GATE 2012 Signals and Systems Questions and Solutions

The input \(x(t)\) and output \(y(t)\) of a system are related by

The system is:

1. time-invariant and stable
2. stable and not time-invariant
3. time-invariant and not stable
4. not time-invariant and not stable

Solution

Time-invariance: for a shifted input \(x(t-t_0)\), substituting \(\tau'=\tau-t_0\) gives

because the explicit \(\cos(3\tau)\) factor is not shifted. The system is not time-invariant.

Stability: take the bounded input \(x(t)=\cos(3t)\). Then \(x(\tau)\cos(3\tau)=\cos^2(3\tau)=\tfrac{1}{2}\big(1+\cos 6\tau\big)\), and

so a bounded input yields an unbounded output. The system is not stable.

Note: this paper's draft key marked (B). The worked analysis above shows the system is neither time-invariant nor stable, so the correct choice is (D), which is also the official answer.

The Fourier transform of a signal \(h(t)\) is \(H(j\omega)=2\cos(\omega)\dfrac{\sin(2\omega)}{\omega}\). The value of \(h(0)\) is:

1. \(\tfrac{1}{4}\)
2. \(\tfrac{1}{2}\)
3. \(1\)
4. \(2\)

Solution

Write \(\cos\omega = \tfrac{1}{2}(e^{j\omega}+e^{-j\omega})\) and let \(X(\omega)=\dfrac{2\sin 2\omega}{\omega}\), which is the transform of the rectangular pulse \(x(t)=1\) for \(-2

Therefore \(h(0) = \tfrac{1}{2}\big[x(1)+x(-1)\big] = \tfrac{1}{2}(1+1) = 1\).

Consider \(\dfrac{d^2y}{dt^2} + 2\dfrac{dy}{dt} + y = \delta(t)\) with \(y(0^-)=-2\) and \(\left.\tfrac{dy}{dt}\right|_{0^-}=0\). Find \(\left.\dfrac{dy}{dt}\right|_{0^+}\).

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

Solution

Taking the Laplace transform with the given initial conditions:

Inverting, \(y(t) = -2e^{-t} - t e^{-t}\), so

Let \(y[n]\) be the convolution of \(h[n]=\left(\tfrac{1}{2}\right)^n u[n]\) and a causal sequence \(g[n]\). If \(y[0]=1\) and \(y[1]=\tfrac{1}{2}\), then \(g[1]\) equals:

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

Solution

With \(h[0]=1,\ h[1]=\tfrac{1}{2}\) and \(g[n]=0\) for \(n<0\):

The unilateral Laplace transform of \(f(t)\) is \(F(s)=\dfrac{1}{s^2+s+1}\). The transform of \(t\,f(t)\) is:

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

Solution

By the frequency-differentiation property \(\mathcal{L}\{t f(t)\} = -\dfrac{d}{ds}F(s)\):

If \(x[n]=\left(\tfrac{1}{3}\right)^{|n|} - \left(\tfrac{1}{2}\right)^n u[n]\), the region of convergence of its Z-transform is:

1. \(\tfrac{1}{3}<|z|<3\)
2. \(\tfrac{1}{3}<|z|<\tfrac{1}{2}\)
3. \(\tfrac{1}{2}<|z|<3\)
4. \(|z|>\tfrac{1}{3}\)

Solution

The two-sided term \(\left(\tfrac{1}{3}\right)^{|n|}\) splits into a causal part (ROC \(|z|>\tfrac{1}{3}\)) and an anti-causal part (ROC \(|z|<3\)), giving \(\tfrac{1}{3}<|z|<3\). The second term \(\left(\tfrac{1}{2}\right)^n u[n]\) has ROC \(|z|>\tfrac{1}{2}\). The overall ROC is the intersection: