# DTFS, DTFT and DFT

 Question 1
A finite duration discrete-time signal x[n] is obtained by sampling the continuous-time signal $x(t) = \cos (200 \pi t)$ at sampling instants t = n/400, n = 0,1,..., 7. The 8-point discrete Fourier transform (DFT) of x[n] is defined as

$x[k]=\sum_{n=0}^{7}x[n]e^{-j\frac{\pi kn}{4}}$, k=0,1,...,7.

Which one of the following statement is TRUE?
 A All X[k] are non-zero B Only X[4] is non-zero C Only X[2] and X [6] are non-zero D Only X[3] and X [5] are non-zero
GATE EC 2020   Signals and Systems
Question 1 Explanation:
\begin{aligned}x(t)&=\cos 200\pi t\\ t&=\frac{n}{400}\\ x(n)&=\cos \left ( 200\pi \frac{n}{400} \right )\\&=\cos \left ( \frac{\pi }{2} n\right ); \; \; n=0,1,...,7\\ &=\left \{ \cos 0,\cos \frac{\pi }{2},\cos \pi ,\cos \frac{3\pi }{2} ,\cos 2\pi ,\cos \frac{5\pi }{2},\cos 3\pi ,\cos \frac{7\pi }{2}\right \} \\ &={1,0,-1,0,1,0,-1,0}\xLeftrightarrow{\text{DFT}} X(K) \\ \text{suppose, } y(n)&={1,-1,1,-1}\xLeftrightarrow{\text{DFT}} Y(k)\end{aligned}
$[Y(K)]=\left.\begin{matrix} [W_{N}] \end{matrix}\right|_{N=4}[y(n)]=\begin{bmatrix} 1 &1 &1 &1 \\ 1 &-j &-1 &j \\ 1&-1 &1 &-1 \\ 1&j& -1 &-j \end{bmatrix}\begin{bmatrix} 1\\ -1\\ 1\\ -1 \end{bmatrix}=\begin{bmatrix} 0 &0 &4 &0 \end{bmatrix}$
Now,as we know,
If for $\, \left \{ a,b,c,d \right \}\xLeftrightarrow{\text{DFT}} \left \{ A,B,C,D \right \}$
Then for, $\left \{ a,0,b,0,c,0,d,0 \right \}\xLeftrightarrow{\text{DFT}}\left \{A,B,C,D,A,B,C,D \right \}$
Similarly for $y(n)={1,-1,1,-1}\xLeftrightarrow{\text{DFT}}Y(k)=\left \{ 0,0,4,0\right \}$
Here, for $x(n)=\left \{ 1,0,-1,0,1,0,-1,0 \right \}$
$X(k)=\left \{ \underset{0}{\uparrow},0,4,0,0,0,4,0 \right \}$
 Question 2
It is desired to find a three-tap causal filter which gives zero signal as an output to an input of the form

$x[n]=c_1 \; exp\left ( -\frac{j\pi n}{2} \right )+c_2 \; exp\left ( \frac{j\pi n}{2} \right )$

where $c_1 \; and \; c_2$ are arbitrary real numbers. The desired three-tap filter is given by

$h[0]=1,\;\; h[1]=a, \;\;h[2]=b$
and
$h[n]=0\;\; for \; n \lt 0 \; or \; n\gt 2$

What are the values of the filter taps a and b if the output is y[n]=0 for all n, when x[n] is as given above?

 A a=1, b=1 B a=0, b=-1 C a=-1, b=1 D a=0, b=1
GATE EC 2019   Signals and Systems
Question 2 Explanation:
\begin{aligned} x(n) &=c_{1} e^{-j \frac{\pi}{2} n}+c_{2} e^{j \frac{\pi}{2} n} \\ \omega_{0} &=\frac{\pi}{2} \mathrm{rad} / \mathrm{s} \\ H\left(e^{i \omega}\right) &=1+a e^{-j \omega}+b e^{-j 2 \omega}\\ \text{To get }y(n)=0& \\ H\left(e^{j \omega_{o}}\right)&=\left.H\left(e^{j \omega}\right)\right|_{\omega=\frac{\pi}{2}}=0\\ 1+a e^{-i \frac{\pi}{2}}+b e^{-i 2 \frac{\pi}{2}}&=0 \\ 1-j a-b&=0 \end{aligned}
From the given options, a=0 and b=1
 Question 3
Consider a six-point decimation-in-time Fast Fourier Transform (FFT) algorithm, for which the signal-flow graph corresponding to X[1] is shown in the figure. Let $W_6=exp\left ( -\frac{j2 \pi}{6} \right )$. In the figure, what should be the values of the coefficients $a_1,a_2,a_3$ in terms of $W_6$ so that X[1] is obtained correctly?

 A $a_1=-1,a_2=W_6,a_3=W_6^2$ B $a_1=1,a_2=W_6^2,a_3=W_6$ C $a_1=1,a_2=W_6,a_3=W_6^2$ D $a_1=-1,a_2=W_6^2,a_3=W_6$
GATE EC 2019   Signals and Systems
Question 3 Explanation:
$\begin{array}{l} X(k)=\sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi}{N} k n} \\ X(1)=\sum_{n=0}^{5} x(n) W_{6}^{n}\\ =x(0)+x(1) W_{6}+x(2) M_{6}^{2}+x(3) n_{6}^{3}+x(4) M_{6}^{4}+x(5) n_{6}^{5} \ldots(i) \\ \end{array}$
From the given flow graph,
$X(k)=[x(0)-x(3)] a_{1}+[x(1)-x(4)] a_{2}+[x(2)-x(5)] a_{3}$
By comparing equations (i) and (ii), we get.
$a_{1}=1, a_{2}=W_{6}, a_{3}=W_{6}^{2}$
 Question 4
Let $X[k]=k+1$, $0\leq k \leq 7$ be 8-point DFT of a sequence $x[n]$,
where $X[k]= \sum_{n=0}^{N-1} x[n]e^{-\frac{j2nk\pi }{N}}$.
The value (correct to two decimal places) of $\sum_{n=0}^{3}x[2n]$ is ______.
 A 1.9 B 3 C 4 D 5
GATE EC 2018   Signals and Systems
Question 4 Explanation:
\begin{aligned} x(k) &=\{1,2,3,4,5,6,7,8\} \\ x(k) &=\sum_{n=0}^{7} x[n] e^{-j \frac{2 \pi}{8} k n} \\ &=\sum_{n=0}^{7} x[n] e^{-i \frac{\pi}{4} k n} \\ \sum_{n=0}^{3} x[2 n] &=\sum_{n=0.2 .4 .6} x[n] \\ &=\frac{1}{2} \sum_{n=0}^{7}\left(x[n]+(-1)^{n} x[n]\right) \\ \sum_{n=0}^{7} x[n] &=x(0)=1 \\ \sum_{n=0}^{7}(-1)^{n} x[n] &=\sum_{n=0}^{7} x[n] e^{-i \frac{\pi}{4} 4 n}=x(4)=5 \\ \sum_{n=0}^{3} x[2 n] &=\frac{1}{2}[1+5]=3 \end{aligned}
 Question 5
An LTI system with unit sample response h[n]=$5\delta [n]-7\delta [n-1]+7\delta [n-3]-5\delta [n-4]$ is a
 A Low - pass filter B high - pass filter C band - pass filter D band - stop filter
GATE EC 2017-SET-2   Signals and Systems
Question 5 Explanation:
\begin{aligned} h[n]=5 \delta[n]&-78[n-1]+7 \delta[n-3]-5 \delta[n-4] & \\ \text { Now, } \quad H\left(e^{j \omega}\right)&=5-7 e^{-j \omega}+7 e^{-3 j \omega}-5 e^{-4 j \omega}\\ \text{Now for }\omega & =0, \\ H\left(e^{j 0}\right)&=5-7+7-5=0\\ \text{and for }\omega & =\pi, &\\ H\left(e^{j \pi}\right) &=5-7(-1)+7(-1)-5(1) \\ &=5+7-7-5=0 \end{aligned}
System is attenuating low and high frequencies whereas passing the mid frequencies. So, its a BPF.
 Question 6
Let h[n] be the impulse response of a discrete-time linear time invariant (LTI) filter. The impulse response is given by

$h[0]=1/3; \; h[1]=1/3; \; h[2]=1/3; \; and \; h[n]=0$ for $n \lt 0 \; and \; n \gt 2.$

Let $H(\omega)$ be the discrete-time Fourier system transform (DTFT) of h[n], where $\omega$ is the normalized angular frequency in radians. Given that $H(\omega_{o})=0$ and $0 \lt \omega _{o} \lt \pi$, the value of $\omega_{0}$ (in radians) is equal to __________.
 A 1.25 B 2.1 C 3.5 D 4.75
GATE EC 2017-SET-1   Signals and Systems
Question 6 Explanation:
\begin{aligned} \text{since }\quad h[n]&=\frac{1}{3} \delta[n]+\frac{1}{3} \delta[n-1]+\frac{1}{3} \delta[n-2] \\ \text{So. }\quad H\left(e^{j \omega}\right)&=\frac{1}{3} e^{-j \omega}[1+2 \cos \omega] \\ H\left(e^{j \omega_{0}}\right)&=0, \quad\text{ for }\left(1+2 \cos \omega_{0}\right)=0 \\ \text{or} \quad \cos \omega_{0}&=-\frac{1}{2} \\ \text{or}\quad \omega_{0}&=\frac{2 \pi}{3}=2.10 \mathrm{radians} \end{aligned}
 Question 7
A continuous-time speech signal $x_{a}(t)$ is sampled at a rate of 8 kHz and the samples are subsequently grouped in blocks, each of size N. The DFT of each block is to be computed in real time using the radix-2 decimation-in-frequency FFT algorithm. If the processor performs all operations sequentially, and takes 20 $\mu$s for computing each complex multiplication (including multiplications by 1 and -1) and the time required for addition/subtraction is negligible, then the maximum value of N is __________
 A 1024 B 2048 C 4096 D 512
GATE EC 2016-SET-3   Signals and Systems
Question 7 Explanation:
\begin{aligned} f_{s} &=8000 \text { samples/sec } \\ \Rightarrow \quad T_{s} &=\frac{1}{f_{s}}=\frac{1}{8000} \sec \end{aligned}
Time for each multiplication $=T_{m}=20 \mu \mathrm{sec}$
$T_{m}=20 \times 10^{-6} \mathrm{sec}$
suppose block is of size N , then time taken to generate each block $=N \times T_{s}=N \times \frac{1}{8000}.$
No. of multiplications that can be performed by processor in the time taken for each block $=\frac{N \times T_{s}}{T_{m}}$
No. of multiplications required to compute DFT by Radix - 2 FFT algorithm $=\frac{N}{2} \log _{2} N$
No. of multiplication required by FFT $\leq$ No. of multiplications that can be performed by the processor in the time taken in each block
[Reason: real time]
\begin{aligned} \frac{N}{2} \log _{2} N & \leq \frac{N \times T_{s}}{T_{m}} \\ \log _{2} N & \leq \frac{2 \times \frac{1}{8000}}{20 \times 10^{-6}}=12.5 \\ N & \leq 2^{125} \\ \therefore \quad N & \leq 2^{12}=4096 \end{aligned}
 Question 8
The Discrete Fourier Transform (DFT) of the 4-point sequence

x[n]={ x[0],x[1],x[2],x[3] } = {3,2,3,4} is
X[k]={ X[0],X[1],X[2],X[3] } = {12,2j,0,-2j}.

If $X_{1}[k]$ is the DFT of the 12-point sequence $x_{1}[k]$={3,0,0,2,0,0,3,0,0,4,0,0}, the value of $|\frac{x_{1}[8]}{x_{1}[11]}|$ is ________
 A 4 B 6 C 8 D 10
GATE EC 2016-SET-2   Signals and Systems
Question 8 Explanation:
From the given question
\begin{aligned} x_{1}[n] &=x\left[\frac{n}{3}\right] \\ x_{1}[k]=\{12,2 j, 0,&-2 j, 12,2 j, 0,-2 j, 12,2 j, 0,-2\} \\ x_{1}[8] &=12 ; \quad x_{1}(11)=-2 j \\ \therefore \quad \frac{x_{1}(8)}{x_{1}(11)} &=\left|\frac{12}{-2 j}\right|=6 \end{aligned}
 Question 9
Consider the signal
$x[n]=6\delta [n+2]+3\delta [n+1]+8\delta [n]+7\delta [n-1]+4\delta [n-2]$.
If $X(e^{j\omega })$ is the discrete-time Fourier transform of x[n],
then $\frac{1}{\pi }\int_{-\pi }^{\pi }X(e^{j\omega })sin^{2}(2\omega )d\omega$ is equal to _________
 A 5 B 6 C 8 D 10
GATE EC 2016-SET-1   Signals and Systems
Question 9 Explanation:
From the definition of DTFT
\begin{aligned} X\left(e^{j \omega}\right) &=\sum_{n=-\infty}^{\infty} x[n] e^{-j \omega n} \\ x[n] &=\frac{1}{2 \pi} \int_{-\pi}^{\pi} X\left(e^{j \omega}\right) e^{j \omega n} d \omega \\ x[0] &=\frac{1}{2 \pi} \int_{-\pi}^{\pi} X\left(e^{j \omega}\right) d \omega \\ \frac{1}{2 \pi} \int_{-\pi}^{\pi} X\left(e^{j \omega}\right) Y\left(e^{j \omega}\right) d\omega &=[x(n) \otimes y(n)]_{n=0} \\ Y\left(e^{i \omega}\right) &=\sin ^{2}(2 \omega) \\ &=\frac{1-\cos 4 \omega}{2} \\ &=\frac{1}{2}-\frac{1}{4} e^{4 / \omega}-\frac{1}{4} e^{-4 j \omega}\\ \end{aligned}
\begin{aligned} y[n]&=\frac{1}{2} \delta[n]-\frac{1}{4} \delta[n+4]-\frac{1}{4} \delta[n-4] \\ y[n]&=\left\{-\frac{1}{4}, 0,0,0, \frac{1}{\underset{\uparrow}{2}}, 0,0,0,-\frac{1}{4}\right\} \\ \Rightarrow \quad y[0]&=\frac{1}{2} \\ x[n]&=\{6,3,\underset{\uparrow}{8},7,4\} ; \quad x[0]=8\\ \frac{1}{\pi} \int_{-\pi}^{\pi} x\left(e^{j \omega}\right) Y\left(e^{j \omega}\right) d \omega &=2 \sum_{n=-\infty}^{\infty} x[0] y[0] \\ &=2 \times 8 \times \frac{1}{2}=8 \end{aligned}
 Question 10
Let $\tilde{x}[n]=1+cos(\frac{\pi n}{8})$ be a periodic signal with period 16. Its DFS coefficients are defined by $a_{k}=\frac{1}{16}\sum_{n=0}^{15}\tilde{x}[n]exp(-j\frac{\pi }{8}kn)$ for all k . The value of the coefficient $a_{31}$ is _______.
 A 0 B 0.5 C 1 D 1.5
GATE EC 2015-SET-3   Signals and Systems
Question 10 Explanation:
\begin{aligned} \tilde{x}[n]&=1+\cos \left(\frac{\pi n}{8}\right)\\ &=1+\frac{e^{\left(\frac{2 \pi}{16}\right)} h}{2}+\frac{e^{-\left(\frac{2 \pi}{16}\right)}}{2} &(N=16)\\ &=1+\frac{e^{\left(\frac{2 \pi}{16}\right)}}{2}+\frac{e^{-\left(\frac{2 \pi}{16}\right)}}{2} \\ \Rightarrow a_{1}&=\frac{1}{2}\text{ and also } a_{31}=a_{15}=a_{-1}=\frac{1}{2} \end{aligned}
There are 10 questions to complete.