GATE 2013 Signals and Systems Questions and Solutions

The impulse response of a continuous-time system is given by

The value of the step response at \(t=2\) is:

1. \(0\)
2. \(1\)
3. \(2\)
4. \(3\)

Solution

The step response is the running integral of the impulse response:

Hence \(y(2)=u(1)+u(-1)=1+0=1\).

A band-limited signal with maximum frequency \(f_m=5\) kHz is to be sampled. According to the sampling theorem, which sampling frequency (kHz) is not valid?

1. \(5\) kHz
2. \(12\) kHz
3. \(15\) kHz
4. \(20\) kHz

Solution

The Nyquist criterion requires \(f_s\ge 2f_m = 10\) kHz. Therefore \(5\) kHz is invalid.

For the periodic signal

the fundamental frequency in rad/s is:

1. \(100\)
2. \(300\)
3. \(500\)
4. \(1500\)

Solution

The component angular frequencies are \(100,\ 300,\ 500\) rad/s. The fundamental frequency is the greatest common divisor of these, so that each component is an integer harmonic of it:

The components are then the 1st, 3rd and 5th harmonics of \(\omega_0\).

Note: the draft key marked (D) 1500. That is incorrect — 1500 is not a common factor of the three frequencies. The fundamental is their GCD, 100 rad/s, which is the official answer (A).

Two systems with impulse responses \(h_1(t)\) and \(h_2(t)\) are connected in cascade. The overall impulse response of the cascaded system is:

1. product of \(h_1(t)\) and \(h_2(t)\)
2. sum of \(h_1(t)\) and \(h_2(t)\)
3. subtraction of \(h_1(t)\) and \(h_2(t)\)
4. convolution of \(h_1(t)\) and \(h_2(t)\)

Solution

A cascade multiplies transfer functions in the \(s\)-domain, \(H(s)=H_1(s)H_2(s)\), which corresponds to convolution in the time domain: \(h(t)=h_1(t)*h_2(t)\).

Which one of the following statements is NOT TRUE for a continuous-time causal and stable LTI system?

1. All the poles of the system must lie on the left side of the \(j\omega\) axis.
2. Zeros of the system can lie anywhere in the \(s\)-plane.
3. All the poles must lie within \(|s|=1\).
4. All the roots of the characteristic equation must be located on the left side of the \(j\omega\) axis.

Solution

For a continuous-time causal and stable LTI system the poles must lie in the open left half \(s\)-plane (\(\operatorname{Re}\{s\}<0\)). The condition \(|s|=1\) is a discrete-time (\(z\)-plane) notion, so statement (C) is not true.

The impulse response of a system is \(h(t)=t\,u(t)\). For an input \(u(t-1)\), the output is:

1. \(\dfrac{t^2}{2}u(t)\)
2. \(\dfrac{t(t-1)}{2}u(t-1)\)
3. \(\dfrac{(t-1)^2}{2}u(t-1)\)
4. \(\dfrac{t^2-1}{2}u(t-1)\)

Solution

Convolving \(h(t)=t\,u(t)\) with \(u(t-1)\):

and \(y(t)=0\) for \(t<1\). Hence \(y(t)=\dfrac{(t-1)^2}{2}u(t-1)\).

Assuming zero initial conditions, the response \(y(t)\) of the system shown below to a unit step input \(u(t)\) is:

1. \(u(t)\)
2. \(t\,u(t)\)
3. \(\dfrac{t^2}{2}u(t)\)
4. \(e^{-t}u(t)\)

Solution

The block is an integrator, \(H(s)=1/s\). With a unit-step input \(U(s)=1/s\):