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

Concept:

The Fourier Series Coefficient of a discrete time Periodic sequence is given by:

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

Where N = Period of the sequence.

Calculation:

Given, x (n) = {1, 1, 0, 0} and N = 4

So, \({{c}_{k}}=\frac{1}{N}~\sum_{n=0}^{N-1}x\left( n \right){{e}^{-jk\omega n}}=\frac{1}{N}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-jK2\pi n}{N}~}}\)

With N = 4,

\(\Rightarrow {{c}_{k}}=\frac{1}{4}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-j2\pi kn}{4}}}=\frac{1}{4}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-j\pi kn}{2}}}\)

\({{c}_{k}}=\frac{1}{4}\left[ x\left( 0 \right)+x\left( 1 \right){{e}^{\frac{-j\pi k}{2}}}+x\left( 2 \right){{e}^{\frac{-j\pi k\left( 2 \right)}{2}}}+x\left( 3 \right){{e}^{\frac{-j\pi k\left( 3 \right)}{2}}} \right]\)

With, x(0) = 1, x(1) = 1, x(2) = 0 and x(3) = 0.

\({{c}_{k}}=\frac{1}{4}\left[ 1+1.{{e}^{\frac{-j\pi k}{2}}}+0+0 \right]\)

\({{c}_{k}}=\frac{1}{4}\left[ 1+{{e}^{\frac{-j\pi k}{2}}} \right]\)

We are asked to calculate \(c_3^*\)

Putting k = 3.

\({{c}_{3}}=\frac{1}{4}\left[ 1+{{e}^{\frac{-j3\pi }{2}}} \right]\)

\(=\frac{1}{4}\left( 1+\cos \frac{3\pi }{2}-jsin\frac{3\pi }{2} \right)\)

\(=\frac{1}{4}\left( 1+j \right)\)

So, \(c_{3}^{*}=\frac{1}{4}\left( 1-j \right)\)

Option (3) is correct.

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

Given, x_in(t) = sin(2π × 4000t) + 0.75 sin (2π × 5000t + π/4)

The above signal is sampled with a frequency F_s, given by, F_s = 16000 Hz

The sampling interval \({T_s} = \frac{1}{{{F_S}}} = \frac{1}{{1600}}sec\)

Since, x(n) = x_in(nT_s)  (Given)

 \( {x_{in}}\left( {n{T_s}} \right) = \sin \left( {2\pi \times 4000.n{T_s}} \right) +\)

\( 0.75\sin \left( {2\pi \times 5000 \times n{T_s} + \frac{\pi }{4}} \right)\)

\(x\left( n \right) = \sin \left( {2\pi \times \frac{{4000\;n}}{{16000}}} \right) + \)

\(0.75\sin \left( {2\pi \times \frac{{5000 \times 1}}{{16000}}.n + \frac{\pi }{4}} \right)\)

\(x\left( n \right) = \sin \left( {\frac{\pi }{2}n} \right) + 0.75\sin \left( {\frac{{5\pi }}{8}n + \frac{\pi }{4}} \right)\)       ----(1)

Now, \(X\left( m \right) = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right){e^{ - j2\pi nm/N}}\), and we are required to find X(0), which can be easily evaluated by putting m = 0 in the expression of X(m)

i.e. \(X\left( 0 \right) = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right).{e^{ - j2\pi n\left( 0 \right)/N}} = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right)\)

With N = 8,  \(X\left( 0 \right) = \mathop \sum \limits_{n = 0}^7 x\left( n \right)\)

From Equation (1),

 \(x\left( n \right) = \sin \left( {\frac{\pi }{2}n} \right) + 0.75\sin \left( {\frac{{5\pi }}{8}n + \frac{\pi }{4}} \right)\)

So, X(0) will be

 \( = \mathop \sum \limits_{n = 0}^7 \left( {\sin \left( {\frac{\pi }{2}n} \right)+0.75\sin \left( {\frac{{5\pi }}{8}n+\frac{\pi }{4}} \right)} \right)\)

Since x(n) is purely real,

So \(\mathop \sum \limits_{n = 0}^7 x\left( n \right)\) will also be purely real.

It is only in option (2), that we have a value whose imaginary part is 0.

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

Given, x_in(t) = sin(2π × 4000t) + 0.75 sin (2π × 5000t + π/4)

The above signal is sampled with a frequency F_s, given by, F_s = 16000 Hz

The sampling interval \({T_s} = \frac{1}{{{F_S}}} = \frac{1}{{1600}}sec\)

Since, x(n) = x_in(nT_s)  (Given)

 \( {x_{in}}\left( {n{T_s}} \right) = \sin \left( {2\pi \times 4000.n{T_s}} \right) +\)

\( 0.75\sin \left( {2\pi \times 5000 \times n{T_s} + \frac{\pi }{4}} \right)\)

\(x\left( n \right) = \sin \left( {2\pi \times \frac{{4000\;n}}{{16000}}} \right) + \)

\(0.75\sin \left( {2\pi \times \frac{{5000 \times 1}}{{16000}}.n + \frac{\pi }{4}} \right)\)

\(x\left( n \right) = \sin \left( {\frac{\pi }{2}n} \right) + 0.75\sin \left( {\frac{{5\pi }}{8}n + \frac{\pi }{4}} \right)\)       ----(1)

Now, \(X\left( m \right) = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right){e^{ - j2\pi nm/N}}\), and we are required to find X(0), which can be easily evaluated by putting m = 0 in the expression of X(m)

i.e. \(X\left( 0 \right) = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right).{e^{ - j2\pi n\left( 0 \right)/N}} = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right)\)

With N = 8,  \(X\left( 0 \right) = \mathop \sum \limits_{n = 0}^7 x\left( n \right)\)

From Equation (1),

 \(x\left( n \right) = \sin \left( {\frac{\pi }{2}n} \right) + 0.75\sin \left( {\frac{{5\pi }}{8}n + \frac{\pi }{4}} \right)\)

So, X(0) will be

 \( = \mathop \sum \limits_{n = 0}^7 \left( {\sin \left( {\frac{\pi }{2}n} \right)+0.75\sin \left( {\frac{{5\pi }}{8}n+\frac{\pi }{4}} \right)} \right)\)

Since x(n) is purely real,

So \(\mathop \sum \limits_{n = 0}^7 x\left( n \right)\) will also be purely real.

It is only in option (2), that we have a value whose imaginary part is 0.

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

Concept:

The Fourier Series Coefficient of a discrete time Periodic sequence is given by:

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

Where N = Period of the sequence.

Calculation:

Given, x (n) = {1, 1, 0, 0} and N = 4

So, \({{c}_{k}}=\frac{1}{N}~\sum_{n=0}^{N-1}x\left( n \right){{e}^{-jk\omega n}}=\frac{1}{N}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-jK2\pi n}{N}~}}\)

With N = 4,

\(\Rightarrow {{c}_{k}}=\frac{1}{4}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-j2\pi kn}{4}}}=\frac{1}{4}\sum_{n=0}^{N-1}x\left( n \right){{e}^{\frac{-j\pi kn}{2}}}\)

\({{c}_{k}}=\frac{1}{4}\left[ x\left( 0 \right)+x\left( 1 \right){{e}^{\frac{-j\pi k}{2}}}+x\left( 2 \right){{e}^{\frac{-j\pi k\left( 2 \right)}{2}}}+x\left( 3 \right){{e}^{\frac{-j\pi k\left( 3 \right)}{2}}} \right]\)

With, x(0) = 1, x(1) = 1, x(2) = 0 and x(3) = 0.

\({{c}_{k}}=\frac{1}{4}\left[ 1+1.{{e}^{\frac{-j\pi k}{2}}}+0+0 \right]\)

\({{c}_{k}}=\frac{1}{4}\left[ 1+{{e}^{\frac{-j\pi k}{2}}} \right]\)

We are asked to calculate \(c_3^*\)

Putting k = 3.

\({{c}_{3}}=\frac{1}{4}\left[ 1+{{e}^{\frac{-j3\pi }{2}}} \right]\)

\(=\frac{1}{4}\left( 1+\cos \frac{3\pi }{2}-jsin\frac{3\pi }{2} \right)\)

\(=\frac{1}{4}\left( 1+j \right)\)

So, \(c_{3}^{*}=\frac{1}{4}\left( 1-j \right)\)

Option (3) is correct.

Let the Fourier series coefficient of the input be \(\rm {a_k}\). Then, the FS coefficient of the output \(\rm {b_k}\) is given by:

\(\rm {b_k} = {a_k} \cdot H\left( {{e^{j{\omega _o}k}}} \right)\)

Now, since the input period is 3 thus, \(\rm {\omega _o} = 2\pi /3\)

Now for

\(\rm \begin{array}{l} k = 0,H\left( {{e^{j0}}} \right) = 1\\ \rm k = 1,H\left( {{e^{j\frac{{2\pi }}{3}}}} \right) = 0\\ \rm k = 2,H\left( {{e^{j\frac{{4\pi }}{3}}}} \right) = 0 \end{array}\)

Thus, there is only 1 non-zero elements in the output

Given, x_in(t) = sin(2π × 4000t) + 0.75 sin (2π × 5000t + π/4)

The above signal is sampled with a frequency F_s, given by, F_s = 16000 Hz

The sampling interval \({T_s} = \frac{1}{{{F_S}}} = \frac{1}{{1600}}sec\)

Since, x(n) = x_in(nT_s)  (Given)

 \( {x_{in}}\left( {n{T_s}} \right) = \sin \left( {2\pi \times 4000.n{T_s}} \right) +\)

\( 0.75\sin \left( {2\pi \times 5000 \times n{T_s} + \frac{\pi }{4}} \right)\)

\(x\left( n \right) = \sin \left( {2\pi \times \frac{{4000\;n}}{{16000}}} \right) + \)

\(0.75\sin \left( {2\pi \times \frac{{5000 \times 1}}{{16000}}.n + \frac{\pi }{4}} \right)\)

\(x\left( n \right) = \sin \left( {\frac{\pi }{2}n} \right) + 0.75\sin \left( {\frac{{5\pi }}{8}n + \frac{\pi }{4}} \right)\)       ----(1)

Now, \(X\left( m \right) = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right){e^{ - j2\pi nm/N}}\), and we are required to find X(0), which can be easily evaluated by putting m = 0 in the expression of X(m)

i.e. \(X\left( 0 \right) = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right).{e^{ - j2\pi n\left( 0 \right)/N}} = \mathop \sum \limits_{n = 0}^{N - 1} x\left( n \right)\)

With N = 8,  \(X\left( 0 \right) = \mathop \sum \limits_{n = 0}^7 x\left( n \right)\)

From Equation (1),

 \(x\left( n \right) = \sin \left( {\frac{\pi }{2}n} \right) + 0.75\sin \left( {\frac{{5\pi }}{8}n + \frac{\pi }{4}} \right)\)

So, X(0) will be

 \( = \mathop \sum \limits_{n = 0}^7 \left( {\sin \left( {\frac{\pi }{2}n} \right)+0.75\sin \left( {\frac{{5\pi }}{8}n+\frac{\pi }{4}} \right)} \right)\)

Since x(n) is purely real,

So \(\mathop \sum \limits_{n = 0}^7 x\left( n \right)\) will also be purely real.

It is only in option (2), that we have a value whose imaginary part is 0.

Given, \(x\left[ n \right] = \sin \left( {\frac{{\pi n}}{5}} \right) + \cos \left( {\frac{\pi }{{10}}n} \right)\). Time period T = 20

It can be written as

\(\begin{array}{l} = \frac{1}{{2j}}{e^{\frac{{j\pi }}{5}n}} - \frac{1}{{2j}}{e^{ - \frac{{j\pi }}{5}n}} + \frac{1}{2}{e^{\frac{{j\pi }}{{10}}n}} + \frac{1}{2}{e^{ - \frac{{j\pi }}{{10}}n}}\\ {\omega _0} = \frac{{2\pi }}{{20}} = \frac{\pi }{{10}} \end{array}\)  

a_k non-zero for k = ±1, ±2

The Fourier series coefficients repeat every period equal to time period of signal.

They are non-zero again after k = 20 ± 1, 20 ± 2

Hence there are fourier series coefficients corresponding to

k=1 which is a₁ 

k= 2 which is a₂

And then again at 20 ± 1, 20 ± 2

which are 

a₁₈ , a₁₉, a₂₀, a₂₁, a₂₂

longest interval is between a2 to a18 which is of 15 samples.

Given \(x\left[ n \right] = \mathop \sum \limits_{k = - \infty }^\infty \delta \left[ {n - 4k} \right]\)

The given sequence is repetitive with period ‘4’

\({c_k} = \frac{1}{4}\mathop \sum \limits_{4 = 0}^3 x\left[ 4 \right]{e^{ - jk\left( {\frac{{2\pi }}{4}} \right)4}}\;\)

(∵ since x[n] = 0 for n ≠ 0 in the period [0, 3])

\(= \frac{1}{4}x\left[ 0 \right] = \frac{1}{4}\;\forall \;k\)

The graph of Fourier coefficients is of form

By definition, Fourier series coefficient

\({{\rm{a}}_{\rm{k}}} = \frac{1}{{\rm{N}}}\mathop \sum \limits_{{\rm{n}} = \left\langle {\rm{N}} \right\rangle }^{} {\rm{x}}\left[ {\rm{n}} \right]{{\rm{e}}^{ - {\rm{jk}}\left( {\frac{{2{\rm{\pi }}}}{{\rm{N}}}} \right){\rm{n}}}}\)

\({{\rm{a}}_{\rm{k}}} = \frac{1}{{\rm{N}}}\mathop \sum \limits_{{\rm{n}} = \left\langle {\rm{N}} \right\rangle }^{} \left(\rm{\mathop \sum \limits_{p= - \infty }^\infty {\rm{\delta }}\left[ {{\rm{n}} - {\rm{pN}}} \right]}\right){{\rm{e}}^{ - {\rm{jk}}\left( {\frac{{2{\rm{\pi }}}}{{\rm{N}}}} \right){\rm{n}}}}\)

\({{\rm{a}}_{\rm{k}}} = \frac{1}{{\rm{N}}}\mathop \sum \limits_{{\rm{n}} = \left\langle {\rm{N}} \right\rangle } \rm{\mathop \sum \limits_{p= - \infty }^\infty {\rm{\left(\delta\left[n-pN\right]e^{-jk\left({2\pi\over N}\right)pN}\right)}}}\)

\({{\rm{a}}_{\rm{k}}} = \frac{1}{{\rm{N}}}\rm{\mathop \sum \limits_{p= - \infty }^\infty}\mathop \sum \limits_{{\rm{n}} = \left\langle {\rm{N}} \right\rangle } \rm{ {\rm{\left(\delta\left[n-pN\right]e^{-jk\left({2\pi\over N}\right)pN}\right)}}}\\ {{\rm{a}}_{\rm{k}}} = \frac{1}{{\rm{N}}}\rm{\mathop \sum \limits_{p= - \infty }^\infty}\mathop \sum \limits_{{\rm{n}} = \left\langle {\rm{N}} \right\rangle } \rm{ {\rm{\left(\delta\left[n-pN\right]e^{-jk\left({2\pi}\right)p}\right)}}}\\ \text{Taking the summation interval to be from 0 to N-1 we have,}\\ {{\rm{a}}_{\rm{k}}} = \frac{1}{{\rm{N}}}\rm{\mathop \sum \limits_{p= - \infty }^\infty e^{-jk\left({2\pi}\right)p}} \rm{ {\rm{}}}\)

Thus,

We have the multiplication property of the Discrete Fourier Transform defined as;

If x₁[n] ↔ a_k

and x₂[n] ↔ b_k, then;

 \({x_1}\left[ n \right]{x_2}\left[ n \right] \to F.S. coefficients ~~c_k = \mathop \sum \limits_{l = 0}^{{N-1}} {a_l}{b_{k - l}}\)

For the given sequence, N = 4. So the Fourier Series Coefficient of the multiplication of the given two sequences will be;

\(c_k = \mathop \sum \limits_{l = 0}^{{3}} {a_l}{b_{k - l}}\)

Thus, \(\rm{\Rightarrow c_k=a_0b_k+a_1b_{k-1}+a_2b_{k-2}+a_3b_{k-3}}\)

Putting the values given, i.e.  \(\rm{a_o=a_3=1, a_1=a_2=2}\) and \(\rm{b_0=b_1=b_2=b_3=1}\), we get;

\( = 1{b_k} + 2{b_{k - 1}} + 2.{b_{k - 2}} + 1.{b_{k - 3}}\)

Since b_k is 1 for all k; therefore the Fourier series coefficients of x₁[n]x₂[n] is 6 for all k;

Hence the given summation evaluates to;  \(\rm{c_0+c_1+c_2+c_3= 24}\).