Fourier Transform of Odd Symmetric Signals MCQ Quiz - Objective Question with Answer for Fourier Transform of Odd Symmetric Signals - Download Free PDF

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:

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:

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

We have 

$\frac{\sin \mathrm{t}}{\mathrm{t}}$is real and even

$\Rightarrow \mathrm{F}\mathrm{T}\left[\frac{\sin \mathrm{t}}{\mathrm{t}}\right]$is also real and even.

Now, convolution in time domain is multiplication in frequency domain.

$\Rightarrow \frac{\sin \mathrm{t}}{\mathrm{t}}\star \frac{\sin \mathrm{t}}{\mathrm{t}}\leftrightarrow \frac{1}{2\pi }\mathrm{F}\mathrm{T}\left[\frac{\sin \mathrm{t}}{\mathrm{t}}\right].\mathrm{F}\mathrm{T}\left[\frac{\sin \mathrm{t}}{\mathrm{t}}\right]$

$\Rightarrow \frac{1}{2\pi }\times \left(\mathrm{E}\mathrm{v}\mathrm{e}\mathrm{n},\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}\right)\times \left(\mathrm{E}\mathrm{v}\mathrm{e}\mathrm{n},\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{l}\right)$

⇒ Even, real 

If Fourier transform is real and even, the inverse Fourier will also be real and even 

Thus $\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{t}}{\mathrm{t}}\ast \frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{t}}{\mathrm{t}}$ is real and even. 

Now $\mathrm{t}.\left[\frac{\sin \mathrm{t}}{\mathrm{t}}\times \frac{\sin \mathrm{t}}{\mathrm{t}}\right]$is real and odd $[\because \mathrm{o}\mathrm{d}\mathrm{d}\times \mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}=\mathrm{o}\mathrm{d}\mathrm{d}]$ 

Now Fourier transform of an odd and real function is imaginary and odd.

Thus, ​$\mathrm{F}\mathrm{T}\left[\mathrm{t}\left[\frac{\sin \mathrm{t}}{\mathrm{t}}\times \frac{\sin \mathrm{t}}{\mathrm{t}}\right]\right]$is imaginary and odd.

Fundamental frequency ${\mathrm{f}}_0=\mathrm{G}\mathrm{C}\mathrm{D}\{12,20,28\}$

$12=2\times 2\times 3$

$20=2\times 2\times 5$

$28=2\times 2\times 7$

$\Rightarrow {\mathrm{f}}_0=4\mathrm{H}\mathrm{z}.$

$\Rightarrow {\omega }_0=2\pi \mathrm{f}=8\pi \mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\mathrm{e}\mathrm{c}$.