Fourier Transform MCQ Quiz in తెలుగు - Objective Question with Answer for Fourier Transform - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

If the Fourier transform of a sequence is F(ω) then the discrete sequence f(n) is given by Inverse Fourier Transform:

\(f\left( n \right)=\frac{1}{2\pi }\underset{-\pi}{\overset{\pi}{\mathop \int }}\,F\left( \omega \right){{e}^{j\omega n}}d\omega\)    ---(1)

Calculation:

We need to find the value of the sequence f(n) at n = 0 i.e. f(0);

Using equation (1), we can write:

\(f\left( 0 \right)=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\mathop \int }}\,F\left( \omega \right)d\omega\)

From the given Fourier transform,

\(\underset{-\pi }{\overset{\pi }{\mathop \int }}\,F\left( \omega \right)d\omega\) is nothing but the area under the curve.

Since the given curve is a triangle, its area is given by:

\(\frac{1}{2}\times base\times height\)

\(\underset{-\pi }{\overset{\pi }{\mathop \int }}\,F\left( \omega \right)d\omega =\frac{1}{2}\times \frac{2\pi }{3}\times 1\)

\(\underset{-\pi }{\overset{\pi }{\mathop \int }}\,F\left( \omega \right)d\omega =\frac{\pi }{3}\)

Concept:

The Fourier Transform of 1 is given by:

1 ↔ 2πδ(ω)

\(\frac{1}{{2\pi }} \leftrightarrow \delta \left( \omega \right)\)

\(\frac{1}{{2\pi }}{e^{j{\omega _0}t}} \leftrightarrow \delta \left( {\omega - {\omega _0}} \right)\)

\(\frac{1}{{2\pi }}{e^{ - j{\omega _0}t}} \leftrightarrow \delta \left( {\omega + {\omega _0}} \right)\)

Analysis:

X(ω) = 2δ(ω + 3ω₀) + 0.5δ (ω + ω₀) + 2δ(ω) + 0.5δ(ω – ω₀) + 2δ (ω - 3ω₀)

Applying Inverse Fourier transform, we get:

\(x\left( t \right) = \frac{2}{{2\pi }}{e^{ - j3{\omega _0}t}} + \frac{{0.5}}{{2\pi }}{e^{ - j{\omega _0}t}} + \frac{2}{{2\pi }}\left( 1 \right) + \frac{{0.5}}{{2\pi }}{e^{j{\omega _0}t}} + \frac{2}{{2\pi }}{e^{3j{\omega _0}t}}\) 

\(x\left( t \right) = \frac{1}{\pi } + \frac{2}{\pi }\cos \left( {3{\omega _0}t} \right) + \frac{1}{{2\pi }}\cos \left( {{\omega _0}t} \right)\)

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:

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

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

The correct answer is 1750 Hz

Concept: 

(1) If the signal is a multiply in the time domain then in the frequency domain we get convolution.

(2) The duration/ extension property of convolution says that if

x₁(ω) is non-zero for ω₁<ω<ω₂ 

x₂(ω) is non-zero for ω3<ω<ω4

Now for Y(ω) = x₁(ω) * x₂(ω)

then

Y(ω) is non-zero for (ω₁ + ω₃)<ω<(ω2 + ω₄)

Solution:

x1(ω) is non zero for -1500π<ω<1500π  

x2(ω) is non zero for -2000π<ω<2000π 

Now for

Y(ω) = x1(ω) * x2(ω)

then

Y(ω) is non-zero for (-1500π-2000π )<ω<(1500π +2000π)

∴ maximum frequency of ωmax  = 3500π 

\(\therefore F_{max} = \frac{3500\pi }{2 \pi}\)

=1750 Hz

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:

Time shifting property of Fourier transform:

If X(ω) is the Fourier transform of x(t),

\(x\left( {t - {t_0}} \right) \leftrightarrow X\left( \omega \right){e^{ - j\omega {t_0}}}\)

Duality property of Fourier transform:

If X(ω) is the Fourier transform of x(t),

\(X\left( t \right) \leftrightarrow 2\pi x\left( { - \omega } \right)\)

Calculation:

\(\begin{array}{l} f\left( t \right)\mathop \leftrightarrow \limits^{F.T} f\left( \omega \right)\\ {e^{ - t}}u\left( t \right)\mathop \leftrightarrow \limits^{F.T} \frac{1}{{1 + j\omega }}\\ {e^{ - 2t}}u\left( t \right)\mathop \leftrightarrow \limits^{F.T} \frac{1}{{2 + j\omega }} \end{array}\)

Using duality property

\(\begin{array}{l} f\left( t \right)\mathop \leftrightarrow \limits^{F.T} f\left( \omega \right)\\ f\left( t \right)\mathop \leftrightarrow \limits^{F.T} 2\pi f\left( { - \omega } \right)\\ f\left[ {\frac{1}{{2 + jt}}} \right]\mathop \leftrightarrow \limits^{F.T} 2\pi {e^{ - \left( { - 2\omega } \right)}}u\left( { - \omega } \right)\\ f\left[ {\frac{1}{{2 + jt}}} \right]\mathop \leftrightarrow \limits^{F.T} 2\pi {e^{2\omega }}u\left( { - \omega } \right) \end{array}\)

Now using time reversal property

f(-t) = f(-ω)

\(\begin{array}{l} f\left[ {\frac{1}{{2 - jt}}} \right]\mathop \leftrightarrow \limits^{F.T} 2\pi {e^{ - 2\omega }}u\left( \omega \right)\\ \therefore {e^{ - 2\omega }}u\left( \omega \right)\mathop \leftrightarrow \limits^{F.T} \frac{1}{{2\pi }}\left[ {\frac{1}{{2 - jt}}} \right] \end{array}\)

The correct answer is option 3)

Concept:

Fourier transform of the exponential signals. 

• e^at u(-t) ⇔ \( \frac{1}{(a -jw)}\)
• e^-at u(t) ⇔ \( \frac{1}{(a + jw)}\)

Solution:

Equation of given graph is

x(t) = e^at u(-t) + e^-at u(t)

Taking Fourier transform both side

F[x(t)] = F[eat u(-t)] + F[e-at u(t)]

 \(X(W) = \frac{1}{(a -jw)} + \frac{1}{(a + jw)}\)

 \(= \frac{(a + jw) + (a - jw)}{a^2 + w^2}\)

 \(= \frac{2a}{a^2 + w^2}\)

At ω = 0 we get IX(w)I = \(\frac{2}{a}\)

At ω = \(\infty\) we get IX(w)I = 0 

These values are satisfied by the only option 3)

Concept:

The Fourier transform of a signal in the time domain is given as:

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

Calculations:

For f(t) = δ(t)

\(F\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty \delta \left( t \right){e^{ - j\omega t}} = 1\)

The spectrum is represented as:

Some common Fourier transform and their spectrum are as shown:

For f(t) = u(t)

\(F\left( \omega \right) = \mathop \smallint \limits_0^\infty u\left( t \right){e^{ - j\omega t}} = \frac{1}{{j\omega }}\)  

For f(t) = e-tu(t)

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

\( \mathop \smallint \limits_0^\infty {e^{ - t}}{e^{ - j\omega t}}dt = \frac{1}{{1 + j\omega }}\)           

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

Concept:

Time-shifting property of Fourier transform:

If X(ω) is the Fourier transform of x(t),

\(x\left( {t - {t_0}} \right) \leftrightarrow X\left( \omega \right){e^{ - j\omega {t_0}}}\)

Duality property of Fourier transform:

If X(ω) is the Fourier transform of x(t),

\(X\left( t \right) \leftrightarrow 2\pi x\left( { - \omega } \right)\)

Calculation:

We know \({\rm{sinc}}t = \frac{{\sin \pi t}}{{\pi t}}\)

Similarly, \({\rm{sinc}}10t = \frac{{\sin 10\;\pi t}}{{10\;\pi t}}\)

By using, duality theorem

\(\frac{{\sin \left( {10\;\pi t} \right)}}{{10\;\pi t}} \to 0.1\;rect\left( {f/10} \right)\)

Using time-shifting property,