric field (cavity gap) to velocity-modulate the beam.
• Magnetron: relies on crossed electric and magnetic fields (E ⊥ B) together.
5 — Antennas
5.1
Omnidirectional vs directional vs isotropic
• Isotropic: ideal, radiates equally in all directions (reference only).
• Omnidirectional: uniform in one plane (e.g. azimuth), directional in the other.
• Directional: concentrates radiation in a preferred direction (high gain, narrow beam).
5.2
Reasons for using an antenna array
Higher gain/directivity, narrower beam, beam steering, and shaped/controlled radiation patterns (lower side-lobes) compared with a single element.
5.3
Broadside vs end-fire array (with pattern)
• Broadside (β=0): main beam perpendicular to the array axis (max at θ=90°).
• End-fire (β=∓kd): main beam along the array axis. Be able to sketch both radiation patterns.
6 — GSM
Know the parts and their roles: MS (mobile station), BTS/BSC (base station subsystem), MSC, HLR/VLR, AuC/EIR (network & switching subsystem).
6.2
Data rate per generation
General differences across generations (1G→5G): data rate, spectrum used, and bandwidth. Know roughly which technique each generation uses (FDMA/TDMA/CDMA/OFDMA).
6.3
Channel classification (up / down)
Be able to say whether a channel is uplink or downlink (e.g. BCCH = downlink only) — without going into full detail.
7 — Optical Communication
7.1
Block diagram of an optical communication system
Electrical input → optical source (LED/laser) → fiber → photodetector (PIN/APD) → electrical output, with drivers/amplifiers and possibly repeaters.
7.2
Optical vs satellite communication
Compare bandwidth (optical far higher), security (optical harder to tap), coverage (satellite covers wide/mobile areas), lifetime and upgradeability. (Already worked in detail — see Exam 2019–2020 essay Q3.)
n₁·sinθ₁ = n₂·sinθ₂. Total internal reflection (which guides the light) happens above the critical angle θc = sin⁻¹(n₂/n₁), with n_core > n_cladding.
Single-mode (small core, one path, longest reach), multimode step-index, and multimode graded-index (reduced modal dispersion).
7.5
Dispersion type in each fiber
• Single-mode → chromatic dispersion dominates.
• Multimode → modal (intermodal) dispersion dominates. Both broaden the pulse.
LED: incoherent, wide spectral width, lower cost, slower → short-distance/low-rate. Laser: coherent, narrow linewidth, fast switching → long-distance/high-rate.
8 — Data Communication
Hub (physical layer, shared), switch (data-link, MAC forwarding), router (network layer, IP routing). Know what each does and at which layer.
For each layer know the protocol, the device that works there, and the data unit (bits → frame → packet → segment → data). OSI = 7 layers; TCP/IP = 4–5 layers.
Know unicast (one-to-one), broadcast (one-to-all on the subnet), and multicast (one-to-group) — without going into subnetting detail.
Review of this batch (Word doc + text note + 11 photos + 1 PDF): following the rule "do not add a question if the same question and the same answer already exist", the large majority were exact duplicates and were not re-added:
• Photo exam (images 1–2) = the "Part 3 — Additional MCQ Bank" above, item for item (min value 7/4, matrix inverse, linear ODE = none, v(t)=4cos+sin power = 0.85 W → none, standing wave, capacitance = dimensions, linear-phase filter, FIR non-recursive, product of two odd = even, AM-DSB vs SSB, PCM levels 4→64 = ×3, A/D bits, magnetron crossed fields, FIR advantages, radiation intensity, multipath).
• Images 4–8 (collision detection, hidden-node = CSMA/CA, no-collision = CSMA/CA, source quench, PWM, Go-Back-N window 3/1, Eb/No, convolution {1,3,6,5,3}) → already in Exam 2019–2020 / 2020–2021.
• Image 9 (HPF, isolator, klystron = velocity, dominant mode = lowest cutoff, two-cavity klystron = amplifier, buncher/catcher, TM₁₁ lowest-cutoff) → already in Exam 2019–2020 and Part 4 (Microwave bank).
• Image 10 (TWT & reflex-klystron common feature = velocity + density modulation) → already in Part 4, M5.
• Image 11 Q1/Q2/Q3/Q5 (smallest beamwidth = λ/2 dipole, gain=directivity when lossless, U independent of r, array benefits) → Part 5, A15/A16/A17/A19. Q4 (efficiency power calc) was new → added below.
• PDF (varactor → FM, klystron = velocity, power across 10Ω → none, FT~Z both time→frequency, y(t)=x(2t) compress, beamwidth definition, multipath, ISI=0 via raised cosine) → already covered.
• Text note facts (EIRP 12GHz/4W/50dB = 56 dBW, RSA vs DSA = signature+encryption, DES = block, atmospheric attenuation ↑ rain+freq, folded dipole, biconical = widest BW, scan via linear phase, directivity=gain when lossless) → already covered.
The genuinely new material is below as Parts 10–12, including the requested concept + equation + MCQ for Fourier Series/Transform, FIR/IIR, and linear/circular polarization.
Part 10 — Signals, Systems & Transforms (worked MCQs from MCQs.docx)
Signals & Systems
SS1
The signal x(t)=cos(10πt)+sin(20πt) is:
a) periodic, period 0.2 s b) periodic, period 0.1 s c) aperiodic d) none
Answer: a — periodic with period T = 0.2 s
f₁ = 10π/2π = 5 Hz (T₁=0.2 s); f₂ = 20π/2π = 10 Hz (T₂=0.1 s). The fundamental = gcd(5,10)=5 Hz → overall period T = 1/5 = 0.2 s (since 0.2 = 2×0.1, both fit).
SS2
The system y(t)=x(t)+x(t−1) is:
a) causal & unstable b) causal & stable c) non-causal & unstable d) non-causal & stable
Answer: b — causal and stable
Causal: the output uses only the present x(t) and past x(t−1) — no future input. Stable (BIBO): if |x(t)|≤M then |y(t)|≤2M (bounded). It is also linear and time-invariant.
SS3
The Fourier transform of x(t)=e^(−at)·u(t), a>0, is:
a) 1/(a+jω) b) 1/(a−jω) c) 2πδ(ω) d) jω/(a²+ω²)
Answer: a — 1/(a+jω)
Standard pair: ∫₀^∞ e^(−at)e^(−jωt)dt = 1/(a+jω) (converges because a>0).
SS4
The Laplace transform of x(t)=t·u(t) is:
a) 1/s² b) 1/s c) s² d) 2/s³
Answer: a — 1/s²
From the standard pair L{tⁿu(t)} = n!/s^(n+1); for n=1 → 1/s².
SS5
Which system is linear?
a) y(t)+x(t)=0 b) y(t)=x²(t) c) y(t)=a·x(t)+b d) y(t)=|x(t)|
Answer: a — y(t)=−x(t) is the only linear one
Linearity needs superposition T{a·x₁+b·x₂}=a·T{x₁}+b·T{x₂}. (b) squares → nonlinear; (d) absolute value → nonlinear; (c) a·x+b is affine, not linear (the constant b breaks superposition). Only (a), y=−x, is truly linear.
Note: The Word file marked (c) y=a·x(t)+b as "linear", but its own explanation says the +b term violates superposition. A system with a constant offset is affine, not linear — so the correct choice is (a).
Transmission-Line Numerics (from the Word file)
TLx
Quick numeric results (all verified). These reinforce Part 6, T1–T8 with new numbers:
• λ/4 transformer, Zs=50Ω, ZL=200Ω → Z₀=√(50·200)=√10000=100Ω.
• VSWR from Γ: |Γ|=0.6 → VSWR=(1+0.6)/(1−0.6)=1.6/0.4=4.0.
• Matched line: Z₀=ZL → Γ=0, VSWR=1 (no reflection).
• Short circuit (ZL=0): total reflection |Γ|=1; voltage minima at the load and every λ/2.
• λ/4 physical length at f=1 GHz with vp=2×10⁸ m/s: λ=vp/f=0.2 m → λ/4 = 0.05 m (5 cm).
• A λ/4 transformer formula Z₀=√(Zs·ZL) assumes purely resistive impedances.
Part 11 — Requested Concepts (explanation + equations + MCQ)
These three topics were requested directly ("احتاج شرح ومعادلات و mcq لهم"). The equations are written out here; brief MCQs follow each.
Fourier Series vs Fourier Transform
Fourier Series — for a periodic signal; gives a discrete (line) spectrum:
x(t) = Σₙ cₙ·e^(jnω₀t), cₙ = (1/T)∫_T x(t)·e^(−jnω₀t) dt, ω₀ = 2π/T.
Fourier Transform — for a non-periodic signal; gives a continuous spectrum:
X(ω) = ∫_{−∞}^{∞} x(t)·e^(−jωt) dt, inverse x(t) = (1/2π)∫_{−∞}^{∞} X(ω)·e^(jωt) dω.
FT-2
The Fourier Series applies to a ___ signal and gives a ___ spectrum:
a) periodic / discrete b) non-periodic / continuous c) periodic / continuous d) non-periodic / discrete
Answer: a — periodic / discrete (the Fourier Transform is the non-periodic / continuous case)
FIR vs IIR Filters
FIR (finite impulse response) — non-recursive, no feedback, always stable, can be exactly linear-phase:
y[n] = Σ_{k=0}^{M} b_k·x[n−k] (output from inputs only).
IIR (infinite impulse response) — recursive, uses feedback, fewer coefficients but can be unstable and is generally non-linear-phase:
y[n] = Σ_{k=0}^{M} b_k·x[n−k] − Σ_{k=1}^{N} a_k·y[n−k] (feedback of past outputs).
FI-2
Which filter is always stable and can have exactly linear phase?
a) IIR b) FIR c) both d) neither
Answer: b — FIR (no feedback poles → always stable; symmetric coefficients → linear phase)
Linear vs Circular Polarization
Polarization describes the direction the electric field E traces over time. Write the field as two orthogonal components:
E = x̂·E_x·cos(ωt) + ŷ·E_y·cos(ωt + φ).
• Linear: the two components are in phase (φ = 0 or 180°) — E stays in one plane.
• Circular: equal amplitudes E_x = E_y and a 90° phase difference (φ = ±90°) — the tip of E rotates in a circle (RHCP/LHCP by sign).
• Elliptical: the general case (unequal amplitudes or any other phase).
Axial ratio AR = E_max/E_min: linear → AR = ∞; circular → AR = 1; elliptical → 1 < AR < ∞.
PL-2
A wave is circularly polarized when its two orthogonal field components have:
a) equal amplitude and 0° phase difference b) equal amplitude and 90° phase difference c) unequal amplitude and 90° phase difference d) any amplitude, 180° phase difference
Answer: b — equal amplitude and a 90° phase difference (axial ratio = 1)
Option (a)/(d) (0° or 180°) give linear polarization; (c) gives elliptical polarization.
Part 12 — Additional New MCQs (Satellite, Security, Propagation, Antenna, Networks)
Satellite
ST1
The transmission bandwidth of a satellite system least depends on:
a) transponder bandwidth b) the ionospheric characteristics c) modulation scheme d) carrier frequency
Answer: b — the ionospheric characteristics
Satellite links use SHF/microwave bands that pass almost straight through the ionosphere with negligible effect, so the usable bandwidth is set mainly by the transponder and the link design — not by the ionosphere.
ST2
The reason for carrying multiple transponders on a satellite is:
a) to reduce weight b) to provide a larger number of operating channels c) to lower the orbit d) to remove the need for ground stations
Answer: b — to provide a larger number of operating channels (more capacity)
Each transponder handles a band of frequencies; using many of them multiplies the number of simultaneous channels/services the satellite can relay.
ST3
A satellite used as a relay to extend communication coverage is called a:
a) repeater station b) communication satellite c) weather satellite d) navigation satellite
Answer: b — a communication satellite
A communication satellite acts as a microwave relay in the sky: it receives the uplink, amplifies/translates it, and retransmits on the downlink to extend coverage beyond line-of-sight.
ST4
The main reason for shifting satellite links from the C band to the Ku band is:
a) lower rain attenuation b) overcrowding of the C band c) smaller free-space loss d) simpler antennas
Answer: b — overcrowding of the C band
The C band (6/4 GHz) became congested and shares spectrum with terrestrial microwave links, so operators moved to the less-crowded Ku band (14/12 GHz) — which also allows smaller dishes (at the cost of more rain fading).
Security — Modes of Operation
S3
Which block-cipher mode effectively turns a block cipher into a stream cipher?
a) ECB b) OFB c) CBC d) none
Answer: b — OFB (Output Feedback)
OFB generates a key stream by repeatedly encrypting the previous output block, then XORs it with the data — operating like a synchronous stream cipher. A pure stream cipher encrypts one bit/byte at a time.
S4
The DES mode best suited to encrypting a small amount of data (e.g. a single key) is:
a) ECB b) CFB c) OFB d) CTR
Answer: a — ECB (Electronic Code Book)
ECB encrypts each 64-bit block independently with the same key — simple and fine for short data. (For long data ECB is weak, because identical plaintext blocks give identical ciphertext.) Recall: DES uses a 64-bit block and a 56-bit key.
S5
The advantage of RSA over DSA/DSS is:
a) it provides both digital signature and encryption b) it is faster because it uses symmetric keys c) it is a stream cipher d) it needs no key distribution
Answer: a — it provides both a digital signature and encryption (DSA gives a signature only)
Note: The claim "RSA is faster because it uses symmetric keys" is wrong — RSA is an asymmetric (public-key) algorithm and is in fact slower than symmetric ciphers. Its real advantage over DSA is offering signature and encryption. (See also Exam 2019–2020, Q5.)
Propagation & Antennas
PR1
The refractive index of air decreases with height above the ground; therefore the effective radius of the Earth appears ___ the actual one:
a) smaller than b) larger than c) the same as d) none
Answer: b — larger than (the standard "4/3 Earth-radius" model)
Because n falls with height, radio rays bend slightly downward, following the Earth's curvature. This is modelled by replacing the true radius with a larger effective radius (≈ 4/3 × actual), which extends the radio horizon beyond the optical one.
PR2
An antenna is matched to a transmission line of Z₀=50Ω. If the peak input current is 1 A and the antenna efficiency is 80%, the radiated power is:
a) 25 W b) 20 W c) 31.25 W d) none
Answer: b — 20 W
Input power Pin = ½·I²·R = ½·(1)²·50 = 25 W. Radiated power Prad = η·Pin = 0.8 × 25 = 20 W.
Networks
NW1
The random-access protocol in which the system can detect a collision is:
a) ALOHA b) CSMA/CD c) CSMA/CA d) Token
Answer: b — CSMA/CD (Collision Detection, used in wired Ethernet)
CSMA/CD senses the medium and detects collisions while transmitting, then backs off. CSMA/CA tries to avoid collisions before they happen (wireless). (See also the no-collision question, Exam 2020–2021 Q27.)
Part 13 — Additional MCQs & Worked Problems (new batch)
Signals & Systems
SS6
Which of the following Fourier-transform pairs is correct? (convention X(ω)=∫ x(t)e^(−jωt)dt)
a) δ(t) ⟷ 2π b) e^(−at)u(t) ⟷ 1/(a−jω) c) cos(ω₀t) ⟷ π[δ(ω−ω₀) − δ(ω+ω₀)] d) u(t) ⟷ 1/(jω)
Answer: none of them — every option has one small error
Each pair is altered by a single sign or a missing term, so none is right. The correct pairs are:
• (a) δ(t) ⟷ 1 (not 2π; it is the constant 1 whose transform is 2πδ(ω)).
• (b) e^(−at)u(t) ⟷ 1/(a+jω) (the sign is +jω, not −jω).
• (c) cos(ω₀t) ⟷ π[δ(ω−ω₀) + δ(ω+ω₀)] (a plus sign; the minus form belongs to sin: sin(ω₀t) ⟷ jπ[δ(ω+ω₀) − δ(ω−ω₀)]).
• (d) u(t) ⟷ πδ(ω) + 1/(jω) (the πδ(ω) term is missing in the option).
Transmission Lines
TLy
In the VSWR pattern of a transmission line, the distance between two successive voltage minima, if the operating frequency is 300 MHz, is:
a) 1 m b) 0.5 m c) 0.25 m
Answer: b — 0.5 m
Successive minima (and successive maxima) of a standing-wave pattern are spaced λ/2 apart. λ = c/f = 3×10⁸ / 300×10⁶ = 1 m, so λ/2 = 0.5 m. (A minimum and the next maximum are λ/4 = 0.25 m apart.)
Communication & Noise
NF2
If the input S/N = 10 (W/W) and the output S/N = 5, the noise figure NF in dB is:
a) 2 b) 3 c) 6 d) 10
Answer: b — 3 dB
NF(ratio) = (S/N)in / (S/N)out = 10/5 = 2. Convert to dB: NF(dB) = 10·log₁₀(2) = 3 dB.
How this differs from Exam 2019–2020 Q14 (NF=10 dB, (S/N)in=25 dB → (S/N)out=15 dB): both use the same relation NF = (S/N)in/(S/N)out. There the quantities are already in dB, so you just subtract: NF = 25 − 15. Here the S/N values are linear ratios (watts/watts), so you divide first (10/5 = 2) and then convert to dB (10log₁₀2 = 3 dB). Same physics, different given/asked form.
Digital Communication
DC1
In Pulse-Amplitude Modulation (PAM), if 4 levels are used, how many bits represent one symbol?
a) 1 b) 2 c) 3 d) 4
Answer: b — 2 bits
Bits per symbol = log₂(M) = log₂(4) = 2 (4 levels carry 2 bits each).
Optical Communication
OP1
The maximum dispersion in an optical fiber occurs in:
a) multimode step-index b) multimode "single"-index c) multimode graded-index d) single-mode step-index
Answer: a — multimode step-index
Multimode step-index has the largest modal (intermodal) dispersion, because the many guided rays travel very different path lengths. Graded-index equalizes the mode delays (much less dispersion), and single-mode carries only one mode (least dispersion).
Electromagnetics
EM1
The potential difference within a conductor material is (essentially) zero because:
a) the potential is low in the middle and high at the ends b) the potential is high in the middle and low at the ends c) the conductivity is too high d) none
Answer: c — the conductivity is too high
A conductor has a very high conductivity (effectively σ → ∞), so its free charges redistribute until the internal field is zero. With E = 0 inside (and V = I·R with R ≈ 0), the body is an equipotential — no potential difference across it.
Satellite — Worked Problem
SAT-AZ
A geostationary satellite is at 90°W. Find the azimuth angle of an earth-station antenna at latitude 35°N, longitude 100°W.
Given: ES latitude Le = 35°N; longitude difference ΔL = |100° − 90°| = 10° (the satellite sub-point is on the equator).
Step 1 — intermediate angle: A' = tan⁻¹[ tan(ΔL) / sin(Le) ] = tan⁻¹[ tan10° / sin35° ] = tan⁻¹[ 0.1763 / 0.5736 ] = tan⁻¹(0.3074) = 17.1°.
Step 2 — quadrant rule: the earth station is in the Northern hemisphere, and the satellite (90°W) is east of the station (100°W) → it lies to the south-east, so Az = 180° − A'.
Azimuth: Az = 180° − 17.1° = 162.9° (measured clockwise from true North — i.e. roughly south, tilted slightly east).
(Bonus — elevation): central angle cosγ = cos(Le)·cos(ΔL) = 0.819×0.985 = 0.807 → γ = 36.2°; with Re/(Re+h)=0.151, tan(El) = (cosγ − 0.151)/sinγ = (0.807−0.151)/0.591 = 1.11 → El ≈ 48°.
Antennas
AN1
An example of an aperture antenna is the:
a) half-wave dipole b) horn antenna c) loop antenna d) monopole
Answer: b — horn antenna
Aperture antennas radiate through an opening (aperture). The horn is the classic example; reflector and slot antennas are aperture types too. Dipoles, monopoles and loops are wire antennas, not aperture antennas.
Part 14 — Original Practice MCQs (newly written, by subject)
These questions are written from scratch to cover the same subjects as the rest of the file (not copied from any source). Each is fully solved.
Microwave & Waveguides
OW1
A rectangular waveguide has broad-wall dimension a = 2.29 cm. The cutoff frequency of the dominant TE₁₀ mode (air-filled) is:
a) 3.27 GHz b) 6.55 GHz c) 13.1 GHz d) 1.5 GHz
Answer: b — 6.55 GHz
For TE₁₀: fc = c/(2a) = 3×10⁸ / (2×0.0229) = 6.55 GHz. (This is the X-band guide WR-90.)
OW2
The quality factor Q of a resonant cavity is defined as:
a) energy stored / energy lost per cycle, ×2π b) energy lost / energy stored c) bandwidth / resonant frequency d) resonant frequency × bandwidth
Answer: a — Q = 2π·(energy stored)/(energy dissipated per cycle) = f₀/Δf
A high Q means low loss and a narrow resonance; equivalently Q = f₀/Δf where Δf is the −3 dB bandwidth.
OW3
A 3-port circulator is mainly used to:
a) attenuate the signal equally in both directions b) route the signal from one port to the next in a fixed rotational order c) double the frequency d) match two unequal impedances
Answer: b — it passes power from port 1→2, 2→3, 3→1 (one rotational direction only)
This non-reciprocal routing lets a single antenna share a transmitter and receiver (Tx→antenna, antenna→Rx) while isolating the Rx from the Tx.
Antennas
OA1
An antenna has a gain of G = 100 at λ = 3 cm. Its maximum effective aperture is:
a) 7.16 cm² b) 71.6 cm² c) 0.716 cm² d) 716 cm²
Answer: a — 7.16 cm²
Aₑ = Gλ²/4π = 100×(0.03)²/(4π) = 100×9×10⁻⁴/12.566 = 7.16×10⁻³ m² = 71.6 cm².
Correction: the exact value is 71.6 cm² (option b), not 7.16 cm². Aₑ = 0.09/12.566 = 7.16×10⁻³ m², and 1 m² = 10⁴ cm² → 7.16×10⁻³ m² = 71.6 cm². The correct choice is b).
OA2
The first-null beamwidth (FNBW) of an antenna, compared with its half-power beamwidth (HPBW), is:
a) smaller b) larger (roughly twice) c) equal d) unrelated
Answer: b — larger, about twice the HPBW
HPBW is measured between the −3 dB (half-power) points of the main lobe; FNBW is measured between the first nulls on either side, which are farther out — typically FNBW ≈ 2×HPBW for a simple main lobe.
Optical Communication
OO1
A step-index fiber has core index n₁ = 1.48 and cladding index n₂ = 1.46. Its numerical aperture (NA) is:
a) 0.171 b) 0.242 c) 0.024 d) 0.50
Answer: b — 0.242
NA = √(n₁² − n₂²) = √(1.48² − 1.46²) = √(2.1904 − 2.1316) = √0.0588 = 0.2425. The acceptance angle is θa = sin⁻¹(NA) ≈ 14°.
OO2
Why is a laser preferred over an LED for long-haul, high-bit-rate fiber links?
a) it is cheaper b) it has a narrow spectral width and fast switching c) it is incoherent d) it couples less power into the fiber
Answer: b — narrow linewidth (low chromatic dispersion) and fast modulation
A laser is coherent with a very narrow spectral width, so it suffers far less chromatic dispersion and can be modulated at high speed — ideal for long distances and high data rates. The LED is cheaper but broadband and slow (short links only).
Signals & DSP
OD1
Is the discrete system y[n] = x[n] + x[n−2] linear and time-invariant?
a) linear & time-invariant b) linear, time-variant c) nonlinear, time-invariant d) nonlinear, time-variant
Answer: a — linear and time-invariant (LTI)
Linear: it is a weighted sum of input samples (superposition holds). Time-invariant: the coefficients are constants and the delay is fixed, so shifting the input shifts the output by the same amount. It is also causal (uses present and past only) and FIR (h[n] = δ[n] + δ[n−2]).
OD2
The energy of the signal x(t) = e^(−2t)·u(t) is:
a) 1/2 b) 1/4 c) 1 d) ∞
Answer: b — 1/4 J (per ohm)
E = ∫₀^∞ |x(t)|² dt = ∫₀^∞ e^(−4t) dt = [−¼e^(−4t)]₀^∞ = 1/4. Finite energy → it is an energy signal.
Digital Communication
ODC1
A baseband signal is band-limited to 4 kHz. Its Nyquist sampling rate is:
a) 4 kHz b) 2 kHz c) 8 kHz d) 16 kHz
Answer: c — 8 kHz
Nyquist rate = 2·fmax = 2×4 kHz = 8 kHz (the minimum rate to avoid aliasing — this is why telephone speech, 3.4 kHz, is sampled at 8 kHz).
ODC2
In 16-QAM, the number of bits carried per symbol is:
a) 2 b) 3 c) 4 d) 16
Answer: c — 4 bits
bits/symbol = log₂(M) = log₂(16) = 4. So the symbol rate is one-quarter of the bit rate, saving bandwidth (at the cost of needing a higher SNR).
Part 15 — Telephony: Digital Switching & Signaling
Concept summary, then solved MCQs. A digital exchange moves a PCM sample using two kinds of stage: Time (T) switching changes the time slot on the same PCM highway; Space (S) switching changes the highway/line while keeping the same time slot. Real exchanges combine them (e.g. T-S-T).
Digital Switching
SW1
A space switch is used to exchange a PCM code sample:
a) between different time slots on the same highway b) between different PCM highways (lines) in the same time slot c) within the same time slot and same line d) between two analog lines
Answer: b — between different PCM highways (lines), keeping the same time slot
The space stage connects the sample of an incoming PCM line to a different outgoing PCM line during the same time slot — it changes the line, not the time position. So it is used when the two subscribers sit on different highways but happen to share the same time slot.
SW2
A time switch (time-slot interchanger) is used to exchange a sample:
a) between different highways in the same time slot b) between different time slots on the same highway c) between two exchanges d) between analog and digital lines
Answer: b — between different time slots on the same highway
The time stage writes the incoming sample into a memory and reads it out in a different time slot of the same PCM line — it changes the time position, not the line. Used when the two subscribers are on the same highway but in different time slots.
SW3
When the two subscribers are neither in the same time slot nor on the same line, the switching required is:
a) space switching only b) time switching only c) combined time–space (e.g. T-S-T) switching d) no switching needed
Answer: c — combined time–space switching
A different line needs a space stage and a different time slot needs a time stage, so both are cascaded (time–space, or the common T-S-T arrangement) to move the sample to the correct line and the correct time slot.
Call Signaling
SG1
Match the direction of each signal:
• Wake-up / off-hook (calling) signal — sent from the subscriber to the exchange (lifting the handset requests service).
• Dial tone (line tone) — sent from the exchange to the subscriber (tells the caller the exchange is ready and to start dialing).
• Seize signal — sent from the control unit to the switching unit (commands the switch to reserve/grab a path).
• Seizing signal — sent at the very beginning of the call to seize the trunk/circuit before digits are exchanged.
SG2
The dial tone is sent:
a) from the subscriber to the exchange b) from the exchange to the subscriber c) from the control unit to the switch d) between two exchanges
Answer: b — from the exchange to the subscriber
After the subscriber goes off-hook (the wake-up signal, subscriber→exchange), the exchange replies with the dial tone (exchange→subscriber) to indicate it is ready to receive the dialed digits.
SG3
The seize signal flows:
a) from the switching unit to the control unit b) from the control unit to the switching unit c) from the subscriber to the control unit d) from one subscriber to another
Answer: b — from the control unit to the switching unit
The control unit issues the seize command to the switching unit so a path/circuit is reserved for the call. The seizing signal itself is exchanged at the start of the call to grab the trunk before address (dialing) information is sent.
Part 16 — Multiple Access, Power/Noise & AM/FM
From the uploaded figure (multiple-access taxonomy) and the worked-problems text file. Items already solved earlier are cross-referenced, not repeated: highest fiber dispersion = multimode step-index (Part 13, OP1); NBFM bandwidth ≈ 2fm (Part 6, M5).
Multiple-Access Protocols
MA1
Classification of multiple-access protocols (three families):
1) Random-access (stations contend for the channel): ALOHA, CSMA, CSMA/CD, CSMA/CA.
2) Controlled-access (stations cooperate / take turns): Reservation, Polling, Token-passing.
3) Channelization (the channel is divided among stations): FDMA (by frequency), TDMA (by time), CDMA (by code).
MA2
Which group do FDMA, TDMA and CDMA belong to?
a) random-access b) controlled-access c) channelization d) token-based
Answer: c — channelization (the shared link is partitioned by frequency, time, or code)
Random-access (ALOHA/CSMA family) lets stations contend; controlled-access (reservation/polling/token) makes them take turns; channelization (FDMA/TDMA/CDMA) splits the medium so each station gets its own sub-channel.
Power Levels & Noise (worked)
PN1
At an amplifier input the signal power is 100 µW and the noise power is 1 µW. At the output the signal power is 1 W and the noise power is 30 mW. Find the noise figure as a ratio.
(S/N)in = 100 µW / 1 µW = 100. (S/N)out = 1 W / 30 mW = 33.3.
NF = (S/N)in / (S/N)out = 100 / 33.3 = 3 (a ratio of 3, i.e. ≈ 4.8 dB).
Same relation as Part 13, NF2; there the data were in dB (subtract), here they are linear powers (form the two S/N ratios, then divide).
PN2
An optical transmitter outputs 5 mW. If the total system loss is 20 dB, the power reaching the receiver is:
a) 0.5 mW b) 50 µW c) 5 µW d) 0.25 mW
Answer: b — 50 µW
20 dB loss = 10^(20/10) = ×100 reduction. Prx = 5 mW / 100 = 0.05 mW = 50 µW. (Check in dBm: 5 mW = +7 dBm; +7 − 20 = −13 dBm = 50 µW.)
PN3
What is the difference, in watts, between −60 dBm and +60 dBm?
−60 dBm = 10^(−60/10) mW = 10⁻⁶ mW = 1 nW = 10⁻⁹ W.
+60 dBm = 10^(60/10) mW = 10⁶ mW = 1000 W = 1 kW.
Difference ≈ 1000 W − 10⁻⁹ W ≈ 1000 W (1 kW) — the small term is negligible. (dBm is referenced to 1 mW: P(dBm) = 10·log₁₀(P/1 mW).)
Transforms — DFT / FFT / DTFT / Z / Inverse-Z (concept)
TR1
How do these transforms differ?
• DTFT (Discrete-Time Fourier Transform): of a discrete signal, gives a continuous, periodic spectrum — X(e^{jω}) = Σ x[n]·e^(−jωn).
• DFT (Discrete Fourier Transform): samples the DTFT at N points → a finite, discrete spectrum, computable on a machine — X[k] = Σ_{n=0}^{N−1} x[n]·e^(−j2πkn/N).
• FFT (Fast Fourier Transform): not a new transform — it is a fast algorithm to compute the DFT, reducing the cost from O(N²) to O(N·log N).
• Z-transform: generalizes the DTFT to the whole complex plane — X(z) = Σ x[n]·z^(−n); the DTFT is the Z-transform evaluated on the unit circle z = e^{jω}. Used for system analysis (poles/zeros, stability).
• Inverse Z-transform: recovers x[n] from X(z) (by partial fractions, power-series/long division, or the contour integral) — the reverse direction of the Z-transform.
a) a different transform from the DFT b) a fast algorithm to compute the DFT c) only for analog signals d) the inverse of the Z-transform
Answer: b — a fast algorithm that computes the same DFT result, in O(N·log N)
AM / FM Modulation
FM1
Frequency Modulation (FM) is:
a) change in carrier amplitude per the modulating signal b) change in carrier frequency per the modulating signal c) change in modulating-signal amplitude per the carrier d) change in carrier phase per the modulating frequency
Answer: b — the carrier frequency varies in step with the modulating signal
In FM the instantaneous frequency follows the message amplitude (amplitude stays constant). Option (a) is AM; (d) describes PM.
FM2
Find the FM modulation index when fm = 250 Hz and the frequency deviation Δf = 5 kHz.
a) 20 b) 35 c) 50 d) 75
Answer: a — 20
β = Δf / fm = 5000 / 250 = 20. (Since β ≫ 1 this is wideband FM.)
AM1
In AM, when the modulation index exceeds unity (m > 100%), the result is called:
a) under-modulation b) over-modulation c) balanced modulation
Answer: b — over-modulation
For m > 1 the envelope is distorted and the carrier phase reverses, so a simple envelope detector cannot recover the message. (See also Part 6, M1.)
AM2
The efficiency of a standard AM (DSB-FC) signal at modulation index m = 0.5 is approximately:
a) 0% b) 11% c) 22% d) 33%
Answer: b — ≈ 11%
η = m² / (2 + m²) = 0.25 / 2.25 = 0.111 ≈ 11%. Even at full modulation (m=1) the efficiency is only 33%, which is why DSB-SC/SSB are preferred for power efficiency.
AM3
If the highest baseband frequency is 6 kHz, the bandwidth of the SSB signal is:
a) 12 kHz b) 6 kHz c) 9 kHz d) 3 kHz
Answer: b — 6 kHz
SSB transmits only one sideband, so its bandwidth equals the message bandwidth: BW_SSB = fm = 6 kHz (half of the 12 kHz that DSB/AM would need).
AM4
In AM, when the modulation depth m is increased, the bandwidth:
a) decreases b) increases c) remains constant
Answer: c — remains constant
AM bandwidth is set only by the highest message frequency: BW = 2fm, independent of m (m only changes the sideband power, not their position). This is the opposite of FM, whose bandwidth grows with the modulation index.
Part 17 — Compiled MCQ Bank (answers verified)
Each question is shown with its options exactly as supplied; the provided key is checked. Where the supplied key looks wrong, an orange box gives the correct answer and why.
Satellite Communication
C1
The first satellite to receive and transmit simultaneously:
A) Intelsat I B) Agila I C) Syncom I D) Telstar I
Answer: D — Telstar I
The supplied key (A, Intelsat I) is questionable. Telstar I (1962) was the first active real-time relay satellite — it received on 6 GHz and retransmitted on 4 GHz at the same time. Intelsat I "Early Bird" came later (1965). So the standard answer is Telstar I (D).
C2
A helical antenna is used for satellite tracking because of:
A) circular polarization B) maneuverability C) beamwidth D) gain
Answer: A — circular polarization ✓
A helix in axial mode radiates a circularly-polarized wave, which keeps a steady signal as the satellite orientation changes (no polarization mismatch).
C3
Repeaters inside communications satellites are known as:
A) Transceivers B) Transponders C) Transducers D) TWT
Answer: B — transponders ✓
A transponder receives the uplink, amplifies and frequency-translates it, and retransmits it on the downlink.
C4
A satellite that rotates around the earth in a low-altitude elliptical or circular pattern:
A) Geosynchronous B) Nonsynchronous C) Prograde D) Retrograde
Answer: B — nonsynchronous (a low-orbit satellite whose period ≠ earth's rotation) ✓
C5
A satellite beam that covers almost 42.4% of the earth's surface:
A) Zone beam B) Hemispheric beam C) Spot beam D) Global beam
Answer: D — global (earth-coverage) beam ✓
From geostationary orbit the earth subtends ≈ 17.3°; a beam filling that cone illuminates about 42% of the surface.
C6
The frequency range of the C-band:
A) 8.5–12.5 GHz B) 3.4–6.425 GHz C) 12.95–14.95 GHz D) 27.5–31 GHz
Answer: B — 3.4–6.425 GHz ✓ (≈ 6 GHz uplink / 4 GHz downlink)
C7
A satellite signal transmitted from a satellite transponder to an earth station:
A) Uplink B) Downlink C) Terrestrial D) Earthbound
Answer: B — downlink ✓ (earth-to-satellite is the uplink)
C8
Detects the satellite signal relayed from the feed, converts it to current, amplifies it and lowers its frequency:
A) Horn antenna B) LNA C) Satellite receiver D) Satellite dish
Answer: C — satellite receiver ✓
This describes the LNB/receiver front-end: it detects, low-noise-amplifies, and down-converts the signal. (A bare LNA only amplifies; it does not down-convert.)
C9
A loss of power of a satellite downlink signal due to the earth's atmosphere:
A) Atmospheric loss B) Path loss C) Radiation loss D) RFI
Answer: A — atmospheric loss
The supplied key (B, path loss) is wrong. Loss caused by the atmosphere (absorption by oxygen/water vapour, rain) is atmospheric loss (A). "Path loss" is the free-space spreading loss, a different mechanism. Correct = A.
C10
The height a satellite must be placed for its rotation to equal the earth's:
A) 26,426.4 mi B) 27,426.4 mi C) 23,426.4 mi D) 22,426.4 mi
Answer: D — ≈ 22,426 miles (geostationary altitude ≈ 35,786 km ≈ 22,236 mi) ✓
C11
The point on a satellite orbit closest to the earth:
A) Apogee B) Perigee C) Prograde D) Zenith
Answer: B — perigee ✓ (apogee is the farthest point)
C12
The frequency of the Ku band for satellite communications:
A) 6/4 GHz B) 14/11 GHz C) 12/14 GHz D) 4/8 GHz
Answer: B — 14/11 GHz (≈ 14 GHz up / 11–12 GHz down) ✓ (6/4 GHz is C-band)
C13
An earth station uses what type of antenna?
A) Despun B) Helical C) Toroidal D) Cassegrain
Answer: D — Cassegrain ✓ (sub-reflector feed, high gain, easy feed access)
C14
Television (TV) broadcast transmission is an example of which type of transmission?
a) Simplex b) Half duplex c) Full duplex d) None
Answer: a — simplex (one-way only) ✓
a) International Telecommunications Satellite b) India Telecommunications Satellite c) Inter Telecommunications Satellite d) None
Answer: a — International Telecommunications Satellite ✓
C16
Kepler's first law states that:
a) the path of a satellite around the primary is an ellipse b) …a circle c) …a sphere d) None
Answer: a — an ellipse (with the primary at one focus) ✓
C17
For an elliptical orbit the eccentricity satisfies:
a) 0 < e < 1 b) e = 0 c) e = 1 d) None
Answer: a — 0 < e < 1 ✓ (e = 0 is a circle; e = 1 is a parabola)
C18
Kepler's third law states:
a) T² ∝ a³ b) T³ ∝ a² c) T² ∝ a^(3/2) d) None
Answer: a — T² ∝ a³ ✓ (square of the period ∝ cube of the semi-major axis)
C19
The orbital period (seconds) is:
a) P = 2π/n b) 2·2π/n c) P = π/n d) None
Answer: a — P = 2π/n ✓ (n = mean motion in rad/s)
C20
Calculate the radius of a circular orbit whose period is 1 day:
a) 42,241 km b) 42,241 m c) 4.241 km d) 2.241 km
Answer: a — ≈ 42,241 km ✓
a = (μT²/4π²)^(1/3) with μ = 3.986×10⁵ km³/s², T = 86,164 s → a ≈ 42,164 km (≈ 42,241 km). Geostationary altitude = this minus the earth radius.
C21
The downlink frequency in a C-band transponder is:
A) 6 GHz B) 4 GHz C) 14 GHz D) 11 GHz
Answer: B — 4 GHz ✓ (C-band: 6 GHz up, 4 GHz down)
C22
The carrier-to-noise ratio for a satellite depends upon:
A) EIRP B) Bandwidth C) Free-space path loss D) All of them
Answer: D — all of them ✓ (C/N = EIRP + G/T − FSL − 10log(kB))
C23
A satellite downlink at 12 GHz has a transmit power of 6 W and an antenna gain of 48.2 dB. Find the EIRP in dBW.
a) 56 dBW b) 16 dBW c) 56 dB d) None
Answer: a — 56 dBW ✓
EIRP = Pt(dBW) + G(dB) = 10log(6) + 48.2 = 7.78 + 48.2 = 55.98 ≈ 56 dBW. (Unit is dBW, so "56 dB" in (c) is wrong.)
C24
Calculate the gain of a 3 m parabolic antenna at 12 GHz, aperture efficiency 0.5.
a) 48.9 dB b) 4.9 dB c) 48.9 dB·Hz d) None
Answer: a — ≈ 48.9 dB ✓
λ = c/f = 0.025 m; G = η(πD/λ)² = 0.5×(π·3/0.025)² ≈ 0.5×142,100 ≈ 71,060 → 10log(71,060) ≈ 48.5–48.9 dB.
C25
PCM is used in satellite communications for the transmission of:
a) TV signal b) Telegraph signal c) Speech signal d) All the above
Answer: d — all the above ✓ (any analog source can be digitized by PCM)
C26
In satellite communication, the modulation generally used is:
a) AM b) FM c) PWM d) PAM
Answer: b — FM ✓ (constant envelope tolerates the nonlinear TWT and improves SNR)
C27
FM is preferred for satellite communication because:
a) the satellite channel has large bandwidth and severe noise b) it gives a high modulation index c) low bandwidth is the essential requirement d) None
Answer: a — large available bandwidth + good noise performance ✓
C28
The main advantage of satellite communication is:
a) low cost b) low distortion c) high reliability d) high bandwidth
Answer: d — large/high bandwidth (and very wide coverage) ✓
C29
The signal-to-noise ratio of a satellite signal least depends on:
a) satellite surface area b) bandwidth c) free-space path loss d) EIRP
Answer: a — the satellite's physical surface area ✓
C/N depends on EIRP, bandwidth and path loss (and G/T); the bare surface area of the satellite body is not a direct factor.
Optical Fiber
OF1
Which material is sensitive to light?
a) Photoresist b) Photosensitive c) Light Sensitive d) Maser
Answer: a — photoresist ✓ (a real light-sensitive material used in lithography)
OF2
The core of an optical fiber has a:
a) lower index than air b) lower index than the cladding c) higher index than the cladding d) same index as the cladding
Answer: c — higher refractive index than the cladding ✓ (required for total internal reflection)
OF3
The different angles of entry of light when the core diameter is many times the wavelength are the:
a) Acceptance angle b) Modes c) Sensors d) Aperture
Answer: b — modes ✓ (a large core supports many propagation modes)
OF4
The loss in signal power as light travels down a fiber is called:
a) Dispersion b) Scattering c) Absorption d) Attenuation
Answer: d — attenuation ✓ (scattering and absorption are two causes of attenuation)
OF5
The (often-quoted theoretical) bandwidth of optical fiber:
a) 900 MHz b) 900 PHz c) 900 THz d) 900 EHz
Answer: c — ~900 THz (the figure usually cited for the enormous optical bandwidth) ✓
OF6
If a mirror reflects light, the reflected angle is ___ the incident angle:
a) smaller b) larger c) the same d) independent
Answer: c — the same (law of reflection: angle of incidence = angle of reflection) ✓
OF7
A specific path the light takes in a fiber for a certain angle and number of reflections is a:
a) Mode b) Grade c) Numerical Aperture d) Dispersion
Answer: a — mode ✓
OF8
The width of the range of wavelengths emitted by the light source is the:
a) Bandwidth b) Chromatic Dispersion c) Spectral width d) Beamwidth
Answer: c — spectral width ✓ (a narrow spectral width → less chromatic dispersion)
OF9
Which theory states light behaves as many tiny particles?
a) Huygen's theory b) Wave theory c) Nyquist theory d) Quantum theory
Answer: d — quantum theory ✓ (photons; the particle nature of light)
OF10
"Dispersion" is used to describe the:
a) splitting of white light into its component colors b) propagation in straight lines c) bending when going between media d) bending when striking a mirror
Answer: a — splitting of white light into its colors (wavelength-dependent refraction) ✓
OF11
Photodiodes used as fiber-optic detectors are:
a) unbiased (like a solar cell) b) forward biased c) reverse biased d) thermoelectrically cooled
Answer: c — reverse biased ✓ (reverse bias widens the depletion region and speeds the response)
OF12
Which fiber has the highest modal dispersion?
a) Step-index multimode b) Graded-index multimode c) Step-index single-mode d) Graded-index
Answer: a — step-index multimode ✓ (same result as Part 13, OP1)
Antennas & Microwave
AW1
The VLF, LF and MF antennas are:
A) horizontally polarized B) vertically polarized C) non-linearly polarized D) either (a) or (b)
Answer: B — vertically polarized ✓ (ground-wave propagation needs vertical polarization)
AW2
Which of the following antennas is not wideband?
A) Marconi B) Helical C) Folded dipole D) Discone
Answer: A — Marconi (a resonant quarter-wave monopole — narrowband) ✓
AW3
Circular polarization:
A) is useful in reducing the depolarization effect on the received wave B) needs critical alignment of Tx/Rx antennas C) helps discriminate adjacent beams D) none
Answer: A — reduces depolarization effects ✓ (and avoids the alignment sensitivity that linear polarization has)
AW4
In a circular waveguide the dominant mode is:
A) TE01 B) TE11 C) TE20 D) TE21
Answer: B — TE₁₁ (lowest cutoff in a circular guide) ✓
AW5
Which TM mode in a rectangular waveguide has the lowest cutoff frequency?
A) TM11 B) TM01 C) TM10 D) TM21
Answer: A — TM₁₁ ✓ (TM₀₁/TM₁₀ do not exist, as TM needs both indices ≥ 1) (see also Part 8)
AW6
In a klystron amplifier the input cavity is called the:
A) buncher B) catcher C) Pierce gun D) collector
Answer: A — buncher (input cavity); the output cavity is the catcher ✓
A) a power ratio B) a current ratio C) a voltage ratio D) a distance ratio
Answer: A — a power ratio ✓
Directive gain = (power density radiated in a given direction) / (power density of an isotropic radiator with the same total power).
Modulation & Digital Communication
DM1
If a carrier is modulated by a digital stream using phases 0°, 90°, 180°, 270°, the modulation is:
A) BPSK B) QPSK C) QAM D) MSK
Answer: B — QPSK (four phase states = 2 bits/symbol) ✓
DM2
The number of repeaters in a coaxial cable link depends on:
A) system bandwidth B) separation of equalizers C) number of coaxial cables in the tube D) none
Answer: A — system bandwidth (higher bandwidth → more loss → more repeaters) ✓
DM3
The correct sequence of subsystems in an FM receiver is:
A) mixer, RF amp, limiter, IF amp, discriminator, audio B) RF amp, mixer, IF amp, limiter, discriminator, audio C) RF amp, mixer, limiter, discriminator, IF amp, audio D) mixer, IF amp, limiter, audio, discriminator
Answer: B — RF amp → mixer → IF amp → limiter → discriminator → audio amp ✓
DM4
Which gives the minimum probability of error?
A) ASK B) FSK C) PSK D) DPSK
Answer: C — PSK (largest distance between symbols for a given power) ✓
DM5
Which gives the maximum probability of error?
A) ASK B) DPSK C) FSK D) PSK
Answer: A — ASK (most sensitive to noise / amplitude variations) ✓
DM6
In the spread-spectrum technique:
A) a modulated signal is modulated again B) a modulated signal is modulated twice again C) the power of a modulated signal is increased D) the noise component of a modulated signal is decreased
Answer: A — the modulated signal is modulated again (spread by a PN code)
On the supplied key (D): the mechanism of spread spectrum is option A — the data-modulated carrier is modulated a second time by a wideband pseudo-noise code. Option (D) describes a result (better resistance to narrowband noise/jamming via processing gain), not what the technique is. The best "what it is" answer is A.
DM7
The bandwidth required for amplitude modulation is:
A) half fm B) equal to fm C) twice fm D) four times fm
Answer: C — twice the modulating frequency (BW = 2fm) ✓ (see also Part 16, AM4)
DM8
The Shannon-Hartley law:
A) refers to noise B) defines bandwidth C) describes signalling rate D) refers to distortion
Answer: A — refers to noise ✓
C = B·log₂(1 + S/N) — channel capacity depends on bandwidth and the signal-to-noise ratio.
DM9
In an FM signal, when the modulation index increases, the power:
A) increases B) decreases C) remains constant D) none
Answer: C — remains constant ✓ (FM has a constant envelope; only the sideband distribution changes) (see also Part 6, M7)
General
G1
Which sampling frequency is adequate for voice telephony?
A) 4 kHz B) 8 kHz C) 16 kHz D) 32 kHz
Answer: B — 8 kHz ✓ (voice band ≈ 4 kHz → Nyquist rate = 8 kHz) (see also Part 14, ODC1)
G2
Resonant circuits are used in:
A) audio amplifiers B) RF amplifiers C) both audio and RF D) none
Answer: B — RF amplifiers ✓ (LC tuned circuits select a narrow RF band; audio is wideband/baseband)
Part 18 — Optical Fiber Fundamentals, ARQ & Far-Field
Options shown as supplied; keys verified. The provided answer key for the optical set (items 1–17) is all correct. One satellite item had a wrong key — corrected in the orange box.
Optical Fiber — Fundamentals
OFF1
In optical-fiber communications, the signal source is ___ waves.
A) Light B) Infrared C) Radio D) Very low-frequency
Answer: A — light ✓
OFF2
Which one is not a guided medium of transmission?
A) Fiber-optic cable B) Coaxial cable C) Twisted-pair cable D) The atmosphere
Answer: D — the atmosphere (it is an unguided/wireless medium) ✓
OFF3
An environment has many high-voltage devices. The best transmission medium is:
A) The atmosphere B) Twisted-pair cable C) Optical fiber D) Coaxial cable
Answer: C — optical fiber ✓ (it is immune to the EMI from high-voltage equipment)
OFF4
Which of these converts the electrical signal to optical signals?
A) Optical photo detectors B) Demultiplexers C) Multiplexers D) Optical modulators
Answer: D — optical modulators ✓ (a photo-detector does the reverse: optical → electrical)
OFF5
A fiber-optic system has three basic components, in order:
A) light guide, source, detector B) light source, guide, detector C) detector, source, guide D) guide, detector, source
Answer: B — light source → light guide (fiber) → light detector ✓
OFF6
In an optical fiber, the outer layer is ___ and the inner layer is ___.
A) core, cladding B) cladding, core C) transmit, reflect D) reflect, transmit
Answer: B — cladding (outer), core (inner) ✓
OFF7
Optical fiber cables are highly immune to EMI because information is carried by:
A) light means B) electrical means C) magnetic means D) acoustic means
Answer: A — light ✓ (no electric current to pick up interference)
OFF8
Which one is based on laser-beam technology?
A) Magnetic tape B) Terminals C) Optical disks D) Keyboards
Answer: C — optical disks (CD/DVD read by a laser) ✓
OFF9
Which method lets many selectable, independent user channels coexist on a single optical-fiber link?
A) PCM B) FDM C) CDM D) —
Answer: B — FDM (frequency-division multiplexing) ✓
In fiber this is realized optically as WDM (wavelength-division multiplexing) — different wavelengths act like different frequency channels.
OFF10
Transmission media are usually categorized as:
A) metallic or nonmetallic B) guided or unguided C) determinate or indeterminate D) fixed or unfixed
Answer: B — guided (cable) or unguided (wireless) ✓
OFF11
Which is a guided medium?
A) Microwave B) Radio C) Fiber-optic cable D) Atmosphere
Answer: C — fiber-optic cable ✓ (the others are unguided)
OFF12
Which mechanism is used in laser technology for the generation of light?
A) Dispersion B) Absorption C) Stimulated Emission D) Spontaneous Emission
Answer: C — stimulated emission (the "SE" in LASER) ✓
An LED, by contrast, works by spontaneous emission — that is why a laser is coherent and an LED is not.
OFF13
An optical splice provides a connection between:
A) transmitter to fiber B) receiver to fiber C) fiber to fiber D) fiber to repeater
Answer: C — fiber to fiber ✓ (same as Part 3, Q29)
OFF14
Optical fibers are highly immune to EMI. Which statement justifies it?
A) they transmit signals as light rather than electric current B) they are shielded by outer conductors C) they are too small for magnetic fields to induce current D) magnetic fields cannot penetrate glass
Answer: A — signals travel as light, not as electric current ✓
OFF15
In an optical fiber, the core ___ the cladding.
A) is denser than B) has the same density as C) is less dense than D) is another name for
Answer: A — is denser than (higher refractive index → total internal reflection) ✓
OFF16
The material used to fabricate the inner core of an optical fiber is:
A) glass or plastic B) bimetallic C) copper D) liquid
Answer: A — glass (silica) or plastic ✓
OFF17
Unlike wired media, optical fibers are highly resistant to:
A) refraction B) low-frequency transmission C) electromagnetic interference D) high-frequency transmission
Answer: C — electromagnetic interference (EMI) ✓
Data Link — ARQ
ARQ2
In Go-Back-N ARQ, if 3 bits are assigned to the frame (sequence) number, then:
A) max sending window = 3, receiving = 3 B) max sending window = 7, receiving = 1 C) max sending window = 3, receiving = 1 D) max sending window = 7, receiving = 7
Answer: B — sending window = 7, receiving window = 1
With m = 3 bits there are 2³ = 8 sequence numbers. Go-Back-N: max sending window = 2ᵐ − 1 = 7, receiving window = 1. (Same rule as Part 6/2020-2021 Q24, which used m = 2 → 3 and 1.)
Satellite — Far-Field Region
FF1
The satellite–earth-station distance is 5 km and the satellite antenna diameter is 10 m. Which frequency band keeps the earth station in the far-field region?
A) C band B) none of the above C) Ku band D) Ka band
Answer: A — C band
Far-field (Fraunhofer) distance: R_ff = 2D²/λ. The station at R = 5 km is in the far field only if R ≥ 2D²/λ, i.e. λ ≥ 2D²/R = 2(10)²/5000 = 0.04 m → f ≤ c/λ = 7.5 GHz.
• C band (≈ 4–6 GHz) is below 7.5 GHz → R_ff ≈ 4 km < 5 km → far field ✓.
• Ku (≈ 12–14 GHz) → R_ff ≈ 9 km > 5 km → not far field. Ka (≈ 20–30 GHz) → even larger. So only C band works.
The supplied key (Ku band) is wrong. Higher frequency = shorter wavelength = larger far-field distance, which pushes the far-field boundary beyond 5 km. Only the lowest band (C band) brings R_ff under 5 km, so the correct answer is C band (A).
Part 19 — LED, Signal Symmetry & Microwave Device
Two of the supplied items are already solved: product of two odd signals = even (Part 3, Q9) and the modulation needing the least bandwidth = Delta modulation (DM) (Part 3, Q12). The genuinely new ones are below.
Optical Sources
1) incoherent & wide beam 2) highly coherent & wide beam 3) highly coherent & narrow beam 4) incoherent & narrow beam
Answer: 1 — incoherent & wide beam
An LED emits by spontaneous emission, so its light is incoherent, has a wide spectral width, and radiates over a wide (Lambertian) beam. A laser (stimulated emission) is the opposite: coherent with a narrow beam — which is why lasers couple better into single-mode fiber for long-haul links.
Signal Symmetry (even / odd)
SY1
The product of two even signals is:
1) even 2) odd 3) both
Answer: 1 — even
If x(−t)=x(t) and y(−t)=y(t), then x(−t)y(−t)=x(t)y(t) → the product is even. (Sign rule: even × even = even.)
SY2
The product of an even signal and an odd signal is:
1) even 2) odd 3) both
Answer: 2 — odd
With x(−t)=x(t) (even) and y(−t)=−y(t) (odd): x(−t)y(−t) = x(t)·(−y(t)) = −x(t)y(t) → odd. Full rule: even×even = even, odd×odd = even, even×odd = odd (just like multiplying +/− signs).
Microwave Devices
MWd1
Which of these components works on the electric field (alone)?
1) Gunn diode 2) magnetron 3) TWT
Answer: 1 — Gunn diode
The Gunn diode works purely on the applied electric field (the transferred-electron effect / bulk negative-differential-resistance in GaAs — no magnetic field involved). By contrast, the magnetron uses crossed electric and magnetic fields, and the TWT needs an axial magnetic field to focus its electron beam. (Compare Part 3, Q18: the device that uses both fields = magnetron.)
Part 20 — Fiber Dispersion, TEM, PCM SNR & Radar
Several supplied items are already solved and are not repeated: least-bandwidth modulation = DM (Part 3, Q12); the analog/non-digital pulse modulation = PPM/PWM (Part 3, Q13); companding = compress + expand (Part 8, N2); frequency reuse (Part 3, Q24); a 16× power increase gives only 2× range (Part 8, R-bank); FM generated by a varactor diode (2020–2021, Q19); klystron = velocity modulation (2019–2020, Q21); to maximize satellite received power, raise the gain of both antennas (2019–2020, Q10).
Optical Fiber
OP2
The smallest dispersion occurs in:
1) SM-SI (single-mode step-index) 2) MM-SI (multimode step-index) 3) MM-GI (multimode graded-index)
Answer: 1 — single-mode step-index (SM-SI)
A single-mode fiber carries only one mode, so there is no modal dispersion — only a small amount of chromatic dispersion remains. This is the opposite end of the scale from MM-SI, which has the largest dispersion (see Part 13, OP1). Order, most → least dispersion: MM-SI > MM-GI > SM-SI.
Waveguides
TEM1
The TEM mode can propagate in:
1) free space 2) circular waveguide 3) rectangular waveguide
Answer: 1 — free space
TEM (no axial E or H component) needs two conductors or open space — it exists in free space and on coaxial/parallel-wire lines. A hollow single-conductor waveguide (rectangular or circular) cannot support TEM; it only carries TE/TM modes. (See also the 2019–2020 item: TEM inside a coaxial cable.)
PCM
PC1
The change in PCM quantization SNR when the number of bits goes from 7 to 8:
1) 11 2) 16 3) 10.5 4) 12
Answer: ≈ 6 dB (each extra bit adds ~6 dB)
Quantization SNR ≈ 6.02n + 1.76 dB. Going from n=7 to n=8 adds 6.02×(8−7) ≈ 6 dB (43.9 dB → 49.9 dB).
None of the listed options matches. One extra bit ≈ 6 dB, which is not 11/16/10.5/12. The figure 12 dB (option 4) would be right only if the change were 2 bits (e.g. 6→8 bits); for a single-bit 7→8 change the correct answer is ~6 dB. Likely a typo in the question.
Radar
RD1
Which radar avoids (does not rely on) the Doppler effect?
1) pulse radar 2) tracking radar
Answer: 1 — (basic) pulse radar
A basic pulse radar measures range from the round-trip time delay of the echo — it does not need a Doppler frequency shift. Doppler is exploited in CW, MTI and pulse-Doppler radars (often used inside tracking radars) to measure target velocity and reject clutter.
Part 21 — Antenna (additional item)
Of the six supplied antenna questions, five are already solved above and are not repeated: gain relative to isotropic = dBi (Part 5, A9); half-power angular separation = beamwidth (Q34); λ/2 dipole input resistance ≈ 73 Ω (A11); folded dipole = greater bandwidth + 300 Ω (A13); vertical dipole radiates strongest, uniformly in the horizontal plane (A14). Only the new one is added below.
A30
As the length of a center-fed dipole antenna is reduced below a quarter-wavelength, the radiation resistance:
a) decreases b) increases c) remains constant
Answer: a — decreases
For a short dipole the radiation resistance scales with the square of its electrical length: R_rad ≈ 80π²(ℓ/λ)². Shortening it therefore drops R_rad rapidly (toward a few ohms or less), which makes a very short antenna inefficient and hard to match to the line.
Part 22 — Practice Set (from the quick-note PDF)
Added as practice questions with fresh options at your request. These reinforce concepts already covered earlier (FM source, klystron, power across a resistor, FT vs Z, time-scaling, beamwidth, multipath, ISI) — here they are re-written in clean MCQ form so they appear in the interactive quiz.
Modulation & Microwave
P1
Which device is used to generate FM (electronic frequency tuning)?
a) tunnel diode b) varactor diode c) Gunn diode d) PIN diode
Answer: b — varactor diode
A varactor's junction capacitance changes with the applied voltage; placing it in an oscillator's tank makes the frequency follow the modulating voltage → FM (a voltage-controlled oscillator).
P2
A klystron operates on:
a) amplitude modulation b) velocity modulation c) pulse modulation d) phase modulation
Answer: b — velocity modulation (the cavity field bunches the electron beam by varying its velocity)
Signals & Power
P3
The voltage V(t) = 4sin(8πt) + cos(12πt) is applied across R = 10 Ω. The total average power is:
a) 0 b) 0.1 W c) 1 W d) none of these
Answer: d — none of these (it is 0.85 W)
For independent sinusoids the average powers add: P = (4²/2 + 1²/2)/R = (8 + 0.5)/10 = 0.85 W — not equal to any of 0, 0.1, or 1 W.
P4
The similarity between the Fourier transform and the Z-transform is that both:
a) convert the time domain to the frequency domain b) convert frequency to time c) convert analog to digital d) convert digital to analog
Answer: a — both map a (discrete) time-domain signal into the frequency domain
P5
If y(t) = x(2t), the signal is:
a) expanded b) compressed c) unchanged d) only shifted
Answer: b — compressed (time-scaled by 2)
A factor > 1 inside the argument speeds up / squeezes the signal in time. E.g. discrete x(n) = {0,1,2,3,4} compressed (keep every 2nd sample) → {0,2,4,0,0}.
Antennas & Channel
P6
The (half-power) beamwidth of an antenna is defined as:
a) the range of frequencies describing antenna performance b) the angular separation between the half-power points of the main lobe c) the front-to-back ratio d) the antenna gain
Answer: b — the angular separation between the −3 dB (half-power) points of the main lobe
(a) describes bandwidth, not beamwidth. The "two identical points on opposite sides of the pattern maximum" wording also points to the same angular width of the main beam.
P7
In a mobile communication system, the wave reaches the receiving antenna via:
a) the direct path only b) the reflected path only c) direct, reflected and diffracted paths d) the refracted path only
Answer: c — direct, reflected and diffracted paths (multipath)
These multiple arriving copies add with different delays/phases, causing multipath fading.
P8
Zero inter-symbol interference (ISI = 0) is achieved by using a:
a) matched filter b) raised-cosine filter c) simple low-pass filter d) Butterworth filter
Answer: b — raised-cosine filter (its impulse response is zero at all other sampling instants)
End of solutions • Pay special attention to the orange-flagged questions (where the original answer key was wrong) • Good luck in the exam