GATE 2017 Signals and Systems Questions and Solutions

The signal energy of the continuous-time signal

is:

1. \(\dfrac{11}{3}\)
2. \(\dfrac{7}{3}\)
3. \(\dfrac{1}{3}\)
4. \(\dfrac{5}{3}\)

Solution

The four ramp terms build a trapezoidal pulse: \(x(t)\) rises as \((t-1)\) on \(1

Note: the draft solution computed a mean-square (power) value of 6, which is not the signal energy and does not match any option. The correct quantity is the energy \(E=5/3\), giving option (D), the official answer.

Consider the system with input-output relation \(y[n]=\big(1+(-1)^{n}\big)x[n]\), where \(x[n]\) is the input and \(y[n]\) the output. The system is:

1. invertible and time-invariant
2. invertible and time-varying
3. non-invertible and time-invariant
4. non-invertible and time-varying

Solution

Time-variance: the coefficient \(\big(1+(-1)^n\big)\) depends explicitly on \(n\). For a shifted input \(x[n-1]\) the response is \(\big(1+(-1)^n\big)x[n-1]\), but the shifted output is \(\big(1+(-1)^{n-1}\big)x[n-1]\); since \(1+(-1)^n\ne 1+(-1)^{n-1}\), the system is time-varying.

Invertibility: for odd \(n\) the gain \(1+(-1)^n=0\), so all odd-index samples of the input are annihilated and cannot be recovered. The system is non-invertible.

Note: the draft concluded "time-invariant" (option C). That is incorrect — the explicit \((-1)^n\) factor makes the system time-varying, so the correct choice is (D), the official answer.

Let \(x(t)=\displaystyle\sum_{k=-\infty}^{+\infty}(-1)^{k}\,\delta\!\left(t-\dfrac{k}{2000}\right)\) be passed through an LTI system with frequency response \(H(\omega)\) (shown below). The Fourier series representation of the output is:

1. \(4000+4000\cos(2000\pi t)+4000\cos(4000\pi t)\)
2. \(2000+2000\cos(2000\pi t)+2000\cos(4000\pi t)\)
3. \(4000\cos(2000\pi t)\)
4. \(2000\cos(2000\pi t)\)

Solution

The alternating impulse train has period \(T_0=2/2000=1/1000\), so \(\omega_0=2\pi/T_0=2000\pi\). The signal has half-wave symmetry, so only odd cosine harmonics appear, with

The harmonics are at \(2000\pi,\ 6000\pi,\dots\). The filter passes only up to \(5000\pi\), so just the fundamental survives: \(y(t)=4000\cos(2000\pi t)\).

Consider \(g(t)=\begin{cases}t-\lfloor t\rfloor, & t\ge0\\ t-\lceil t\rceil, & \text{otherwise}\end{cases}\), \(t\in\mathbb{R}\), where \(\lfloor t\rfloor\) is the floor and \(\lceil t\rceil\) the ceiling. Find the coefficient of the second-harmonic component of the Fourier series of \(g(t)\).

Solution

The function \(t-\lfloor t\rfloor\) (and its odd extension via the ceiling for \(t<0\)) is the familiar sawtooth, which is periodic. For the standard unit-period sawtooth \(g(t)=t-\lfloor t\rfloor\), the complex Fourier coefficients are \(c_k=\dfrac{j}{2\pi k}\) (\(k\ne0\)), so the magnitude of the second harmonic is \(|c_2|=\dfrac{1}{4\pi}\approx 0.08\).

Note: the draft answer ("\(g(t)\) is non-periodic, hence no Fourier series") is incorrect — the sawtooth is periodic. The exact numeric value of the "second-harmonic coefficient" depends on the precise definition, which the OCR garbled (both branches printed identically). The value above assumes the standard sawtooth; please confirm against the original paper.

Let \(z(t)=x(t)*y(t)\), where \(*\) denotes convolution, and let \(c\) be a positive real constant. The correct expression for \(z(ct)\) is:

1. \(c\,x(ct)*y(ct)\)
2. \(x(ct)*y(ct)\)
3. \(c\,x(t)*y(ct)\)
4. \(c\,x(ct)*y(t)\)

Solution

By the time-scaling property of convolution, if \(x(t)*y(t)=z(t)\) then \(x(ct)*y(ct)=\dfrac{1}{c}z(ct)\). Rearranging, \(z(ct)=c\,x(ct)*y(ct)\).

The output \(y(t)\) of the system shown is to be sampled so it can be reconstructed uniquely from its samples. The required minimum sampling rate is:

1. 1000 samples/s
2. 1500 samples/s
3. 2000 samples/s
4. 3000 samples/s

Solution

The modulator gives \(z(t)=x(t)\cos(1000\pi t)\), shifting the spectrum to \(\pm1000\pi\). The lowpass filter \(h(t)=\dfrac{\sin1500\pi t}{\pi t}\) passes components up to \(1500\pi\) rad/s, so the highest frequency in \(y(t)\) is

Hence \(f_{s,\min}=2f_m=1500\) samples/s.

Note: the draft mislabelled \(f_m\) as "150 Hz" — a typo; \(1500\pi\) rad/s equals 750 Hz, and \(2\times750=1500\) samples/s, consistent with the stated answer (B).

A cascade system has impulse responses \(h_1[n]=\{\underset{\uparrow}{1},-1\}\) and \(h_2[n]=\{\underset{\uparrow}{1},1\}\), where \(\uparrow\) marks the time origin. The input sequence \(x[n]\) producing the output \(y[n]=\{\underset{\uparrow}{1},2,1,-1,-2,-1\}\) is:

1. \(x[n]=\{1,2,1,1\}\)
2. \(x[n]=\{1,1,2,2\}\)
3. \(x[n]=\{1,1,1,1\}\)
4. \(x[n]=\{1,2,2,1\}\)

Solution

The overall impulse response is \(h[n]=h_1[n]*h_2[n]=\{1,-1\}*\{1,1\}=\{1,0,-1\}\), so \(H(z)=1-z^{-2}\). Dividing \(Y(z)\) by \(H(z)\):

so \(x[n]=\{1,2,2,1\}\). (Check: \(\{1,2,2,1\}*\{1,0,-1\}=\{1,2,1,-1,-2,-1\}\).)

The pole-zero plots of three discrete-time systems P, Q and R on the z-plane are shown. Which statement about their frequency selectivity is TRUE?

1. All three are high-pass filters.
2. All three are band-pass filters.
3. All three are low-pass filters.
4. P is low-pass, Q is band-pass and R is high-pass.

Solution

All three plots place zeros at \(z=1\) (\(\omega=0\)) and \(z=-1\) (\(\omega=\pi\)), forcing the magnitude response to zero at both DC and the Nyquist frequency. A response that vanishes at both extremes and peaks in between is band-pass, so all three are band-pass filters.

Consider a causal, stable LTI system with rational \(H(z)\) whose impulse response begins at \(n=0\), with \(H(1)=\tfrac54\). The poles are \(P_k=\tfrac{1}{\sqrt2}\exp\!\big(j\tfrac{(2k-1)\pi}{4}\big)\), \(k=1,2,3,4\), and all zeros are at \(z=0\). Let \(g[n]=j^{n}h[n]\). Find \(g[8]\).

Solution

The four poles give \(z^4+\tfrac14\) in the denominator, and four zeros at the origin give \(z^4\) in the numerator:

From \(H(1)=\dfrac{K}{1+\tfrac14}=\dfrac54\) we get \(K=\dfrac{25}{16}\). Expanding,

Since \(j^{8}=1\), \(g[8]=h[8]\approx0.098\).