Definition of Discrete Fourier Transform (DFT) MCQ Quiz - Objective Question with Answer for Definition of Discrete Fourier Transform (DFT) - Download Free PDF

Analog signal recorded:​ The process of capturing and storing an analog signal, which is a continuous-time signal. The minimum length of the analog signal recorded. It means capturing a continuous analog signal for at least a specified duration.

To find the minimum length of the analog signal recorded, considering the given condition B = 6kHz, DFT points N = 2^m and resolution $\le 200Hz.$

Resolution = $\frac{Bandwidth}{N}$

In this case:

$\frac{6kHz}{2^m}\le 200Hz$

Solving for m.

2^m $\ge \frac{6kHz}{200Hz}$

m $\ge lo{g}_2[\frac{6kHz}{200Hz}]$

m $\ge lo{g}_2[30]$

m$\ge 4.9069$

Since m must be an integer, we round up to the next integer, so m = 5.

Now, the minimum length of the analog signal recorded is given by:

Signal length = $\frac{N}{Bandwidth}$

Signal length = $\frac{2^5}{6kHz}$

Signal length = $\frac{32}{6kHz}$

Signal length $\approx$ 0.00533 seconds.

This is very closest to option 3 0.005 seconds.

Here, option 3 is correct.

We know that, $a_k=\frac{1}{N}\sum _{n=0}^{N-1}x(n)e^{\frac{-j2\pi }{N}kn}$

$a_k={\frac{1}{N}X(\mathrm{\Omega })|}_{\mathrm{\Omega }=\frac{2\pi k}{N}}$

$a_k={\frac{1}{N}(1+\cos \mathrm{\Omega })e^{-j\mathrm{\Omega }}|}_{\mathrm{\Omega }=\frac{2\pi k}{N}}$ = $\frac{1}{N}(1+\cos \frac{2\pi k}{N})e^{\frac{-j2\pi k}{N}}$

$a_3=\frac{1}{5}(1+\cos \frac{2\pi \times 3}{5})e^{\frac{-j2\pi \times 3}{5}}$

$a_3=\frac{1}{5}(1+\cos \frac{6\pi }{5})e^{\frac{-j6\pi }{5}}$

$\left|a_3\right|=\frac{1}{5}(1+\cos \frac{6\pi }{5})=0.038$

Hence, the correct answer is 0.038

Given:

 $X(\omega )=\{\begin{array}{ll}1,& \text{ for }|\omega |<{\omega }_0\\ 0,& \text{ for }|\omega |>{\omega }_0\end{array}$

By taking inverse Fourier transform,

$x(t)=\frac{\sin {\omega }_{0}t}{\pi t}$

$x\left(\frac{\pi }{2{\omega }_0}\right)=\frac{2{\omega }_0}{\pi \times \pi }\sin {\omega }_0\times \frac{\pi }{2{\omega }_0}$ $=\frac{2{\omega }_0}{{\pi }^2}\sin \frac{\pi }{2}=\frac{2{\omega }_0}{{\pi }^2}$

So, option (C) and (D) are wrong.

$x(0)=\underset{t\to 0}{Lt}\frac{\sin {\omega }_{0}t}{\pi t}=$ $\underset{t\to 0}{Lt}\frac{{\omega }_0\cos {\omega }_{0}t}{\pi }=\frac{{\omega }_0}{\pi }$

So, x(0) ∝ ω₀ Option (B) is wrong.

When ω0​ → ∞, X(ω) will be a D.C signal and inverse Fourier transform of a D.C signal will be impulse signal.

So, option (A) is correct.

Hence, the correct option is (A).

Concept:

Expression for calculating IDFT

$X(K)=\sum _0^{N-1}x(n)e^{\frac{2\pi nK}{N}}$

Where

X(K) → DFT of sequence x(n) 

N → Period of the sequence

x(n) → sequence in the time domain

Solution:

Given;

$x(n)=3+{\cos }^2(\frac{2\pi n}{N})for;n=0,1,.....N-1$

The DFT of the sequence can be evaluated by expanding the cosine as a sum of complex exponentials.

Then;

    $x(n)=3+{\cos }^2(\frac{2\pi n}{N})$

$\Rightarrow x(n)=3+\frac{1}{4}{(e^{\frac{j2\pi n}{N}}+e^{\frac{-j2\pi n}{N}})}^2$

$\Rightarrow x(n)=3+\frac{1}{2}+\frac{1}{4}{e}^{\frac{j4\pi n}{N}}+\frac{1}{4}{e}^{\frac{-j4\pi n}{N}}$

$\Rightarrow x(n)=\frac{7}{2}+\frac{1}{4}{e}^{j\frac{2\pi }{N}(2n)}+\frac{1}{4}{e}^{j\frac{2\pi }{N}(N-2)n}$

The DFT coefficients are;

X(K) = 7/2N      for; K = 0 

        = 1/4N      for; K =  2 and K = N - 2

         = 0           else

Concept:

DFT of delayed sequence:

Let x(n) be a discrete sequence, and x'(n) is a delayed or shifted sequence of x(n) by n0.

If;

DFT{x(n)} ↔ X(K)

Then;

DFT{x'(n)} = DFT{x[(n - n0),mod N]} =  X(K)e-(j2πn0k/N)

Solution:

Given;

y(n) = x(n)+x(n-N)

Length of y(n) = 2N

Perform 2N point DFT

If;

x(n) ↔ X(k)

x(n-N) ↔ X(K)e^-(j2πNk/2N)

Then;

2N point DFT of y(n);

    Y(K) =  X(k) + X(K)e-(j2πNk/2N)

⇒ Y(K) = (1 + e-(j2πNk/2N))X(K)

⇒ Y(K) = [1 + (-1)^k] X(K)

∴ Y(K) =  2X(K)   ;  for K = even

           = 0            ;  for K = odd 

Concept:

$X\left(K\right)=\sum _{n=0}^{N-1}x\left(n\right)e^{\frac{-j2\pi kn}{N}}$

Analysis:

$X\left(0\right)=\sum _{n=0}^{N-1}x\left(n\right)$

= x(0) + x(1) + x(2) + x(3) + x(4) + x(5) + x(6) + x(7) …1)

$X\left(4\right)=\sum _{n=0}^{N-1}x\left(n\right){\left(-1\right)}^n$

= x(0) - x(1) + x(2) - x(3) + x(4) - x(5) + x(6) - x(7) …2)

Adding equation 1) and 2)

X(0) + X(4) = 2[x(0) + x(2) + x(4) + x(6)]     …3)

From equation 3),

[X(0) + X(4)] /2 = $\sum _{n=0}^{3}x\left(2n\right)$

$=(1+5)/2=3$

$\sum _{n=0}^{3}x\left(2n\right)=3$

Concept:

1). The N point DFT sequence is circularly even if it is symmetric about a point on circle i.e.

x[n] = x[N - n] for 1 ≤ n ≤ N -1

2). The N point DFT sequence is circularly odd if it is anti-symmetric about a point on circle i.e.

x[n] = -x[N- n] for 1 ≤ n ≤ N -1

Analysis:

Given:

DFT sequence x[n] = {2, 3, 4, 3}  and N = 4

Checking if x[n] = x[N - n], we can write:

x[1] = x[4 - 1] = x[3] = 3

x[2] = x[4 - 2] = x[2] = 4

x[3] = x[4 - 3] = x[1] = 3

Hence, it is 4 point circularly even.

Concept:

Sampling theorem is the bridge between continuous-time signals and discrete-time signals i.e., we convert continuous signals to discrete signals using sampling theorem.

y[n]= x(nTs)

Ts = Sampling interval

Calculation:

It is given that a finite duration discrete signal x(n) is obtained by sampling x(t) = cos(200πt) with T_s = n/400

x(t) = cos 200 π t

Since the sampling instant is t = n/400, x(t) can be written as:

$x\left(n\right)=\cos \left[200\pi \cdot \frac{n}{400}\right]$

$x\left(n\right)=\cos \left(\frac{\pi }{2}n\right)$ n = 0, 1, 2 …7

$x\left(n\right)=\left\{\cos 0,\cos \frac{\pi }{2},\cos \pi ,\cos \frac{3\pi }{2},\dots ,\cos \frac{7\pi }{2}\right\}$

x(n) = {+1, 0, -1, 0, +1, 0, -1, 0}

The discrete Fourier Transform of x[n] will be X[k].

Assuming, y(n) = [1, -1, 1, -1], i.e.

$x\left(n\right)=y\left(\frac{n}{2}\right)$

Then its 4-point DFT will be:

$Y\left(K\right)=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -j& -1& j\\ 1& -1& 1& -1\\ 1& j& -1& -j\end{array}\right]\left[\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right]$

After solving we’ll get,

Y(K) $=\left[\begin{array}{cccc}0& 0& 4& 0\end{array}\right]$

We know that,

If y(n) ↔ Y(K)

Then $y\left(\frac{n}{m}\right)$ ↔ [Y(K), Y(K), Y(K) … m times]

So, $y\left(\frac{n}{2}\right)$ ↔ [Y(K), Y(K)]

Since, Y(K) = [0, 0, 4, 0]

$x\left(n\right)=y\left(\frac{n}{2}\right)\leftrightarrow \left[\begin{array}{cccccccc}0,& 0,& \underset{X(2)}{\underbrace{4,}}& 0,& 0,& 0,& \underset{X(6)}{\underbrace{4,}}& 0\end{array}\right]$

Concept:

The DFT of a sequence is given as:

$\left[\begin{array}{c}X\left[0\right]\\ \mathrm{X}\left[1\right]\\ \mathrm{X}\left[2\right]\\ \mathrm{X}\left[3\right]\end{array}\right]=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -\mathrm{j}& -1& \mathrm{j}\\ 1& -1& 1& -1\\ 1& \mathrm{j}& -1& -\mathrm{j}\end{array}\right]\left[\begin{array}{c}\mathrm{x}\left[0\right]\\ \mathrm{x}\left[1\right]\\ \mathrm{x}\left[2\right]\\ \mathrm{x}\left[3\right]\end{array}\right]$

Analysis:

Substituting value we have

$\left[\begin{array}{c}X\left[0\right]\\ \mathrm{X}\left[1\right]\\ \mathrm{X}\left[2\right]\\ \mathrm{X}\left[3\right]\end{array}\right]=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -\mathrm{j}& -1& \mathrm{j}\\ 1& -1& 1& -1\\ 1& \mathrm{j}& -1& -\mathrm{j}\end{array}\right]\left[\begin{array}{r}1\\ 2\\ 3\\ 4\end{array}\right]$

$=\left[\begin{array}{r}10\\ 1-2\mathrm{j}-3+4\mathrm{j}\\ 1-2+3-4\\ 1+2\mathrm{j}-3-4\mathrm{j}\end{array}\right]$

X[K] = [10, -2+2j, -2, -2-2j]

option(3) is the correct answer

Given:

 $X(\omega )=\{\begin{array}{ll}1,& \text{ for }|\omega |<{\omega }_0\\ 0,& \text{ for }|\omega |>{\omega }_0\end{array}$

By taking inverse Fourier transform,

$x(t)=\frac{\sin {\omega }_{0}t}{\pi t}$

$x\left(\frac{\pi }{2{\omega }_0}\right)=\frac{2{\omega }_0}{\pi \times \pi }\sin {\omega }_0\times \frac{\pi }{2{\omega }_0}$ $=\frac{2{\omega }_0}{{\pi }^2}\sin \frac{\pi }{2}=\frac{2{\omega }_0}{{\pi }^2}$

So, option (C) and (D) are wrong.

$x(0)=\underset{t\to 0}{Lt}\frac{\sin {\omega }_{0}t}{\pi t}=$ $\underset{t\to 0}{Lt}\frac{{\omega }_0\cos {\omega }_{0}t}{\pi }=\frac{{\omega }_0}{\pi }$

So, x(0) ∝ ω₀ Option (B) is wrong.

When ω0​ → ∞, X(ω) will be a D.C signal and inverse Fourier transform of a D.C signal will be impulse signal.

So, option (A) is correct.

Hence, the correct option is (A).

Concept:

By duality property

 $x(n)\underset{}{\overset{DFT}{\to }}\stackrel{DFT}{\to }Nx(-k)$

Calculation:

Given 

X̅  = DFT x(n) =  X[k] = [1, 0, 0, 0, 2, 0, 0, 0] 

$\mathrm{X}[\mathrm{k}]=\sum _{\mathrm{n}=0}^7\mathrm{x}[\mathrm{n}]\mathrm{e}\mathrm{x}\mathrm{p}(-\mathrm{j}\frac{2\pi }{8}\mathrm{n}\mathrm{k})$

Therefore we hane to find 

y̅ = (DFT (x̅ )) = DFT (DFT x(n))

By duality property

$x(n)\underset{}{\overset{DFT}{\to }}\stackrel{DFT}{\to }Nx(-k)$ 

DFT (DFT x(n)) = y̅  = N x(-k)

X(k) = [1, 0, 0, 0, 2, 0, 0, 0] 

X(-k) = [1, 0, 0, 0, 2, 0, 0, 0] 

y(n) =  N X(-k) = 8 [1, 0, 0, 0, 2, 0, 0, 0] 

y(0) = 8

Calculation:

$DTFTX\left(e^{j\omega }\right)=\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}x\left(n\right)e^{-j\omega n}$ 

Given x(n) = aⁿu(n),  -1 < a < 1

$X\left(e^{j\omega }\right)=\underset{0}{\sum ^{\mathrm{\infty }}}{a}^n{e}^{-j\omega n}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(a{e}^{-j\omega }\right)}^n$ 

$\Rightarrow \text{ Knowing }\text{ that},\underset{K=0}{\sum ^{\mathrm{\infty }}}{r}^n=\frac{1}{1-r}$ 

Hence, x((n + k))_N Sequence x(n) shifted clockwise by k samples.

The keyword to pick in the question is “real sequence”

For real sequence we have

X[k] = X^*[N -k]

Here N = 5

X[1] = X^*[4]

X[4] = X^*[1] = 1+ j1

X[2] = X^*[3]

X[2] = 2 – j2

The zeroth value of the sequence is

$x\left(0\right)=\frac{1}{N}\sum _{k=0}^{k=4}X\left[k\right]$

$\frac{X\left[0\right]+X\left[1\right]+X\left[2\right]+X\left[3\right]+X\left[4\right]}{5}$

$=\frac{4+1+2+2+1}{5}$

x(0) = 2

Given matrix product 

[p q r s] = [a b c d] $\left[\begin{array}{cccc}a& b& c& d\\ d& a& b& c\\ c& d& a& b\\ b& c& d& c\end{array}\right]$

take transpose on both sides

$\left[\begin{array}{c}p\\ q\\ r\\ s\end{array}\right]=\left[\begin{array}{cccc}a& d& c& b\\ b& a& d& c\\ c& b& a& d\\ d& c& b& a\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]$

This is a circulation convolution 

[p q r s] = [a b c d] [a b c d]

take DFT both sides

$DFT[\mathrm{p}\mathrm{q}\mathrm{r}\mathrm{s}]=[\alpha \beta \gamma \delta ]\cdot [\alpha \beta \gamma \delta ]$

$DFT[\mathrm{p}\mathrm{q}\mathrm{r}\mathrm{s}]=[{\alpha }^2{\beta }^2{\gamma }^2{\delta }^2]$