Properties of Fourier Transform MCQ Quiz in मराठी - Objective Question with Answer for Properties of Fourier Transform - मोफत PDF डाउनलोड करा

Concept:

\(\delta \left( t \right) \overset{FT}{\longleftrightarrow } 1\)

\(\delta \left( {t - {t_0}} \right)\overset{FT}{\longleftrightarrow } \;{e^{ - j\omega {t_0}}}\)

Analysis:

The given signal can be written as:

y(t) = δ(t + 1) + (-δ(t – 1))

y(t) = δ(t + 1) – δ(t – 1)

Taking the Fourier transform, we get:

y(jω) = e^jω – e^-jω

= cos ω + j sin ω – (cos ω – j sin ω)

= 2j sin ω

∴ k = 2        

Concept:

The energy of a discrete time signal using Parsevel’s theorem is given by:

\(\mathop \sum \limits_{n = - \infty }^\infty {\left| {x\left( n \right)} \right|^2} = \frac{1}{{2\pi }}\mathop \smallint \nolimits_{ - \pi }^\pi {\left| {X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega \)

Calculation:

Given discrete time signal is:

\(x\left( n \right) = 2\left[ {\mathop \sum \limits_{k = 0}^\infty \delta \left( {n - 1 - k} \right) - \delta \left( {n - 6 - k} \right)} \right]\)

\(u\left( n \right) = \mathop \sum \limits_{k = 0}^\infty \delta \left( {n - k} \right)\)

\(u\left( {n - 1} \right) = \mathop \sum \limits_{k = 0}^\infty \delta \left( {n - 1 - k} \right)\)

\(u\left( {n - 6} \right) = \mathop \sum \limits_{k = 0}^\infty \delta \left( {n - 6 - k} \right)\)

∴ x(n) = 2[u(n - 1) – u(n - 6)]

= 2, 1 ≤ n ≤ 5

\(\frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi {\left| {X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = 2\mathop \sum \limits_{n = - \infty }^\infty {\left| {x\left( n \right)} \right|^2}\) 

\(\frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi {\left| {X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = 2\left[ {{2^2} + {2^2} + {2^2} + {2^2} + {2^2}} \right]\) 

\(\frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi {\left| {X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = 40\) 

Also,

\(\mathop \sum \limits_{n = - \infty }^\infty {\left| {nx\left( n \right)} \right|^2} = \frac{1}{{2\pi }}\mathop \smallint \nolimits_{ - \pi }^\pi {\left| {\frac{d}{{d\omega }}X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega \) 

\(\mathop \smallint \nolimits_{ - \pi }^\pi {\left| {\frac{d}{{d\omega }}X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = 2\pi \mathop \sum \limits_{n = - \infty }^\infty {\left| {nx\left( n \right)} \right|^2}\) 

\(\mathop \smallint \nolimits_{ - \infty }^\infty {\left| {\frac{d}{{d\omega }}X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = 2\pi \left[ {{1^2} \times {2^2} + {2^2} \times {2^2} + {3^2} \times {2^2} + {4^2} \times {2^2} + {5^2} \times {2^2}} \right]\)

\(\mathop \smallint \nolimits_{ - \pi }^\pi {\left| {\frac{d}{{d\omega }}X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = 440\pi \) 

Hence, the correct options are (1) and (3).

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) 

Concept:

\(\delta \left( t \right) \overset{FT}{\longleftrightarrow } 1\)

\(\delta \left( {t - {t_0}} \right)\overset{FT}{\longleftrightarrow } \;{e^{ - j\omega {t_0}}}\)

Analysis:

The given signal can be written as:

y(t) = δ(t + 1) + (-δ(t – 1))

y(t) = δ(t + 1) – δ(t – 1)

Taking the Fourier transform, we get:

y(jω) = e^jω – e^-jω

= cos ω + j sin ω – (cos ω – j sin ω)

= 2j sin ω

∴ k = 2        

Concept:

The transfer function of a system is defined as:

\(H\left( {j\omega } \right) = \frac{{Y\left( {j\omega } \right)}}{{X\left( {j\omega } \right)}}\)

Y(jω) = H(jω) × X(jω)

Taking the inverse Fourier transform of the above, we get the equation of the system in the time domain.

Also,

\(If\;x\left( t \right)\overset{F.T.}{\longleftrightarrow } X\left( {j{\rm{\omega }}} \right)\)

\(then\frac{{{d^n}x\left( t \right)}}{{dt}} \leftrightarrow {\left( {j{\rm{\omega }}} \right)^n}.X\left( {j{\rm{\omega }}} \right)\)

Calculation:

\(Given\;H\left( {j{\rm{\omega }}} \right) = \frac{{1 + j{\rm{\omega }}}}{{\left( {2 + j{\rm{\omega }}} \right)\left( {3 + j{\rm{\omega }}} \right)}}\)

This can be written as:

\( \Rightarrow \frac{{Y\left( {j{\rm{\omega }}} \right)}}{{X\left( {j{\rm{\omega }}} \right)}} = \frac{{1 + j{\rm{\omega }}}}{{\left( {2 + j{\rm{\omega }}} \right)\left( {3 + j{\rm{\omega }}} \right)}}\)

\( \Rightarrow \frac{{Y\left( {j{\rm{\omega }}} \right)}}{{X\left( {j{\rm{\omega }}} \right)}} = \frac{{1 + j{\rm{\omega }}}}{{6 + 2\left( {j{\rm{\omega }}} \right) + 3\left( {j{\rm{\omega }}} \right) + {{\left( {j{\rm{\omega }}} \right)}^2}}}\;\)

\( \Rightarrow \frac{{Y\left( {j{\rm{\omega }}} \right)}}{{X\left( {j{\rm{\omega }}} \right)}} = \frac{{1 + j{\rm{\omega }}}}{{{{\left( {j{\rm{\omega }}} \right)}^2} + 5j{\rm{\omega }} + 6}}\)

⇒ (jω)² Y(jω) + 5(jω) Y(jω) + 6Y(jω) = X(jω) + (jω) × (jω)

⇒ Taking the inverse Fourier Transform of the above, we get:

\( \Rightarrow \frac{{{d^2}y\left( t \right)}}{{d{t^2}}} + \frac{{5dy\left( t \right)}}{{dt}} + 6y\left( t \right) = x\left( t \right) + \frac{{dx\left( t \right)}}{{dt}}\)

\(\Rightarrow \overset{\ddot{\ }}{\mathop{y}}\,\left( t \right)+5\dot{y}\left( t \right)+6y\left( t \right)=x\left( t \right)+\dot{x}\left( t \right)\)

So, Option (4) is correct.

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

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 \(\mathop \smallint \limits_{ - \infty }^\infty {\left| {{\rm{x}}\left( {\rm{t}} \right)} \right|^2}{\rm{dt}} = \frac{1}{4}\)

Concept:

The Fourier transform of a rectangular pulse is a sinc pulse, i.e.

\(rect\left( t \right)\mathop \leftrightarrow \limits^{F.T} \frac{{2\sin \omega }}{\omega }\) 

Considering another signal y(t) = y₁ (t + 2), its Fourier transform will be:

\({y_1}\left( {t + 2} \right) \leftrightarrow \;\frac{{2\sin \omega }}{\omega }{e^{2j\omega }}\) 

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

We can write:

\(x\left( t \right)*y\left( t \right)\mathop \leftrightarrow \limits^{F.T} \;X\left( {j\omega } \right)Y\left( {j\omega } \right)\) 

From the definition Fourier transform, we have:

\(x\left( t \right)*y\left( t \right) = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^\infty X\left( {j\omega } \right)Y\left( {j\omega } \right){e^{j\omega t}}d\omega \) 

For t = 0, the above expression can be written as:

\(\mathop \smallint \limits_{ - \infty }^\infty X\left( {j\omega } \right)Y\left( {j\omega } \right)d\omega = 2\pi {\left[ {x\left( t \right)*y\left( t \right)} \right]_{t = 0}}\) 

Using Equation (1):

\(\mathop \smallint \limits_{ - \infty }^\infty X\left( {j\omega } \right) \cdot \frac{{2\sin \omega }}{\omega }{e^{2j\omega }}d\omega = 2\pi {\left( {x\left( t \right)*y\left( t \right)} \right)_{t = 0}}\) 

∴ We need to evaluate the value of x(t) * y(t) at t = 0, to get the required integral value.

The convolution in the time domain is defined as:

\(x\left( t \right)*y\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( \tau \right)y\left( {t - \tau } \right)d\tau \) 

\(x\left( t \right)*y{\left. {\left( t \right)} \right|_{t = 0}} = \mathop \smallint \limits_{ - \infty }^\infty x\left( \tau \right)y\left( { - \tau } \right)d\tau \) 

x(τ) and y(-τ) are represented as shown:

The required integral is nothing but the shaded portion, i.e.

\(x\left( t \right)*y({\left. {t)} \right|_{t = 0}} = \left( {1 \times 1} \right) + \frac{1}{2} \times 1 \times 1 + \left( {1 \times 2} \right)\) 

= 3.5

\(\therefore \;\mathop \smallint \limits_{ - \infty }^\infty X\left( {j\omega } \right) \cdot \frac{{2\sin \omega }}{\omega }{e^{j2\omega }}d\omega = 2\pi \times 3.5\) 

= 7π

= 21.991

Concept:

The normalized Gaussian pulse is given as:

\(f\left( t \right) = K{e^{ - a{t^2}}}\)

And its Fourier transform is given as:

\(F.T.\;\left\{ {f\left( t \right)} \right\} = \sqrt {\frac{\pi }{a}} \;{e^{ - \frac{{{\pi ^2}{f^2}}}{a}}}\)

\( = \sqrt {\frac{\pi }{a}} \;{e^{ - {{\left( {\frac{\pi^2 \omega^2 }{4\pi^2a}} \right)}}}}\)

Analysis:

\({e^{ - p{t^2}}} \leftrightarrow \sqrt {\frac{\pi }{p}} {e^{ - {\omega ^2}/4p}} = X\left( \omega \right)\)

\(\frac{d}{{d\omega }}X\left( \omega \right) = \sqrt {\frac{\pi }{p}} \frac{d}{{d\omega }}{e^{ - \frac{{{\omega ^2}}}{{4p}}\;}}\)

\(= \sqrt {\frac{\pi }{p}} \;{e^{ - \frac{{{\omega ^2}}}{{4p}}}}\left[ { - \frac{{2\omega \;}}{{4p}}} \right]\)

Substituting the values of ω and X(ω) in the given equation, the equation evaluates to

Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left.

For example: