Signals and Systems Test 3

Concept:

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

\(1\;\mathop \leftrightarrow \limits^{F.T} 2\pi \;\delta \left( \omega \right)\)

\({e^{j{\omega _0}t}} \cdot 1\;\mathop \leftrightarrow \limits^{FT} 2\pi \;\delta \left( {\omega - {\omega _0}} \right)\)

\({e^{ - j{\omega _0}t}} \cdot 1\;\mathop \leftrightarrow \limits^{FT} 2\pi \;\delta \;\left( {\omega + {\omega _0}} \right)\)

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  

\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1\;for}&{ - 1 \le t \le 1}\\ 0&{otherwise} \end{array}} \right.\)

Fourier transform:

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

\(=\mathop \smallint \limits_{ - 1 }^1 {e^{ - j\omega t}}(1)dt\)

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

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

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

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

Calculation:

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

If f(n) ↔ F(ω)

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

So, the inverse DTFT of F(ω - π) is

e^jπn f(n), i.e

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

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

Let,

\({x_1}\left( t \right)\mathop \leftrightarrow \limits^{F.T} {x_1}\left( {j\omega } \right)\)

\({x_1}\left( {3t} \right)\mathop \leftrightarrow \limits^{F.T} \frac{1}{3}{x_1}\left( {\frac{{j\omega }}{3}} \right)\)

Similarly

\({x_2}\left( t \right)\mathop \leftrightarrow \limits^{F.T} \;{x_2}\left( {j\omega } \right)\)

\({x_2}\left( {3t} \right)\mathop \leftrightarrow \limits^{F.T} \;\frac{1}{3}{x_2}\left( {\frac{{j\omega }}{3}} \right)\)

y₂(t) = x₁(3t) * x₂(3t)

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

\({y_2}\left( t \right)\mathop \leftrightarrow \limits^{F.T} \;{y_2}\left( {j\omega } \right) = \frac{1}{9}{x_1}\left( {\frac{{j\omega }}{3}} \right){x_2}\left( {\frac{{j\omega }}{3}} \right)\)         ---(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)

\(H\left( j\omega \right)=\underset{-\infty }{\overset{\infty }{\mathop \int }}\,h\left( t \right){{e}^{-j\omega t}}dt\)

\(=\underset{-\infty }{\overset{\infty }{\mathop \int }}\,{{e}^{-2t}}{{e}^{-j\omega t}}dt\)

\(=\underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{-\left( 2+j\omega \right)t}}dt\)

\(=\frac{1}{2+j\omega }\)

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

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

x(t) = 2 cos 2t

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

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

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

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

Output \(=\frac{1}{2\sqrt{2}}\text{ }\!\!~\!\!\text{ }2\cos \left( 2t-0.25~\pi \right)\)

\(=\frac{1}{\sqrt{2}}\cos \left( 2t-0.25~\pi \right)\)

= 2^-0.5 cos (2t – 0.25 π)

Concept:

\(f\left( n \right)\overset{DTFT}{\mathop{\leftrightarrow }}\,F\left( \omega \right)\)

\(n~f\left( n \right)\leftrightarrow \frac{jdF\left( \omega \right)}{d\omega }\)

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

\(F\left( \omega \right) = \mathop \sum \limits_{ - \infty }^\infty f\left( n \right){e^{ - j\omega n}}\)

Now, the DTFT of ‘nf(n)’ will be:

\(j\frac{d~F\left( \omega \right)}{d\omega }= \mathop \sum \limits_{ - \infty }^\infty nf\left( n \right){e^{ - j\omega n}}\)

At, ω = 0

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

Calculation:

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

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

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

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

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

From Equation (1),

\(\mathop \sum \limits_{ - \infty }^\infty\,nf\left( n \right)u\left( n \right)=\mathop \sum \limits_{ 0 }^\infty\,nf\left( n \right)\)

\(=\mathop \sum \limits_{ 0 }^\infty\,n{{\left( \frac{1}{3} \right)}^{n}}u\left( n \right)=j\frac{dF\left( 0 \right)}{d\omega }\)

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

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

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

\(=\frac{3}{4}=0.75\)