Concept:

The Fourier transform of a constant signal is an impulse function, i.e.

$1\stackrel{F.T}{\leftrightarrow }2\pi \delta \left(\omega \right)$

$e^{j{\omega }_{0}t}\cdot 1\stackrel{FT}{\leftrightarrow }2\pi \delta \left(\omega -{\omega }_0\right)$

$e^{-j{\omega }_{0}t}\cdot 1\stackrel{FT}{\leftrightarrow }2\pi \delta \left(\omega +{\omega }_0\right)$

Application:

Acos ω₀ t can be written as:

$A\cos {\omega }_{0}t=\frac{A}{2}{e}^{j{\omega }_{0}t}+\frac{A}{2}{e}^{-j{\omega }_{0}t}$

∴ The Fourier Transform of cos ω₀ t will be:

$A\cos {\omega }_{0}t\leftrightarrow \frac{A}{2}\cdot 2\pi \delta \left(\omega -{\omega }_0\right)+\frac{A}{2}\cdot 2\pi \delta \left(\omega +{\omega }_0\right)$

$\cos {\omega }_{0}t\leftrightarrow A\pi \left[\delta \left(\omega -{\omega }_0\right)+\delta \left(\omega +{\omega }_0\right)\right]$       ---(1)

Given:

X(jω₀) = 10 [δ(ω – 4π) + δ(ω + 4π)]        ---(2)

Comparing equation (1) and (2), we get:

ω₀ = 4π

2πf₀ = 4π

f₀ = 2 Hz  

$x\left(t\right)=\{\begin{array}{cc}1for& -1\le t\le 1\\ 0& otherwise\end{array}$

Fourier transform:

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

$=\underset{-1}{\overset{1}{\int }}{e}^{-j\omega t}(1)dt$

$=\frac{1}{-j\omega }{\left[e^{-j\omega }\right]}_{-1}^1$

$X(j\omega )=\frac{1}{-j\omega }\left(e^{-j\omega }-e^{j\omega }\right)$

X(jω) is zero at ω = π and ω = 2π.

Concept:

If the DTFT of a signal f(n) ↔ F(ω), then from the frequency shifting property

 $e^{j{\omega }_{o}n}f\left(n\right)\leftrightarrow F(\omega -{\omega }_o)$

Calculation:

Given, f(n) = {a, b, c, d}

If f(n) ↔ F(ω)

Then, $e^{j\pi n}f\left(n\right)\leftrightarrow F(\omega -\pi )$

So, the inverse DTFT of F(ω - π) is

e^jπn f(n), i.e

$\Rightarrow {(-1)}^{n}f\left(n\right)=\{f\left(0\right),(-1)f\left(1\right),f\left(2\right),(-1)f\left(3\right)\}$

The frequency response

$\begin{array}{l}H\left(e^{j\omega }\right)=\frac{y\left(e^{j\omega }\right)}{x\left(e^{j\omega }\right)}\\ =\frac{2}{1-\frac{3}{4}{e}^{-j\omega }+\frac{1}{8}{e}^{-j2\omega }}\\ =\frac{2}{\left(1-\frac{1}{2}{e}^{-j\omega }\right)\left(1-\frac{1}{4}{e}^{-j\omega }\right)}\\ =\frac{4}{1-\frac{1}{2}{e}^{-j\omega }}-\frac{2}{1-\frac{1}{4}{e}^{-j\omega }}\end{array}$

Taking inverse Fourier transform

Concept:

The Fourier Transform of a signal x(t) is defined as:

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

Application:

For the given signal, the Fourier Transform is defined as:

$X\left(j\omega \right)=\underset{0}{\overset{1}{\int }}4\cdot e^{-j\omega t}dt+\underset{1}{\overset{2}{\int }}2\cdot e^{-j\omega t}dt$ 

$={\frac{4e^{-j\omega t}}{\left(-j\omega \right)}|}_0^1+{\frac{2e^{-j\omega t}}{\left(-j\omega \right)}|}_1^2$ 

$=\frac{-4}{j\omega }\left[e^{-j\omega }-1\right]-\frac{2}{j\omega }\left[e^{-2j\omega }-e^{-j\omega }\right]$ 

$=\frac{4-4e^{-j\omega }-2e^{-2j\omega }+2e^{-j\omega }}{j\omega }$ 

$=\frac{4-2e^{-j\omega }-2e^{-j2\omega }}{j\omega }$ 

At ω = π,

$X\left(j\pi \right)=\frac{4-2e^{-j\pi }-2e^{-j2\pi }}{2\pi }$ 

$=\frac{4-2\left(-1\right)-2}{j\pi }=\frac{4}{j\pi }$ 

∴ The required value of jπ × X(jπ) will be:

$j\pi \times X\left(j\pi \right)=\frac{4}{j\pi }\times j\pi =4$ 

Concept:

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

$f\left(n\right)=\frac{1}{2\pi }\underset{-\frac{T}{2}}{\stackrel{\frac{T}{2}}{\int }}F\left(\omega \right)e^{j\omega n}d\omega$    ---(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 a rectangular pulse is a sinc pulse, i.e.

$rect\left(t\right)\stackrel{F.T}{\leftrightarrow }\frac{2\sin \omega }{\omega }$ 

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:

$x\left(t\right)\ast y\left(t\right)\stackrel{F.T}{\leftrightarrow }X\left(j\omega \right)Y\left(j\omega \right)$ 

From the definition Fourier transform, we have:

$x\left(t\right)\ast y\left(t\right)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X\left(j\omega \right)Y\left(j\omega \right)e^{j\omega t}d\omega$ 

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

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X\left(j\omega \right)Y\left(j\omega \right)d\omega =2\pi {\left[x\left(t\right)\ast y\left(t\right)\right]}_{t=0}$ 

Using Equation (1):

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X\left(j\omega \right)\cdot \frac{2\sin \omega }{\omega }{e}^{2j\omega }d\omega =2\pi {\left(x\left(t\right)\ast y\left(t\right)\right)}_{t=0}$ 

∴ 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:

$x\left(t\right)\ast y\left(t\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(\tau \right)y\left(t-\tau \right)d\tau$ 

$x\left(t\right)\ast y{\left(t\right)|}_{t=0}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\left(\tau \right)y\left(-\tau \right)d\tau$ 

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

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

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

= 3.5

$\therefore \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X\left(j\omega \right)\cdot \frac{2\sin \omega }{\omega }{e}^{j2\omega }d\omega =2\pi \times 3.5$ 

= 7π

= 21.991

Let,

$x_1\left(t\right)\stackrel{F.T}{\leftrightarrow }{x}_1\left(j\omega \right)$

$x_1\left(3t\right)\stackrel{F.T}{\leftrightarrow }\frac{1}{3}{x}_1\left(\frac{j\omega }{3}\right)$

Similarly

$x_2\left(t\right)\stackrel{F.T}{\leftrightarrow }{x}_2\left(j\omega \right)$

$x_2\left(3t\right)\stackrel{F.T}{\leftrightarrow }\frac{1}{3}{x}_2\left(\frac{j\omega }{3}\right)$

y₂(t) = x₁(3t) * x₂(3t)

Since the convolution in the time domain is the multiplication in the frequency domain, we can write:

$y_2\left(t\right)\stackrel{F.T}{\leftrightarrow }{y}_2\left(j\omega \right)=\frac{1}{9}{x}_1\left(\frac{j\omega }{3}\right)x_2\left(\frac{j\omega }{3}\right)$         ---(1)

Given: y₁(t) = x₁(t) * x₂(t)

$y_1\left(j\omega \right)=x_1\left(j\omega \right)x_2\left(j\omega \right)$

Substituting ω → ω/3 in the above, we can write:

$y_1\left(\frac{j\omega }{3}\right)=x_1\left(\frac{j\omega }{3}\right)x_2\left(\frac{j\omega }{3}\right)$       ---(2)

Using equation (2), Equation (1) can now be written as:

$y_2\left(j\omega \right)=\frac{1}{9}{y}_1\left(\frac{j\omega }{3}\right)$

Taking the inverse Fourier transform, we get:

$y_2\left(t\right)=\frac{1}{3}{y}_1\left(3t\right)$

$\therefore \frac{y_2\left(t\right)}{y_1\left(3t\right)}=\frac{1}{3}=0.333$

h(t) = e^-2t u(t)

$H\left(j\omega \right)=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}h\left(t\right)e^{-j\omega t}dt$

$=\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{e}^{-2t}{e}^{-j\omega t}dt$

$=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{e}^{-(2+j\omega )t}dt$

$=\frac{1}{2+j\omega }$

$\left|H\left(j\omega \right)\right|=\sqrt{\frac{1}{4+{\omega }^2}}$

$\mathrm{\angle }H\left(j\omega \right)=-{\tan }^{-1}\left(\frac{\omega }{2}\right)$

x(t) = 2 cos 2t

The phase response at ω = 2 rad/sec is:

$\mathrm{\angle }H(j\omega )=-\frac{\pi }{4}=-0.25\pi$

The magnitude response at ω = 2 rad/sec is:

$|H(j\omega ){|}_{\omega =2}=\frac{1}{2\sqrt{2}}$

Output $=\frac{1}{2\sqrt{2}}2\cos (2t-0.25\pi )$

$=\frac{1}{\sqrt{2}}\cos (2t-0.25\pi )$

= 2^-0.5 cos (2t – 0.25 π)

Concept:

$f\left(n\right)\stackrel{DTFT}{\leftrightarrow }F\left(\omega \right)$

$nf\left(n\right)\leftrightarrow \frac{jdF\left(\omega \right)}{d\omega }$

Also, the discrete-time Fourier Transform of a sequence f(n) is given by:

$F\left(\omega \right)=\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(n\right)e^{-j\omega n}$

Now, the DTFT of ‘nf(n)’ will be:

$j\frac{dF\left(\omega \right)}{d\omega }=\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}nf\left(n\right)e^{-j\omega n}$

At, ω = 0

$j\frac{dF\left(0\right)}{d\omega }=\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}nf\left(n\right)$     ---(1)

Calculation:

$f\left(x\right)={\left(\frac{1}{3}\right)}^{n}u\left(n\right)$

$F\left(\omega \right)=\frac{1}{1-\left(\frac{1}{3}\right)e^{-j\omega }}$

$n{\left(\frac{1}{3}\right)}^{n}u\left(n\right)\leftrightarrow j\frac{dF\left(\omega \right)}{d\omega }$

$j\frac{dF\left(\omega \right)}{d\omega }=j\left[\frac{\left(\frac{1}{3}\right)\left(e^{-j\omega }\right)(-j)}{{(1-\left(\frac{1}{3}\right)e^{-j\omega })}^2}\right]$

$=\frac{\frac{1}{3}{e}^{-j\omega }}{{(1-\frac{1}{3}{e}^{-j\omega })}^2}$

From Equation (1),

$\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}nf\left(n\right)u\left(n\right)=\sum _0^{\mathrm{\infty }}nf\left(n\right)$

$=\sum _0^{\mathrm{\infty }}n{\left(\frac{1}{3}\right)}^{n}u\left(n\right)=j\frac{dF\left(0\right)}{d\omega }$

$f\left(n\right)={\left(\frac{1}{3}\right)}^{n}u\left(n\right)$

$j\frac{dF\left(0\right)}{d\omega }=\frac{\frac{1}{3}}{{(1-\frac{1}{3})}^2}$

$=\frac{1}{3\times {\left(\frac{2}{3}\right)}^2}=\frac{1}{3}\times \frac{9}{4}$

$=\frac{3}{4}=0.75$