# Analog Communication Systems

 Question 1
Consider a superheterodyne receiver tuned to $600 \text{ kHz}$. If the local oscillator feeds a $1000 \text{ kHz}$ signal to the mixer, the image frequency (in integer) is ____________________ $\text{kHz}$ .
 A 850 B 1825 C 1400 D 1250
GATE EC 2021   Communication Systems
Question 1 Explanation:
\begin{aligned} f_{s} &=600 \mathrm{kHz} \\ f_{l} &=1000 \mathrm{kHz} \\ f_{s i} &=? \\ \text { Default } f_{l}>f_{s} & \\ \mathrm{IF} &=f_{l}-f_{s}=400 \mathrm{kHz} \\ f_{\mathrm{si}} &=f_{s}+2 \mathrm{IF}=600 \mathrm{~K}+800 \mathrm{~K}=1400 \mathrm{kHz} \end{aligned}
 Question 2
A sinusoidal message signal having root mean square value of $4\:V$ and frequency of $\text{1 kHz}$ is fed to a phase modulator with phase deviation constant 2 rad/ volt. If the carrier signal is $c\left ( t \right )=2\cos\left ( 2\pi 10^{6} t\right )$, the maximum instantaneous frequency of the phase modulated signal (rounded off to one decimal place) is ______ $\text{Hz}$ .
 A 1.01131e+06 B 12543.3 C 124534 D 1.02536e+06
GATE EC 2021   Communication Systems
Question 2 Explanation:
Given sinusoidal message signal $\rightarrow V_{\mathrm{rms}}=4 \mathrm{~V}$
\begin{aligned} f_{m} &=1 \mathrm{kHz} \\ V_{\text {peak }} &=\sqrt{2} \times V_{\mathrm{rms}}=\sqrt{2} \times 4 \mathrm{~V} \\ m(t) &=4 \sqrt{2} \sin 2 \pi \times 10^{3} t\\ K_{p}&=2 \mathrm{rad} / \mathrm{volt}\\ c(t) &=2 \cos 2 \pi \times 10^{6} \mathrm{t} \\ \left(f_{i}\right)_{\max } &=f_{c}+\frac{K_{p}}{2 \pi}\left[\frac{d}{d t} m(t)\right]_{\max } \\ \frac{d}{d t} m(t) &=4 \sqrt{2} \times\left(2 \pi \times 10^{3}\right) \cos 2 \pi \times 10^{3} t \\ \left[\frac{d}{d t} m(t)\right]_{\max } &=4 \sqrt{2} \times 2 \pi \times 10^{3} \\ \left(f_{i}\right)_{\max } &=10^{6}+\frac{2}{2 \pi}\left(4 \sqrt{2} \times 2 \pi \times 10^{3}\right)=1011313.7 \mathrm{~Hz} \end{aligned}
 Question 3
A $4\:kHz$ sinusoidal message signal having amplitude $4\:V$ is fed to a delta modulator (DM) operating at a sampling rate of $32\:\text{kHz}$ . The minimum step size required to avoid slope overload noise in the DM (rounded off to two decimal places) is _____
 A 2.54 B 5.45 C 7.82 D 3.14
GATE EC 2021   Communication Systems
Question 3 Explanation:
\begin{aligned} f_{m}&=4 \mathrm{kHz} \\ A_{m}&=4 \mathrm{~V} \\ f_{s}&=32 \mathrm{kHz}\\ \text { To avoid } \text{SOE} \rightarrow &\frac{\Delta}{T_{s}} \geq 2 \pi f_{m} A_{m} \\ \Delta \times 32 k &\geq 2 \pi \times 4 k \times 4 \mathrm{~V} \\ \Delta &\geq \pi\\ (\Delta)_{\min }&=\pi volts\\ (\Delta)_{\min }&=3.14 volts \end{aligned}
 Question 4
Consider a carrier signal which is amplitude modulated by a single-tone sinusoidal message signal with a modulation index of $50\%$. If the carrier and one of the sidebands are suppressed in the modulated signal, the percentage of power saved (rounded off to one decimal place) is ______________
 A 54.2 B 78.6 C 98.2 D 94.4
GATE EC 2021   Communication Systems
Question 4 Explanation:
If carrier and one of the sidebands are suppressed, then % of power saved
\begin{aligned} &=\frac{\text { Power saved }}{\text { Total power }}=\frac{P_{c}+\frac{P_{c} \mu^{2}}{4}}{P_{c}\left[1+\frac{\mu^{2}}{2}\right]}=\frac{4+\mu^{2}}{4+2 \mu^{2}} \\ &=\frac{4+(0.5)^{2}}{4+2 \times(0.5)^{2}}=0.944=94.4 \% \end{aligned}
 Question 5
$S_{PM}(t)\; and \; S_{FM}(t)$ as defined below, are the phase modulated and the frequency modulated waveforms, respectively, corresponding to the message signal m(t) shown in the figure.

$S_{PM}(t)= \cos (1000 \pi t+K_p m(t))$
and $S_{FM}(t)= \cos (1000 \pi t+K_f \int_{-\infty }^{t} m(\tau )d\tau )$

where $K_p$ is the phase deviation constant in radians/volt and $K_f$ is the frequency deviation constant in radians/second/volt. If the highest instantaneous frequencies of $S_{PM}(t)\; and \; S_{FM}(t)$ are same, then the value of the ratio $\frac{K_p}{K_f}$ is ______ seconds.
 A 0.5 B 1 C 2 D 2.5
GATE EC 2020   Communication Systems
Question 5 Explanation:
\begin{aligned}S(t)_{pm}&=A_c \cos [2\pi f_{c}t+k_pm(t)]\\ S(t)_{Fm}&=A_c \cos [2\pi f_{c}t+k_{f}\int_{0}^{\infty }m(t)dt]\end{aligned}
Instantaneous frequency are equal
\begin{aligned}f_{i}&=\frac{1}{2\pi }\frac{\mathrm{d} }{\mathrm{d} t}\theta (t)\\f_{iPM}&=f_{c}+\frac{K_p}{2\pi }\frac{\mathrm{d} }{\mathrm{d} t} m(t)\\ f_{iFM}&=f_{c}+\frac{K_f}{2\pi }m(t)\\ \text{Given that, } (f_{iPM})_{max}&=(f_{iFM})_{max} \\ f_{c}+\frac{k_P}{2\pi}[\frac{\mathrm{d} }{\mathrm{d} t}m(t)]_{max}&=f_{c}+\frac{K_{f}}{2\pi }[m(t)]_{max}\\ \left.\begin{matrix} \frac{k_P}{2\pi}\frac{\mathrm{d} }{\mathrm{d} t}m(t) \end{matrix}\right|_{max} &= \left. \begin{matrix} \frac{K_{f}}{2\pi }[m(t)] \end{matrix}\right|_{max}\\ \left.\begin{matrix} \frac{\mathrm{d} }{\mathrm{d} t}m(t) \end{matrix}\right|_{max} &=5, \left.\begin{matrix} m(t) \end{matrix}\right|_{max}=10 \\ 5k_{p}&=10k_{f} \\ \frac{k_{p}}{k_{f}}&=2\end{aligned}
 Question 6
For the modulated signal $x(t)= m(t) \cos(2 \pi f_c t)$, the message signal $m(t) = 4 \cos (1000 \pi t)$ and the carrier frequency $f_c$ is 1 MHz. The signal $x(t)$ is passed through a demodulator, as shown in the figure below. The output $y(t)$ of the demodulator is
 A $\cos (460 \pi t)$ B $\cos (920 \pi t)$ C $\cos (1000 \pi t)$ D $\cos (540 \pi t)$
GATE EC 2020   Communication Systems
Question 6 Explanation:
Output of multiplier
$=x(t)\cos 2\pi (f_{c}+40)t=m(t)\cos 2\pi f_{c}t\cdot \cos2\pi (f_{c}+40)t$
$=\frac{m(t)}{2}[\cos 2\pi (2f_{c}+40)t+\cos 2\pi (40)t]$
Given, $m(t)=4\cos 1000\pi t$
So, output of multiplier
$=2\cos 2\pi (500)t[\cos 2\pi (2f_{c}+40)t+\cos 2\pi (40)t]$
$=\cos 2\pi (2f_{c}+540)t+\cos 2\pi (2f_{c}-460)t+\cos 2\pi (540)t+\cos 2\pi (460)t$
Output of Low pass filter
$=\cos [2\pi (460)]t$
$=\cos 920\pi t$
 Question 7
The baseband signal m(t) shown in the figure is phase-modulated to generate the PM signal $\varphi (t)=2cos(2 \pi f_c t+km(t))$. The time t on the x-axis in the figure is in milliseconds. If the carrier frequency is $f_c$=50kHz and $k=10 \pi$, then the ratio of the minimum instantaneous frequency (in kHz) to the maximum instantaneous frequency (in kHz) is __________ (rounded off to 2 decimal places).
 A 0.25 B 0.5 C 0.75 D 0.85
GATE EC 2019   Communication Systems
Question 7 Explanation:
\begin{aligned} f_i&=\frac{1}{2 \pi}\frac{d\theta (t)}{dt}= \\ &=f_c+\frac{k}{2 \pi}\frac{dm (t)}{dt} \\ &=f_c+5\frac{dm (t)}{dt} \end{aligned}

\begin{aligned} f_{i(min)}&=50kHz-(5 \times 1000Hz)=45kHz\\ f_{i(max)}&=50kHz+(5 \times 2000Hz)=60kHz\\ \frac{f_{i(min)}}{f_{i(max)}}&=\frac{45}{60}=0.75 \end{aligned}
 Question 8
Let $c(t)=A_{c} cos(2\pi f_{c}t) \; and \; m(t)=cos(2\pi f_{m}t)$. It is given that $f_{c}>>5f_{m}$. The signal $c(t) +m(t)$ is applied to the input of a non-linear device, whose output $v_{o}(t)$ is related to the input $v_{i}(t)+bv_{i}(t)$,where a and b are positive constants. The output of the non-linear device is passed through an ideal band-pass filter with center frequency $f_{c}$ and bandwidth $3f_{m}$ ,to produce an amplitude modulated (AM) wave. If it is desired to have the sideband power of the AM wave to be half of the carrier power, then $a/b$ is
 A 0.25 B 0.5 C 1 D 2
GATE EC 2018   Communication Systems
Question 8 Explanation:
\begin{aligned} v_{f t}^{(t)} &=A_{c} \cos \left(2 \pi f_{c} t\right)+\cos \left(2 \pi t_{m} t\right) \\ v_{d}(t) &=a v_{i}(t)+b v_{i}^{2}(t) \\ =&\left[a A_{c} \cos \left(2 \pi t_{c} t\right)+a \cos \left(2 \pi f_{m} t\right)\right] \\ \quad &+b\left[\begin{array}{c}A_{c}^{2} \cos ^{2}\left(2 \pi f_{c} t\right)+\cos ^{2}\left(2 \pi f_{m} t\right) \\ +2 A_{c} \cos \left(2 \pi f_{c} t\right) \cos \left(2 \pi t_{m} t\right)\end{array}\right] \end{aligned}
After passing through the given BPF,
$y(t)=a A_{c} \cos 2 \pi f_{c} t+2 b A_{c} \cos \left(2 \pi f_{c} t\right) \cos \left(2 \pi f_{m} t\right)$
$=a A_{c}\left[1+\frac{2 b}{a} \cos \left(2 \pi f_{m} t\right)\right] \cos \left(2 \pi f_{c} t\right)$
Modulation index,
\begin{aligned} \mu &=\begin{array}{l}2 b \\ a\end{array} \\ P_{S B} &=\frac{\mu^{2}}{2} P_{C}=\frac{1}{2} P_{C} \\ \mu^{2} &=1 \Rightarrow \mu=1 \\ \frac{2 b}{a} &=1 \\ \frac{a}{b} &=2 \end{aligned}
 Question 9
Consider the following amplitude modulated signal: $s(t)=cos(2000\pi t)+4cos(2400\pi t)+cos(2800\pi t).$ The ratio (accurate to three decimal places) of the power of the message signal to the power of the carrier signal is __________.
 A 0.12 B 0.2 C 0.8 D 1
GATE EC 2018   Communication Systems
Question 9 Explanation:
The power of the carrier signal is,
$P_{c}=\frac{(4)^{2}}{2}=8 \mathrm{W}$
The power of message signal can be taken as the power carried by the sidebands.
So, the power of the message signal is,
$P_{m}=2\left[\frac{(1)^{2}}{2}\right]=1 \mathrm{W}$
Required ratio,
$\frac{P_{m}}{P_{c}}=\frac{1}{8}=0.125$
 Question 10
A modulating signal given By $x(t) =5sin(4 \pi 10^{3}t-10\pi cos2\pi 10^{3}t$)V is fed to a phase modulator with phase deviation constant $k_{p}$=0.5rad/V. If the carrier frequency is 20 kHz, the instantaneous frequency (in kHz) at t = 0.5 ms is __________
 A 65 B 70 C 75 D 80
GATE EC 2017-SET-2   Communication Systems
Question 10 Explanation:
Given that, the modulating signal is.
$x(t)=5 \sin (4000 \pi t-10 \pi \cos 2000 \pi t) \mathrm{V}$
The standard phase modulated signal can be given as,
$s(t)=A_{c} \cos \left(\omega_{c} t+k_{p} x(t)\right)$
Instantaneous angle of the modulated signal is.
$\theta(t)=\omega_{c} t+k_{p} x(t)$
Instantaneous frequency is,
\begin{array}{l} \omega_{1}(t)=\frac{d \theta(t)}{d t}\\ =\omega_{c}+k_{p} \frac{d x(t)}{d t} \\ f_{i}(t)=f_{c}+\frac{25}{2 \pi}[\cos (4000 \pi t-10 \pi \cos 2000 \pi t) \\ \quad \left.\left(4000 \pi+20000 \pi^{2} \sin 2000 \pi t\right)\right] \\ A t=0.5 \mathrm{ms} \\ \begin{aligned} f_{1}(t)=& f_{c}+\frac{25}{2 \pi}[\cos (2 \pi-10 \pi \cos (\pi))] \\ &\quad\left(4000 \pi+20000 \pi^{2} \sin (\pi)\right) & \\ &=f_{c}+\frac{25}{2 \pi}[\cos (12 \pi)](4000 \pi) \\ &=f_{c}+50 \mathrm{kHz} \\ =& 70 \mathrm{kHz} \; \; ( \because \text { Given that, } f_{c}=20 \mathrm{kHz}) \end{aligned} \end{array}
There are 10 questions to complete.