GATE 2010 Signals and Systems Questions and Solutions

Given \(f(t)\) and \(g(t)\) as shown below, the Laplace transform of \(g(t)\) is:

1. \(\dfrac{1}{s}(e^{3s}-e^{5s})\)
2. \(\dfrac{1}{s}(e^{-5s}-e^{-3s})\)
3. \(\dfrac{e^{-3s}}{s}(1-e^{-2s})\)
4. \(\dfrac{1}{s}(e^{5s}-e^{3s})\)

Solution

The pulse \(g(t)\) equals 1 for \(3

For the same \(f(t)\) and \(g(t)\), \(g(t)\) can be expressed as:

1. \(g(t)=f(2t-3)\)
2. \(g(t)=f\!\left(\tfrac{t}{2}-3\right)\)
3. \(g(t)=f\!\left(2t-\tfrac{3}{2}\right)\)
4. \(g(t)=f\!\left(\tfrac{t}{2}-\tfrac{3}{2}\right)\)

Solution

Since \(g(t)\) has width 2 while \(f(t)\) has width 1, first expand \(f\) by a factor of 2 (scale time by \(\tfrac{1}{2}\)): \(f_1(t)=f\!\left(\tfrac{t}{2}\right)\). Then shift by 3:

Given the finite-length input \(x[n]\) and output \(y[n]\) of an LTI system as shown, the impulse response \(h[n]\) is:

1. \(h[n]=\{1,0,0,1\}\)
2. \(h[n]=\{1,0,1\}\)
3. \(h[n]=\{1,1,1,1\}\)
4. \(h[n]=\{1,1,1\}\)

Solution

From the figure, \(x[n]=\{1,-1\}\) and \(y[n]=\{1,0,0,0,-1\}\). Using \(Y(z)=X(z)H(z)\) with \(X(z)=1-z^{-1}\) and \(Y(z)=1-z^{-4}\):

Therefore \(h[n]=\{1,1,1,1\}\). (Equivalently, the output length \(5 = M+N-1 = 2+N-1\) gives \(N=4\) taps.)

Let \(x(t)\) be a rectangular pulse of unit height from \(t=-1\) to \(t=+1\). Find \(\displaystyle\int_{-\infty}^{\infty}|X(\omega)|^2\,d\omega\), where \(X(\omega)\) is the Fourier transform of \(x(t)\).

1. 2
2. \(2\pi\)
3. 4
4. \(4\pi\)

Solution

By Parseval's theorem,

The second-harmonic component of the periodic waveform shown has an amplitude of:

1. 0
2. 1
3. \(\tfrac{2}{\pi}\)
4. 5

Solution

The waveform is odd and possesses half-wave symmetry. Such a signal contains only sine terms at odd harmonics; all even harmonics are absent. Hence the second-harmonic amplitude is \(0\).

The system represented by \(\displaystyle y(t)=\int_{-\infty}^{5t} x(\tau)\,d\tau,\ t>0\), is:

1. Linear and causal
2. Linear but not causal
3. Causal but not linear
4. Neither causal nor linear

Solution

Integration is a linear operation, so the system is linear. However, the upper limit \(5t\) exceeds \(t\) for \(t>0\), so the output depends on future inputs — the system is not causal.