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

Explanation:

Discrete Fourier Transform (DFT) and Symmetry Properties

Introduction: The Discrete Fourier Transform (DFT) is a mathematical technique used to analyze the frequency content of discrete signals. When a signal is real-valued, its DFT exhibits specific symmetry properties that simplify computations and provide insights into the relationship between various frequency components. In this problem, we are tasked with finding the value of X(6) of an 8-point DFT, given the values of X(0), X(1), X(2), and X(3). The key to solving this lies in exploiting the symmetry properties of the DFT.

Symmetry Properties of DFT for Real-Valued Sequences:

• Conjugate Symmetry: If the input sequence is real-valued, the DFT coefficients satisfy the following property:

  X(N-k) = conjugate(X(k)) for k = 1, 2, ..., N-1.

  Here, N is the number of points in the DFT (in this case, N = 8).

• Special Cases:
  • X(0) is always real because it represents the sum of all input values (DC component).
  • If N is even, X(N/2) (middle frequency) is also real.

In this problem, the input sequence is real-valued, so we can apply the conjugate symmetry property to find X(6).

Given:

• X(0) = 5
• X(1) = 1 + j
• X(2) = 3 + j
• X(3) = 2 + 3j

We need to determine the value of X(6). Using the conjugate symmetry property of the DFT:

X(N-k) = conjugate(X(k))

For N = 8, the relationship for X(6) becomes:

X(6) = conjugate(X(2))

From the given data:

X(2) = 3 + j

Taking the complex conjugate:

conjugate(X(2)) = 3 - j

Therefore:

X(6) = 3 - j

Correct Answer: Option 2

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).

Let $I=\sum _{n=0}^{4}x(n)\sin \frac{4\pi n}{5}$

$=\frac{1}{2j}\sum _{n=0}^{4}x(n)\cdot [e^{j\frac{4\pi n}{5}}-e^{-j\frac{4\pi n}{5}}]$

$=\frac{1}{2j}[\sum _{n=0}^{4}x(n)e^{j\frac{4\pi n}{5}}-\sum _{n=0}^{4}x(n)\cdot e^{-j\frac{4\pi n}{5}}]$  ....(1)

As we know, $a_k=\frac{1}{N}\sum _{n=0}^{4}x(n)\cdot e^{-jk\cdot \frac{2\pi }{N}n}$

$=\frac{1}{5}\sum _{n=0}^{4}x(n)\cdot e^{-j\frac{2\pi }{N}kn}$

Put K = 2;  $a_2=\frac{1}{5}\sum _{n=0}^{4}x(n)\cdot e^{-j\frac{4\pi n}{5}}$

Put K = -2;  $a_{-2}=\frac{1}{5}\sum _{n=0}^{4}x(n)\cdot e^{j\frac{4\pi n}{5}}$

From equation (i),  $I=\frac{1}{2j}[5a_{-2}-5a_2]$

$=\frac{5}{2j}[a_3-a_2]$            $\left[\begin{array}{rl}{a}_{-2}& =a_{-2+N}\\ & =a_{-2+5}\\ & =a_3\end{array}\right]$

⇒ $I=\frac{5}{2j}[-2j-2j]=-10$

Concept:

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

It is defined as:

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

where n = 0, 1, …, N – 1

Calculation:

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

Length of the sequence, N = 4

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

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

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

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

Concept:

A square wave is represented as:

The fundamental frequency for a square wave is the inverse of the time period, i.e.

$f=\frac{1}{T}$

Calculation:

With T = 4 msec, the fundamental frequency will be:

$f=\frac{1}{4\times {10}^{-3}}=250Hz$

Important Note: A square wave is a combination of many frequencies, i.e.

f_square = f0 + f₁ + f₂ + ... + f_n

f₀ = fundamental frequency

f₁, f₂, ...f_n are the harmonics, i.e. multiples of the fundamental frequency

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:

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

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:

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:

$$\begin{gathered} {x}_1\left(n\right)=\left\{\begin{array}{c}2\\ \uparrow \end{array}\begin{array}{c}1\\ \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}1\\ \end{array}\right\}; \\ {x}_2\left(n\right)=\left\{\begin{array}{c}1\\ \uparrow \end{array}\begin{array}{c}2\\ \end{array}\begin{array}{c}3\\ \end{array}\begin{array}{c}4\\ \end{array}\right\} \end{gathered}$$

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

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

$y\left(n\right)=\left[2121\right]\left[\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\\ 3& 4& 1& 2\\ 2& 3& 4& 1\end{array}\right]$

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

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)

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

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

z[3] = y[3]

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

${\left[x\left(n\right)\ast h\left(n\right)\right]}_{atn=N-1}={\left[x\left(n\right)h\left[n\right]\right]}_{atn=N-1}$

Application:

The given sequence is a 16 point sequence.

Hence for N = 16 – 1 = 15:

Z[k] = y [k]

Concept:

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

$x\left(n\right)=\sum _{k=0}^{N-1}{a}_k{e}^{jk{\omega }_{0}n}$

$x\left(n\right)=\dots +a_{-1}{e}^{-j{\omega }_{0}n}+a_1{e}^{j{\omega }_{0}n}+\dots$

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

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

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

$x\left(n\right)\stackrel{DTFT}{\leftrightarrow }X\left(\omega \right)$

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

At n = 0,

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

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

$\therefore {\int }_{-\pi }^{\pi }X\left(\omega \right)d\omega =2\pi x\left(0\right)$

Concept:

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

$x\left(n\right)=\sum _{k=0}^{N-1}{a}_k{e}^{jk{\omega }_{0}n}$

x(n) = … + a_–1 e^-jω_oⁿ + 1 + a₁ e^jω_oⁿ + …   ---(1) 

a_k = 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,

a_k = a_k+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 = a_1+10+10 = a_1+10+10+10 ---- = a_1+m(10)

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

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

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.