GATE ECE · Signals And Systems

Dtfs Dtft And Dft Practice Questions

Generate GATE-level questions on DTFS, DTFT, and DFT. Focus on: 1. Discrete-time Fourier series and transforms. 2. DFT and FFT algorithms. 3. Frequency domain analysis of discrete signals.

39 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025MSQ

Let $x[n]$ be a discrete-time signal whose $z$ -transform is $X(z)$ . Which of the following statements is/are TRUE?

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📖 Explanation

(A) The DTFT does not always exist. It exists only if the signal is absolutely summable: $\sum_{n=-\infty }^{\infty } |x[n]| < \infty$ So this statement is False. (B) The region of convergence (RoC) contains no poles, but it can include zeros (since zeros do not affect convergence). So this statement is False. (C) This is True. The DTFT of $x[n]$ exists iff the RoC of $X(z)$ includes the unit circle $|z| = 1$ . (D) For $x[n] = \alpha \delta[n]$ , the $z$ -transform is $X(z) = \alpha$ (a constant), which is defined for all $z \in \mathbb{C}$ . Hence, the RoC is the entire z-plane, so this is True.

Q3GATE 2023MCQ

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

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📖 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; $y[n]=\delta[n]-\delta[n-1]-0.5 \delta[n-2]+2.5 \delta[n-3]-\delta[n-5]$

Q4GATE 2023MCQ

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

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📖 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}

Q8GATE 2021NAT

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 ______

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📖 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}

Q9GATE 2020MCQ

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?

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📖 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 $x(n)=\left \{ 1,0,-1,0,1,0,-1,0 \right \}$ $X(k)=\left \{ \underset{0}{\uparrow},0,4,0,0,0,4,0 \right \}$

Q16GATE 2016NAT

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 _________

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📖 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}