# Signals and Systems

 Question 1
For a vector $\bar{x}=\left [x[0],x[1],...,x[7] \right ]$, the 8-point discrete Fourier transform (DFT) is denoted by $\bar{X}=DFT(\bar{x})=\left [X[0],X[1],...,X[7] \right ]$, where
$X[k]=\sum_{n=0}^{7}x[n]exp\left ( -j\frac{2 \pi}{8}nk \right )$
Here, $j=\sqrt{-1}$. If $\bar{x}==[1,0,0,0,2,0,0,0]$ and $\bar{y}=DFT(DFT(\bar{x}))$, then the value of $y[0]$ is _________ (rounded off to one decimal place).
 A 3.2 B 6.8 C 4 D 8
GATE EC 2022      DTFS, DTFT and DFT
Question 1 Explanation:
\begin{aligned} X[K]&=\sum_{n=0}^{7}x[n]exp\left ( -j\frac{2\pi}{8}mk \right )\\ \bar{y}&=DFT(DFT(\bar{x})) \; \; where\\ \bar{x}&=[x[0],x[1],...,x[7]]\\ y(0)&=? \;\;x[n]=[1,0,0,0,2,0,0,0]\\ x[n]&\overset{DFT}{\rightarrow}\overset{DFT}{\rightarrow}N\cdot x(-k)=\bar{y}(n)\\ \bar{y}(n)&=N\cdot x(-k)|_{mod.N}=N\cdot x(N.K),N=8\\ \bar{y}(n)&=8[1,0,0,0,2,0,0,0]\\ \bar{y}(n)&=[8,0,0,0,16,0,0,0]\\ y(0)&=8 \end{aligned}
 Question 2
The outputs of four systems $(S_1,S_2,S_3,S_4)$ corresponding to the input signal $\sin (t)$, for all time $t$, are shown in the figure.
Based on the given information, which of the four systems is/are definitely NOT LTI (linear and time-invariant)?

 A $S_1$ B $S_2$ C $S_3$ D $S_4$
GATE EC 2022      LTI Systems Continuous and Discrete
Question 2 Explanation:

Since, LTI systems does not change the frequency of sinusoidal input. So S3 and S4 are definitely not LTI as input and output sinusoidal frequencies are different.
 Question 3
Let $x_1(t)=e^{-t}u(t)$ and $x_2(t)=u(t)-u(t-2)$, where $u(\cdot)$ denotes the unit step function. If $y(t)$ denotes the convolution of $x_1(t)$ and $x_2(t)$, then $\lim_{t\rightarrow \infty }y(t)=$ _________ (rounded off to one decimal place).
 A 0 B 0.4 C 0.6 D 0.8
GATE EC 2022      LTI Systems Continuous and Discrete
Question 3 Explanation:
\begin{aligned} x_1(t)=e^{-1}U(t)&\underleftrightarrow{L.T.}\frac{1}{(s+1)}\\ x_2(t)=U(t)-U(t-2)&\underleftrightarrow{L.T.}\frac{1}{s}-\frac{e^{-2s}}{s}=\frac{1-e^{-2s}}{s}\\ y(t)=x_1(t)\otimes x_2(t)&\underleftrightarrow{L.T.} \frac{(1-e^{-2s})}{s(s+1)} \\ y(t)&\underleftrightarrow{L.T.}\frac{1-e^{-2s}}{s(s+1)}\\ \lim_{t \to \infty }y(t)&=\lim_{s \to 0 }\frac{s(1-e^{-2s})}{s(s+1)}=0 \end{aligned}
 Question 4
The Fourier transform $X(j\omega )$ of the signal $x(t)=\frac{t}{(1+t^2)^2}$ is _________.
 A $\frac{\pi}{2j}\omega e^{-|\omega|}$ B $\frac{\pi}{2}\omega e^{-|\omega|}$ C $\frac{\pi}{2j} e^{-|\omega|}$ D $\frac{\pi}{2} e^{-|\omega|}$
GATE EC 2022      DTFS, DTFT and DFT
Question 4 Explanation:
$x(t)=\frac{t}{(1+t^2)^2}$
As we know that FT of $te^{-|t|} \; \underleftrightarrow{FT} \;\frac{-j4\omega }{(1+\omega ^2)^2}$
Duality $\frac{-j4\omega }{(1+t ^2)^2} \leftrightarrow 2 \pi(-\omega )e^{-|-\omega |}$
$\Rightarrow \frac{t}{(1+t^2)^2} \underrightarrow{FT} \frac{-2\pi}{-j4}\omega e^{-|\omega |}$
$\Rightarrow \;\;\;\rightarrow\frac{\pi}{j2} \omega e^{-|\omega |}$
 Question 5
Consider the signals $x\left [ n \right ]=2^{n-1}u\left [ -n+2 \right ]$ and $y\left [ n \right ]=2^{-n+2}u\left [ n+1 \right ]$, where $u[n]$ is the unit step sequence. Let $X(e^{jw})$ and $Y(e^{jw})$ be the discrete-time Fourier transform of $x[n]$ and $y[n]$, respectively. The value of the integral
$\frac{1}{2\pi } \int_{0}^{2\pi }X\left ( e^{j\omega } \right )Y\left ( e^{-j\omega } \right )d\omega$
(rounded off to one decimal place) is ______
 A 2.4 B 12.5 C 8 D 10.8
GATE EC 2021      DTFS, DTFT and DFT
Question 5 Explanation:
\begin{aligned} x[n] &=2^{n-1} u[-n+2] \\ y[n] &=2^{-n+2} u[n+1] \\ y[-n] &=2^{n+2} u[-n+1] \\ V &=\frac{1}{2 \pi} \int_{0}^{2 \pi} X\left(e^{j \omega}\right) Y\left(e^{-j \omega}\right) d \omega &\ldots(i)\\ z[n] &=x[n] * y[-n] \\ z[n] & \rightarrow Z\left(e^{j \omega}\right) \\ z[n] &=\frac{1}{2 \pi} \int_{-\pi}^{\pi} Z\left(e^{j \omega}\right) e^{j \omega n} d \omega \\ \text { Put } n=0, \qquad z[0] &=\frac{1}{2 \pi} \int_{0}^{2 \pi} Z\left(e^{j \omega}\right) d \omega &\ldots(ii) \end{aligned}
Compare equations (i) and (ii)
\begin{aligned} z[0] &=V \\ Z\left(e^{j \omega}\right) &=X\left(e^{j \omega}\right) Y\left(e^{-j \omega}\right) \\ \text{Apply IDTFT},\qquad Z[n] &=x[n] * y[-n]=x[n] * p[n]\\ \because \qquad p[n]&=y[-n]=2^{n+2} u[-n+1]\\ z[n] &=\sum_{k=-\infty}^{+\infty} 2^{k-1} u[-k+2] 2^{n-k+2} u[-n+k+1] \\ &=\sum_{k=-\infty}^{2} 2^{k-1} 1 \cdot 2^{n-k+2} u[k+1-n] \\ &=\sum_{k=-\infty}^{2} 2^{k-1+n-k+2} u[k+1-n] \\ z[n] &=\sum_{k=-\infty}^{2} 2^{n+1} u[k-n+1] \\ \text { Put } n=0,\qquad \qquad V &=z[0]=\sum_{k=-\infty}^{2} 2^{1} \cdot u[k+1]=2 \sum_{k=-1}^{2} 1=2[1+1+1+1] \\ V &=8 \end{aligned}
 Question 6
For a unit step input u[n], a discrete-time $\text{LTI}$ system produces an output signal $\left ( 2\delta \left [ n+1 \right ] +\delta \left [ n \right ]+\delta \left [ n-1 \right ]\right )$. Let $y[n]$ be the output of the system for an input $\left ( \left ( \dfrac{1}{2} \right )^n u\left [ n \right ]\right )$. The value of $y[0]$ is ______
 A 0 B 0.5 C 1 D 1.5
GATE EC 2021      LTI Systems Continuous and Discrete
Question 6 Explanation:

The impulse response $h(n)=s(n)-s(n-1)$
\begin{aligned} h(n)=2 \delta(n+1)+\delta(n)&+ \delta(n-1)-2 \delta(n)-\delta(n-1)-\delta(n-1-1) \\ &=2 \delta(n+1)+\delta(n)+\delta(n-1)-2 \delta(n)-\delta(n-1)-\delta(n-2) \\ h(n) &=2 \delta(n+1)-\delta(n)-\delta(n-2) \\ \text{For input }\qquad x_{1}(n)&=(1 / 2)^{n} u(n) \text{ then output } y_{1}(n)=x_{1}(n) * h(n)\\ y_{1}(n)&=\left(\frac{1}{2}\right)^{n} u(n) *[2 \delta(n+1)-\delta(n)-\delta(n-2)]\\ y_{1}(n)&=\left(\frac{1}{2}\right)^{n} u(n) * 2 \delta(n+1)-\left(\frac{1}{2}\right)^{n} u(n) * \delta(n)-\left(\frac{1}{2}\right)^{n} u(n) * \delta(n-2) \\ y_{1}(n)&=2\left(\frac{1}{2}\right)^{n+1} u(n+1)-\left(\frac{1}{2}\right)^{n} u(n)-\left(\frac{1}{2}\right)^{n-2} u(n-2) \\ \left.y_{1}(n)\right|_{n=0} &=2\left(\frac{1}{2}\right)^{1} u(1)-\left(\frac{1}{2}\right)^{0} u(0)-\left(\frac{1}{2}\right)^{-2} u(-2) \\ &=1-1 \\ y_{1}(0) &=0 \end{aligned}
 Question 7
The exponential Fourier series representation of a continuous-time periodic signal x(t) is defined as
$x\left ( t \right )=\sum_{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}$
where $\omega _0$ is the fundamental angular frequency of x(t) and the coefficients of the series are $a_{k}$. The following information is given about $x(t)$ and $a_{k}$.

I. x(t) is real and even, having a fundamental period of 6
II. The average value of x(t) is 2
III. $a_{k}=\left\{\begin{matrix} k, & 1\leq k\leq 3\\ 0,& k> 3 \end{matrix}\right.$

The average power of the signal x(t) (rounded off to one decimal place) is ____________
 A 14 B 45 C 63 D 32
GATE EC 2021      Fourier Transforms, Frequency Response and Correlation
Question 7 Explanation:
$1. x(t)$ is real and even so $a_{k}$ is also real and even $a_{k}=a_{-k}$
2. Average of x(t) is 2 i.e., $a_{0}=2.$
$\begin{array}{llll} 3. \;x(t) \rightarrow a_{k}=k &1 \leq k \leq 3 \quad& a_{1}=1 \quad& a_{-1}=1 \\ \quad \quad \quad \quad \;\;\;\;\;\;\;\;0 & k>3 & a_{2}=2 & a_{-2}=2 \\ && a_{3}=3 & a_{-3}=3 \end{array}$
$4. \quad T_{0}=6$
Parsval's Power Theorem
\begin{aligned} \frac{1}{T} \int_{0}^{T}\left|x(t)^{2}\right| d t &=\sum_{n=-\infty}^{+\infty}\left|a_{k}\right|^{2} \\ P(t) &=\sum_{n=-\infty}^{+\infty}\left|a_{k}\right|^{2}=\left|a_{-3}\right|^{2}+\left|a_{-2}\right|^{2}+\left|a_{-1}\right|^{2}+\left|a_{0}\right|^{2}+\left|a_{1}\right|^{2}+\left|a_{2}\right|^{2}+\left|a_{3}\right|^{2} \\ &=2\left|a_{1}\right|^{2}+2\left|a_{2}\right|^{2}+2\left|a_{3}\right|^{2}+\left|a_{0}\right|^{2} \\ &=2 \times 1^{2}+2 \times(2)^{2}+2(3)^{2}+(2)^{2} \\ &=2+8+18+4 \\ P_{x(t)} &=32 \end{aligned}
 Question 8
Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k]=y[k] is/are
 A k=0,1,2,,15 B k=0 C k=15 D k=0 and k=15
GATE EC 2021      DTFS, DTFT and DFT
Question 8 Explanation:
If two' N' point signals x(n) and h(n) are convolving with each other linearly and circularly
then
$y(k)=z(k)$ at $k=N-1$
where, y(n)= Linear convolution of x(n) and h(n)
z(n)= Circular convolution of x(n) and h(n)
Since, $N=16$ (Given)
Therefore, $\quad y(k)=z(k)$ at $k=N-1=15$
 Question 9
Consider a real-valued base-band signal x(t), band limited to $\text{10 kHz}$. The Nyquist rate for the signal $y\left ( t \right )=x\left ( t \right )x\left ( 1+\dfrac{t}{2} \right )$ is
 A $\text{15 kHz}$ B $\text{30 kHz}$ C $\text{60 kHz}$ D $\text{20 kHz}$
GATE EC 2021      Sampling
Question 9 Explanation:

$\mathrm{NR}=2 \times f_{\mathrm{max}}=2 \times 15=30 \mathrm{kHz}$
 Question 10
The transfer function of a stable discrete-time LTI system is $H(z)=\frac{K(z-\alpha )}{z+0.5}$, where K and $\alpha$ are real numbers. The value of $\alpha$ (rounded off to one decimal place) with $|\alpha| \gt 1$, for which the magnitude response of the system is constant over all frequencies, is _____.
 A -1 B 1 C 2 D -2
GATE EC 2020      Z-Transform
Question 10 Explanation:
System is all-pass filter.
For digital all-pass filter, condition is
$\text{zero}=\frac{1}{\text{pole}^{*}}\; \; ...(i)$
By given transfer function,
$\text{zero}=\alpha$
$\text{Pole}=-0.5$
using condition(i),
$\alpha =\frac{1}{-0.5}=-2$

There are 10 questions to complete.