GATE ECE · Communication Systems

Analog Communication Systems Practice Questions

Practice problems for Analog Communication Systems

62 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025MCQ

Consider a frequency-modulated (FM) signal $f(t) = A_c \cos\left(2\pi f_c t + 3\sin(2\pi f_1 t) + 4\sin(6\pi f_1 t)\right)$ , where $A_c$ and $f_c$ are, respectively, the amplitude and frequency (in Hz) of the carrier waveform. The frequency $f_1$ is in Hz, and assume that $f_c > 100f_1$ . The peak frequency deviation of the FM signal in Hz is _______.

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📖 Explanation

The general form of an FM signal is: $f(t) = A_c \cos\left(2\pi f_c t + \phi(t)\right)$ , where $\phi(t)$ is the instantaneous phase deviation. Here, $\phi(t) = 3\sin(2\pi f_1 t) + 4\sin(6\pi f_1 t)$ To find instantaneous frequency deviation, we take the derivative of $\phi(t)$ : $\frac{d\phi(t)}{dt} = 3 \cdot 2\pi f_1 \cos(2\pi f_1 t) + 4 \cdot 6\pi f_1 \cos(6\pi f_1 t)$ $= 6\pi f_1 \cos(2\pi f_1 t) + 24\pi f_1 \cos(6\pi f_1 t)$ So, instantaneous frequency deviation: $\Delta f(t) = \frac{1}{2\pi} \frac{d\phi(t)}{dt} = 3f_1 \cos(2\pi f_1 t) + 12f_1 \cos(6\pi f_1 t)$ Peak frequency deviation is the maximum absolute value of $\Delta f(t)$ : $\Delta f_{\text{peak}} = \max|\Delta f(t)| = 3f_1 + 12f_1 = 15f_1$

Q10GATE 2020NAT

$S_{PM}(t)\; and \; S_{FM}(t)$ 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.

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📖 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}

Q13GATE 2018MCQ

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

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📖 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}

Q16GATE 2017NAT

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 __________

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📖 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}