GATE 2021 Signals and Systems Questions and Solutions

Two discrete-time LTI systems with impulse responses \(h_1[n]=\delta[n-1]+\delta[n+1]\) and \(h_2[n]=\delta[n]+\delta[n-1]\) are connected in cascade (\(\delta[n]\) is the Kronecker delta). The impulse response of the cascaded system is:

1. \(\delta[n-2]+\delta[n+1]\)
2. \(\delta[n-1]\,\delta[n]+\delta[n+1]\,\delta[n-1]\)
3. \(\delta[n-2]+\delta[n-1]+\delta[n]+\delta[n+1]\)
4. \(\delta[n]\,\delta[n-1]+\delta[n-2]\,\delta[n+1]\)

Solution

The cascade impulse response is the convolution \(h[n]=h_1[n]*h_2[n]\), which is the product of the two \(z\)-transforms:

Taking the inverse \(z\)-transform term by term gives

A continuous-time signal is \(x(t)=1-|t|\) for \(|t|\le 1\) and \(x(t)=0\) for \(|t|>1\) (a unit triangular pulse). With \(X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt\), the maximum magnitude of \(X(\omega)\) is ____.

Solution

The triangular pulse of width \(2\) and height \(1\) has the transform

Since \(\operatorname{Sa}^2(\cdot)\le 1\) with its maximum at the origin, \(|X(\omega)|\) is largest at \(\omega=0\), where \(X(0)=\int_{-\infty}^{\infty}x(t)\,dt\) equals the pulse area \(=1\).

Let \(f(t)\) be even, with Fourier transform \(F(\omega)\). Suppose \(\dfrac{dF(\omega)}{d\omega}=-\omega F(\omega)\) for all \(\omega\), and \(F(0)=1\). Then:

1. \(f(0)<1\)
2. \(f(0)>1\)
3. \(f(0)=1\)
4. \(f(0)=0\)

Solution

Solving the first-order ODE \(F'(\omega)+\omega F(\omega)=0\) with \(F(0)=1\) gives a Gaussian:

Hence \(f(0)=\dfrac{1}{\sqrt{2\pi}}\approx0.399<1\).

If the input \(x(t)\) and output \(y(t)\) of a system are related by \(y(t)=\max\!\big(0,\,x(t)\big)\), then the system is:

1. linear and time-variant
2. linear and time-invariant
3. non-linear and time-variant
4. non-linear and time-invariant

Solution

Half-wave rectification violates additivity: for \(x_1=-2\Rightarrow y_1=0\) and \(x_2=1\Rightarrow y_2=1\), the response to \(x_1+x_2=-1\) is \(0\), not \(y_1+y_2=1\). So the system is non-linear. The rule contains no explicit time dependence, so delaying the input merely delays the output — the system is time-invariant.

The causal signal with \(z\)-transform \(z^{2}(z-a)^{-2}\) is (\(u[n]\) is the unit step):

1. \(a^{2n}u[n]\)
2. \((n+1)a^{n}u[n]\)
3. \(n^{-1}a^{n}u[n]\)
4. \(n^{2}a^{n}u[n]\)

Solution

Start from the standard pair \(n\,a^{n}u[n]\leftrightarrow \dfrac{az}{(z-a)^2}\), so that \(n\,a^{n-1}u[n]\leftrightarrow \dfrac{z}{(z-a)^2}\). The extra factor of \(z\) is a one-step advance:

The last equality holds because the \((n+1)\) factor kills the \(n=-1\) term.