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 · 19 PYQs · 0 AI practice · GATE ECE 2027

Q21GATE 2015MCQ

Two sequences [a,b,c] and [A,B,C] are related as

\begin{bmatrix} A\\ B\\ C \end{bmatrix}=\begin{bmatrix} 1 & 1& 1\\ 1& W_{3}^{-1} & W_{3}^{-2}\\ 1&W_{3}^{-2} & W_{3}^{-4} \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix} \; where\; W_{3}=e^{j\frac{2\pi }{3}}

If another sequence [ p, q, r] is derived as,

\begin{bmatrix} p\\ q\\ r \end{bmatrix}=\begin{bmatrix} 1 & 1& 1\\ 1& W_{3}^{1} & W_{3}^{2}\\ 1&W_{3}^{2} & W_{3}^{4} \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & W_{3}^{2} &0 \\ 0&0 & W_{3}^{4} \end{bmatrix} \begin{bmatrix} A/3\\ B/3\\ C/3\end{bmatrix}

then the relationship between the sequences [ p, q, r] and [a, b, c] is

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📖 Explanation

\begin{aligned} \left[\begin{array}{l}A \\ B \\ C\end{array}\right]&=\left[\begin{array}{l}a+b+c \\ a+b W_{3}^{-1}+c W_{3}^{-2} \\ a+b W_{3}^{-2}+c W_{3}^{-1}\end{array}\right]\$ \\ \left[\begin{array}{l}p \\ q \\ r\end{array}\right]&=\left[\begin{array}{ccc}1 & w_{3}^{2} & w_{3}^{1} \\ 1 & w_{3}^{1} w_{3}^{2} & w_{3}^{2} w_{3}^{1} \\ 1 & w_{3}^{2} w_{3}^{2} & W_{3}^{1} W_{3}^{1}\end{array}\right]\$\left[\begin{array}{c}\frac{a+b+c}{3} \\ \frac{a+b W_{3}^{-1}+c W_{3}^{2}}{3} \\ \frac{a+b W_{3}^{-2}+c W_{3}^{-1}}{3}\end{array}\right]\$ \\ \text{where }&\quad W_{3}^{4}=e^{i \frac{2 \pi}{3}} \\ &\quad W_{3}^{2}=e^{i \frac{4 \pi}{3}}\\ W_{3}^{-1}&=e^{-j \frac{2 \pi}{3}} \\ W_{3}^{-2}&=e^{-j \frac{4 \pi}{3}} \\ W_{3}^{-4}&=W_{3}^{-1}=e^{-j \frac{2 \pi}{3}} \\ \left[\begin{array}{l}p \\q \\r\end{array}\right]&=\left[\begin{array}{l} c \\a \\b\end{array}\right] \end{aligned}

Q22GATE 2014MCQ

The N -point DFT X of a sequence x[n], $0\leq n\leq N-1$ is given by, $X[k]=\frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi }{N}nk}$ , $0\leq k\leq N-1$ Denote this relation as $X = DFT(x)$ . For N=4, which one of the following sequences satisfies $DFT(DFT(x)) = x$ ?

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📖 Explanation

$\text { DFT }(\text{DFT}(x))=x$

\begin{aligned} \text { Let, } x &=[123] \\ \text { DFT }(x) &=\frac{1}{\sqrt{4}}\left[\begin{array}{cccc}1 & 1 & 1 & 1 \\ 1 & -j & -1 & j \\ 1 & -1 & 1 & -1 \\ 1 & j & -1 & -j\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 2\end{array}\right] \\ &=\frac{1}{2}\left[\begin{array}{r}8 \\ -2 \\ 0 \\ -2\end{array}\right]=\left[\begin{array}{r}4 \\ -1 \\ 0 \\ -1\end{array}\right] \end{aligned}

\begin{array}{l} \text { DFT } (\text { DFT }(x))=\frac{1}{\sqrt{4}}\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -j & -1 & j \\ 1 & -1 & 1 & -1 \\ 1 & j & -1 & -j \end{array}\right]\left[\begin{array}{c} 4 \\ -1 \\ 0 \\ -1 \end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{l} 2 \\ 4 \\ 6 \\ 4 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \\ 2 \end{array}\right] \end{array}

Q26GATE 2014NAT

A Fourier transform pair is given by $(\frac{2}{3})^{n} \; u[n+3]\overset{FT}{\leftrightarrow}\frac{Ae^{-j6\pi f}}{1-(\frac{2}{3})e^{-j2 \pi f}}$ where u [n] denotes the unit step sequence. The values of A is _____.

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📖 Explanation

$\left(\frac{2}{3}\right)^{n} u(n+3) \stackrel{F.T.}{\longleftrightarrow} \frac{Ae^{j6\pi f}}{1-\left(\frac{2}{3}\right) e^{-i 2 \pi f}}$ Let, $x[n]=\left(\frac{2}{3}\right)^{n} u(n+3)$ Discrete Fourier transform is given as

\begin{aligned} X\left(e^{j \omega}\right)&=\sum_{n=-\infty }^{\infty } x[n] e^{-j \omega n} \\ &=\sum_{n=-3}^{\infty }\left(\frac{2}{3}\right)^{n} e^{-j \omega n} \\ \text { Put, } n+3 &=m \\ \text { So, } x\left(e^{i \omega}\right) &=\sum_{m=0}^{\infty }\left(\frac{2}{3}\right)^{m-3} e^{-j \omega(m-3)} \\ &=\left(\frac{2}{3}\right)^{-3} \sum_{m=0}^{\infty }\left(\frac{2}{3}\right)^{m} e^{-j \omega m \cdot e^{j \omega 3}} \\ &=\left(\frac{3}{2}\right)^{3} \sum_{m=0}^{\infty }\left(\frac{2}{3} e^{-j \omega}\right)^{m} \cdot e^{j \omega 3} \\ &=\frac{3.375 \cdot e^{i 2 \pi / 3}}{1-\frac{2}{3} e^{-i 2 \pi t}} \\ &=\frac{3.375 e^{j \pi \pi f}}{1-\frac{2}{3} e^{-i 2 \pi t}} \\ \text{So,}\quad A&=3.375 \end{aligned}

Q39GATE 2004MCQ

The impulse response h[n] of a linear time invariant system is given as

\left\{\begin{matrix} -2\sqrt{2} & n=1,-1\\ 4\sqrt{2} & n=2,-2\\ 0& otherwise \end{matrix}\right.

If the input to the above system is the sequence $e^{j\pi n/4}$ , then the output is

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📖 Explanation

\begin{aligned} h(n)&= 4 \sqrt{2} \delta(n+2)-2 \sqrt{2} \delta(n+1) \\ &-2 \sqrt{2} \delta(n-1)+4 \sqrt{2} \delta(n-2) \\ x(n)&=e^{j \pi n / 4} \\ y(n)&= x(n)^{*} h(n) \\ &=4 \sqrt{2} \delta(n+2)^{*} e^{j n \pi / 4}+4 \sqrt{2} \delta(n-2)^{*} e^{i \pi / 4} \\ &-2 \sqrt{2}\left[\delta(n+1)^{*} e^{j m / 4}+\delta(n-1) e^{j m / 4}\right]\$ \\ &\quad x(n)^{\star} \delta(n-a) \rightarrow x(n-a) \\ &y(n)=4 \sqrt{2}\left[e^{j \pi / 4(n+2)}+e^{j \pi / 4(n-2)}\right]\$ \\ &\quad -2 \sqrt{2}\left[e^{j \pi / 4(n+1)}+e^{j \pi / 4(n-0)}\right]\$ \\ &=e^{j n \pi / 4}\left[4 \sqrt{2}\left(e^{j \pi / 2}+e^{-j \pi / 2}\right)\right]\$ \\ &\quad -2 \sqrt{2}\left(e^{i \pi / 4}+e^{-j \pi / 4}\right) \\ &=e^{i n / 4}[0-2 \sqrt{2} \times 2 \cos \pi / 4] \\ y(n)&=-4 e^{i n \pi / 4} \end{aligned}