GATE 2016 Signals and Systems Questions and Solutions

The value of \(\displaystyle\int_{-\infty}^{+\infty} e^{-t}\,\delta(2t-2)\,dt\), where \(\delta(t)\) is the Dirac delta function, is:

1. \(\dfrac{1}{2e}\)
2. \(\dfrac{2}{e}\)
3. \(\dfrac{1}{e^{2}}\)
4. \(\dfrac{1}{2e^{2}}\)

Solution

Using the scaling property \(\delta(2t-2)=\tfrac{1}{2}\delta(t-1)\):

Let \(f(x)\) be a real, periodic function satisfying \(f(-x)=-f(x)\). The general form of its Fourier series representation is:

1. \(f(x)=a_0+\sum_{k=1}^{\infty}a_k\cos(kx)\)
2. \(f(x)=\sum_{k=1}^{\infty}b_k\sin(kx)\)
3. \(f(x)=a_0+\sum_{k=1}^{\infty}a_{2k}\cos(kx)\)
4. \(f(x)=\sum_{k=0}^{\infty}a_{2k+1}\sin((2k+1)x)\)

Solution

Since \(f(-x)=-f(x)\), the function is odd. An odd function has only sine terms (no DC, no cosine terms), so \(f(x)=\sum_{k=1}^{\infty}b_k\sin(kx)\).

The Fourier transform of a continuous-time signal \(x(t)\) is \(X(\omega)=\dfrac{1}{(10+j\omega)^{2}}\), \(-\infty<\omega<\infty\). Find the value of \(x(t)\) at \(t=1\) (up to 1 decimal place).

Solution

The standard pair \(t\,e^{-at}u(t)\leftrightarrow\dfrac{1}{(a+j\omega)^{2}}\) with \(a=10\) gives

Note: the OCR of this problem corrupted the question text (it printed a stray "value of \(|\ln t|\) at \(t=1\)" and the worked draft drifted into an unrelated sinc-integral). The well-posed question that matches the given transform asks for \(x(1)\); inverting \(X(\omega)\) yields \(x(t)=t\,e^{-10t}u(t)\), so \(x(1)\approx 0.0\), which is the official answer. Please verify against the original paper.

Suppose the maximum frequency in a band-limited signal \(x(t)\) is 5 kHz. Then the maximum frequency in \(x(t)\cos(2000\pi t)\), in kHz, is (numerical):

Solution

Multiplication by \(\cos(2000\pi t)\) (i.e. \(f_c=1\) kHz) shifts the spectrum, so the highest frequency becomes \(f_m+f_c=5+1=6\) kHz.

Suppose \(x_1(t)\) and \(x_2(t)\) have the Fourier transforms shown below. Which statement is TRUE?

1. \(x_1(t)\) and \(x_2(t)\) are complex and \(x_1(t)x_2(t)\) is also complex with nonzero imaginary part
2. \(x_1(t)\) and \(x_2(t)\) are real and \(x_1(t)x_2(t)\) is also real
3. \(x_1(t)\) and \(x_2(t)\) are complex but \(x_1(t)x_2(t)\) is real
4. \(x_1(t)\) and \(x_2(t)\) are imaginary but \(x_1(t)x_2(t)\) is real

Solution

The spectra are not conjugate-symmetric, so \(x_1(t)\) and \(x_2(t)\) are complex. From the figures \(X_2(\omega)=X_1(-\omega)\), so \(x_2(t)=x_1(-t)\). Since \(X_1(\omega)\) is real, \(x_1(t)\) is conjugate-symmetric, \(x_1(-t)=x_1^*(t)\). Hence

which is real. So the signals are complex but their product is real.

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

Solution

With \(\omega=1\): \(|H(j1)|=\dfrac{1}{\sqrt{1^2+1}}=\dfrac{1}{\sqrt2}\approx0.707\).

The transfer function of a system is \(\dfrac{Y(s)}{R(s)}=\dfrac{s}{s+2}\). The steady-state output is \(A\cos(2t+\varphi)\) for the input \(\cos(2t)\). The values of \(A\) and \(\varphi\) respectively are:

Solution

At \(\omega=2\): \(|H(j2)|=\dfrac{2}{\sqrt{4+4}}=\dfrac{1}{\sqrt2}\approx0.707\) and \(\angle H(j2)=90^\circ-\tan^{-1}(1)=45^\circ\).

The Laplace transform of \(f(t)=e^{2t}\sin(5t)u(t)\) is:

1. \(\dfrac{5}{s^2-4s+29}\)
2. \(\dfrac{5}{s^2+5}\)
3. \(\dfrac{s-2}{s^2-4s+29}\)
4. \(\dfrac{5}{s+5}\)

Solution

\(\mathcal{L}\{\sin5t\,u(t)\}=\dfrac{5}{s^2+25}\). Applying the frequency shift \(s\to s-2\):

Consider a causal stable LTI system characterized by \(\dfrac{dy(t)}{dt}+\dfrac{1}{6}y(t)=3x(t)\). Find the response to the input \(x(t)=3e^{-t/3}u(t)\).

Solution

Here \(H(s)=\dfrac{3}{s+1/6}\) and \(X(s)=\dfrac{9}{s+1/3}\), so

A signal \(z(t)\) is an eigensignal of an LTI system \(T\) if \(T\{z(t)\}=\gamma z(t)\). Suppose the impulse response of \(T\) is real and even. Which statement is TRUE?

1. \(\cos(t)\) is an eigensignal but \(\sin(t)\) is not
2. \(\cos(t)\) and \(\sin(t)\) are both eigensignals but with different eigenvalues
3. \(\sin(t)\) is an eigensignal but \(\cos(t)\) is not
4. \(\cos(t)\) and \(\sin(t)\) are both eigensignals with identical eigenvalues

Solution

A real, even impulse response gives a real, even \(H(j\omega)\), so \(H(j1)=H(-j1)\). Feeding \(\cos t=\tfrac12(e^{jt}+e^{-jt})\) and \(\sin t=\tfrac1{2j}(e^{jt}-e^{-jt})\) through the system, both emerge scaled by the same real \(H(j1)\). Hence both are eigensignals with identical eigenvalue \(H(j1)\).

For the state-space system \(\dot{x}(t)=\begin{bmatrix}1&0\\0&2\end{bmatrix}x(t)\), \(y(t)=c^Tx(t)\), \(c=\begin{bmatrix}1\\1\end{bmatrix}\), \(x(0)=\begin{bmatrix}1\\1\end{bmatrix}\), the value of \(y(t)\) at \(t=\log_e 2\) is (numerical):

Solution

The diagonal system gives \(x(t)=\begin{bmatrix}e^{t}\\e^{2t}\end{bmatrix}\), so \(y(t)=e^{t}+e^{2t}\). At \(t=\ln2\):

Consider a continuous-time system with input \(x(t)\) and output \(y(t)=x(t)\cos(t)\). This system is:

1. linear and time-invariant
2. non-linear and time-invariant
3. linear and time-varying
4. non-linear and time-varying

Solution

The output depends linearly on \(x(t)\), so the system is linear. For a shifted input the response is \(x(t-t_1)\cos(t)\), whereas the shifted output is \(x(t-t_1)\cos(t-t_1)\); these differ, so the system is time-varying.

Signals \(x_1(t)\leftrightarrow X_1(\omega)\) (bandwidth \(B_1\)) and \(x_2(t)\leftrightarrow X_2(\omega)\) (bandwidth \(B_2\)) drive a system with impulse response \(h(t)=e^{-2|t|}\); the input is \(x_1(t)\,x_2(t)\). The minimum sampling rate to uniquely reconstruct \(y(t)\) is:

1. \(2B_1\)
2. \(2(B_1+B_2)\)
3. \(4(B_1+B_2)\)
4. \(\infty\)

Solution

Multiplication in time convolves the spectra, so \(x_1(t)x_2(t)\) has bandwidth \(B_1+B_2\). The LTI block \(h(t)\) cannot widen the bandwidth, so the output bandwidth is \(B_1+B_2\) and the Nyquist rate is \(2(B_1+B_2)\).

Let \(S=\displaystyle\sum_{n=0}^{\infty}n\alpha^{n}\) where \(|\alpha|<1\). The value of \(\alpha\) in \(0<\alpha<1\) such that \(S=2\alpha\) is (numerical):

Solution

Using \(\sum_{n=0}^{\infty}n\alpha^{n}=\dfrac{\alpha}{(1-\alpha)^2}\) and setting it equal to \(2\alpha\):