Definition of Discrete Fourier Transform (DFT) MCQ Quiz in मल्याळम - Objective Question with Answer for Definition of Discrete Fourier Transform (DFT) - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Given x[n] = {4, 5, 6, 7}

$x_1\left[n\right]=x\left[\frac{n}{3}\right]=\left\{4,0,0,5,0,0,6,0,0,7,0,0\right\}$

From the definition of N-point DFT:

If $x_n\left[n\right]\begin{array}{c}\stackrel{N-Point}{\to }\\ DFT\end{array}{X}_1\left(k\right)$ 

$X_1\left(k\right)-{\sum }_{n=0}^{N-1}{x}_1\left[n\right]e^k$

$X_1\left(0\right)={\sum }_{n=0}^{N-1}{x}_1\left[n\right]e^0$

$\therefore X_1\left(0\right)={\sum }_{n=0}^{N-1}{x}_1\left[n\right]=x_1\left(0\right)+x_1\left(1\right)+\dots +x_1\left(11\right)$   (As N = 12)   ---(1)

$X_1\left(\frac{N}{2}\right)={\sum }_{n=0}^{N-1}{x}_1\left[n\right]e^{-\frac{j2\pi }{N}\times \frac{N}{2}n}$

$X_1\left(\frac{N}{2}\right)={\sum }_{n=0}^{N-1}{x}_1\left[n\right]e^{-j\pi n}$

$X_1\left(\frac{N}{2}\right)={\sum }_{n=0}^{N-1}{\left(-1\right)}^n{x}_1\left[n\right]$

For N = 12:

$X_1\left(6\right)={\sum }_{n=0}^{N-1}{\left(-1\right)}^n{x}_1\left[n\right]=x_1\left(0\right)-x_1\left(1\right)+x_1\left(2\right)-x_1\left(3\right)\dots -x_1\left(11\right)$     ---(2)

x₁(n) = [4, 0, 0, 5, 0, 0, 6, 0, 0, 7, 0, 0]

From equation (1):

X₁(0) = 4 + 0 + 4 + 5 + 0 + 0 + 6 + 0 + 0 + 7 + 0 + 0

X₁(0) = 22

From equation (2):

X₁(6) = 4 – 0 + 0 – 5 + 0 – 0 + 6 – 0 + 0 – 7 + 0 – 0

X₁(6) = -2

∴ |X₁(0) – X₁(6)| = |22 – (-2)| = 24

$\begin{array}{l}x\left(k\right)=\sum _{n=0}^{7}x\left(n\right)e^{\frac{-j2\pi nk}{8}}\\ =\sum _{n=0}^{3}x\left(n\right)e^{-j\frac{2\pi }{8}nk}+\sum _{n=4}^{7}x\left(n\right)e^{-j\frac{2\pi }{8}nk}\end{array}$

let n – 4 = l → n = 4 + l

Second summation becomes

$\sum _{l=0}^{3}x\left(4+l\right)e^{-j\frac{2\pi }{8}\left(4+l\right)k}$

$\sum _{l=0}^{3}x\left(4+l\right)e^{-j\frac{2\pi }{8}lk}{e}^{-j\pi k}$

Changing the variable

l → n, 2^nd summation becomes

$\sum _{n=0}^{3}x\left(n+4\right)e^{-j\frac{2\pi }{8}nk}{e}^{-j\pi k}$      ______(1)

Symmetry property in Question

$x\left(n+\frac{N}{2}\right)=-x\left(n\right)$

x(n + 4) = -x(n)          _____(2)

Substitute in (1)

$-\sum _{n=0}^{3}x\left(n\right)e^{-j\frac{2\pi }{8}nk}{e}^{-j\pi k}$

e^-jπk = (-1)^k

The total summation is

$X\left[k\right]=\sum _{n=0}^{3}x\left(n\right)e^{-j\frac{2\pi }{8}nk}-\sum _{n=0}^{3}x\left(n\right)e^{-j\frac{2\pi }{8}nk}{\left(-1\right)}^k$

For k = even (0, 2, 4, 6)

X[0] = X[2] = X[4] = X[6] = 0

Taking D.F.T of x[n]

$X\left[k\right]=\sum _{n=0}^{3}x\left[n\right]W_4^{kn}$

$X\left[k\right]=1-W_4^{2k}$

Taking DFT of h[n]

$H\left[k\right]=\sum _{n=0}^{3}h\left[n\right]W_4^{kn}$

$=1+\frac{1}{2}{W}_4^k+\frac{1}{4}{W}_4^{2k}+\frac{1}{8}{W}_4^{3k}$

Y[k] = X[k] H[k]

$=\left(1-W_4^{2k}\right)\left(1+\frac{1}{2}{W}_4^k+\frac{1}{4}{W}_4^{2k}+\frac{1}{8}{W}_4^{3k}\right)$

$=1+\frac{1}{2}{W}_4^k-\frac{3}{4}{W}_4^{2k}-\frac{3}{8}{W}_4^{3k}-\frac{1}{4}{W}_4^{4k}-\frac{1}{8}{W}_4^{5k}$

Since,

 $W_4^{4k}=1^k$

And $W_4^{5k}=W_4^k$

We obtain

$Y\left[k\right]=\frac{3}{4}+\frac{3}{8}{W}_4^k-\frac{3}{4}{W}_4^{2k}-\frac{3}{8}{W}_4^{3k};k=0,1,2,3$

Thus by definition of DFT we get

$y\left[n\right]=\{\frac{3}{4},\frac{3}{8},\frac{-3}{4},\frac{-3}{8}]$

Hence y[2] = - 3/4

= - 0.75

Concept:

N-point discrete Fourier transform (DFT) is given by

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

Calculation:

Given $\mathrm{x}\left[\mathrm{n}\right]=\left\{3,2,3,4\right\}$

$\begin{array}{l}\mathrm{X}\left[\mathrm{k}\right]=\left(12,2\mathrm{j},01-2\mathrm{j}\right)\\ {\mathrm{x}}_1\left[\mathrm{n}\right]=\left\{3,0,0,2,0,0,3,0,0,4,0,0\right\}\\ \therefore {\mathrm{x}}_1\left[\mathrm{n}\right]=\mathrm{x}\left(\frac{\mathrm{n}}{3}\right)\end{array}$

Apply DTFT on both sides we get

$\begin{array}{l}{\mathrm{X}}_1\left[\mathrm{k}\right]=\sum _{\mathrm{n}=0}^{\mathrm{N}-1}{\mathrm{x}}_1\left(\mathrm{n}\right).{\mathrm{e}}^{-\frac{\mathrm{j}2\pi }{\mathrm{N}}\mathrm{n}\mathrm{k}}\\ {\mathrm{X}}_1\left(8\right)=\sum _{\mathrm{n}=0}^{11}{\mathrm{x}}_1\left(\mathrm{n}\right).{\mathrm{e}}^{\frac{\mathrm{j}2\pi }{12}\times 8\mathrm{n}}=\sum _{\mathrm{n}=0}^{11}{\mathrm{x}}_1\left(\mathrm{n}\right){\mathrm{e}}^{-\frac{\mathrm{j}4\pi }{3}\mathrm{n}}\end{array}$

${\mathrm{X}}_1(8)=3+2+3+4=12$

$\begin{array}{l}{\mathrm{X}}_1\left(11\right)={\mathrm{x}}_1^{\ast }\left(1\right)\\ {\mathrm{X}}_1\left(1\right)=\sum _{\mathrm{n}=0}^{11}{\mathrm{x}}_1\left(\mathrm{n}\right).{\mathrm{e}}^{\frac{\mathrm{j}2\pi }{12}\times \mathrm{n}.1}=\sum _{\mathrm{n}=0}^{11}{\mathrm{x}}_1\left(\mathrm{n}\right){\mathrm{e}}^{\frac{\mathrm{j}\pi }{6}.\mathrm{n}}\\ {\mathrm{X}}_1\left(1\right)=3+2.{\mathrm{e}}^{-\frac{\mathrm{j}\pi }{6}.3}+3.{\mathrm{e}}^{-\frac{\mathrm{j}\pi }{6}.6}+4{\mathrm{e}}^{-\frac{\mathrm{j}\pi }{6}.9}\end{array}$

$=3–2\mathrm{j}+3(-1)+4(\mathrm{j})=2\mathrm{j}$

$\begin{array}{l}{\mathrm{X}}_1^{\ast }\left(1\right)=-2\mathrm{j}\\ \therefore \left|\frac{{\mathrm{X}}_1\left(8\right)}{{\mathrm{X}}_1\left(11\right)}\right|=\frac{12}{2}=6\end{array}$

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

Concept:

N-point discrete Fourier transform (DFT) is given by

$x\left(k\right)=\sum _{n=0}^{N-1}x\left(n\right)e^{-\left(\frac{J2\pi }{N}\right)}\times kn\$$ 

Solution:

x(n) = {1, 2, 3, 4, 5}

$x\left(k\right)=\sum _{n=0}^{N-1}x\left(n\right)e^{-\frac{J2\pi }{N}\times kn}$ 

$x\left(k\right)=\sum _{n=0}^{N-1}x\left(n\right)e^{-\frac{J2\pi }{N}\times kn}$ 

$=\sum _{n=0}^{4}x\left(n\right)=x\left(0\right)+x\left(1\right)+x\left(2\right)+x\left(3\right)+x\left(4\right)$ 

= 1 + 2 + 3 + 4 + 5

= 15

$x\left(1\right)=\sum _{n=0}^{4}x\left(n\right)e^{-\frac{J2\pi }{5}\times 1\times n}$ 

$=\sum _{n=0}^{4}x\left(n\right)e^{-\frac{J2\pi n}{5}}$ 

$=x\left(0\right)+x\left(1\right)e^{-\frac{J2\pi }{5}}+x\left(2\right)e^{-\frac{J4\pi }{5}}+x\left(3\right)e^{-\frac{J6\pi }{5}}+x\left(4\right)e^{-\frac{J8\pi }{5}}$ 

$=x\left(0\right)+x\left(1\right)\left[\cos \frac{2\pi }{5}-Jsin\frac{2\pi }{5}\right]+x\left(2\right)\left[\cos \frac{4\pi }{5}-Jsin\frac{4\pi }{5}\right]+x\left(3\right)\left[\cos \frac{6\pi }{5}-Jsin\frac{6\pi }{5}\right]+x\left(4\right)\left[\cos \frac{8\pi }{5}-Jsin\frac{8\pi }{5}\right]$

X(1) = -2.5 + 3.45

Similarly $x\left(2\right)=\sum _{n=0}^{4}x\left(n\right)e^{-\frac{J4\pi n}{5}}$

On calculating x(2) = -2.5 + 813

Using property of DFT, x(k) = x * (N – k)

x(3) = x * (5 – 3) = x * (2) = -2.5 - .81J

x(4) = x * ( 5 – 4) = x * (1) = -2.5 – 3.4J

x(k) = [15m -2.5 + 3.4J, -2.5 + .81J, -2.5 – 0.815, -2.5 – 3.45]

Note → we can calculate x(0) directly by using –

$x\left(0\right)=\sum _{n=0}^{4}x\left(n\right)=x\left(0\right)+x\left(1\right)+x\left(2\right)+\dots \dots x\left(N-1\right)$ 

In the given question we calculate x(0), which is 15 and we can eliminate other options directly. No need to calculate x(1), x(2), x(3), x(4) further.

$X\left(k\right)={\sum }_{n=0}^{7}x\left(n\right)e^{\frac{-j2\pi nk}{8}}$

$X\left(4\right)={\sum }_{n=0}^{7}x\left(n\right)e^{\frac{-j2\pi n}{2}}$

$X(4)={\sum }_{n=0}^{7}x\left(n\right)e^{-j\pi n}$

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

$X_1\left(k\right)={\sum }_{n=0}^{3}x\left(2n\right)e^{\frac{-j2\pi \left(2n\right)k}{4}}$

$X_1\left(4\right)={\sum }_{n=0}^{3}x\left(2n\right)e^{-j4\pi n}$

X₁(4) = x(0) + x(2) + x(4) + x(6)        ---(2)

$X_2\left(k\right)={\sum }_{n=0}^{3}x\left(2n+1\right)e^{-j2\pi \frac{\left(2n+1\right)}{4}k}$

$X_2\left(4\right)={\sum }_{n=0}^{3}x\left(2n+1\right)e^{-j2\pi \frac{\left(2n+1\right)4}{4}}$

$X_2\left(4\right)={\sum }_n^{3}x\left(2n+1\right)e^{-j4\pi n}.e^{-j2\pi }$

X₂(4) = x(1) + x(3) + x(5) + x(7)      ---(3)

From equation (1), (2), and (3):

X(4) = X₁(4) – X₂(4)

X(4) = X₁(4) – X₂(4)

X₁(4) = X(4) + X₂(4)

Concept:

${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$

DFT of signal x[n] is:

$X[K]=\sum _{K=0}^{N-1}x\left[n\right]e^{-j\left(\frac{2\pi }{N}\right)Kn}$

For 8 point DFT N = 8

Inverse DFT $x\left[n\right]=\frac{1}{N}\underset{K=0}{\sum ^{N-1}}X\left[K\right]e^{-j\frac{2\pi }{N}Kn}$        …1)

Calculations:

$6{\cos }^2\left(\frac{\pi }{4}n\right)=6\left[\frac{(1+\cos \left(\frac{\pi }{2}n\right))}{2}\right]$

$=3+3\cos \left(\frac{\pi }{2}n\right)$

$=3+\frac{3}{2}[e^{j{\left(\frac{2\pi }{8}\right)}^{2n}}+e^{-j{\left(\frac{2\pi }{8}\right)}^{2n}}]$

= Converting to equation 1) form

$x\left[n\right]=\frac{1}{8}[24+12e^{j{\left(\frac{2\pi }{8}\right)}^{2n}}+12e^{j\frac{2\pi }{8}(8-2)n}]$

By coefficient matching

X[K] = {24, 0, 12, 0, 0, 0, 12, 0}

$\underset{K=0}{\sum ^3}X[2K+1]=X\left(1\right)+X\left(3\right)+X\left(5\right)+X\left(7\right)$

= 0

Analysis:

Given,

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

$X\left(k\right)=\sum _{h=0}^{3}x\left(n\right)e^{-j}\frac{2\pi }{4}nk$

We can now write:

$X\left(0\right)=\sum _{n=0}^{3}x\left(n\right)\cdot 1$

$=x\left(0\right)+x\left(1\right)+x\left(2\right)+x\left(3\right)$

X(0) = 0

$X\left(1\right)=\sum _{n=0}^{3}x\left(n\right){\left(-j\right)}^n$

$=1+j-1-j$

X(1) = 0

$X\left(2\right)=\sum _{n=0}^{3}x\left(n\right){\left(-1\right)}^n$

$=1+1+1+1$

X(2) = 4

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

$=1-j-1+j$

X(3) = 0

Hence,

X(k) = [0, 0, 4, 0]

We observe that the term for k = 2 is not zero.

Short Trick:

X(k) = X* (N – K) where N = 4

x(0) = x(0) + x(1) + x(2) + x(3) = 0

x(N/2) = x(2) = x(0) – x(1) + x(2) – x(3) = 4

Hence for k = 2, X(k) ≠ 0

Concept:

Convolution in time domain results in multiplication in the frequency domain.

To find circular convolution of two signals we can follow the following steps:

• First, make the length of the signals equal to N by adding extra zeros if needed.         
• Form two matrices, 1^st matrix using the cyclic rotation of one of the signals and 2^nd matrix with another signal.
• Multiply the two matrices.

Calculation:

Given that:

$x_1\left[n\right]=\left\{1,2,3,0\right\}$ 

$x_2\left[n\right]=\left\{1,3,2,1\right\}$  

$X_3\left(k\right)=X_1\left(k\right)X_2\left(k\right)$ 

$x_3\left[n\right]=x_1\left[n\right]⊛x_2\left[n\right]$ 

$=\left\{1,2,3,0\right\}⊛\left\{1,3,2,1\right\}$ 

$x_3\left[n\right]=\left[\begin{array}{c}\begin{array}{cc}1& 0\\ 2& 1\end{array}\begin{array}{cc}3& 2\\ 0& 3\end{array}\\ \begin{array}{cc}3& 2\\ 0& 3\end{array}\begin{array}{cc}1& 0\\ 2& 1\end{array}\end{array}\right]\left[\begin{array}{c}1\\ 3\\ 2\\ 1\end{array}\right]$ 

$=\left[\begin{array}{c}1\times 1+0\times 3+3\times 2+2\times 1\\ 2\times 1+1\times 3+0\times 2+3\times 1\\ 3\times 1+2\times 3+1\times 2+0\times 1\\ 0\times 1+3\times 3+2\times 2+1\times 1\end{array}\right]=\left[\begin{array}{c}9\\ 8\\ 11\\ 14\end{array}\right]$ 

x₃ [2] = 11