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

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:

\( \mathcal{L}\{e^{-at}x(t)\} = X(s + a) \), where \( X(s) = \mathcal{L}\{x(t)\} \)

Given:

\( \mathcal{L}\{x(t)\} = \sqrt{\frac{2}{s - 3}} \)

Calculation:

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

Using the time-shifting property:

\( \mathcal{L}\{e^{-6t}x(t)\} = \sqrt{\frac{2}{(s + 6) - 3}} = \sqrt{\frac{2}{s + 3}} \)

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:

\[ x(t) = \begin{cases} e^{at} & \text{for } t < 0 \\ e^{-at} & \text{for } t \ge 0 \end{cases} \]

Step 2: Split the Integral

We can split the integral into two parts: one for \(t < 0\) and one for \(t \ge 0\):

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

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:

Fourier Transform:

• The Fourier transform of a function converts it from time domain to frequency domain.
• If f(t) is a time-domain function, its Fourier Transform is defined as:
• \( F(p) = \int_{-\infty}^{\infty} f(t)e^{-ipt} dt \)
• For Gaussian functions like \(( e^{-t^2/2} )\) , their Fourier transform is also Gaussian.
• The Fourier transform of  \(e^{-t^2/2}\)  is  \(\sqrt{2\pi} e^{-p^2/2} \).
• Multiplying by t in time domain corresponds to taking derivative with respect to p in frequency domain:
• \( \mathcal{F}[t f(t)] = i \frac{d}{dp}F(p) \)

Calculation:

Given,

Let f(t) = t e^−t²⁄2

Let F(p) = Fourier transform of e^−t²⁄2 = e^−p²⁄2

⇒ Fourier transform of t f(t) = i × d/dp (e^−p²⁄2)

⇒ = i × (−p e^−p²⁄2) = −i p e^−p²⁄2

⇒ Now t e^−t²⁄2 = f(t),

so full FT is i × d/dp (F(p))

⇒ Add F(p) itself:

Final result = (1 + i p) e^{−(p² − 1)/2}

∴ The correct Fourier transform is: \(\rm (1-ip)e^{-ip}e^{-(p^2-1)/2}\)  

Concept:

Some common Fourier Transform properties are as shown:

If X(ω) is the Fourier transform of x(t) i.e. x(t) ↔ X(ω), then

x₁(t) = u(t + 1.5) - u(t - 1.5)

\(\Rightarrow \quad x_1(t)=\operatorname{rect}\left(\frac{t}{3}\right)\)

\( x_1(t)=\operatorname{rect}\left(\frac{t}{3}\right) \stackrel{F T}{\longleftrightarrow} 3 S a(1.5 \omega) \)

Now, \(x_2(t)=\delta(t+3)+\operatorname{rect}\left(\frac{t}{2}\right)+2 \delta(t-2) \)

Taking Fourier transform

\( X_2(\omega) =e^{3 j \omega}+2 S a(\omega)+2 e^{-2 k \omega} \)

∵  \(y(t) =x_1(t) * x_2(t) \)

\(Y(\omega) =X_1(\omega) \cdot X_2(\omega) \)

We know, \(Y(\omega) =\int_{-\infty}^{\infty} y(t) \cdot e^{-j \omega t} \cdot d t \)

∴ \(\int_{-\infty}^{\infty} y(t) =Y(0)\)

∴ Y(0) = X₁(0).X₂(0)

= 3[1 + 2 + 2] = 15

Concept:

Fourier Transform:

\(F.T\left[ {x\left( t \right)} \right] = X\left( ω \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - jω t}}dt\)

\(I.F.T\left[ {X\left( ω \right)} \right] = x\left( t \right)\)

\(= \frac{1}{{2π }}\mathop \smallint \limits_{ - \infty }^\infty X\left( ω \right){e^{jω t}}dt\)

Some properties of fourier transform:

Analysis:

We know that:

\({x^*}\left[ { + n} \right]\mathop \leftrightarrow \limits^{F.T} {X^*}\left( {{e^{ - j\omega }}} \right) = {X_1}\left( \omega \right)\)

Then,

 \({x^*}\left[ { + n} \right]\mathop \leftrightarrow \limits^{F.T} {X^*}\left( { - \omega } \right) = {X^*}\left( {{e^{ - j\left( { - \omega } \right)}}} \right)\)

\(= {X^*}\left( {{e^{j\omega }}} \right)\)

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.

Concept:

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

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt\) 

Let ω = 0

\(X\left( 0 \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)dt\) 

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:

\(\sin ct = \frac{{\sin \pi t}}{{\pi t}}\) 

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

\({x_1}\left( t \right) \cdot {x_2}\left( t \right) \leftrightarrow \frac{1}{{2\pi }}\left[ {\frac{1}{1}\left( \omega \right)*\frac{1}{2}\left( \omega \right)} \right]\)

Calculation:

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

\(x\left( t \right) = \left( {\frac{{\sin 5\pi t}}{{5\pi t}}} \right) \cdot \left( {\frac{{\sin 5\pi t}}{{5\pi t}}} \right)\) 

\(\mathop \smallint \limits_{ - \infty }^\infty \sin {c^2}\left( {5t} \right)dt = X\left( 0 \right)\)

X(0) = 0.2

Concept:

If x(t) has Fourier transform X(ω)

\(x(t) ↔X(ω) \)

Then,

\(e^{-at} u(t) ↔\frac {1}{a+jω} \)

Also, the Fourier transform of a discrete-time signal is 1, i.e.

δ(t) ↔ 1

Calculation:

\(H(f)=\frac{j3πf}{1+jπf} \)

Since ω = 2πf, the above can be written as:

\(H(ω)=\frac{j1.5ω}{1+j0.5ω}=\frac{3jω}{2+jω}\)

\(H(ω)=3-\frac{6}{2+jω}\)

Taking the inverse Fourier transform of the above, we get:

h(t) = 3δ(t) – 6e^-2t 4(t)

Concept:

Some common Fourier Transform properties are as shown:

If X(ω) is the Fourier transform of x(t) i.e. x(t) ↔ X(ω), then

Concept:

Time-shifting property of Fourier Transform:

Shiting in time domain results in a phase shift in the frequency domain, i.e.

\(If\;x\left( t \right)\mathop \leftrightarrow \limits^{FT} X\left( f \right)\)

\(x\left( {t - {t_0}} \right)\mathop \leftrightarrow \limits^{FT} X\left( f \right){e^{ - j2\pi f{t_0}}}\)

If a signal x(t) is passed through a system having transfer function H(f), then the output of the system is:

\(Y\left( f \right) = X\left( f \right)H\left( f \right)\)

where \(x\left( t \right)\mathop \leftrightarrow \limits^{FT} X\left( f \right)\) and

\(y\left( t \right)\mathop \leftrightarrow \limits^{FT} Y\left( f \right)\)

Calculation:

Since both X(f) and H(f) are band-limited to the same frequency band, we can write:

\(Y\left( f \right) = X\left( f \right).H\left( f \right) = X\left( f \right){e^{ - j4\pi f}} = X\left( f \right).{e^{ - i2\pi f \times 2}}\)

Using time shifting property of the Fourier transform, we can write:

\(X\left( f \right).{e^{ - i2\pi f \times 2}}\xrightarrow{IFT} x\left( {t - 2} \right)\)

Concept:

According to Parseval's Theorem, the energy of a signal is related to the Fourier Transform as:

\(E\left( x\left( t \right) \right)=\frac{1}{2\pi }\underset{-\infty }{\overset{0}{\mathop \int }}\,{{\left| X\left( j\omega \right) \right|}^{2}}d\omega \)

Where X(jω) is the Fourier Transform of signal x(t)

Also, since the energy of the signal is defined as:

\(E\left( x\left( t \right) \right)=\underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{x}^{2}}\left( t \right)d\left( t \right)\)

Parseval’s Theorem can be written as:

\(\underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{x}^{2}}\left( t \right)d=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{\left| X\left( j\omega \right) \right|}^{2}}d\omega \)      ---(1)

Application:

From equation (1) we can write:

\(\underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{\left| X\left( j\omega \right) \right|}^{2}}d\omega =2\pi \underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{x}^{2}}\left( t \right)dt\)         ---(2)

Since time-shifting does not affect the energy of a signal, we can evaluate the energy of x(t + 1) as well which will be the same as the energy of the signal x(t).

Now the energy of the above signal s(t) will be the same as the energy of the original signal x(t).

s(t) is defined as:

s(t) = t + 2, -2 < t < -1

s(t) = 2t + 3, -1 < t < 0

Since the signal s(t) is also symmetrical about the origin, equation (2) can be written as:
\(\underset{-\infty }{\overset{\infty }{\mathop \int }}\,X{{\left( j\omega \right)}^{2}}=2\pi \times 2\underset{-\infty }{\overset{0}{\mathop \int }}\,{{s}^{2}}\left( t \right)dt\)

\(=2\times 2\pi \left[ \underset{-\infty ~}{\overset{-1}{\mathop \int }}\,{{\left( t+2 \right)}^{2}}dt+\underset{-1}{\overset{0}{\mathop \int }}\,{{\left( 2t+3 \right)}^{2}}dt \right]\)

\(=4\pi \left[ \left. \frac{{{\left( t+2 \right)}^{2}}}{3} \right|_{-2}^{-1}+\left. \frac{{{\left( 2t+3 \right)}^{3}}}{3\times 2} \right|_{-1}^{0} \right]\)

\(=4\pi \left[ \frac{1-0}{3}+\frac{{{3}^{3}}-1}{6} \right]\)

\(=4\pi \left( \frac{1}{3}+\frac{26}{3} \right)\)

\(=4\pi \times \frac{14}{3}\)

\(\therefore ~\underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{\left| X\left( j\omega \right) \right|}^{2}}=\frac{56\pi }{3}=58.61\)

Concept:

\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {A;\;\;\;\left| t \right| \le \frac{\tau }{2}}\\ {0;\;\;\;\;\left| t \right| > \frac{\tau }{2}} \end{array}} \right.\)

\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {A; - \frac{\tau }{2} \le t \le \frac{\tau }{2}}\\ {0;\;\;\;\;otherwise} \end{array}} \right.\)

This signal can be represented as:

\(x\left( t \right) = A\left[ {u\left( {t - \frac{\tau }{2}} \right) - u\left( {\tau + \frac{1}{2}} \right)} \right]\)

By applying the Fourier transform,

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \frac{\tau }{2}}^{\frac{\tau }{2}} A.{e^{ - j\omega t}}dt\)

\( = - \frac{A}{{j\omega }}\left[ {{e^{ - \frac{{j\omega }}{2}}} - {e^{\frac{{j\omega }}{2}}}} \right]\)

\( = 2A\frac{{{e^{\frac{{j\omega \tau }}{2}}} - {e^{ - \frac{{j\omega \tau }}{2}}}}}{{2j\omega }}\)

\( = \frac{{2A}}{\omega }\sin \frac{{\omega \tau }}{2}\)

\( = \frac{{2A\tau }}{{\omega \tau }}\sin \frac{{\omega \tau }}{2}\)

\( = A\tau \frac{{\sin \frac{{\omega \tau }}{2}}}{{\frac{{\omega \tau }}{2}}}\)

By putting ω = 2πF

\( = A\tau \frac{{\sin \pi F\tau }}{{\pi F\tau }}\)

Calculation:

\({\rm{x}}\left( {\rm{t}} \right) = \frac{{\sin {\rm{t}}}}{{{\rm{\pi t}}}}{\rm{*}}\frac{{\sin {\rm{t}}}}{{{\rm{\pi t}}}}\)

The Fourier transform of the given since function will be a rectangular wave as shown:

Since the convolution time domain is the multiplication in the frequency domain, the Fourier transform (Spectrum) of x(t) will be:

Now inverse Fourier, we get the time-domain expression of x(t) as:

Thus, \(\frac{{{\rm{sint}}}}{{{\rm{\pi t}}}}{\rm{*}}\frac{{{\rm{sint}}}}{{{\rm{\pi t}}}} = \frac{{{\rm{sint}}}}{{{\rm{\pi t}}}}\)

Analysis:

When two blocks are cascaded their impulse responses are to be convolved.

Here two blocks are cascaded, then equivalent impulse response of the whole system will be:

b₁(t) ⊗ b₂(t)

In the above figure:

z(t) = x(t) ⊗ b₁(t)

y(t) = z(t) ⊗ b₂(t)

The expression for y(t) in terms of x(t) will be:

y(t) =x(t) ⊗ (b1(t)⊗b₂(t))

Important Points

Properties of convolution:

Associativity: f1 * (f2 * f3) = (f1 * f2) * f3

Commutativity: f1 * f2 = f2 * f1 

Distribitivity: f1 * (f2 + f3) = f1 * f2 + f1 * f3

Multilinearity: a(f1 * f2) = (af1) * f2 = f1(af2) 

\(y\left( t \right) = x\left( t \right) + \frac{1}{2}x\left( {t - 1} \right)\)

By applying Fourier transform,

\(Y\left( \omega \right) = X\left( \omega \right) + \frac{1}{2}{e^{ - j\omega }}X\left( \omega \right)\)

\(Y\left( \omega \right) = \left( {1 + \frac{1}{2}{e^{ - j\omega }}} \right)X\left( \omega \right)\)

\(\Rightarrow \frac{{Y\left( \omega \right)}}{{X\left( \omega \right)}} = 1 + \frac{1}{2}{e^{ - j\omega }}\)

\(\Rightarrow H\left( \omega \right) = 1 + \frac{1}{2}{e^{ - j\omega }}\)

\(\Rightarrow H\left( 0 \right) = 1 + \frac{1}{2} = \frac{3}{2}\)

Y(ω) = X(ω).H(ω)

⇒ |Y(ω)|² = |X(ω)|².|H(ω)|²

⇒ |Y(0)|² = |X(0)|².|H(0)|²

We know,

\({\rm{x}}\left( {\rm{t}} \right) = \frac{{{\rm{sin}}4{\rm{\pi t}}}}{{4{\rm{\pi t}}}}\)

Integrating mod square of x(t) to find the energy will be a tedious process. 

Fourier transform of  \(\frac{{{\rm{sin}}4{\rm{\pi t}}}}{{4{\rm{\pi t}}}}\)

From Parseval’s theorem

\(\mathop \smallint \limits_{ - \infty }^\infty {\left| {{\rm{x}}\left( {\rm{t}} \right)} \right|^2}{\rm{dt}} = \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - \infty }^\infty {\left| {{\rm{X}}\left( {\rm{\omega }} \right)} \right|^2}{\rm{d\omega }}\)

we have,

\(\begin{array}{l} \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - 4{\rm{\pi }}}^{4{\rm{\pi }}} {\left( {\frac{1}{4}} \right)^2}{\rm{d\omega }} = \frac{1}{{2{\rm{\pi }}}}\mathop \smallint \limits_{ - 4{\rm{\pi }}}^{4{\rm{\pi }}} \frac{1}{{16}}{\rm{d\omega }}\\ \Rightarrow = \left. {\frac{1}{{2{\rm{\pi }}}}.\frac{1}{{16}}{\rm{\omega }}} \right|_{ - 4{\rm{\pi }}}^{4{\rm{\pi }}}\\ = \frac{1}{{2{\rm{\pi }}}}.\frac{1}{{16}}.8{\rm{\pi }} = \frac{1}{4} \end{array}\)

Thus,

signal's energy \(\mathop \smallint \limits_{ - \infty }^\infty {\left| {{\rm{x}}\left( {\rm{t}} \right)} \right|^2}{\rm{dt}} = \frac{1}{4}\)