GATE 2020 Signals and Systems Questions and Solutions

For input \(x(t)\) and impulse response \(h(t)\), an LTI system gives \(x(t)*h(t)=y(t)\). Define \(|x(t)|*|h(t)|=z(t)\). Which statement is true?

1. For all \(t\in(-\infty,\infty)\), \(z(t)\le y(t)\)
2. For some but not all \(t\), \(z(t)\le y(t)\)
3. For all \(t\in(-\infty,\infty)\), \(z(t)\ge y(t)\)
4. For some but not all \(t\), \(z(t)\ge y(t)\)

Solution

By the triangle inequality applied to the convolution integral,

Since \(z(t)\) is itself non-negative and \(z(t)\ge|y(t)|\ge y(t)\), we have \(z(t)\ge y(t)\) for all \(t\).

Which statement is true about the two-sided Laplace transform?

1. It exists for every signal that may or may not have a Fourier transform.
2. It has no poles for any bounded signal that is non-zero only inside a finite time interval.
3. The number of finite poles and finite zeros must be equal.
4. If a signal is a weighted sum of shifted one-sided exponentials, its Laplace transform has no poles.

Solution

A bounded signal of finite duration has a region of convergence that is the entire \(s\)-plane. Since the ROC can never contain a pole, such a signal has no finite poles. Hence (B) is correct.

For a discrete-time LTI system obeying \(y[n]-a\,y[n-1]=b_0 x[n]-b_1 x[n-1]\), which statement is true?

1. \(y[n]\) is unaffected by the values of \(x[n-k]\), \(k>2\)
2. The system is necessarily causal
3. The system impulse response is non-zero at infinitely many instants
4. When \(x[n]=0\) for \(n<0\), \(y[n]\) for \(n>0\) is determined by \(x[n]\)

Solution

The transfer function is \(H(z)=\dfrac{b_0-b_1 z^{-1}}{1-a z^{-1}}\), a single pole at \(z=a\). Its inverse can be taken right-sided (causal) or left-sided (anti-causal) depending on the ROC, so the system is not necessarily causal. Either way the pole at \(z=a\) makes \(h[n]\propto a^{n}\) non-zero at infinitely many instants. Hence (C).

The causal realization of a transfer function \(H(s)\) with poles at \(2-j,\,-2+j\) and zeros at \(2+j,\,-2-j\) will be:

1. stable, real, all-pass
2. unstable, complex, all-pass
3. unstable, real, high-pass
4. stable, complex, low-pass

Solution

Each zero is the mirror image (through the origin) of a pole, so \(|H(j\omega)|\) is constant — an all-pass system. One pole (\(2-j\)) lies in the right half-plane, so the causal realization is unstable. The poles and zeros do not occur in complex-conjugate pairs, so the coefficients are complex (the system is not real).

Consider \(x[n]=\left(\tfrac12\right)^{n}u[n]\) (with \(u[n]=1\) for \(n\ge0\), else \(0\)). The Z-transform of \(x[n-k]\), \(k>0\), is \(\dfrac{z^{-k}}{1-\tfrac12 z^{-1}}\), with region of convergence:

1. \(|z|<2\)
2. \(|z|>2\)
3. \(|z|<\tfrac12\)
4. \(|z|>\tfrac12\)

Solution

The ROC of \(x[n]\) is \(|z|>\tfrac12\). A time shift multiplies by \(z^{-k}\), which does not change the ROC (apart from possibly the point \(z=0\)). Hence the ROC remains \(|z|>\tfrac12\).