# DTFS, DTFT and DFT

 Question 1
Let an input $x[n]$ having discrete time Fourier transform.
$X\left(e^{j \Omega}\right)=1-e^{-j \Omega}+2 e^{-3 j \Omega}$ be passed through an LTI system. The frequency response of the LTI system is $H\left(e^{j \Omega}\right)=1-\frac{1}{2} e^{-j 2 \Omega}$. The output $y[n]$ of the system is
 A $\delta[n]+\delta[n-1]-\frac{1}{2} \delta[n-2]-\frac{5}{2} \delta[n-3]+\delta[n-5]$ B $\delta[n]-\delta[n-1]-\frac{1}{2} \delta[n-2]-\frac{5}{2} \delta[n-3]+\delta[n-5]$ C $\delta[n]-\delta[n-1]-\frac{1}{2} \delta[n-2]+\frac{5}{2} \delta[n-3]-\delta[n-5]$ D $\delta[n]+\delta[n-1]+\frac{1}{2} \delta[n-2]+\frac{5}{2} \delta[n-3]+\delta[n-5]$
GATE EC 2023   Signals and Systems
Question 1 Explanation:
\begin{aligned} y[n] & =x[n] * h[n] \\ Y\left(e^{j \Omega}\right) & =X\left(e^{j \Omega}\right) H\left(e^{j \Omega}\right) \\ & =\left[1-e^{-j \Omega}+2 e^{-3 j \Omega}\right]\left[1-\frac{1}{2} e^{-2 j \Omega}\right] \\ & =1-e^{-j \Omega}+2.5 e^{-3 j \Omega}-0.5 e^{-j 2 \Omega}-e^{-j 5 \Omega} \end{aligned}

Taking IDTFT;
$y[n]=\delta[n]-\delta[n-1]-0.5 \delta[n-2]+2.5 \delta[n-3]-\delta[n-5]$
 Question 2
Consider a discrete-time periodic signal with period $N=5$. Let the discrete-time Fourier series (DTFS) representation be $x[n]=\sum_{k=0}^{4} a_{k} e^{\frac{j k 2 \pi n}{5}}$, where $a_{0}=1, a_{1}=3 j, a_{2}=2 j$, $a_{3}=-2 j$ and $a_{4}=-3 j$. The value of the sum $\sum_{n=0}^{4} x[n] \sin \frac{4 \pi n}{5}$ is
 A -10 B 10 C -2 D 2
GATE EC 2023   Signals and Systems
Question 2 Explanation:
Let, $I=\sum_{n=0}^{4} x(n) \sin \frac{4 \pi n}{5}$

\begin{aligned} & =\frac{1}{2 j} \sum_{n=0}^{4} x(n) \cdot\left[e^{j \frac{4 \pi n}{5}}-e^{-j \frac{4 \pi n}{5}}\right] \\ & =\frac{1}{2 j}\left[\sum_{n=0}^{4} x(n) e^{j \frac{4 \pi n}{5}}-\sum_{n=0}^{4} x(n) \cdot e^{-j \frac{4 \pi n}{5}}\right] \quad (i) \end{aligned}
As we know, $\quad a_{k}=\frac{1}{N} \sum_{n=0}^{4} x(n) \cdot e^{-j k \cdot \frac{2 \pi}{N} n}$ $=\frac{1}{5} \sum_{n=0}^{4} x(n) \cdot e^{-j \frac{2 \pi}{N} k n}$

Put $K=2 ; \quad a_{2}=\frac{1}{5} \sum_{n=0}^{4} x(n) \cdot e^{-j \frac{4 \pi n}{5}}$

Put $K=-2 ; \quad a_{-2}=\frac{1}{5} \sum_{n=0}^{4} x(n) \cdot e^{j \frac{4 \pi n}{5}}$

From equation (i),

\begin{aligned} I&=\frac{1}{2 j}\left[5 a_{-2}-5 a_{2}\right] & =\frac{5}{2 j}\left[a_{3}-a_{2}\right] \\ I & =\frac{5}{2 j}[-2 j-2 j]=-10 \end{aligned}
$\begin{array}{rl} a_{-2} & =a_{-2+N} \\ \because \quad & =a_{-2+5} \\ & =a_{3} \end{array}$

 Question 3
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   Signals and Systems
Question 3 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 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   Signals and Systems
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   Signals and Systems
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}

There are 5 questions to complete.