GATE 2007 Signals and Systems Questions and Solutions

Let a signal \(a_1 F(\omega_1 t + \phi_1)\) be applied to a stable linear time-invariant system. Let the corresponding steady-state output be represented as \(a_2 F(\omega_2 t + \phi_2)\). Then which of the following statements is true?

1. \(F\) is not necessarily a sine or cosine function but must be periodic with \(\omega_1 = \omega_2\)
2. \(F\) must be a sine or cosine function with \(a_1 = a_2\)
3. \(F\) must be a sine function with \(\omega_1 = \omega_2\) and \(\phi_1 = \phi_2\)
4. \(F\) must be a sine or cosine function with \(\omega_1 = \omega_2\)

Solution

For a linear time-invariant system, the only signals that pass through with their shape preserved (eigenfunctions) are the complex exponentials, i.e. sines and cosines. At any input frequency \(\omega_1\), the steady-state output is sinusoidal at the same frequency, with amplitude scaled by \(|H(j\omega_1)|\) and phase shifted by \(\phi_2 = \phi_H + \phi_1\), where \(|H|\) and \(\phi_H\) are the magnitude and phase of the frequency response.

Hence \(F\) must be a sine or cosine and \(\omega_1 = \omega_2\).

The frequency spectrum of a signal is shown in the figure. If this is ideally sampled at intervals of 1 ms, then the frequency spectrum of the sampled signal will be:

Solution

The sampling interval is \(T_s = 1\,\mathrm{ms} = 10^{-3}\,\mathrm{s}\), so the sampling frequency is

After ideal sampling, the spectrum becomes a scaled, periodic replication of the original baseband spectrum at integer multiples of \(f_s\):

Consider the discrete-time system shown in the figure where the impulse response of \(G(z)\) is \(g(0)=0,\ g(1)=g(2)=1,\ g(3)=g(4)=\dots=0\). This system is stable for the range of values of \(K\):

1. \([-1,\ \tfrac{1}{2}]\)
2. \([-1,\ 1]\)
3. \([-\tfrac{1}{2},\ 1]\)
4. \([-\tfrac{1}{2},\ 2]\)

Solution

Since \(g(1)=g(2)=1\), we have \(g[n]=\delta[n-1]+\delta[n-2]\), so

The closed-loop transfer function is

The poles are the roots of the characteristic equation \(z^{2} - Kz - K = 0\). Applying the Jury stability test (poles inside the unit circle) to \(z^2 + a_1 z + a_0\) with \(a_1=-K,\ a_0=-K\):

• \(|a_0| < 1 \Rightarrow |K| < 1\).
• \(1 + a_1 + a_0 > 0 \Rightarrow 1 - K - K > 0 \Rightarrow K < \tfrac{1}{2}\).
• \(1 - a_1 + a_0 > 0 \Rightarrow 1 + K - K > 0\), always true.

Combining these gives \(-1 < K < \tfrac{1}{2}\).

Note: the closed-loop denominator is \(z^{2}-Kz-K\) (an earlier draft printed \(z^{2}-Kz+1\)); the corrected form is used here. The final range is unchanged.

A signal \(x(t)\) is given by

Which among the following gives the fundamental Fourier term of \(x(t)\)?

1. \(\dfrac{4}{\pi}\cos\!\left(\dfrac{\pi t}{T} - \dfrac{\pi}{4}\right)\)
2. \(\dfrac{4}{\pi}\cos\!\left(\dfrac{\pi t}{2T} + \dfrac{\pi}{4}\right)\)
3. \(\dfrac{4}{\pi}\sin\!\left(\dfrac{\pi t}{T} - \dfrac{\pi}{4}\right)\)
4. \(\dfrac{4}{\pi}\sin\!\left(\dfrac{\pi t}{2T} + \dfrac{\pi}{4}\right)\)

Solution

The signal is periodic with period \(T_0 = \tfrac{7T}{4} - (-\tfrac{T}{4}) = 2T\), giving fundamental frequency

The exponential Fourier coefficient for \(k=1\) is

Comparing with \(C_1 = \tfrac{a_1}{2} - j\tfrac{b_1}{2}\) gives \(a_1 = b_1 = \tfrac{4}{\sqrt2\,\pi}\), so

If \(u(t)\) and \(r(t)\) denote the unit step and unit ramp functions respectively, and \(u(t)*r(t)\) their convolution, then the function \(u(t+1)*r(t-2)\) is given by:

1. \(\tfrac{1}{2}(t-1)\,u(t-1)\)
2. \(\tfrac{1}{2}(t-1)\,u(t-2)\)
3. \(\tfrac{1}{2}(t-1)^2\,u(t-1)\)
4. None of these

Solution

Convolution is shift-invariant, so the time shifts simply add. First note that \(u(t)*r(t) = \int_0^t \tau\,d\tau = \tfrac{t^2}{2}\,u(t)\). Applying the shifts \(+1\) and \(-2\) (net \(-1\)):

Given \(X(z) = 1 - 3z^{-1}\) and \(Y(z) = 1 + 2z^{-2}\) are the Z-transforms of two signals \(x[n]\) and \(y[n]\). An LTI system has impulse response \(h[n] = x[n-1]*y[n]\). The output of the system for the input \(\delta[n-1]\) is:

1. has Z-transform \(z^{-1}X(z)Y(z)\)
2. equals \(\delta[n-2] - 3\delta[n-3] + 2\delta[n-4] - 6\delta[n-5]\)
3. has Z-transform \(1 - 3z^{-1} + 2z^{-2} - 6z^{-3}\)
4. does not satisfy any of the above three

Solution

Since \(h[n] = x[n-1]*y[n]\),

For input \(\delta[n-1]\), i.e. \(I(z)=z^{-1}\), the output is

A signal is processed by a causal filter with transfer function \(G(s)\). For a distortion-free output signal waveform, \(G(s)\) must:

1. provide zero phase shift for all frequencies
2. provide constant phase shift for all frequencies
3. provide linear phase shift that is proportional to frequency
4. provide a phase shift that is inversely proportional to frequency

Solution

Distortionless transmission requires a constant magnitude response and a phase that is linear in frequency, \(\phi(\omega) = -\omega t_d\). A linear phase corresponds to a pure time delay \(t_d\), which preserves the waveform shape.

With \(G(z) = \alpha z^{-1} + \beta z^{-3}\), the digital filter has the linear-phase characteristic of the previous question if:

1. \(\alpha = \beta\)
2. \(\alpha = -\beta\)
3. \(\alpha = \beta^{1/3}\)
4. \(\alpha = \beta^{-1/3}\)

Solution

The impulse response is \(g[n] = \{0,\ \alpha,\ 0,\ \beta\}\) for \(n = 0,1,2,3\), i.e. nonzero taps at \(n=1\) and \(n=3\), centred about \(n=2\). A finite-impulse-response filter has exactly linear phase when its coefficients are symmetric about the centre tap, which requires the \(n=1\) and \(n=3\) taps to be equal:

(The anti-symmetric choice \(\alpha=-\beta\) also gives linear phase but produces a band-pass/high-pass response, not the low-pass behaviour required here.)