Q21GATE 2015MCQ

Consider the signal $s(t)=m(t)cos(2\pi f_{c}t)+\hat{m}(t)sin(2\pi f_{c}t) \; where\; \hat{m}t$ denotes the Hilbert transform of m(t) and the bandwidth of m(t) is very small compared to $f_{c}$ . The signal s(t) is a

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Q22GATE 2015MCQ

A message signal $m(t)=A_{m} sin(2\pi f_{m}f)$ is used to modulate the phase of a carrier $A_{c}cos(2\pi f_{c}t)$ to get the modulated signal $y(t)=A_{c} cos(2\pi f_{c}t+m(t))$ . The bandwidth of y(t).

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📖 Explanation

The bandwidth of a PM signal is given by $B=2\left(\Delta f+f_{m}\right)$ Now, for single tone PM signal

\begin{aligned} \Delta f&=A_{m} K_{\rho} f_{m} \\ \therefore \quad B&=2\left(K_{p} A_{m}+1\right) f_{m} \end{aligned}

Q23GATE 2015NAT

In the system shown in Figure (a), m(t) is a low-pass signal with bandwidth W Hz. The frequency response of the band-pass filter H(f) is shown in Figure (b). If it is desired that the output signal z(t) = 10x(t), the maximum value of W (in Hz) should be strictly less than ________.

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📖 Explanation

\begin{array}{l} \quad x(t)=m(t) \cos 2400 \pi t \\ \text { thus } \quad y(t)=10 x(t)+x^{2}(t) \\ \therefore y(t)=10 m(t) \cos (2400 \pi t)+m^{2}(t) \cos ^{2}(2400 \pi t) \\ =10 m(t) \cos (2400 \pi t)+\frac{1}{2} m^{2}(t) \frac{1}{2} \cos (4800 \pi t) \end{array}

Drawing the spectrum set spectrum of x(t) be $y(t) = 10x(t) + x^{2}(t)$ To recover z(t)

\begin{aligned} 1200-W \gt 700 & \Rightarrow W \lt 500 \\ 1200+W \lt 1700 & \Rightarrow W \lt 500 \\ 2 W \lt 700 & \Rightarrow W \lt 350 \\ 2400-2 W \gt 1700 & \Rightarrow W \lt 350 \end{aligned}

So, that maximum value of W should be less than 350 Hz

Q24GATE 2015MCQ

The complex envelope of the bandpass signal $x(t)=-\sqrt{2}(\frac{sin(\pi t/ 5)}{\pi t/ 5})sin(\pi -\frac{\pi}{4})$ , centered about $f=\frac{1}{2}$ Hz, is

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📖 Explanation

\begin{aligned} x(t) &=-\sqrt{2}\left[\frac{\sin \left(\frac{\pi t}{5}\right)}{\left(\frac{\pi t}{5}\right)}\right] \sin \left(\pi t-\frac{\pi}{4}\right) \\ x(t) &=\text{Re}\left[\tilde{x}(t) e^{i 2 \pi t_{c} t}\right]\$ \\ \tilde{x}(t) &=\text { Complex envelope } \\ &=\sqrt{2}\left[\frac{\sin \left(\frac{\pi t}{5}\right)}{\left(\frac{\pi t}{5}\right)}\right] e^{i \pi / 4} \end{aligned}

Q25GATE 2014NAT

In a double side-band (DSB) full carrier AM transmission system, if the modulation index is doubled, then the ratio of total sideband power to the carrier power increases by a factor of ______.

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📖 Explanation

Ratio of $P_{S B}$ to $P_{C}$ is

\begin{array}{l} \quad P_{S B}={P}_{C}{\mu}^{2} \\ \begin{array}{l} \therefore \quad \frac{P_{S B}}{P_{C}}=\frac{\mu^{2}}{2} \\ \text { and }\left(\frac{P_{S B}}{P_{C}}\right)_{1}=\frac{\mu_{1}^{2}}{2} \\ \left(\frac{P_{S B}}{P_{C}}\right)_{2}=\frac{\mu_{2}^{2}}{2} \\ \text { where, } \quad\left(\frac{P_{S B}}{P_{C}}\right)_{2}=\frac{\mu_{2}^{2}}{2} \\ \text { hence } \frac{\left(P_{S B} / P_{C}\right)_{1}}{\left(P_{S B} / P_{C}\right)_{2}}=\frac{1}{4} \\ \left(\frac{P_{S B}}{P_{C}}\right)_{2}=4\left(\frac{P_{S B}}{P_{C}}\right)_{1} \end{array} \end{array}

Four times.

Q26GATE 2014NAT

Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index ( $\mu$ ) when the maximum and minimum values of the envelope, respectively, are 3 V and 1 V, is ______.

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📖 Explanation

$\mu=\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }}=\frac{3-1}{3+1}=\frac{2}{4}=0.50$

Q27GATE 2014MCQ

Consider an FM signal $f(t)=cos[2\pi f_{c}+\beta _{1}sin 2 \pi f_{1}t+\beta _{2}sin2\pi f_{2}t]$ . The maximum deviation of instantaneous frequency from the carrier frequency $f_{c}$ is

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📖 Explanation

Maximum frequency deviation is

\begin{array}{l} =-\frac{1}{2 \pi}\left|\frac{d \phi}{d t}\right|_{\max }\\ \phi \rightarrow \text{ Phase deviation}\\ =\frac{1}{2 \pi}\left|\beta_{1} 2 \pi f_{1} \cos 2 \pi f_{1} t+\beta_{2} 2 \pi f_{2} \cos 2 \pi t_{2} t\right|_{\max } \\ \quad=\frac{1}{2 \pi}\left(2 \pi \beta_{1} f_{1}+2 \pi \beta_{2} f_{2}\right) \\ \quad=\beta_{1} f_{1}+\beta_{2} f_{2} \end{array}

Q28GATE 2014NAT

A modulated signal is $y(t)=m(t)cos(40000\pi t)$ , where the baseband signal m(t) has frequency components less than 5 kHz only. The minimum required rate (in kHz) at which y(t) should be sampled to recover m(t) is _____.

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📖 Explanation

The minimum sampling rate is $f_{s}=2 f_{m}=2 \times 5 \mathrm{kHz}=10 \mathrm{kHz}$

Q29GATE 2014NAT

If the figure, M(f) is the Fourier transform of the message signal m(t) where A = 100 Hz and B = 40 Hz. Given $v(t)= cos(2 \phi f_{c}t)$ and $w(t)=cos(2 \pi (f_{c}+A)t)$ , where $f_{c} \gt A$ . The cutoff frequencies of both the filters are $f_{c}$ . The bandwidth of the signal at the output of the modulator (in Hz) is__

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📖 Explanation

After passing through first multiplier the output is $m(t) x v(t) = x(t)$ Therefore, The Fourier transform of output y(t) of the high pass filter centred at $f_{c}$ is Now, the output of second multiplier is $y(t) \cdot w(t) = z( t)$ Therefore, After passing through the low pass filter centerd at $f_{c}$ the Fourier transform of output s(t) is Thus, bandwidth $= (A - B) - 0$ $= 100 Hz - 40 Hz = 60 Hz$

Q30GATE 2012MCQ

The signal m(t) as shown is applied both to a phase modulator (with $k_{p}$ as the phase constant) and a frequency modulator (with $k_{f}$ as the frequency constant) having the same carrier frequency The ratio $k_{p}/k_{f}$ (in rad/Hz) for the same maximum phase deviation is

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📖 Explanation

For phase modulator $\phi(t)=2 \pi f_{c} t+k_{p} m(t)$ maximum phase deviation is $\left(\phi_{D}\right)_{\max }=k_{p} \max [m(t)]=2 k_{p} \quad\ldots(1)$ for frequency modulator

\begin{aligned} \phi(t) &=2 \pi f_{c} t+2 \pi k_{f} \int_{0}^{t} m(t) d t \\ \left(\phi_{D}^{\prime}\right)_{\max } &=2 \pi k_{t} t\left[\int_{0}^{t} m(t) d t\right]_{\max } \\ \left(\phi_{D}^{\prime}\right)_{\max } &=2 \pi k_{t}\left[\int_{0}^{2} m(t) d t\right]\$ \\ \left(\phi_{D}^{\prime}\right)_{\max } &=2 \pi k_{f}\left[\int_{0}^{2} 2 d t\right]\$ \\ \left(\phi_{D}^{\prime}\right)_{\max } &=8 \pi k_{f} &\ldots(2)\\ \text{given}\\ \left(\phi_{D}^{\prime}\right)_{\max }&=\left(\phi_{D}\right)_{\max }\\ 8 \pi k_{f}&=2 k_{p} \\ \mid \frac{k_{p}}{k_{f}}&=4 \pi \end{aligned}

Q31GATE 2011MCQ

A message signal $m(t) = \cos 2000 \pi t + 4 \cos 4000 \pi t$ modulates the carrier $c(t) = \cos 2 \pi f_{c} t$ where $f_{c}$ =1MHz to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy

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📖 Explanation

\begin{aligned} \frac{I}{f_{c}}& \lt \lt R C\lt \lt \frac{1}{f_{m}} \\ \frac{I}{1 \mathrm{MHz}}&\lt \lt R C\lt \lt \frac{1}{2 \mathrm{kHz}} \\ \Rightarrow \quad \quad 1 \mu \mathrm{s}&\lt \lt R C\lt \lt 0.5 \mathrm{ms} \end{aligned}

Q32GATE 2011MCQ

The Column-I lists the attributes and the Column-II lists the modulation systems. Match the attribute to the modulation system that best meets it

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Q33GATE 2010MCQ

Consider an angle modulated signal $x(t)=6cos[2\pi \times 10^{6}t+2sin(8000\pi t)+$ $4cos(8000\pi t)]$ V. The average power of x(t) is

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📖 Explanation

Average power of angle modulated signal x(t) is $A_{c}^{2} / 2. Here, A_{c}=6$ Therefore Average power $=\frac{6^{2}}{2}=18 \mathrm{W}$

Q34GATE 2010MCQ

Suppose that the modulating signal is $m(t)=2cos(2\pi f_{m}t)$ and the carrier signal is $X_{c}(t)=A_{c}cos(2\pi f_{c}t)$ , which one of the following is a conventional AM signal without over-modulation

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📖 Explanation

Conventional AM signal is

\begin{aligned} x(t) &=A_{c}[1+m(t)] \cos \left(2 \pi f_{c} t\right) \\ &=A_{c} \cos \left(2 \pi f_{c} t\right)+A_{c} m(t) \cos \left(2 \pi f_{c} t\right) \end{aligned}

In option (B), Modulation index $=\frac{2}{1}=2$ So, it is conventional AM signal but with overmodulation. In option (C) $x(t)=A_{c} \cos \left(2 \pi f_{c} t\right)+\frac{A_{c}}{4} m(t) \cos \left(2 \pi f_{c} t\right)$ Here,Modulation index $=\frac{2}{4}=\frac{1}{2}$ Therefore, it is conventional AM signal without over-modulation.

Q35GATE 2009MCQ

A message signal given by $m(t)=(\frac{1}{2})cos\omega _{1}t-(\frac{1}{2})sin\omega _{2}t$ is amplitude-modulated with a carrier of frequency $\omega_{c}$ to generate $s(t)=[1 + m(t)]cos\omega _{c} t$ What is the power efficiency achieved by this modulation scheme ?

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📖 Explanation

\begin{aligned} \eta &=\frac{m_{a}^{2}}{2+m_{a}^{2}} \times 100 \% \\ m_{a_{1}} &=\frac{V_{m}}{V_{c}}=\frac{1}{2}=m_{a_{2}} \\ m_{a} &=\sqrt{m_{a_{1}}^{2}+m_{a_{2}}^{2}} \\ \eta &=\frac{0.25 \times 2}{2+0.25 \times 2} \times 100 \%=20 \% \end{aligned}

Q36GATE 2009MCQ

For a message signal $m(t)=cos(2\pi f_{m}t)$ and carrier of frequency $f_{c}$ , which of the following represents a single side-band (SSB) signal ?

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📖 Explanation

$\cos \left[2 \pi\left(f_{c}+f_{m}\right) t\right]$ represents only USB of AMSSB/SC signal.

Q37GATE 2008MCQ

The signal $\cos\omega _{c}t+0.5 \cos\omega _{m}t \sin \omega_{c} t$ is

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📖 Explanation

The signal $\cos \omega_{c} t+0.5 \cos \omega_{m} t \sin \omega_{c} t$ is either AM or narrow-band FM signal.

Q38GATE 2008MCQ

Consider the amplitude modulated (AM) signal $A_{c}cos\omega _{c}t+2cos\omega _{m}tcos\omega _{c}t$ . For demodulating the signal using envelope detector, the minimum value of $A_{c}$ should be

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📖 Explanation

Modulated signal,

\begin{aligned} \phi_{\text {AU }}(t) &=A_{c} \cos \omega_{c} t+2 \cos \omega_{m} t \cos \omega_{c} t \\ &=A_{c}\left[1+\frac{2}{A_{c}}-\cos \omega_{m} t\right] \cos \omega_{c} t \\ \mu &=\frac{2}{A_{c}} \end{aligned}

Conditionfor envelope detection of an AM signal is

\begin{aligned} \mu &\leq 1\\ \Rightarrow \quad \frac{2}{A_{c}} &\leq 1 or A_{c} \geq 2 \\ \end{aligned}

Therefore minimum value of $A_{C}$ should be 2

Q39GATE 2008MCQ

Consider the frequency modulated signal $10cos[2\pi \times 10^{5}t+5sin(2 \pi \times 1500t)$ $+ 7.5sin(2 \pi \times 1000t)]$ with carrier frequency of $10^{5}$ Hz. The modulation index is

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📖 Explanation

Modulation index, $m_{t}=\frac{\delta}{f_{m}}$ Where, $\quad \delta=$ maximum frequency deviation $f_{m}=$ maximum frequency component Given that. $f_{m}=1500 \mathrm{Hz}$ Deviation,

\begin{array}{l} \Delta \theta=5 \sin (2 \pi \times 1500 t)+7.5 \sin (2 \pi \times 1000 t) \\ \Delta \omega=\frac{d \theta}{d t}=5 \times 2 \pi \times 1500 \cos (2 \pi \times 500 t) \\ +7.5 \times 2 \pi \times 1000 \cos (2 \pi \times 1000 t) \\ \Delta w_{\max }=2 \pi(7500+7500) \\ \delta=\frac{\Delta \omega_{\max }}{2 \pi}=15000 \mathrm{Hz} \\ m_{f}=\frac{\delta}{f_{m}}=\frac{15000}{1500}=10 \end{array}

Q40GATE 2007MCQ

In the following scheme, if the spectrum M(f) of m(t) is as shown, then the spectrum Y(f) of y(t) will be

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📖 Explanation

In nonlinear or switching modulation schemes, mixing a baseband signal $M(f)$ with a carrier introduces spectral shifts. When passed through a bandpass filter centered at the carrier frequency, the resulting spectrum $Y(f)$ consists of the original baseband spectrum translated to $\pm f_c$ . The correct amplitude and shape depend on the gating/switching function's Fourier series, but the basic form is a double-sideband spectrum at the carrier. Option B correctly identifies this translated spectrum diagram.