Question

(b) Fos the pesiedic Siannl Showl ind the ketch the Omplitude and phase spedisa. la T -loT ge) -T -1 -1 Subject: Signal and system Book: signal & system by Alan V. Oppenheim Explain Every step of question

Please explain every step and writing must be accurate.

Answer 1

A signal $f(t)$ having a period $T_o$ can be expressed as

$f(t)=\sum_{k=-\infty}^\infty C_ke^{j\omega_ot}$ for $\omega_o=\frac{2\pi}{T_o}$

where $C_k$ is the complex Fourier series coefficient and is given by

$C_k=\frac{1}{T_o}\int_{T_o}f(t)e^{-jk\omega_ot}dt$ for $k\neq 0$

$C_0=\frac{1}{T_o}\int_{T_o}f(t)dt$

Part-1

For the first signal, the period is $T_0=10\pi$

$C_0=\frac{1}{10\pi}\int_{-5\pi}^{5\pi}f(t)dt=\frac{1}{10\pi}\int_{-\pi}^{\pi}dt=\frac{\pi-(-\pi)}{10\pi}=\frac{1}{5}$

The exponential Fourier series coefficient is given by

$C_k=\frac{1}{10\pi}\int_{-5\pi}^{5\pi}f(t)e^{-jk\frac{2\pi}{10\pi}t}dt$

$\Rightarrow C_k=\frac{1}{10\pi}\int_{-5\pi}^{5\pi}f(t)e^{-jk\frac{t}{5}}dt$

$\Rightarrow C_k=\frac{1}{10\pi}\left[\int_{-5\pi}^{\pi}0\cdot e^{-jk\frac{t}{5}}dt+\int_{-\pi}^{\pi}1\cdot e^{-jk\frac{t}{5}}dt+\int_{\pi}^{5\pi}0\cdot e^{-jk\frac{t}{5}}dt \right ]$

$\Rightarrow C_k=\frac{1}{10\pi}\left[0+\int_{-\pi}^{\pi}e^{\frac{-jk}{5}t}dt+0 \right ]$

$\Rightarrow C_k=\frac{1}{10\pi}\left[\frac{e^{\frac{-jk}{5}t}}{\frac{-jk}{5}} \right ]_{-\pi}^{\pi}$

$\Rightarrow C_k=\frac{1}{10\pi}\left[\frac{5j}{k} e^{\frac{-jk}{5}t} \right ]_{-\pi}^{\pi}$

$\Rightarrow C_k=\frac{j}{2k\pi}\left[ e^{\frac{-jk}{5}\pi}-e^{\frac{-jk}{5}(-\pi)} \right ]$

$\Rightarrow C_k=\frac{j}{2k\pi}\left[ e^{-j(\frac{k\pi}{5})}-e^{j(\frac{k\pi}{5})} \right ]$

$\Rightarrow C_k=\frac{j}{2k\pi}\left[ -2j\sin\left(\frac{k\pi}{5} \right ) \right ]$

$\Rightarrow C_k=\frac{1}{k\pi}\sin\left(\frac{k\pi}{5} \right )$

The amplitude and phase spectra are generated using MATLAB for $-50\leq k\leq50$

Below is the MATLAB code

k=-50:50;
C=zeros(1,length(k));
for m=1:length(k)
C(m)=1/(k(m)*pi)*sin(k(m)*pi/5);
end
C(k==0)=1/5;
figure(1)
subplot(2,1,1)
plot(k,abs(C))
xlabel('k')
ylabel('|C_k|')
title('Amplitude Spectrum')
subplot(2,1,2)
plot(k,angle(C)*180/pi)
xlabel('k')
ylabel('\angleC_k(deg)')
title('Phase Spectrum')

the Output plot is

Part-2

For the second signal, the time period is $T_o=\pi$

$C_0=\frac{1}{\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(t)dt=\frac{1}{\pi}\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{4}{\pi}dt=\frac{4}{\pi^2}\left[\frac{\pi}{4}-\frac{-\pi}{4} \right ]=\frac{2}{\pi}$

The exponential Fourier series coefficient is given by

$C_k=\frac{1}{\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(t)e^{-jk\frac{2\pi}{\pi}t}dt$

$\Rightarrow C_k=\frac{1}{\pi}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(t)e^{-2jkt}dt$

$\Rightarrow C_k=\frac{1}{\pi}\left[\int_{\frac{-\pi}{2}}^{-\frac{\pi}{4}}0\cdot e^{-2jkt}dt+\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{4}{\pi}te^{-2jkt}dt+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}0\cdot e^{-2jkt}dt \right ]$

$\Rightarrow C_k=\frac{1}{\pi}\left[0+\frac{4}{\pi}\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}te^{-2jkt}dt+0 \right ]$

$\Rightarrow C_k=\frac{4}{\pi^2}\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}te^{-2jkt}dt$

Now lets find $\int te^{-2jkt}dt$

Integrating by parts

$\int fg'=fg-\int f'g$

$f=t\rightarrow f'=1$

$g'=e^{-2jkt}\rightarrow g=-\frac{e^{-2jkt}}{2jk}$

$\int te^{-2jkt}dt=-\frac{te^{-2jkt}}{2jk}-\int-\frac{e^{-2jkt}}{2jkt}dt$

$\Rightarrow \int te^{-2jkt}dt=-\frac{te^{-2jkt}}{2jk}+\frac{1}{2jk}\int e^{-2jkt}dt$

$\Rightarrow \int te^{-2jkt}dt=-\frac{te^{-2jkt}}{2jk}-\frac{e^{-2jkt}}{4j^2k^2}$

$\Rightarrow \int te^{-2jkt}dt=\frac{jte^{-2jkt}}{2k}+\frac{e^{-2jkt}}{4k^2}$

$\Rightarrow \int te^{-2jkt}dt=\frac{(2jkt+1)e^{-2jkt}}{4k^2}$

So,

$C_k=\frac{4}{\pi^2}\left[\frac{(2jkt+1)e^{-2jkt}}{4k^2} \right ]_{\frac{-\pi}{4}}^{\frac{\pi}{4}}$

$\Rightarrow C_k=\frac{4}{\pi^2}\left[\frac{(2jk(\frac{\pi}{4})+1)e^{-2jk(\frac{\pi}{4})}}{4k^2}-\frac{(2jk(-\frac{\pi}{4})+1)e^{-2jk(-\frac{\pi}{4})}}{4k^2} \right ]$

$\Rightarrow C_k=\frac{4}{\pi^2}\left[\frac{(jk(\frac{\pi}{2})+1)e^{-jk(\frac{\pi}{2})}}{4k^2}-\frac{(-jk(\frac{\pi}{2})+1)e^{jk(\frac{\pi}{2})}}{4k^2} \right ]$

$\Rightarrow C_k=\frac{4}{\pi^2}\left[\frac{(jk(\frac{\pi}{2})+1)(-j)}{4k^2}-\frac{(-jk(\frac{\pi}{2})+1)(j)}{4k^2} \right ]$

$\Rightarrow C_k=\frac{4}{k^2\pi^2}\left[(jk(\frac{\pi}{2})+1)(-j)-(-jk(\frac{\pi}{2})+1)(j) \right ]$

$\Rightarrow C_k=\frac{4}{k^2\pi^2}\left[(k(\frac{\pi}{2})-j)-(k(\frac{\pi}{2})+j) \right ]$

$\Rightarrow C_k=\frac{4}{k^2\pi^2}\left[-2j \right ]$

$\Rightarrow C_k=\frac{-8j}{k^2\pi^2}$

The amplitude and phase spectra are generated using MATLAB for $-10\leq k\leq10$

k=-10:10;
C=zeros(1,length(k));
for m=1:length(k)
C(m)=-8*j/(k(m)^2*pi^2);
end
C(k==0)=2/pi;
figure(1)
subplot(2,1,1)
plot(k,abs(C))
xlabel('k')
ylabel('|C_k|')
title('Amplitude Spectrum')
subplot(2,1,2)
plot(k,angle(C)*180/pi)
xlabel('k')
ylabel('\angleC_k(deg)')
title('Phase Spectrum')

Note: For the sake of proper legibility in the graphs, they are generated using MATLAB. You can refer them and create your hand-drawn plots if required. You can also change the range of 'k' if required just by modifying line-1 in the above given codes.