# Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) MCQ Quiz - Objective Question with Answer for Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) - Download Free PDF

Last updated on Feb 22, 2023

## Latest Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) MCQ Objective Questions

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 1:

.Find the N- point DFT of the sequence

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

The DFT coefficients of the DFT sequence will be _______

1. X(K) = 5/2N      for; K = 0

= 1/4N      for; K =  2 and K = N - 2

= 0           else

2. X(K) = 7/2N      for; K = 0

= 1/4N      for; K =  2 and K = N

= 0           else

3. X(K) = 7/2N      for; K = 0

= 1/4N      for; K =  2 and K = N-2

= 0           else

4. X(K) = 7/2N      for; K = 0

= 1/4N      for; K =  1 and K = N-2

= 0           else

Option 3 :

X(K) = 7/2N      for; K = 0

= 1/4N      for; K =  2 and K = N-2

= 0           else

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 1 Detailed Solution

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;

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

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

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

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

The DFT coefficients are;

X(K) = 7/2N      for; K = 0

= 1/4N      for; K =  2 and K = N - 2

= 0           else

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 2:

Let x(n) be a sequence that is zero outside the interval [0,N-1] with a DFT X(K).

Derive the expression for the DFT of y(n);

If y(n) = x(n) + x(n - N) and choose the correct answer/answers from below.

1. Y(K) =  2X(K)   ;  for K = odd
2. Y(K) =  0   ;  for K = odd
3. Y(K) =  2X(K)   ;  for K = even
4. Y(K) =  X(K)   ;  for K = even

Option :

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 2 Detailed Solution

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)kX(K)

∴ Y(K) =  2X(K)   ;  for K = even

= 0            ;  for K = odd

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 3:

Consider two signals x(n) = {0,1,1,2} and h(n) = {a,b,c}. The signal y(n) = x(n) ⊗ h(n). The DFT of y(n) is given as Y(e). If Y(ej0is 36; then which of the following relation is correct?

1. a × b × c = 27
2. a + b + c = 9
3. ab + c = 9
4. a + b - c = 12

Option 2 : a + b + c = 9

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 3 Detailed Solution

Concept:

If x(n) and y(n) are two discrete sequence in time domain and they DFT of x(n) and y(n) and X(K) and Y(K) respectively.

Then;

x(n) ⊗ y(n) ↔ X(K).Y(K)

Summation property

if y(n ) is a  discrete sequence in the time domain.

$$Y(e^{j0}) = \sum_{0}^{N-1}y(n)$$

Calculation:

Given;

x(n) = {0,1,1,2}

h(n) = {a,b,c}

y(n) = x(n) ⊗ h(n)

y(n) = {0,a,(a+b),(2a+b+c),(2b+c),(2c)}

$$Y(e^{j0}) = \sum_{0}^{5}y(n) = (0+a+2a+b+c+2b+c+2c) = 4(a+b+c)$$

Then;

$$4(a+b+c) = 36$$

$$\therefore (a+b+c)=9$$

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 4:

Select the figure from among the given options that can replace the question mark (?) in the following series.

Option 3 :

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 4 Detailed Solution

The logic followed here is:

The images are rotated in the next boxes.

Hence, the correct answer is "Option 3".

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 5:

DFT of a sequence x(n) is given as

X(K) = {64, 32, 80, 32}

The value of IDFT [IDFT [IDFT [IDFT {X(K))}]]] is ______

1. [4, -2, 5, -2]
2. [-4, 2, -5, 2]
3. [4, 2, -5, 2]
4. [4, 2, 5, 2]

Option 4 : [4, 2, 5, 2]

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 5 Detailed Solution

Concept:

Expression for calculating IDFT

$$x(n) = \frac{1}{N} \sum_{k=0}^{N-1}X(k)e^{-j\left ( \frac{2\pi}{N} \right )kn}$$

Where

X(K) → DFT of sequence x(n)

N → Period of the sequence

x(n) → sequence in the time domain

As per the property of IDFT;

If; x(n) = x(-n) [circularly even]

then

IDFT [IDFT [IDFT [IDFT [IDFT.....m times {X(n)}]]]]  ↔ $$\frac{1}{( \sqrt{N})^{m}} [X(K)]$$

Here, m should be even

Calculation:

Given;

X(K) = {64, 32, 80, 32} = 16{4, 2, 5, 2}

Here

X(K) is circularly even [except X(0) if laterally invert the remaining sequence, the sequence will be the same]

Then using the property of IDFT

IDFT [IDFT [IDFT [IDFT {X(K))}]]]:

$$\frac{1}{( \sqrt{4})^{4}} [X(K)] = \frac{1}{16} [16(4, 2, 5, 2)] = [4, 2, 5, 2]$$

## Top Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) MCQ Objective Questions

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 6

Inverse discrete Fourier transform of Y(k) = {1, 0, 1, 0} is

1. y(n) = {0, 0.5, 0, 0.5}
2. y(n) = {0.5, 0, 0.5, 0}
3. y(n) = {0.5, 0.5, 0, 0}
4. y(n) = {0, 0, 0.5, 0.5}

Option 2 : y(n) = {0.5, 0, 0.5, 0}

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 6 Detailed Solution

Concept:

The finite-length sequence can be obtained from the discrete Fourier transform by performing inverse discrete Fourier transform.

It is defined as:

$$x\left( n \right) = \frac{1}{N}\mathop \sum \limits_{k = 0}^{N - 1} X\left( k \right){e^{\frac{{j2\pi nk}}{N}}}$$

where n = 0, 1, …, N – 1

Calculation:

Given sequence is Y(k) = {1, 0, 1, 0}

Length of the sequence, N = 4

$$y\left( 0 \right) = \frac{1}{4}\mathop \sum \limits_{k = 0}^3 X\left( k \right){e^{\frac{{j2\pi nk}}{4}}} = \frac{1}{4}\left( {1 + 0 + 1 + 0} \right) = 0.5$$

$$y\left( 1 \right) = \frac{1}{4}\mathop \sum \limits_{k = 0}^3 X\left( k \right){e^{\frac{{j2\pi nk}}{4}}} = \frac{1}{4}\left( {1 + 0 + {e^{i\pi }} + 0} \right) = 0$$

$$y\left( 2 \right) = \frac{1}{4}\mathop \sum \limits_{k = 0}^3 X\left( k \right){e^{\frac{{j2\pi nk}}{4}}} = \frac{1}{4}\left( {1 + 0 + {e^{i2\pi }} + 0} \right) = 0.5$$

$$y\left( 3 \right) = \frac{1}{4}\mathop \sum \limits_{k = 0}^3 X\left( k \right){e^{\frac{{j2\pi nk}}{4}}} = \frac{1}{4}\left( {1 + 0 + {e^{i3\pi }} + 0} \right) = 0$$

y(n) = {0.5, 0, 0.5, 0}

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 7

The output of a circular convolution performed on two signals x1(n) = {2, 1, 2, 1} and x2(n) = {1, 2, 3, 4} is:

1. {16, 14, 16, 14}
2. {14, 16, 14, 16}
3. {12, 14, 12, 14}
4. {14, 12, 14, 12}

Option 2 : {14, 16, 14, 16}

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 7 Detailed Solution

Concept:

Convolution in the 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, 1st matrix using the cyclic rotation of one of the signals and 2nd matrix with another signal.
• Multiply the two matrices.

Calculation:

Given:

$$\\{x_1}\left( n \right)=\left\{ {\begin{array}{*{20}{c}} 2\\ \uparrow \end{array}\;\begin{array}{*{20}{c}} 1\\ {} \end{array}\;\begin{array}{*{20}{c}} 2\\ {} \end{array}\;\begin{array}{*{20}{c}} 1\\ {} \end{array}} \right\};\\{x_2}\left( n \right)=\left\{ {\begin{array}{*{20}{c}} 1\\ \uparrow \end{array}\;\begin{array}{*{20}{c}} 2\\ {} \end{array}\;\begin{array}{*{20}{c}} 3\\ {} \end{array}\;\begin{array}{*{20}{c}} 4\\ {} \end{array}} \right\}$$

$$y\left( n \right) = {x_1}\left( n \right)\;⊛{x_2}\;\left( n \right)$$

$$= \left\{ {\begin{array}{*{20}{c}} 2\\ \uparrow \end{array},\;\begin{array}{*{20}{c}} 1\\ {} \end{array}\;,\begin{array}{*{20}{c}} 2\\ {} \end{array}\;,\begin{array}{*{20}{c}} 1\\ {} \end{array}\;} \right\} ⊛\;\;\left\{ {\begin{array}{*{20}{c}} 1\\ \uparrow \end{array},\;\begin{array}{*{20}{c}} 2\\ {} \end{array},\begin{array}{*{20}{c}} 3\\ {} \end{array},\begin{array}{*{20}{c}} 4\\ {} \end{array}\;} \right\}$$

$$y\left( n \right) = \left[ {2\;1\;2\;1} \right]\left[ {\begin{array}{*{20}{c}} 1&2&3&4\\ 4&1&2&3\\ 3&4&1&2\\ 2&3&4&1 \end{array}} \right]$$

$$y\left( n \right) = \left\{ {\;\begin{array}{*{20}{c}} {14}\\ \uparrow \end{array},\;\begin{array}{*{20}{c}} {16}\\ {} \end{array},\begin{array}{*{20}{c}} 14\\ {} \end{array},\begin{array}{*{20}{c}} {16}\\ {} \end{array}} \right\}$$

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 8

The difference in the number of complex multipliers required for 16-point DFT and 16-point radix-2 FFT is:

1. 30
2. 63
3. 224
4. 256

Option 3 : 224

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 8 Detailed Solution

Concept:

For an N-point DFT as shown, the number of multiplication:

(M)DFT = N(rows) × [N multiplication per Row]

M(DFT) = N2

$$\left[ {\begin{array}{*{20}{c}} 1&2&3&.. &N\\ 1& ..&.. & .. & .. \\ 1& .. & .. & ..& .. \\ .. &.. & .. & .. & .. \\ N& .. & .. & ..&N \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ .. \\ .. \\ .. \\ N \end{array}} \right]$$

And for an N-point FFT, the number of multiplication equals the number of stages × Multiplications per stage, i.e.

$${\left( M \right)_{FFT}} = {\log _2}N \times \frac{N}{2}$$

Calculation:

(M)DFT = N2 = 256

$${\left( M \right)_{FFT}} = \frac{{16}}{2}{\log _2}\left( {16} \right)$$

$$= \frac{{16}}{2} \times 4 = 32$$

(M)DFT – (M)FFT = 256 – 32 = 224

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 9

Consider a signal $$x\left( n \right) = \left\{ { - 1,\;2,\;\begin{array}{*{20}{c}} 4\\ \uparrow \end{array},\;2,\; - 1,\;3} \right\}$$ . What is $$\mathop \smallint \nolimits_{ - \pi }^\pi X\left( \omega \right)d\omega ?$$

1. 7
2. 14π
3. 9

Option 3 : 8π

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 9 Detailed Solution

$$x\left( n \right) = \left\{ { - 1,\;2,\;\begin{array}{*{20}{c}} 4\\ \uparrow \end{array},\;2,\; - 1,\;3} \right\}$$

$$x\left( n \right)\mathop \leftrightarrow \limits^{DTFT} X\left( \omega \right)$$

$$x\left( n \right) = \frac{1}{{2\pi }}\mathop \smallint \nolimits_{ - \pi }^\pi X\left( \omega \right){e^{ - j\omega n}}d\omega$$

At n = 0,

$$x\left( 0 \right) = \frac{1}{{2\pi }}\mathop \smallint \nolimits_{ - \pi }^\pi X\left( \omega \right){e^{ - j\left( 0 \right)n}}d\omega$$

$$2\pi x\left( 0 \right) = \mathop \smallint \nolimits_{ - \pi }^\pi X\left( \omega \right){e^{ - j\left( 0 \right)n}}d\omega$$

$$\therefore \mathop \smallint \nolimits_{ - \pi }^\pi X\left( \omega \right)d\omega = 2\pi x\left( 0 \right)$$

= 2 π (4) = 8 π

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 10

Let X[k] = k + 1, 0 ≤ k ≤ 7 be 8-point DFT of a sequence x[n], where

$$X\left[ k \right] = \mathop \sum \limits_{n = 0}^{N - 1} x\left[ n \right]{e^{ - j2\pi nk/N}}$$

The value (correct to two decimal places) of $$\mathop \sum \limits_{n = 0}^3 x\left[ {2n} \right]$$ is ______

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 10 Detailed Solution

Concept:

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

Analysis:

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

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

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

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

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

From equation 3),

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

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

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

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 11

Consider a discrete-time periodic signal $$\rm x\left[ n \right] = sin\left( {\frac{{\pi n}}{5}} \right)$$. Let ak be the complex Fourier Series Coefficient (FSC) of x[n]. ais non-zero when k = Bm ± 1 , where 'm' is any integer. The value of B is_______.

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 11 Detailed Solution

Concept:

The Fourier series representation of discrete time periodic sequence in given by:

$$x\left( n \right) = \mathop \sum \limits_{k = 0}^{N - 1} {a_k}{e^{jk{\omega _0}n}}$$

x(n) = … + a–1 e-jωon + 1 + a1 eon + …   ---(1)

ak = Fourier series coefficient periodic by N.

ω0 = Fundamental frequency.

Calculation:

$$x\left( n \right) = \sin \left( {\frac{{\pi n}}{5}} \right)$$

The period of the given sequence will be:

$$N = \frac{{2\pi }}{{\pi /5}} = 10$$

So $${\omega _0} = \frac{{2\pi }}{{10}} = \frac{\pi }{5}$$

x(n) can be written as:

$$x\left( n \right) = \frac{1}{{2j}}\left( {{e^{j\frac{\pi }{5} \cdot n\;}} - {e^{ - j\frac{\pi }{5}n}}} \right)$$ (Euler’s identity)

$$\Rightarrow \frac{1}{{2j}} \cdot {e^{j\frac{\pi }{5} \cdot n}} - \frac{1}{{2j}} \cdot {e^{ - j\frac{\pi }{5}n}} = \frac{1}{{2j}}{e^{j{\omega _0}n}} = \frac{{ - 1}}{{2j}}\;{e^{ - j{\omega _0}n}}$$

Comparing this with Equation-(1), we get:

$$\Rightarrow {a_{ - 1}} = \frac{{ - 1}}{{2j}}\;\;and\;\;{a_1} = \frac{1}{{2j}}$$

Since the fourier coefficient are periodic,

ak = ak+N

So, a-1 = a-1+10 = a-1+10+10 = a-1+10+10+10 … = a-1+m(10)

For any integer m.

Similarly,

a1 = a1+10 = a1+10+10 = a1+10+10+10 ---- = a1+m(10)

So, ak will be non-zero for k = ± 1 + m(10)

For any other value of k, ak will be zero.

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 12

The sequence x(n) = {2, 3, 4, 3} is:

1. Circularly odd
2. Circularly even
3. Partly circularly odd and partly circularly even
4. Neither circularly odd nor circularly even

Option 2 : Circularly even

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 12 Detailed Solution

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.

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 13

The x[n] = {4, 3, 2, 1, 2, 3} signal is:

1. 4 point circularly even
2. 6 Point circularly even
3. 2 point circularly even
4. 3 point circularly even

Option 2 : 6 Point circularly even

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 13 Detailed Solution

Concept:

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

Analysis:

Given: DFT sequence x[n] = {4, 3, 2, 1, 2, 3} and N = 6

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

x[1] = x[6 - 1] = x[5] = 3

x[2] = x[6 - 2] = x[4] = 2

x[3] = x[6 - 3] = x[3] = 1

Hence it is 6 point circularly even.

Special Note:

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

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 14

The discrete Fourier series representation for the following sequence:

$$x\left( n \right) = \cos \frac{\pi }{4}n$$ is

1. $$\frac{1}{2}{e^{j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{ - j{{\rm{\Omega }}_0}n}}$$ and $${{\rm{\Omega }}_0} = \frac{\pi }{8}$$
2. $$\frac{1}{2}{e^{ - j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{ - j2{{\rm{\Omega }}_0}n}}$$ and $${{\rm{\Omega }}_0} = \frac{\pi }{4}$$
3. $$\frac{1}{2}{e^{ - j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{ - j{{\rm{\Omega }}_0}n}}$$ and $${{\rm{\Omega }}_0} = \frac{\pi }{6}$$
4. $$\frac{1}{2}{e^{j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{j7{{\rm{\Omega }}_0}n}}$$ and $${{\rm{\Omega }}_0} = \frac{\pi }{4}$$

Option 4 : $$\frac{1}{2}{e^{j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{j7{{\rm{\Omega }}_0}n}}$$ and $${{\rm{\Omega }}_0} = \frac{\pi }{4}$$

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 14 Detailed Solution

Concept:

The Fourier series representation of the discrete-time periodic sequence is given by:

$$x\left( n \right) = \mathop \sum \limits_{k = 0}^{N - 1} {a_k}{e^{jk{\omega _0}n}}$$

$$x\left( n \right) = \ldots + {a_{ - 1}}{e^{ - j{\omega _0}n}} + {a_1}{e^{j{\omega _0}n}} + \ldots$$

ak = Fourier series coefficient periodic by N.

ω0 = Fundamental frequency.

Application:

Given sequence is: $$x\left( n \right) = \cos \frac{\pi }{4}n$$

The fundamental frequency of the sequence $${\omega _0} = \frac{\pi }{4}$$

We know that, $$\cos \theta = \frac{{{e^{j\theta }} + {e^{ - j\theta }}}}{2}$$

Now, we can rewrite the given sequence as

$$x\left( n \right) = \frac{{{e^{\frac{{j\pi }}{4}n}} + {e^{ - \frac{{j\pi }}{4}n}}}}{2}$$

$$= \frac{1}{2}{e^{\frac{{j\pi }}{4}n}} + \frac{1}{2}{e^{\frac{{ - j\pi }}{4}n}}$$

We can write $${e^{\frac{{ - j\pi }}{4}n}} = {e^{\frac{{j7\pi }}{4}n}}$$

Now, x(n) becomes

$$x\left( n \right) = \frac{1}{2}{e^{\frac{{j\pi }}{4}n}} + \frac{1}{2}{e^{\frac{{j7\pi }}{4}n}}$$

$$= \frac{1}{2}{e^{j{{\rm{\Omega }}_0}n}} + \frac{1}{2}{e^{j7{{\rm{\Omega }}_0}n}}$$ and $${{\rm{\Omega }}_0} = \frac{\pi }{4}$$

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 15

Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k] = y[k] is/are

1. k = 0, 1, 2, ..., 15
2. k = 0 and k = 15
3. k = 15
4. k = 0

Option 3 : k = 15

#### Discrete Fourier Transform (DFT) and Discrete Fourier Series (DFS) Question 15 Detailed Solution

Concept:

Linear convolution = x[n] * h[n]

y[n] = x[n] * h[n]

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

Also, the function which is circularly convoled can be written as:

z[n] = IDFT [x(k) y(k)]

i.e. z[n] = x[n] h(n)

→ circular convolution

We have 16 point sequence x(n) and h(n) but for the case of understanding and solving, we will

consider 4-point sequence as:

x(n) = {1, 1, 2, 2}

h(n) = {1, 2, 3, 4}

Linear convolution:

y(n) = {1, 3, 7, 13, 16, 14, 8}

Circular convolution:

z[n] = x[n] h(n)

$$\left[ {\begin{array}{*{20}{c}} 1&2&2&1\\ 1&1&2&2\\ 2&1&1&2\\ 2&2&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ 3\\ 4 \end{array}} \right]$$

z[n] = [15, 17, 15, 13]

z[3] = y[3]

Clearly, if N point sequence is given for x(n) and h(n), then:

$${\left[ {x\left( n \right)*h\left( n \right)} \right]_{at\;n = N - 1}} = {\left[ {x\left( n \right)h\left[ n \right]} \right]_{at\;n = N - 1}}$$

Application:

The given sequence is a 16 point sequence.

Hence for N = 16 – 1 = 15:

Z[k] = y [k]
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