Signals and Systems Test 3

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

Output, $y\left(t\right)=\left(e^{-2t}-e^{-4t}\right)u\left(t\right)$

By applying the Fourier transform,

$Y\left(\omega \right)=\frac{1}{j\omega +2}-\frac{1}{j\omega +4}=\frac{j\omega +4-j\omega -2}{\left(j\omega +2\right)\left(j\omega +4\right)}$ 

$=\frac{2}{\left(j\omega +2\right)\left(j\omega +4\right)}$ 

We know Y(ω) = X(ω) H(ω)

$X\left(\omega \right)=\frac{Y\left(\omega \right)}{H\left(\omega \right)}=\frac{\frac{2}{\left(j\omega +2\right)\left(j\omega +4\right)}}{1/\left(j\omega +2\right)}$ 

$=\frac{2}{j\omega +4}=2e^{-4t}u\left(t\right)$ 

Given that, x(t) = e^-2|t| sgn(t)

$sgn\left(t\right)=\begin{array}{c}1,t>0\\ -1,t<0\end{array}$

x(t) = e^-2t t > 0

= -e^-2t t < 0

$x\left(t\right)=e^{-2t}u\left(t\right)-e^{-2t}u\left(-t\right)$

By applying Fourier transform,

$X\left(\omega \right)=\frac{1}{2+j\omega }-\frac{1}{2-j\omega }$

$X\left(\omega \right)=\frac{-j2\omega }{{\omega }^2+4}$ 

Step 1:

f(t) = e^-t u(t)

$E_f\left(t\right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left|f\left(t\right)\right|}^{2}dt=\underset{0}{\overset{\mathrm{\infty }}{\int }}{\left(e^{-t}\right)}^{2}dt$

$=\frac{1}{2}=0.5$

$E_f\left(\omega \right)for\left|\omega \right|\le 5=\frac{1}{2\pi }\underset{-5}{\overset{5}{\int }}\frac{1}{1+{\omega }^2}d\omega$

$=\frac{1}{2\pi }{\left({\tan }^{-1}\omega \right)}_{-5}^5=\frac{2}{2\pi }{\left({\tan }^{-1}\omega \right)}_0^5$

$=\frac{1}{\pi }\left({\tan }^{-1}5-{\tan }^{-1}0\right)$

$=\frac{1.37}{\pi }=0.437$

Percentage of energy contained $=\frac{0.437}{0.5}\times 100$ 

= 87.433%