Overview of Fourier Transform MCQ Quiz - Objective Question with Answer for Overview of Fourier Transform - Download Free PDF

• x(n) → X(e^jω): Denotes that the Fourier transform of x(n) is X(e^jω).
• y(n) → Y(e^jω): Denotes that the Fourier transform of y(n) is Y(e^jω).

The question asks us to identify which of the provided statements is NOT TRUE.

Correct Option: Option 2

Detailed Explanation:

Let us analyze the given options one by one to determine why Option 2 is the correct answer.

Option 2: x(n−nd)→ejωnd(ejω)" id="MathJax-Element-238-Frame" role="presentation" style="position: relative;" tabindex="0">x(n−nd)→ejωnd(ejω)x(n−nd)→ejωnd(ejω) $x\left(n-n_{d}\right) → e^{j \omega n_{d}} \left(e^{j \omega}\right)$

This statement is NOT TRUE. Let us break it down:

• When a signal is delayed in the time domain by n_d (i.e., x(n - n_d)), the Fourier transform of the delayed signal introduces a phase shift in the frequency domain.
• The correct Fourier transform for x(n - n_d) is given by:
  x(n - n_d) → X(e^jω) × e^-jωn_d
• In the given statement, the phase factor e^jωn_d is incorrectly written as e^jωn_d (positive exponent), which is incorrect. The correct phase factor must have a negative exponent.

Therefore, Option 2 is incorrect because it does not accurately represent the Fourier transform of a delayed signal. The correct representation should involve the term e^-jωn_d, not e^jωn_d.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: a × x(n) + b × y(n) → a × X(e^jω) + b × Y(e^jω)

This statement is TRUE. The Fourier transform is a linear operation, which means that the Fourier transform of a linear combination of signals is equal to the same linear combination of their respective Fourier transforms. Hence, this property is valid.

Option 3: ejω0×x(n)→X(ej(ω−ω0))" id="MathJax-Element-239-Frame" role="presentation" style="position: relative;" tabindex="0">ejω0×x(n)→X(ej(ω−ω0))ejω0×x(n)→X(ej(ω−ω0)) $e^{jω_{0}} × x(n) → X\left(e^{j\left(ω - ω_{0}\right)}\right)$

This statement is TRUE. When a signal is modulated in the time domain by e^jω₀ (complex exponential multiplication), the effect in the frequency domain is a shift in the frequency spectrum by ω₀. The Fourier transform accurately reflects this frequency shift, making this statement valid.

Option 4: x(n) × y(n) → X(e^jω) × Y(e^jω)

This statement is TRUE. Convolution in the time domain corresponds to multiplication in the frequency domain. Hence, the Fourier transform of the convolution of x(n) and y(n) is the product of their respective Fourier transforms. This property is fundamental to the Fourier transform and is valid.

Explanation:

Let F(ω) is the Fourier transform of f(t).

Hence an odd and imaginary signal always has an odd and real Fourier transform

Hence statement (1) is wrong.

Convolution:

Convolution is a mathematical operation used to express the relation between input and output of an LTI system. It relates the input, output, and impulse response of an LTI system as

y(t) = x(t) * h(t)

Where y (t) = output of LTI

x (t) = input of LTI

h (t) = impulse response of LTI

1. Convolution of two even signals or two odd signals always results in an even signal.

2. Convolution of odd signal and even signal always results in the odd signal.

Hence statement (2) is also false.

So option (1) is the correct answer.

f(t) = 3e^-4t u(t)

\(F\left( \omega \right) = \mathop \smallint \nolimits_{ - \infty }^\infty f\left( t \right){e^{ - j\omega t}}dt\)

\({e^{ - at}}\mathop \leftrightarrow \limits^{F.T} \frac{1}{{a + j\omega }}\)

∴ \(3{e^{ - 4t}}\;\mathop \leftrightarrow \limits^{FT} \frac{3}{{4 + j\omega }}\)

At ω = 4,

\(F\left( \omega \right){\left. \right|_{\omega = 4}} = \frac{3}{{4 + j4}} = \left( {\frac{{\frac{3}{4}}}{{1 + j}}} \right)\)

Explanation:

Let F(ω) is the Fourier transform of f(t).

Hence an odd and imaginary signal always has an odd and real Fourier transform

Hence statement (1) is wrong.

Convolution:

Convolution is a mathematical operation used to express the relation between input and output of an LTI system. It relates the input, output, and impulse response of an LTI system as

y(t) = x(t) * h(t)

Where y (t) = output of LTI

x (t) = input of LTI

h (t) = impulse response of LTI

1. Convolution of two even signals or two odd signals always results in an even signal.

2. Convolution of odd signal and even signal always results in the odd signal.

Hence statement (2) is also false.

So option (1) is the correct answer.

f(t) = 3e^-4t u(t)

\(F\left( \omega \right) = \mathop \smallint \nolimits_{ - \infty }^\infty f\left( t \right){e^{ - j\omega t}}dt\)

\({e^{ - at}}\mathop \leftrightarrow \limits^{F.T} \frac{1}{{a + j\omega }}\)

∴ \(3{e^{ - 4t}}\;\mathop \leftrightarrow \limits^{FT} \frac{3}{{4 + j\omega }}\)

At ω = 4,

\(F\left( \omega \right){\left. \right|_{\omega = 4}} = \frac{3}{{4 + j4}} = \left( {\frac{{\frac{3}{4}}}{{1 + j}}} \right)\)

Explanation:

Let F(ω) is the Fourier transform of f(t).

Hence an odd and imaginary signal always has an odd and real Fourier transform

Hence statement (1) is wrong.

Convolution:

Convolution is a mathematical operation used to express the relation between input and output of an LTI system. It relates the input, output, and impulse response of an LTI system as

y(t) = x(t) * h(t)

Where y (t) = output of LTI

x (t) = input of LTI

h (t) = impulse response of LTI

1. Convolution of two even signals or two odd signals always results in an even signal.

2. Convolution of odd signal and even signal always results in the odd signal.

Hence statement (2) is also false.

So option (1) is the correct answer.

f(t) = 3e^-4t u(t)

\(F\left( \omega \right) = \mathop \smallint \nolimits_{ - \infty }^\infty f\left( t \right){e^{ - j\omega t}}dt\)

\({e^{ - at}}\mathop \leftrightarrow \limits^{F.T} \frac{1}{{a + j\omega }}\)

∴ \(3{e^{ - 4t}}\;\mathop \leftrightarrow \limits^{FT} \frac{3}{{4 + j\omega }}\)

At ω = 4,

\(F\left( \omega \right){\left. \right|_{\omega = 4}} = \frac{3}{{4 + j4}} = \left( {\frac{{\frac{3}{4}}}{{1 + j}}} \right)\)

• x(n) → X(e^jω): Denotes that the Fourier transform of x(n) is X(e^jω).
• y(n) → Y(e^jω): Denotes that the Fourier transform of y(n) is Y(e^jω).

The question asks us to identify which of the provided statements is NOT TRUE.

Correct Option: Option 2

Detailed Explanation:

Let us analyze the given options one by one to determine why Option 2 is the correct answer.

Option 2: x(n−nd)→ejωnd(ejω)" id="MathJax-Element-238-Frame" role="presentation" style="position: relative;" tabindex="0">x(n−nd)→ejωnd(ejω)x(n−nd)→ejωnd(ejω) $x\left(n-n_{d}\right) → e^{j \omega n_{d}} \left(e^{j \omega}\right)$

This statement is NOT TRUE. Let us break it down:

• When a signal is delayed in the time domain by n_d (i.e., x(n - n_d)), the Fourier transform of the delayed signal introduces a phase shift in the frequency domain.
• The correct Fourier transform for x(n - n_d) is given by:
  x(n - n_d) → X(e^jω) × e^-jωn_d
• In the given statement, the phase factor e^jωn_d is incorrectly written as e^jωn_d (positive exponent), which is incorrect. The correct phase factor must have a negative exponent.

Therefore, Option 2 is incorrect because it does not accurately represent the Fourier transform of a delayed signal. The correct representation should involve the term e^-jωn_d, not e^jωn_d.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: a × x(n) + b × y(n) → a × X(e^jω) + b × Y(e^jω)

This statement is TRUE. The Fourier transform is a linear operation, which means that the Fourier transform of a linear combination of signals is equal to the same linear combination of their respective Fourier transforms. Hence, this property is valid.

Option 3: ejω0×x(n)→X(ej(ω−ω0))" id="MathJax-Element-239-Frame" role="presentation" style="position: relative;" tabindex="0">ejω0×x(n)→X(ej(ω−ω0))ejω0×x(n)→X(ej(ω−ω0)) $e^{jω_{0}} × x(n) → X\left(e^{j\left(ω - ω_{0}\right)}\right)$

This statement is TRUE. When a signal is modulated in the time domain by e^jω₀ (complex exponential multiplication), the effect in the frequency domain is a shift in the frequency spectrum by ω₀. The Fourier transform accurately reflects this frequency shift, making this statement valid.

Option 4: x(n) × y(n) → X(e^jω) × Y(e^jω)

This statement is TRUE. Convolution in the time domain corresponds to multiplication in the frequency domain. Hence, the Fourier transform of the convolution of x(n) and y(n) is the product of their respective Fourier transforms. This property is fundamental to the Fourier transform and is valid.

Analysis:

\(h\left( t \right) = \frac{{2\cos \left( {4\pi t} \right)\sin \left( {\pi t} \right)}}{{\pi t}}\)

\(h\left( t \right) = 2\cos \left( {4\pi t} \right)\times \frac{{\sin \left( {\pi t} \right)}}{{\pi t}}\)

Since \(cos~at=\frac{e^{jat}+e^{-jat}}{2}\), the above can be written as:

\(h\left( t \right) = 2(\frac{e^{j4\pi t}+e^{-j4\pi t}}{2})\times \frac{{\sin \left( {\pi t} \right)}}{{\pi t}}\)

\(h\left( t \right) = (e^{j4\pi t}+e^{-j4\pi t})\times \frac{{\sin \left( {\pi t} \right)}}{{\pi t}}\)

Let:

\(h\left( t \right) = {e^{ {j4\pi t}}}{h_1}\left( t \right) + {e^{ - {j4\pi t} }}{h_1}\left( t \right)\)

Taking the Fourier Transform of the above, we can write:

H(ω) = H₁ (ω – 4π) + H₁ (ω + 4ω)

The spectrum is drawn as:

The input signal given is:

x(t) = 1 + cos (πt) + sin (4πt)

The input signal frequencies are 0, π, 4π.

∴ Only ‘4π’ frequency will be allowed to pass through.

∴ The output is y(t) = sin (4πt)