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
1011313.7
B
12543.3
C
124534.2
D
1025361.2
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.