# Analog Communication Systems

 Question 1
$S_{PM}(t)\; and \; S_{FM}(t)$ 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 1 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 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 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 6
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 6 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}
 Question 7
The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________
 A 4.1 B 4.8 C 5.2 D 5.8
GATE EC 2017-SET-2   Communication Systems
Question 7 Explanation:
\begin{aligned} \text { Given that, } P_{C}&=5 \mathrm{kW} \text { for } \mu_{(\mathrm{max})}=0.5\\ \text { So, } \quad P_{t(\max )}&=P_{c}\left[1+\frac{(0.5)^{2}}{2}\right]\\ &=5\left[1+\frac{0.25}{2}\right] \mathrm{kW}=5.625 \mathrm{kW} \\ \text { For } \mu & =0.4,\\ &\begin{aligned} P_{ \text {Qrex } )}\left(1+\frac{\mu^{2}}{2}\right)=&P_{t(\max )} \\ P_{C(\text{rax})}=& \frac{5.625}{1+\frac{(0.4)^{2}}{2}} \mathrm{kW} \\ & =5.208 \mathrm{kW} \end{aligned} \end{aligned}
 Question 8
For a superheterodyne receiver, the intermediate frequency is 15 MHz and the local oscillator frequency is 3.5 GHz. If the frequency of the received signal is greater than the local oscillator frequency, then the image frequency (in MHz) is __________
 A 3240 B 6422 C 3485 D 8432
GATE EC 2016-SET-3   Communication Systems
Question 8 Explanation:
\begin{aligned} I F &=15 \mathrm{MHz} \\ f_{l} &=3.5 \mathrm{GHz} \\ f_{\mathrm{si}} &=f_{l}-\mathrm{IF} \\ &=3500-15 \mathrm{MHz}=3485 \mathrm{MHz} \end{aligned}
 Question 9
The amplitude of a sinusoidal carrier is modulated by a single sinusoid to obtain the amplitude modulated signal $s(t)=5 cos 1600\pi t+20 cos 1800\pi t +5cos 2000 \pi t$. The value of the modulation index is __________
 A 0.1 B 0.2 C 0.5 D 0.8
GATE EC 2016-SET-1   Communication Systems
Question 9 Explanation:
\begin{aligned} s(t)=& 5 \cos 1600 \pi t+20 \cos 1800 \pi t+5 \cos 2000 \pi t \\ s(t)=& 20 \cos 1800 \pi t+5 \cos 1600 \pi t+5 \cos 200 \pi t \\ s(t)=& A_{c} \cos 2 \pi f_{c} t+\frac{A_{c} \mu}{2} \cos 2 \pi\left(f_{c}-t_{m}\right) t \\ &+\frac{A_{c} \mu}{2} \cos 2 \pi\left(f_{c}+f_{m}\right) t \end{aligned}
comparing, we get
\begin{aligned} A_{c}&=20 \mathrm{v} ; \frac{A_{c} \mu}{2}=5 \mathrm{V}\\ \mu&=\frac{10}{20}=0.5 \end{aligned}
 Question 10
A superheterodyne receiver operates in the frequency range of 58 MHz - 68 MHz. The intermediate frequency $f_{IF}$ and local oscillator frequency $f_{LO}$ are chosen such that $f_{IF}$ $\leq f_{LO}$ . It is required that the image frequencies fall outside t he 58 MHz - 68 MHz band. The minimum required $f_{IF}$ (in MHz) is ________
 A 2 B 3 C 4 D 5
GATE EC 2016-SET-1   Communication Systems
Question 10 Explanation:
$f_{s}=58 \mathrm{MHz}-68 \mathrm{MHz}$
$f_{s l}$ should fall outside the range $58 \mathrm{MHz}-68 \mathrm{MHz}$
\begin{aligned} \text { Hence } \quad f_{\text {smin }} &=58 \mathrm{MHz} \\ f_{\mathrm{si}} &=f_{s}+2 I F \gt 68 \mathrm{MHz} \\ 58 \mathrm{MHz}+2 I F & \gt 68 \mathrm{MHz} \\ I F & \gt 5 \mathrm{MHz} \\ \Rightarrow (I F)_{\min }&=5 \mathrm{MHz} \end{aligned}
There are 10 questions to complete. 