GATE 2014 Signals and Systems Questions and Solutions

\(x(t)\) is nonzero only for \(T_x

1. \(z(t)\) can be nonzero over an unbounded interval
2. \(z(t)\) is nonzero for \(t
3. \(z(t)\) is zero outside of \(T_x+T_y
4. \(z(t)\) is nonzero for \(t>T_x'+T_y'\)

Solution

Convolution adds the supports: if \(x\) lives on \((T_x,T_x')\) and \(y\) on \((T_y,T_y')\), then \(z=x*y\) is confined to \((T_x+T_y,\;T_x'+T_y')\) and is zero outside that interval.

Let \(X(s)=\dfrac{3s+5}{s^2+10s+21}\) be the Laplace transform of a signal \(x(t)\). Then \(x(0^+)\) is:

1. \(0\)
2. \(3\)
3. \(5\)
4. \(21\)

Solution

By the initial value theorem,

For a periodic square wave, which one of the following statements is TRUE?

1. The Fourier series coefficients do not exist.
2. The Fourier series coefficients exist but the reconstruction converges at no point.
3. The Fourier series coefficients exist but the reconstruction converges at most points.
4. The Fourier series coefficients exist and the reconstruction converges at every point.

Solution

A square wave satisfies the Dirichlet conditions, so its Fourier coefficients exist. However, at the jump discontinuities the series converges to the midpoint value (Gibbs phenomenon), so reconstruction fails exactly at those isolated points but converges everywhere else — i.e. at most points.

Let \(g:[0,\infty)\to[0,\infty)\) be defined by \(g(x)=x-\lfloor x\rfloor\), where \(\lfloor x\rfloor\) denotes the integer part of \(x\). The value of the constant (DC) term in the Fourier series expansion of \(g(x)\) is (numerical):

Solution

\(g(x)\) is the periodic sawtooth of period 1. The constant term is its average over one period:

The function shown in the figure can be represented as:

1. \(u(t)-u(t-T)+\dfrac{(t-T)}{T}u(t-T)-\dfrac{(t-2T)}{T}u(t-2T)\)
2. \(u(t)+\dfrac{t}{T}u(t-T)-\dfrac{t}{T}u(t-2T)\)
3. \(u(t)-u(t-T)+\dfrac{(t-T)}{T}u(t-T)-\dfrac{(t-2T)}{T}u(t)\)
4. \(u(t)-u(t-T)+\dfrac{(t-T)}{T}u(t-T)-\dfrac{(t-2T)}{T}u(t-2T)\)

Solution

Decompose the waveform into a unit step on \((0,T)\) plus a unit-slope ramp that starts at \(t=T\) and is cancelled by an equal negative ramp starting at \(t=2T\):

Note: in this paper's draft, options (A) and (D) are printed with identical text — an OCR/typesetting duplication. The intended correct expression is the one given above; the official key is (A).

Let \(X(z)=\dfrac{1}{1-z^{-3}}\) be the Z-transform of a causal signal \(x[n]\). Then the values of \(x[2]\) and \(x[3]\) are:

1. \(0\) and \(0\)
2. \(0\) and \(1\)
3. \(1\) and \(0\)
4. \(1\) and \(1\)

Solution

For a causal sequence, expand as a power series in \(z^{-1}\):

So \(x[n]=1\) for \(n=0,3,6,\dots\) and \(0\) otherwise. Hence \(x[2]=0\) and \(x[3]=1\).

Let \(f(t)\) be a continuous-time signal with Fourier transform \(F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}\,dt\). Define \(g(t)=\int_{-\infty}^{\infty}F(u)e^{-jut}\,du\). What is the relationship between \(f(t)\) and \(g(t)\)?

1. \(g(t)\) would always be proportional to \(f(t)\)
2. \(g(t)\) would be proportional to \(f(t)\) if \(f(t)\) is an even function
3. \(g(t)\) would be proportional to \(f(t)\) only if \(f(t)\) is a sinusoidal function
4. \(g(t)\) would never be proportional to \(f(t)\)

Solution

From the inverse transform, \(2\pi f(t)=\int_{-\infty}^{\infty}F(u)e^{jut}\,du\). Replacing \(t\to-t\):

This is proportional to \(f(t)\) only when \(f(-t)=f(t)\), i.e. when \(f\) is even.

Consider an LTI system with transfer function \(H(s)=\dfrac{1}{s(s+4)}\). If the input is \(\cos(3t)\) and the steady-state output is \(A\sin(3t+\alpha)\), the value of \(A\) is:

1. \(\dfrac{1}{30}\)
2. \(\dfrac{1}{15}\)
3. \(\dfrac{3}{4}\)
4. \(\dfrac{4}{3}\)

Solution

The steady-state amplitude is \(|H(j\omega)|\) at \(\omega=3\):

Hence \(A=\dfrac{1}{15}\) (with phase \(\angle H(j3)=-90^\circ-\tan^{-1}\tfrac34\)).

Consider an LTI system with impulse response \(h(t)=e^{-5t}u(t)\). If the output is \(y(t)=e^{-3t}u(t)-e^{-5t}u(t)\), then the input \(x(t)\) is:

1. \(e^{-3t}u(t)\)
2. \(2e^{-3t}u(t)\)
3. \(e^{-5t}u(t)\)
4. \(2e^{-5t}u(t)\)

Solution

In the \(s\)-domain \(Y(s)=H(s)X(s)\), with \(H(s)=\dfrac{1}{s+5}\) and

Therefore \(X(s)=\dfrac{Y(s)}{H(s)}=\dfrac{2}{s+3}\), giving \(x(t)=2e^{-3t}u(t)\).

A discrete system is represented by

with \(X_1(0)=1,\ X_2(0)=0\). The pole locations for \(a=1\) are:

1. \(1\pm j0\)
2. \(-1\pm j0\)
3. \(\pm 1+j0\)
4. \(0\pm j1\)

Solution

For \(a=1\) the state matrix is \(\begin{bmatrix}1&0\\2&1\end{bmatrix}\). Its eigenvalues are the roots of \((1-\lambda)^2=0\), so \(\lambda=1\) (repeated). The poles are at \(z=1\), i.e. \(1\pm j0\).

An input \(x(t)=2+5\sin(100\pi t)\) is sampled at \(400\) Hz and applied to the system

where \(N\) is the number of samples per cycle. The steady-state output \(y[n]\) is:

1. \(0\)
2. \(1\)
3. \(2\)
4. \(5\)

Solution

With \(f_s=400\) Hz, the sampled signal is \(x[n]=2+5\sin(\tfrac{\pi}{4}n)\), so \(\omega_0=\pi/4\) and \(N=2\pi/\omega_0=8\). The averaging filter has

At \(\omega=\omega_0=\pi/4\): \(H=\tfrac18\dfrac{1-e^{-j2\pi}}{1-e^{-j\pi/4}}=0\), so the sinusoid is rejected. At DC (\(\omega=0\)), \(H=1\), so the constant passes unchanged. Steady-state output \(=1\times2=2\).

A function \(f(t)\) is shown in the figure. Its Fourier transform \(F(\omega)\) is:

1. real and even function of \(\omega\)
2. real and odd function of \(\omega\)
3. imaginary and odd function of \(\omega\)
4. imaginary and even function of \(\omega\)

Solution

The waveform is real and odd (its right half is the negative of its left half). The Fourier transform of a real odd signal is purely imaginary and odd in \(\omega\).

A signal is \(x(t)=1\) for \(|t|<1\) and \(0\) for \(|t|>1\). The Fourier transform of \(y(t)=x(2t)*x(t/2)\) is:

1. \(\dfrac{4}{\omega^{2}}\sin\!\big(\tfrac{\omega}{2}\big)\sin(2\omega)\)
2. \(\dfrac{4}{\omega^{2}}\sin\!\big(\tfrac{\omega}{2}\big)\)
3. \(\dfrac{4}{\omega^{2}}\sin(2\omega)\)
4. \(\dfrac{4}{\omega^{2}}\sin^{2}\omega\)

Solution

The rectangular pulse transforms as \(X(\omega)=\dfrac{2\sin\omega}{\omega}\). Using time-scaling, \(x(2t)\leftrightarrow\dfrac{\sin(\omega/2)}{\omega/2}\) and \(x(t/2)\leftrightarrow\dfrac{4\sin 2\omega}{2\omega}\). Convolution in time means multiplication in frequency:

For the signal \(f(t)=3\sin(8\pi t)+6\sin(12\pi t)+\sin(14\pi t)\), the minimum sampling frequency (Hz) satisfying the Nyquist criterion is (numerical):

Solution

The highest frequency component is \(14\pi=2\pi f\Rightarrow f_{\max}=7\) Hz. Minimum sampling rate \(=2f_{\max}=14\) Hz.

A continuous-time LTI system with system function \(H(\omega)\) has the pole-zero plot shown. For this system, which alternative is TRUE?

1. \(|H(0)|>|H(\omega)|,\ |\omega|>0\)
2. \(|H(\omega)|\) has multiple maxima, at \(\omega_1\) and \(\omega_2\)
3. \(|H(0)|<|H(\omega)|,\ |\omega|>0\)
4. \(|H(\omega)|=\) constant, \(-\infty<\omega<\infty\)

Solution

The magnitude response is \(|H(\omega)|=\dfrac{\prod(\text{distances from zeros})}{\prod(\text{distances from poles})}\). With a zero located at (or close to) the origin, the numerator distance is smallest at \(\omega=0\), driving \(|H|\) to its minimum there; as \(|\omega|\) increases the distance from the zero grows and \(|H(\omega)|\) rises. Hence \(|H(0)|<|H(\omega)|\) for \(|\omega|>0\) — a high-pass-type magnitude characteristic.

Note: this paper's draft solution text described an "all-pass" configuration (which would give a constant magnitude, option D). That reasoning is inconsistent with the plotted zero near the origin. The correct conclusion is (C), the official key.

A sinusoid \(x(t)\) of unknown frequency is sampled by an impulse train of period 20 ms. The samples are applied to an ideal lowpass filter with cutoff 25 Hz. The output is a sinusoid of frequency 20 Hz. Then \(x(t)\) has frequency:

1. 10 Hz
2. 60 Hz
3. 30 Hz
4. 90 Hz

Solution

The sampling rate is \(f_s=1/(20\text{ ms})=50\) Hz.

An input at \(f_m\) appears after sampling at the aliased frequency \(|f_s-f_m|\). Setting this equal to the observed 20 Hz:

A differentiable non-constant even function \(x(t)\) has derivative \(y(t)\); their Fourier transforms are \(X(\omega)\) and \(Y(\omega)\). Which statement is TRUE?

1. \(X(\omega)\) and \(Y(\omega)\) are both real
2. \(X(\omega)\) is real and \(Y(\omega)\) is imaginary
3. \(X(\omega)\) and \(Y(\omega)\) are both imaginary
4. \(X(\omega)\) is imaginary and \(Y(\omega)\) is real

Solution

A real even \(x(t)\) has a real (and even) transform \(X(\omega)\). Its derivative \(y(t)=x'(t)\) is real and odd, so \(Y(\omega)=j\omega X(\omega)\) is imaginary (and odd).