Fourier Transforms, Frequency Response and Correlation

 Question 1
$X(\omega )$ is the Fourier transform of $x(t)$ shown below. The value of $\int_{-\infty }^{\infty }|X(\omega )|^2 d\omega$ (rounded off to two decimal places) is ______.
 A 58.61 B 42.45 C 26.12 D 78.92
GATE EC 2020   Signals and Systems
Question 1 Explanation:
\begin{aligned} \int_{-\infty }^{\infty }\left | X(\omega ) \right |^{2}d\omega &=2\pi \int_{-\infty }^{\infty }\left | x(t) \right |^{2}dt \\ &=2\pi \int_{-\infty }^{\infty }\left | y(t) \right |^{2}\\ &=2\times 2\pi \int_{-2}^{0}\left | y(t) \right |^{2}dt\\ &=2\times 2\pi\left [ \int_{-2}^{-1}(t+2)^{2}dt+\int_{-1}^{0}(2t+3)^{2}dt \right ]\\&=4\pi \left [\left \{ \frac{(t+2)^{3}}{3} \right \} _{-1}^{-2}+\left \{ \frac{(2t+3)^3}{3\times 2} \right \}_{0}^{-1} \right ]\\&=4\pi \left [ \frac{1-0}{3}+\frac{3^{3}-1}{6} \right ]\\&=4\pi \left [ \frac{1}{3}+\frac{26}{6} \right ]\\&= 4\pi \times \left [ \frac{1}{3}+\frac{26}{6} \right ] \\ &=4\pi \times \left [ \frac{1}{3}+\frac{13}{3} \right ]\\&=4\pi \times \frac{14}{3} \\&=\frac{56\pi }{3}=58.61\end{aligned}
 Question 2
The input 4sinc(2t) is fed to a Hilbert transformer to obtain y(t), as shown in the figure below:

Here $sinc(x)=\frac{sin(\pi x)}{\pi x}$. The value (accurate to two decimal places) of $\int_{-\infty }^{\infty } |y(t)|^{2}dt$ is _______.
 A 7 B 8 C 9 D 10
GATE EC 2018   Signals and Systems
Question 2 Explanation:
Hilbert transform does not alter the amplitude spectrum of the signal.
\begin{aligned} \text{So, }\quad \int_{-\infty}^{\infty}|y(t)|^{2} d t&=\int_{-\infty}^{\infty}|x(t)|^{2} d t=\int_{-\infty}^{\infty}|X(f)|^{2} d f \\ x(t)&=4 \sin \mathrm{c}(2 t) \\ \text{sinc}(t) &\stackrel{CTFT}{\longleftrightarrow} \text{rect}(f) \\ 4 \text{sinc}(2 t) &\stackrel{CTFT}{\longleftrightarrow}+\frac{4}{2} \text{rect}\left(\frac{f}{2}\right)=2 \text{rect}\left(\frac{f}{2}\right) \end{aligned}\\ \begin{aligned} \int_{-\infty}^{\infty}|X(f)|^{2} d f &=2 \times(2)^{2}=8 \\ \text { So. } \int_{-\infty}^{\infty}|y(t)|^{2} d t &=8 \end{aligned}

 Question 3
Consider an LTI system with magnitude response
$|H(f)|=\left\{\begin{matrix} 1-\frac{|f|}{20} & |f|\leq 20\\ 0 & |f|> 20 \end{matrix}\right.$
And phase response Arg {H(f)}=-2f
If the input to the system is
$x(t)=8cos(20\pi t+\frac{\pi }{4})+16sin(40\pi t+\frac{\pi }{8})+24cos(80\pi t+\frac{\pi }{16})$
Then the average power of the output signal y(t) is _________.
 A 6 B 7 C 8 D 9
GATE EC 2017-SET-2   Signals and Systems
Question 3 Explanation:
Since,

\begin{aligned} \text{So,}\quad y(t)&=\frac{1}{2} \times 8 \cos (20 \pi t+\phi) \\ \text{So,}\quad &=4 \cos (20 \pi t+\phi) ; \phi=\frac{\pi}{4}-20^{\circ} \\ P_{y}&=\frac{4^{2}}{2}=8 \mathrm{W} \end{aligned}
 Question 4
A continuous time signal $x(t)=4cos(200\pi t)+8cos(400 \pi t)$, where t is in seconds, is the input to a linear time invariant (LTI) filter with the impulse response

$h(t)=\left\{\begin{matrix} \frac{2sin(300\pi t)}{\pi t} & t\neq 0\\ 600 & t=0 \end{matrix}\right.$

Let y(t) be the output of this filter. The maximum value of |y(t)| is ________.
 A 8 B 6.5 C 7.3 D 8.9
GATE EC 2017-SET-1   Signals and Systems
Question 4 Explanation:
$x(t)=4 \cos 200 \pi t+8 \cos 400 \pi t$

\begin{aligned} \text{So.}\quad y(t)&=8 \cos 200 \pi t \\ \therefore \quad|y(t)|_{\max }&=8 \end{aligned}
 Question 5
If the signal $x(t)=\frac{sin(t)}{\pi t}*\frac{sin(t)}{\pi t}$ with * denoting the convolution operation, then x(t) is equal to
 A $\frac{sin(t)}{\pi t}$ B $\frac{sin(2t)}{2\pi t}$ C $\frac{2sin(t)}{\pi t}$ D $(\frac{sin(t)}{\pi t})^{2}$
GATE EC 2016-SET-3   Signals and Systems
Question 5 Explanation:

\begin{aligned} x(t) &=x_{1}(t) * x_{1}(t) \\ X(\omega) &=X_{1}(\omega) \cdot X_{1}(\omega) \end{aligned}

$\therefore \quad x(t)=\frac{\sin t}{\pi t}$
 Question 6
The energy of the signal $x(t)=\frac{sin(4\pi t)}{4\pi t}$ is ________
 A 0 B 0.25 C 0.5 D 0.75
GATE EC 2016-SET-2   Signals and Systems
Question 6 Explanation:

\begin{aligned} \text { Energy, } E_{x(i)} &=\frac{1}{2 \pi} \int_{-\infty}^{\infty}|X(\omega)|^{2} d \omega \\ &=\frac{1}{2 \pi} \int_{-4 \pi}^{4 \pi}\left(\frac{1}{4}\right)^{2} d \omega \\ &=\frac{1}{2 \pi} \times \frac{1}{16}[8 \pi]=\frac{1}{4}=0.25 \mathrm{J} \end{aligned}
 Question 7
A network consisting of a finite number of linear resistor (R), inductor (L), and capacitor (C) elements, connected all in series or all in parallel, is excited with a source of the form

$\sum_{k=1}^{3}a_{k}cos(k\omega _{0}t)$, where $a_{k}\neq 0, \omega _{0}\neq 0$.

The source has nonzero impedance. Which one of the following is a possible form of the output measured across a resistor in the network?
 A $\sum_{k=1}^{3}b_{k}cos(k\omega _{0}t+\varnothing _{k})$ , where $b_{k}\neq a_{k}, \forall k$ B $\sum_{k=1}^{4}b_{k}cos(k\omega _{0}t+\varnothing _{k})$, where $b_{k}\neq 0, \forall k$ C $\sum_{k=1}^{3}a_{k} cos(k\omega _{0}t+\varnothing _{k})$ D $\sum_{k=1}^{2}a_{k} cos(k\omega _{0}t+\varnothing _{k})$
GATE EC 2016-SET-1   Signals and Systems
Question 7 Explanation:

When a sinusoidal input is given to LTI system, the output is also a sinusoid with change in magnitude and the phase shift offered by LTI system.
 Question 8
Consider the function $g(t)=e^{-t}sin(2\pi t)u(t)$ where u(t) is the unit step function. The area under g(t) is _______.
 A 0.15 B 0.34 C 0.5 D 1
GATE EC 2015-SET-3   Signals and Systems
Question 8 Explanation:
\begin{aligned} g(t) &=e^{-t} \sin (2 \pi t) u(t) \\ \text { Area } &=\int_{-\infty}^{\infty} g(t) d t \\ &=\int_{-\infty}^{\infty} \sin (2 \pi t) u(t) e^{-t} d t=\left.G(s)\right|_{s=0} \\ G(s) &=\frac{2 \pi}{(s+1)^{2}+4 \pi^{2}} \\ \text { Area } &=\left.\frac{2 \pi}{(s+1)^{2}+4 \pi^{2}}\right|_{S=0}=0.155 \end{aligned}
 Question 9
A real-valued signal x(t) limited to the frequency band $|f|\leq \frac{W}{2}$ is passed through a linear time invariant system whose frequency response is
$H(f)=\left\{\begin{matrix} e^{-j4\pi f}, &|f|\leq \frac{W}{2} \\ 0, & |f|> \frac{W}{2} \end{matrix}\right.$
The output of the system is
 A x(t+4) B x(t-4) C x(t+2) D x(t-2)
GATE EC 2014-SET-4   Signals and Systems
Question 9 Explanation:
\begin{aligned} \text{Given:} \quad H(f)&=\left\{\begin{array}{cc}e^{-j 4 \pi t}, & |f| \leq \frac{\omega}{2} \\ 0, & |f|>\frac{\omega}{2}\end{array}\right. \\ \text{So, } \quad|H(t)|&=1 \\ H(f)&=e^{-j 2.2 \pi t} \\ H(\omega)&=e^{j 2 \omega} &|f| \leq \frac{\omega}{2} \\ h(t)&=\delta(t-2)\\ \text{As we know that}\\ y(t)&=x(t) * h(t)\\ \text{or}\quad y(t)&=X(\omega) \cdot H(\omega) \\ y(t)&=x(t) * \delta(t-2) \\ y(t)&=x(t-2) \end{aligned}
 Question 10
The phase response of a passband waveform at the receiver is given by
$\varphi (f)=-2\pi \alpha (f-f_{c})-2\pi \beta f_{c}$
where $f_{c}$
is the centre frequency, and $\alpha$ and $\beta$ are positive constants. The actual signal propagation delay from the transmitter to receiver is
 A $\frac{\alpha - \beta}{\alpha +\beta }$ B $\frac{\alpha \beta}{\alpha +\beta }$ C $\alpha$ D $\beta$
GATE EC 2014-SET-3   Signals and Systems
Question 10 Explanation:
Group delay is given by
$=\frac{-1}{2 \pi} \frac{d \phi(f)}{d f}=\frac{-1}{2 \pi}[-2 \pi \alpha]$
Group delay $=\alpha$
There are 10 questions to complete.