GATE 2024 Signals and Systems Questions and Solutions

If the energy of a continuous-time signal \(x(t)\) is \(E\), and the energy of \(2x(2t-1)\) is \(cE\), then \(c\) is ____ (1 decimal place).

Solution

Time shifting leaves energy unchanged. Time scaling by \(2\) gives energy \(E/2\), and amplitude scaling by \(2\) multiplies energy by \(4\):

Consider the discrete-time systems \(\{T_1x\}[n]=x[0]+x[1]+\dots+x[n]\) and \(\{T_2x\}[n]=x[0]+\tfrac12x[1]+\dots+\tfrac{1}{2^{n}}x[n]\). Which statement is true?

1. \(T_1\) and \(T_2\) are BIBO stable.
2. \(T_1\) and \(T_2\) are not BIBO stable.
3. \(T_1\) is BIBO stable but \(T_2\) is not.
4. \(T_1\) is not BIBO stable but \(T_2\) is.

Solution

\(T_1\) is an accumulator: the bounded input \(x[n]=u[n]\) gives \(\{T_1x\}[n]=n+1\to\infty\), so \(T_1\) is unstable. For \(T_2\), any bounded input \(|x[n]|\le M\) yields

so \(T_2\) is BIBO stable.

A signal \(y(t)\) is the time-reversal of \(x(t)\), i.e. \(y(t)=x(-t)\). Which statement is always true for the convolution of \(x(t)\) and \(y(t)\)?

1. It is an even signal.
2. It is an odd signal.
3. It is a causal signal.
4. It is an anti-causal signal.

Solution

Let \(z(t)=x(t)*x(-t)\) (the autocorrelation). Replacing \(t\to-t\) and using the commutativity of convolution gives \(z(-t)=x(-t)*x(t)=z(t)\), so \(z\) is even.

Let \(X(\omega)\) be the Fourier transform of \(x(t)=e^{-t^{4}}\cos t\). The value of \(\dfrac{dX(\omega)}{d\omega}\) at \(\omega=0\) is ____ (1 decimal place).

Solution

Using the property \(t\,x(t)\leftrightarrow j\dfrac{dX(\omega)}{d\omega}\),

Here \(x(t)=e^{-t^{4}}\cos t\) is even, so \(t\,x(t)\) is odd and its integral over \((-\infty,\infty)\) is zero.

For the signal \(x(t)=e^{t^{2}}\big[u(t-1)-u(t-10)\big]\), the ROC of its Laplace transform is:

1. \(-\infty<\operatorname{Re}(s)<\infty\)
2. \(\operatorname{Re}(s)\ge 10\)
3. \(\operatorname{Re}(s)\le 1\)
4. \(1\le\operatorname{Re}(s)\le 10\)

Solution

The bracket restricts \(x(t)\) to the finite interval \(1\le t\le 10\). A finite-duration signal has a Laplace transform that converges for the entire \(s\)-plane.

The input \(x(t)\) and output \(y(t)\) of a system are related by \(y(t)=e^{-t}\displaystyle\int_{-\infty}^{t}e^{\tau}x(\tau)\,d\tau\). The system is:

1. nonlinear
2. linear and time-invariant
3. linear but not time-invariant
4. noncausal

Solution

Rewriting, \(y(t)=\int_{-\infty}^{t}e^{-(t-\tau)}x(\tau)\,d\tau=x(t)*h(t)\) with \(h(t)=e^{-t}u(t)\). A convolution with a fixed kernel is both linear and time-invariant (and causal, since \(h(t)=0\) for \(t<0\)).

If the \(z\)-transform of a finite-duration signal \(x[n]\) is \(X(z)\), then the \(z\)-transform of \(y[n]=x[2n]\) is:

1. \(Y(z)=X(z^{2})\)
2. \(Y(z)=\tfrac12\big[X(z^{-1/2})+X(-z^{-1/2})\big]\)
3. \(Y(z)=\tfrac12\big[X(z^{1/2})+X(-z^{1/2})\big]\)
4. \(Y(z)=\tfrac12\big[X(z^{2})+X(-z^{2})\big]\)

Solution

Writing \(Y(z)=\sum_n x(2n)z^{-n}\) and substituting \(m=2n\), the even-index selector \(\tfrac12[1+(-1)^m]\) gives