Fourier Transform MCQ Quiz in मल्याळम - Objective Question with Answer for Fourier Transform - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

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

$f\left(n\right)=\frac{1}{2\pi }\underset{-\pi }{\stackrel{\pi }{\int }}F\left(\omega \right)e^{j\omega n}d\omega$    ---(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:

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

From the given Fourier transform,

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

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

$\frac{1}{2}\times base\times height$

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

$\underset{-\pi }{\stackrel{\pi }{\int }}F\left(\omega \right)d\omega =\frac{\pi }{3}$

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:

$X\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-j\omega t}dt$ 

Let ω = 0

$X\left(0\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)dt$ 

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:

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

Calculation:

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

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

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\sin c^2\left(5t\right)dt=X\left(0\right)$

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)

$x(t)\leftrightarrow X(\omega )$

Then,

$e^{-at}u(t)\leftrightarrow \frac{1}{a+j\omega }$

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

δ(t) ↔ 1

Calculation:

$H(f)=\frac{j3\pi f}{1+j\pi f}$

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

$H(\omega )=\frac{j1.5\omega }{1+j0.5\omega }=\frac{3j\omega }{2+j\omega }$

$H(\omega )=3-\frac{6}{2+j\omega }$

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:

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

Using duality property

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

Now using time reversal property

f(-t) = f(-ω)

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

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 = $\mathrm{\infty }$ 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:

$X\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(t\right)e^{-j\omega t}$

Calculations:

For f(t) = δ(t)

$F\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\delta \left(t\right)e^{-j\omega t}=1$

The spectrum is represented as:

Some common Fourier transform and their spectrum are as shown:

For f(t) = u(t)

$F\left(\omega \right)=\underset{0}{\overset{\mathrm{\infty }}{\int }}u\left(t\right)e^{-j\omega t}=\frac{1}{j\omega }$  

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

$F\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-t}u\left(t\right)e^{-j\omega t}dt$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-t}{e}^{-j\omega t}dt=\frac{1}{1+j\omega }$           

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 $\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}t=\frac{\sin \pi t}{\pi t}$

Similarly, $\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}10t=\frac{\sin 10\pi t}{10\pi t}$

By using, duality theorem

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

Using time-shifting property,