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
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
\delta[n]+\delta[n-1]-\frac{1}{2} \delta[n-2]-\frac{5}{2} \delta[n-3]+\delta[n-5] | |
\delta[n]-\delta[n-1]-\frac{1}{2} \delta[n-2]-\frac{5}{2} \delta[n-3]+\delta[n-5] | |
\delta[n]-\delta[n-1]-\frac{1}{2} \delta[n-2]+\frac{5}{2} \delta[n-3]-\delta[n-5] | |
\delta[n]+\delta[n-1]+\frac{1}{2} \delta[n-2]+\frac{5}{2} \delta[n-3]+\delta[n-5] |
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]
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
-10 | |
10 | |
-2 | |
2 |
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}
\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).
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).
3.2 | |
6.8 | |
4 | |
8 |
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 _________.
\frac{\pi}{2j}\omega e^{-|\omega|} | |
\frac{\pi}{2}\omega e^{-|\omega|} | |
\frac{\pi}{2j} e^{-|\omega|} | |
\frac{\pi}{2} e^{-|\omega|} |
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 |}
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 ______
\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 ______
2.4 | |
12.5 | |
8 | |
10.8 |
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}
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.