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,

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

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

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

$c_k=\frac{1}{4}[1+1.e^{\frac{-j\pi k}{2}}+0+0]$

$c_k=\frac{1}{4}[1+e^{\frac{-j\pi k}{2}}]$

We are asked to calculate $c_3^{\ast }$

Putting k = 3.

$c_3=\frac{1}{4}[1+e^{\frac{-j3\pi }{2}}]$

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

$=\frac{1}{4}(1+j)$

So, $c_3^{\ast }=\frac{1}{4}(1-j)$

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

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

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)

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

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

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

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

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

Now, $X\left(m\right)=\sum _{n=0}^{N-1}x\left(n\right)e^{-j2\pi nm/N}$, 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. $X\left(0\right)=\sum _{n=0}^{N-1}x\left(n\right).e^{-j2\pi n\left(0\right)/N}=\sum _{n=0}^{N-1}x\left(n\right)$

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

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

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

Since x(n) is purely real,

So $\sum _{n=0}^{7}x\left(n\right)$ 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

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

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)

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

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

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

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

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

Now, $X\left(m\right)=\sum _{n=0}^{N-1}x\left(n\right)e^{-j2\pi nm/N}$, 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. $X\left(0\right)=\sum _{n=0}^{N-1}x\left(n\right).e^{-j2\pi n\left(0\right)/N}=\sum _{n=0}^{N-1}x\left(n\right)$

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

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

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

Since x(n) is purely real,

So $\sum _{n=0}^{7}x\left(n\right)$ 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

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

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,

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

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

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

$c_k=\frac{1}{4}[1+1.e^{\frac{-j\pi k}{2}}+0+0]$

$c_k=\frac{1}{4}[1+e^{\frac{-j\pi k}{2}}]$

We are asked to calculate $c_3^{\ast }$

Putting k = 3.

$c_3=\frac{1}{4}[1+e^{\frac{-j3\pi }{2}}]$

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

$=\frac{1}{4}(1+j)$

So, $c_3^{\ast }=\frac{1}{4}(1-j)$

Option (3) is correct.

Let the Fourier series coefficient of the input be ${\mathrm{a}}_{\mathrm{k}}$. Then, the FS coefficient of the output ${\mathrm{b}}_{\mathrm{k}}$ is given by:

${\mathrm{b}}_{\mathrm{k}}={\mathrm{a}}_{\mathrm{k}}\cdot \mathrm{H}\left({\mathrm{e}}^{\mathrm{j}{\omega }_{\mathrm{o}}\mathrm{k}}\right)$

Now, since the input period is 3 thus, ${\omega }_{\mathrm{o}}=2\pi /3$

Now for

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

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)

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

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

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

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

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

Now, $X\left(m\right)=\sum _{n=0}^{N-1}x\left(n\right)e^{-j2\pi nm/N}$, 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. $X\left(0\right)=\sum _{n=0}^{N-1}x\left(n\right).e^{-j2\pi n\left(0\right)/N}=\sum _{n=0}^{N-1}x\left(n\right)$

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

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

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

Since x(n) is purely real,

So $\sum _{n=0}^{7}x\left(n\right)$ 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 $x\left[n\right]=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\delta \left[n-4k\right]$

The given sequence is repetitive with period ‘4’

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

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

$=\frac{1}{4}x\left[0\right]=\frac{1}{4}\mathrm{\forall }k$

The graph of Fourier coefficients is of form

By definition, Fourier series coefficient

${\mathrm{a}}_{\mathrm{k}}=\frac{1}{\mathrm{N}}\sum _{\mathrm{n}=⟨\mathrm{N}⟩}^{}\mathrm{x}\left[\mathrm{n}\right]{\mathrm{e}}^{-\mathrm{j}\mathrm{k}\left(\frac{2\pi }{\mathrm{N}}\right)\mathrm{n}}$

${\mathrm{a}}_{\mathrm{k}}=\frac{1}{\mathrm{N}}\sum _{\mathrm{n}=⟨\mathrm{N}⟩}^{}\left(\sum _{\mathrm{p}=-\mathrm{\infty }}^{\mathrm{\infty }}\delta \left[\mathrm{n}-\mathrm{p}\mathrm{N}\right]\right){\mathrm{e}}^{-\mathrm{j}\mathrm{k}\left(\frac{2\pi }{\mathrm{N}}\right)\mathrm{n}}$

${\mathrm{a}}_{\mathrm{k}}=\frac{1}{\mathrm{N}}\sum _{\mathrm{n}=⟨\mathrm{N}⟩}\sum _{\mathrm{p}=-\mathrm{\infty }}^{\mathrm{\infty }}\left(\delta [\mathrm{n}-\mathrm{p}\mathrm{N}]{\mathrm{e}}^{-\mathrm{j}\mathrm{k}\left(\frac{2\pi }{\mathrm{N}}\right)\mathrm{p}\mathrm{N}}\right)$

$$\begin{gathered} {\mathrm{a}}_{\mathrm{k}}=\frac{1}{\mathrm{N}}\sum _{\mathrm{p}=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{\mathrm{n}=⟨\mathrm{N}⟩}\left(\delta [\mathrm{n}-\mathrm{p}\mathrm{N}]{\mathrm{e}}^{-\mathrm{j}\mathrm{k}\left(\frac{2\pi }{\mathrm{N}}\right)\mathrm{p}\mathrm{N}}\right) \\ {\mathrm{a}}_{\mathrm{k}}=\frac{1}{\mathrm{N}}\sum _{\mathrm{p}=-\mathrm{\infty }}^{\mathrm{\infty }}\sum _{\mathrm{n}=⟨\mathrm{N}⟩}\left(\delta [\mathrm{n}-\mathrm{p}\mathrm{N}]{\mathrm{e}}^{-\mathrm{j}\mathrm{k}\left(2\pi \right)\mathrm{p}}\right) \\ \text{Taking the summation interval to be from 0 to N-1 we have,} \\ {\mathrm{a}}_{\mathrm{k}}=\frac{1}{\mathrm{N}}\sum _{\mathrm{p}=-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{j}\mathrm{k}\left(2\pi \right)\mathrm{p}} \end{gathered}$$

Thus,

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

If x₁[n] ↔ a_k

and x₂[n] ↔ b_k, then;

 $x_1\left[n\right]x_2\left[n\right]\to F.S.coefficients{c}_k=\sum _{l=0}^{N-1}{a}_l{b}_{k-l}$

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

$c_k=\sum _{l=0}^3{a}_l{b}_{k-l}$

Thus, $\Rightarrow {\mathrm{c}}_{\mathrm{k}}={\mathrm{a}}_0{\mathrm{b}}_{\mathrm{k}}+{\mathrm{a}}_1{\mathrm{b}}_{\mathrm{k}-1}+{\mathrm{a}}_2{\mathrm{b}}_{\mathrm{k}-2}+{\mathrm{a}}_3{\mathrm{b}}_{\mathrm{k}-3}$

Putting the values given, i.e.  ${\mathrm{a}}_{\mathrm{o}}={\mathrm{a}}_3=1,{\mathrm{a}}_1={\mathrm{a}}_2=2$ and ${\mathrm{b}}_0={\mathrm{b}}_1={\mathrm{b}}_2={\mathrm{b}}_3=1$, we get;

$=1b_k+2b_{k-1}+2.b_{k-2}+1.b_{k-3}$

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;  ${\mathrm{c}}_0+{\mathrm{c}}_1+{\mathrm{c}}_2+{\mathrm{c}}_3=24$.