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

Concept: Fourier Transform Symmetry Property

If a time-domain function $f(t)$ is:

• Real: It has no imaginary part

• Even: f(t) = f(-t)" id="MathJax-Element-103-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t) = f(-t)" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0"> f(t) = f(-t) $f(t) = f(-t)$ $f(t) = f(-t)$ f(t)=f(−t)

Then it's  Fourier Transform $F(\omega )$ will be:

• Purely real

• Even: F(ω)=F(−ω)" id="MathJax-Element-105-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">

🔍 Example: Cosine Wave

Let’s take: f(t) = cos(ω0​t)

• It is real.

• It is even, because " id="MathJax-Element-107-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-4-Frame" role="presentation" style="position: relative;" tabindex="0">  cos(−ω_0​t)=cos(ω_0​t)

Fourier Transform of " id="MathJax-Element-108-Frame" role="presentation" style="position: relative;" tabindex="0">" id="MathJax-Element-5-Frame" role="presentation" style="position: relative;" tabindex="0"> cos(ω_0​t) is: F(ω)=π [δ(ω−ω_0​)+δ(ω+ω_0​)]

This result is:

• Purely real (involves delta functions, no imaginary component)

• Even since it's symmetric around $\omega =0$

Concept:

Sinc pulse shaping is commonly used in digital communication systems to achieve ideal Nyquist pulse shaping, minimizing intersymbol interference (ISI).

The sinc function is defined as:

Explanation:

The sinc function arises as the inverse Fourier Transform of a rectangular function in the frequency domain.

In other words, if the frequency spectrum is a perfect rectangular shape (ideal low-pass filter), its time-domain representation is a sinc function.

Conclusion:

Correct Answer: Option 4) rectangular

The correct option is 1

Concept:

If the Laplace transform of a function \( x(t) \) is known, then the Laplace transform of \( e^{-at}x(t) \) can be obtained using the time-shifting property of the Laplace transform.

This property states that:

, where \( X(s) = \mathcal{L}\{x(t)\} \)

Given:

Calculation:

We are asked to find the Laplace transform of \( e^{-6t}x(t) \).

Using the time-shifting property:

The correct answer is option 4

Concept:

The time scaling property of Fourier transform states that:

If

Then:

‘a’ any real constant.

Proof:

For a > 0:

Let at = τ. Thus adt = dτ

The above expression can now be written as:

For a

Let -at = ui – adt = du

Note:

The scaling property states that time compression of signal results in its spectral expansion and vice-versa.

Explanation:

The Fourier transform of the signal \(x(t) = e^{-a|t|}\) can be calculated as follows:

Step 1: Definition of Fourier Transform

The Fourier transform of a function \(x(t)\) is defined as:

\[ X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \]

For the given signal \(x(t) = e^{-a|t|}\), we need to consider the absolute value function. Therefore, the signal can be written as:

\[ x(t) = \begin{cases} e^{at} & \text{for } t

Step 2: Split the Integral

We can split the integral into two parts: one for \(t

\[ X(j\omega) = \int_{-\infty}^{0} e^{at} e^{-j\omega t} dt + \int_{0}^{\infty} e^{-at} e^{-j\omega t} dt \]

Simplifying the exponents inside the integrals, we get:

\[ X(j\omega) = \int_{-\infty}^{0} e^{(a - j\omega)t} dt + \int_{0}^{\infty} e^{-(a + j\omega)t} dt \]

Step 3: Evaluate the Integrals

Let's evaluate each integral separately.

For the first integral \( \int_{-\infty}^{0} e^{(a - j\omega)t} dt \):

\[ \int_{-\infty}^{0} e^{(a - j\omega)t} dt = \left[ \frac{e^{(a - j\omega)t}}{a - j\omega} \right]_{-\infty}^{0} \]

Evaluating the limits:

\[ \left[ \frac{e^{(a - j\omega)t}}{a - j\omega} \right]_{-\infty}^{0} = \frac{1}{a - j\omega} - \lim_{t \to -\infty} \frac{e^{(a - j\omega)t}}{a - j\omega} \]

Since \(a > 0\), \(e^{(a - j\omega)t}\) approaches 0 as \(t\) approaches \(-\infty\):

\[ \frac{1}{a - j\omega} - 0 = \frac{1}{a - j\omega} \]

For the second integral \( \int_{0}^{\infty} e^{-(a + j\omega)t} dt \):

\[ \int_{0}^{\infty} e^{-(a + j\omega)t} dt = \left[ \frac{-e^{-(a + j\omega)t}}{a + j\omega} \right]_{0}^{\infty} \]

Evaluating the limits:

\[ \left[ \frac{-e^{-(a + j\omega)t}}{a + j\omega} \right]_{0}^{\infty} = 0 - \left( -\frac{1}{a + j\omega} \right) = \frac{1}{a + j\omega} \]

Step 4: Add the Results

Now, combining both integrals, we get:

\[ X(j\omega) = \frac{1}{a - j\omega} + \frac{1}{a + j\omega} \]

To simplify, find a common denominator:

\[ X(j\omega) = \frac{a + j\omega + a - j\omega}{(a - j\omega)(a + j\omega)} = \frac{2a}{a^2 + \omega^2} \]

Therefore, the Fourier transform of \(x(t) = e^{-a|t|}\) is:

\[ X(j\omega) = \frac{2a}{a^2 + \omega^2} \]

Important Information:

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

Option 1: \(X(j\omega) = \frac{2a}{a+\omega}\)

This option is incorrect because the denominator should be \(a^2 + \omega^2\), not \(a + \omega\).

Option 3: \(X(j\omega) = \frac{a}{a - j\omega}\)

This option is incorrect because it does not account for the full expression obtained from the Fourier transform, and it lacks the correct form of the denominator.

Option 4: \(X(j\omega) = \frac{2a}{a + j\omega}\)

This option is incorrect because it only partially matches one of the terms derived from the Fourier transform calculation and does not include the complete expression.

Conclusion:

Understanding the Fourier transform process and carefully evaluating each integral's limits and results are essential for correctly identifying the transform of a given signal. In this case, the correct Fourier transform of \(x(t) = e^{-a|t|}\) is \(X(j\omega) = \frac{2a}{a^2 + \omega^2}\), making Option 2 the correct choice.

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

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

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

Concept:

Fourier Transform:

It is used for frequency analysis of any Bounded Input and Bounded Output (BIBO) signal.

Fourier Transform for any function x(t) is given by

Inverse Fourier Transform for any function X(ω) is given by

Frequency shifting

If X(ω) has the inverse Fourier transform as x(t).

Calculation:

δ (ω) is impulse function that exists only at t = 0

so its inverse Fourier transform will be

Given signal is δ(ω – ω0)

Inverse Fourier transform is

Impulse function exists at ω0.

We know that the area of the impulse is 1.

Important Points

The delta function is a normalized impulse, that is, sample number zero has a value of one, while all other samples have a value of zero.

For this reason, the delta function is frequently called the unit impulse. 

Fourier Transform of δ (t) is 1.

Concept:

Fourier Transform:

Some properties of fourier transform:

Analysis:

We know that:

Then,

Option (2) correct.

More information:

• Fourier transform of the real signal is always even conjugate in nature.
• F.T [Real & even signal] = purely real and even.
• F.T [Real & odd signal] = purely imaginary and odd.
• Shifting in the time domain only changes the phase spectrum of the signal.

Given,

x(t) = 0 for |t| > 1

 x(t) = 1 - |t| for |t| ≤ 1

The waveform of function can be drawn as,

Hence, given function is triangular function,

From standered Fourier transform,

If 

Fourier transform of x(t)  is given by

A = T = 1

From the waveform of function,

Therefore, the peak value of the sampling function occurs at (ω = 0).

The maximum magnitude (peak) of X(ω) will be 1.

Hilbert Transformer:

Hilbert Transform of a signal x(t) is defined as:

The impulse response of Hilbert Transform is given by

h(t) ≠ 0 for t

Hence, the System is Non – causal.

• Hilbert transform does not change the domain of the signal.

• The magnitude of the frequency component present in x(t) remains unchanged when it is passed through the system.
• The phase of the positive frequency components is shifted by -π/2 and that of negative frequency components is shifted by +π/2.
• A signal x(t) and its Hilbert transform  are orthogonal to each other.

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

Concept:

A function is odd, if the function on one side of x-axis">t-axisx-axis is sign inverted with respect to the other side or graphically, symmetric about the origin.

f(t) = - f(-t)

Symmetry condition of Fourier Transform

Calculation:

We have the wave form of function f (t) as

From the wave form, f(t) is an odd function

∴ f (t) = - f (- t)

⇒ Fourier transform of the function is imaginary and odd function of ω

Concept:

The Fourier transform of a signal x(t) is defined as:

Let ω = 0

Analysis:

For x(t) = sin c² (5t), we find it’s Fourier transform X(ω) and then find X(0) using the above concept.

Sinc function is defined as:

Also, the Fourier transform of a sinc function is a rectangular pulse as shown:

Calculation:

x(t) = sin c²5t = (sin c 5t) (sin c 5t)

X(0) = 0.2