GATE 2008 Signals and Systems Questions and Solutions

A signal \(e^{-\alpha t}\sin(\omega t)\) is the input to a real linear time-invariant system. Given \(K\) and \(\phi\) are constants, the output of the system will be of the form \(K e^{-\beta t}\sin(\nu t + \phi)\) where:

1. \(\beta\) need not be equal to \(\alpha\) but \(\nu\) equals \(\omega\)
2. \(\nu\) need not be equal to \(\omega\) but \(\beta\) equals \(\alpha\)
3. \(\beta\) equals \(\alpha\) and \(\nu\) equals \(\omega\)
4. \(\beta\) need not be equal to \(\alpha\) and \(\nu\) need not be equal to \(\omega\)

Solution

For a real LTI system, the input \(e^{-\alpha t}\sin(\omega t)\) is a damped sinusoid built from the complex exponentials \(e^{(-\alpha \pm j\omega)t}\). These exponentials are eigenfunctions of the system, so the output contains exactly the same complex frequencies \(-\alpha \pm j\omega\). The system can change the amplitude \(K\) and phase \(\phi\), but the decay rate and oscillation frequency are preserved: \(\beta = \alpha\) and \(\nu = \omega\).

The impulse response of a causal LTI system is \(h(t)\). Consider:
Statement (I): Principle of superposition holds.
Statement (II): \(h(t)=0\) for \(t<0\).
Which one of the following is correct?

1. Statement (I) is correct and Statement (II) is wrong
2. Statement (II) is correct and Statement (I) is wrong
3. Both Statement (I) and Statement (II) are wrong
4. Both Statement (I) and Statement (II) are correct

Solution

A linear system obeys superposition, so Statement (I) is correct. A causal system cannot respond before the input is applied, so \(h(t)=0\) for \(t<0\), making Statement (II) correct as well.

Given a sequence \(x[n]\), to generate the sequence \(y[n]=x[3-4n]\), which one of the following procedures is correct?

1. First delay \(x[n]\) by 3 samples, then pick every 4th sample, then time-reverse.
2. First advance \(x[n]\) by 3 samples, then pick every 4th sample, then time-reverse.
3. First pick every 4th sample, time-reverse, then advance by 3 samples.
4. First pick every 4th sample, time-reverse, then delay by 3 samples.

Solution

We need \(y[n] = x[-4n+3]\). Build it in three steps:

1. Advance by 3: \(z_1[n] = x[n+3]\).
2. Decimate (every 4th sample): \(z_2[n] = z_1[4n] = x[4n+3]\).
3. Time-reverse: \(y[n] = z_2[-n] = x[-4n+3]\). ✓

A system with input \(x(t)\) and output \(y(t)\) is defined by

The system will be:

1. Causal, time-invariant and unstable
2. Causal, time-invariant and stable
3. Non-causal, time-invariant and unstable
4. Non-causal, time-variant and unstable

Solution

Causality: at \(t=-2\), \(y(-2)=\int_{-\infty}^{4}x(\tau)\,d\tau\) depends on future input values, so the system is non-causal.

Time-invariance: the upper limit \(-2t\) scales time, so a shifted input does not produce a correspondingly shifted output: \(y'(t)\ne y(t-t_0)\). The system is time-variant.

Stability: integrating over an unbounded interval gives an unbounded output for a bounded input, so the system is unstable.

A signal \(x(t)=\mathrm{sinc}(\alpha t)\), where \(\mathrm{sinc}(x)=\dfrac{\sin(\pi x)}{\pi x}\), is the input to an LTI system with impulse response \(h(t)=\mathrm{sinc}(\beta t)\). With \(K\) a constant, which statement about the output is true?

1. It is of the form \(K\,\mathrm{sinc}(\gamma t)\) with \(\gamma=\min(\alpha,\beta)\)
2. It is of the form \(K\,\mathrm{sinc}(\gamma t)\) with \(\gamma=\max(\alpha,\beta)\)
3. It is of the form \(K\,\mathrm{sinc}(\alpha t)\)
4. It is of the form \(K\,\mathrm{sinc}(\beta t)\)

Solution

A sinc in time corresponds to a rectangular (ideal low-pass) spectrum. The output spectrum is the product \(Y(j\omega)=X(j\omega)H(j\omega)\), i.e. the product of two rectangles, which equals the narrower rectangle (lower cutoff).

The narrower band corresponds to the smaller of \(\alpha,\beta\). Hence the output is a sinc with \(\gamma=\min(\alpha,\beta)\).

Let \(x(t)\) be periodic with period \(T\) and Fourier coefficients \(a_k\). Let \(y(t)=x(t-t_0)+x(t+t_0)\) have Fourier coefficients \(b_k\). If \(b_k=0\) for all odd \(k\), a possible value of \(t_0\) is:

1. \(T/8\)
2. \(T/4\)
3. \(T/2\)
4. \(2T\)

Solution

Using the time-shift property,

For \(b_k=0\) at all odd \(k\), we need \(\cos(k\omega_0 t_0)=0\), i.e. \(\omega_0 t_0 = \tfrac{\pi}{2}\). With \(\omega_0 = \tfrac{2\pi}{T}\),

\(H(z)\) is the transfer function of a real system. When \(x[n]=(1+j)^n\) is input, the output is zero. Further, the ROC of \(\left(1-\tfrac{1}{2}z^{-1}\right)H(z)\) is the entire \(z\)-plane (except \(z=0\)). It can be inferred that \(H(z)\) has a minimum of:

1. one pole and one zero
2. one pole and two zeros
3. two poles and one zero
4. two poles and two zeros

Solution

The output vanishes for \(x[n]=(1+j)^n\), so \(H(z)\) has a zero at \(z=1+j\). Because the system is real, zeros occur in conjugate pairs, so there is also a zero at \(z=1-j\): that is two zeros.

The ROC of \(\left(1-\tfrac{1}{2}z^{-1}\right)H(z)\) being the whole plane (except the origin) means this product is FIR (no poles). Therefore the only pole of \(H(z)\) is at \(z=\tfrac{1}{2}\), cancelled by the factor: that is one pole.

Given \(X(z)=\dfrac{z}{(z-a)^2},\ |z|>a\). The residue of \(X(z)\,z^{n-1}\) at \(z=a\) for \(n\ge 0\) is:

1. \(a^{n-1}\)
2. \(a^{n}\)
3. \(n\,a^{n}\)
4. \(n\,a^{n-1}\)

Solution

We need the residue at the double pole \(z=a\) of

For a second-order pole,

Let \(x(t)=\operatorname{rect}\!\left[t-\tfrac{1}{2}\right]\) (where \(\operatorname{rect}(x)=1\) for \(-\tfrac{1}{2}\le x\le\tfrac{1}{2}\) and \(0\) otherwise). With \(\mathrm{sinc}(x)=\dfrac{\sin(\pi x)}{\pi x}\), the Fourier transform of \(x(t)+x(-t)\) is:

1. \(\mathrm{sinc}\!\left(\tfrac{\omega}{2\pi}\right)\)
2. \(2\,\mathrm{sinc}\!\left(\tfrac{\omega}{2\pi}\right)\)
3. \(2\,\mathrm{sinc}\!\left(\tfrac{\omega}{2\pi}\right)\cos\!\left(\tfrac{\omega}{2}\right)\)
4. \(\mathrm{sinc}\!\left(\tfrac{\omega}{2\pi}\right)\sin\!\left(\tfrac{\omega}{2}\right)\)

Solution

Here \(x(t)=1\) for \(0\le t\le 1\). Its transform is

Then