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

$X_1\left(j\omega \right)$ and $X_2\left(j\omega \right)$ are not conjugate symmetric, hence $x_1\left(t\right),x_2\left(t\right)$ are not real

Concept:

Duality Property

If $\mathrm{x}(\mathrm{t})\stackrel{\mathrm{F}.\mathrm{T}.}{\leftrightarrow }\mathrm{x}(\omega )$

then $\mathrm{x}(\mathrm{t})\stackrel{\mathrm{F}.\mathrm{T}.}{\leftrightarrow }2\pi \mathrm{x}(-\omega )$

Fourier time of signal x(t) is defined as,

$x(\omega )={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t).e^{-j\omega t}dt}$

putting t = 0;

$x(0)={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)dt}$

Explanation:

We know that $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{x}\left(\mathrm{t}\right)\cdot \mathrm{d}\mathrm{t}=\mathrm{X}\left(0\right)$

Also we know

Where $X\left(0\right)=1$

$\therefore 2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left(\frac{\mathrm{s}\mathrm{i}\mathrm{n}2\pi \mathrm{t}}{\pi \mathrm{t}}\right)\cdot \mathrm{d}\mathrm{t}=2\mathrm{X}\left(0\right)=2\left(1\right)=2$

​

We have,

              $y\left(t\right)=\frac{dx\left(t\right)}{dt}$........eq(i)

Since, x(t) is a Differentiable non - constant even function.

so, x(t) = x(-t) and its Fourier transform X(ω) = X(-ω)

Taking Fourier transform of eq(i)

⇒ Y(ω) = jω X(ω)

Y(-ω) = j(-ω)X(-ω)

Y(-ω) = -jω X(ω)

Y(-ω) = -Y(ω) 

So, Y(ω) is odd and imaginary.

Since, x(t) is differentiable, non-constant,  even function. So, its frequency response will be real i.e., X(ω) is real.

Hence, the correct option is (2)

We have,

              $y\left(t\right)=\frac{dx\left(t\right)}{dt}$........eq(i)

Since, x(t) is a Differentiable non - constant even function.

so, x(t) = x(-t) and its Fourier transform X(ω) = X(-ω)

Taking Fourier transform of eq(i)

⇒ Y(ω) = jω X(ω)

Y(-ω) = j(-ω)X(-ω)

Y(-ω) = -jω X(ω)

Y(-ω) = -Y(ω) 

So, Y(ω) is odd and imaginary.

Since, x(t) is differentiable, non-constant,  even function. So, its frequency response will be real i.e., X(ω) is real.

Hence, the correct option is (2)

Concept:

Duality Property

If $\mathrm{x}(\mathrm{t})\stackrel{\mathrm{F}.\mathrm{T}.}{\leftrightarrow }\mathrm{x}(\omega )$

then $\mathrm{x}(\mathrm{t})\stackrel{\mathrm{F}.\mathrm{T}.}{\leftrightarrow }2\pi \mathrm{x}(-\omega )$

Fourier time of signal x(t) is defined as,

$x(\omega )={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t).e^{-j\omega t}dt}$

putting t = 0;

$x(0)={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)dt}$

Explanation:

We know that $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{x}\left(\mathrm{t}\right)\cdot \mathrm{d}\mathrm{t}=\mathrm{X}\left(0\right)$

Also we know

Where $X\left(0\right)=1$

$\therefore 2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left(\frac{\mathrm{s}\mathrm{i}\mathrm{n}2\pi \mathrm{t}}{\pi \mathrm{t}}\right)\cdot \mathrm{d}\mathrm{t}=2\mathrm{X}\left(0\right)=2\left(1\right)=2$

​

$X_1\left(j\omega \right)$ and $X_2\left(j\omega \right)$ are not conjugate symmetric, hence $x_1\left(t\right),x_2\left(t\right)$ are not real

We have,

              $y\left(t\right)=\frac{dx\left(t\right)}{dt}$........eq(i)

Since, x(t) is a Differentiable non - constant even function.

so, x(t) = x(-t) and its Fourier transform X(ω) = X(-ω)

Taking Fourier transform of eq(i)

⇒ Y(ω) = jω X(ω)

Y(-ω) = j(-ω)X(-ω)

Y(-ω) = -jω X(ω)

Y(-ω) = -Y(ω) 

So, Y(ω) is odd and imaginary.

Since, x(t) is differentiable, non-constant,  even function. So, its frequency response will be real i.e., X(ω) is real.

Hence, the correct option is (2)

Concept:

Duality Property

If $\mathrm{x}(\mathrm{t})\stackrel{\mathrm{F}.\mathrm{T}.}{\leftrightarrow }\mathrm{x}(\omega )$

then $\mathrm{x}(\mathrm{t})\stackrel{\mathrm{F}.\mathrm{T}.}{\leftrightarrow }2\pi \mathrm{x}(-\omega )$

Fourier time of signal x(t) is defined as,

$x(\omega )={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t).e^{-j\omega t}dt}$

putting t = 0;

$x(0)={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)dt}$

Explanation:

We know that $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{x}\left(\mathrm{t}\right)\cdot \mathrm{d}\mathrm{t}=\mathrm{X}\left(0\right)$

Also we know

Where $X\left(0\right)=1$

$\therefore 2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\left(\frac{\mathrm{s}\mathrm{i}\mathrm{n}2\pi \mathrm{t}}{\pi \mathrm{t}}\right)\cdot \mathrm{d}\mathrm{t}=2\mathrm{X}\left(0\right)=2\left(1\right)=2$

​

$X_1\left(j\omega \right)$ and $X_2\left(j\omega \right)$ are not conjugate symmetric, hence $x_1\left(t\right),x_2\left(t\right)$ are not real

x(t) can be expressed as the product of complex sinusoid e^j10t and rectangular pulse z(t) defined as:

$z\left(t\right)=\{\begin{array}{cc}1& \left|t\right|\le \pi \\ 0& 0,otherwise\end{array}$

$z\left(t\right)\stackrel{F.T.}{\leftrightarrow }Z\left(j\omega \right)=\frac{2}{\omega }\sin \left(\pi \omega \right)$

Since x(t) = ej10t z(t)

Using the frequency shift property

$e^{j10t}z\left(t\right)\stackrel{F.T.}{\leftrightarrow }z\left(j^{\left(\omega -10\right)}\right)$

$x\left(t\right)\stackrel{F.T.}{\leftrightarrow }\left(\frac{2}{\omega -10}\right)\sin \left(\left(\omega -10\right)\pi \right)$

at ω = 13

$\left(\frac{2}{3}\right)\left(\sin 3\pi \right)$

$\Rightarrow \left(-1\right)\left(0\right)$

From the symmetry properties of Fourier transform, we know that a real valued signal has magnitude spectrum that is even and a phase spectrum that is odd. This is true for the given $\left|\mathrm{X}{\mathrm{e}}^{\mathrm{j}\omega }\right|\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{\angle }\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)$. Thus, $\mathrm{x}\left[\mathrm{n}\right]$ is real valued.

Now, if $\mathrm{x}\left[\mathrm{n}\right]$ is an even function, then by symmetry properties $\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)$ must be real and even for a real $\mathrm{x}\left[\mathrm{n}\right]$. However, since $\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)=\left|\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)\right|{\mathrm{e}}^{-\mathrm{j}2\omega },\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)$ is not a real valued function for all $\omega$. Consequently, $\mathrm{x}\left[\mathrm{n}\right]$ is not even.

Finally, for a periodic $\mathrm{x}\left[\mathrm{n}\right]$, the Fourier transform is zero except for impulses located at integer multiples of fundamental frequency. This not true for $\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)$. Thus, $\mathrm{x}\left[\mathrm{n}\right]$ is aperiodic. For aperiodic discrete signal, the Parseval’s theorem is

$\sum _{\mathrm{n}=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|\mathrm{x}\left[\mathrm{n}\right]\right|}^2=\frac{1}{2\pi }\underset{2\pi }{\overset{}{\int }}{\left|\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)\right|}^2\mathrm{d}\omega$ 

We have

$\begin{array}{l}\mathrm{X}\left(-\mathrm{j}\omega \right)=\sum _{\mathrm{k}=-\mathrm{\infty }}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^{\left|\mathrm{k}\right|}\delta \left(-\omega -\frac{\mathrm{k}\pi }{4}\right)\\ \sum _{\mathrm{k}=-\mathrm{\infty }}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^{\left|\mathrm{k}\right|}\delta \left(\omega +\frac{\mathrm{k}\pi }{4}\right)\end{array}$

$=\mathrm{X}\left(\mathrm{j}\omega \right)$ as $\mathrm{k}$ varies from $-\mathrm{\infty }\mathrm{t}\mathrm{o}\mathrm{\infty }$ 

$\Rightarrow \mathrm{X}\left(-\mathrm{j}\omega \right)=\mathrm{X}\left(\mathrm{j}\omega \right)$ 

Thus $\mathrm{X}\left(\mathrm{j}\omega \right)$ is even

And $\mathrm{X}\left(\mathrm{j}\omega \right)$ is real

$\Rightarrow \mathrm{X}\left(\mathrm{j}\omega \right)$ is real and even

$\Rightarrow \mathrm{x}\left(\mathrm{t}\right)$ is real and even

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.