signalsignal processingsignal-theory
I have the following problem: according to my exercise text I have to calculate a signal in the frequency domain. I found out that I could use a the folding rule for time-discrete LTI systems and have to calculate:
$$Y(e^{j\omega}) = \int_0^{2\pi}{X(e^{j\alpha})*X(e^{j\alpha+\omega})\frac{d\alpha}{2\pi}}$$
$$X(e^{j\omega}) = \sqrt{2}\sum_{-\infty}^\infty rect(\frac{\omega + 2\pi k}{\pi/2})$$
The result should be \$1-\frac{|\omega|}{\pi}\$. But why?
So, in general case, the electric field should be kind of: $$E_3=A\sqrt{\frac{1+b\cos(2\omega_h t)}{2}}\text{e}^{j\omega_c t}$$ You may worry about the sign '\$\pm\$'before the square root. That would be a very good consideration and we need to extend our solution into a generic expression. We may notice the amplitude factor could be extended as: $$\pm\sqrt{\frac{1+b\cos(2\omega_h t)}{2}}\\ =\pm\sqrt{\frac{1+b[2\cos^2(\omega_h t)-1]}{2}}\\ =\cos(\omega_h t)\cdot\sqrt{b+\frac{1-b}{2\cos^2(\omega_h t)}} $$ It is easy to verify that for \$b=1\$ the above expression is just \$\cos(\omega_ht)\$ as we desired. So we can rewrite \$E_3\$ as: $$ E_3=\frac{A\cdot\beta(t)}{2}\cdot\bigg[\text{e}^{j(\omega_c-\omega_h)t}+\text{e}^{j(\omega_c+\omega_h)t}\bigg]\;, $$ where \$\beta(t)=\sqrt{b+\frac{1-b}{2\cos^2(\omega_h t)}}\$. Maybe...Maybe... You also worry about the singularities of \$\beta(t)\$ $$\cos(\omega_ht)\rightarrow0\quad\Rightarrow\quad\beta(t)\rightarrow\infty\;.$$ To cure this problem, I introduce a prescriptive parameter \$\epsilon\$, it could be an arbitrary small positive number (like 0.01,0.001) in the denominator for avoiding the divergence. $$ \beta(t)\rightarrow\beta'(t)=\sqrt{b+\frac{1-b}{2\cos^2(\omega_h t)+\epsilon}} $$ My next step is to find the Fourier spectrum of '\$\cos(\omega_h t)\sqrt{b+\frac{1-b}{2\cos^2(\omega_h t)}}\$' between \$ -\omega_h\$ to \$\omega_h\$--because we already have the phase '\$\text{e}^{j\omega_c t}\$'in our expression. This will be useful for getting monochrome modes.
Here are the pictures I made
for b=1
for b=0.5
$$-------------------------------------$$ Here is the code for Matlab
Th=20; %% period of omega_h
omegah=2*pi/Th;%%% we define \omegah via defining Th
N=90; %% divide the time interval (one period) into N equal parts.
L=100; %% Length of time interval interms of numbers of periods.
tt=Th/N*[1:L*N]; %% we generate a time vector including 3 periods.
b=0.5;
amph=cos(omegah*tt).*sqrt(b+(1-b)*power((2*cos(omegah*tt).*cos(omegah*tt)),-1)); %% this is our factor in the amplitude.
plot(tt,amph) %%% the picture of the amplitude including 'L' periods.
ttrial=Th/N*[1:3*N];
amphtrial=cos(omegah*ttrial).*sqrt(b+(1-b)*power((2*cos(omegah*ttrial).*cos(omegah*ttrial)),-1));
plot(ttrial,amphtrial) %%% the picture of the amplitude including '3' periods, this picture just helps you tell the pattern of the mode easily.
%%
FThh=fft(amph);
Homg=fftshift(FThh); %%% if you were not familiar of usage of commands 'fft' and 'fftshift' check them in the matlab help, they are extremely useful!!!
angleomg=-pi*N/Th:2*pi/(L*Th):pi*N/Th-2*pi/(L*Th);
plot(angleomg,abs(Homg))
title('b=0.5')
xlabel('\omega')
ylabel('E')
Apply the continuous to discrete time equation you've already given. The discrete time domain is itself a sampled version of the continuous time, sampled by the function \$x_d[n] = x(nT_s)\$
Best Answer
With the given \$X(e^{jw})\$ you won't get that answer.
Graphical proof:
If we try to plot \$X(e^{jw})\$, it will be a train of pulses with pulse width = \$\pi/2\$ and amplitude = \$\sqrt2\$ . And pulse will be centered at integer multiple of \$2\pi\$ as shown in the figure.
Calculating the value of \$Y\$ at \$w=0\$,
$$Y(e^{j\omega})|_{w=0} = \int_0^{2\pi}{X^2(e^{j\alpha})\frac{d\alpha}{2\pi}}$$
This is the \$\frac{1}{2\pi}\times\$ area under \$X^2(e^{j\omega})\$ from \$0\$ to \$2\pi\$.
$$Y(e^{j\omega})|_{w=0} = \frac{1}{2\pi}\times (2\times\frac{\pi}{4} + 2\times\frac{\pi}{4}) = \frac{1}{2}$$
Which does not satisfy \$1-\frac{|\omega|}{\pi}\$. But a slight change in \$X(e^{jw})\$ can give you the result.
Changing the question:
If \$X(e^{jw})\$ was defined as follows, $$X(e^{j\omega}) = \sqrt{2}\sum_{-\infty}^\infty rect(\frac{\omega + 2\pi k}{\pi})$$
Then, \$X(e^{jw})\$ will be a pulse train similar to previous one but the pulse width would be \$\pi\$. See the figure given below.
The black represents the \$X(e^{j\alpha})\$ and blue represents the \$X(e^{j\alpha+w})\$. (assume that 0 < w < \pi)
So the product
$${X(e^{j\alpha})\times X(e^{j\alpha+\omega})}$$
will be zero everywhere (0 to \$2\pi\$) except in the region marked by gray color. And the amplitude of this product will be \$\sqrt2\times \sqrt2 =2\$.
So the integral $$\int_0^{2\pi}{X(e^{j\alpha})*X(e^{j\alpha+\omega})d\alpha}$$
will be given by the area under this product curve. Which will be summation of area under two gray-rectangles:
$$\int_0^{2\pi}{X(e^{j\alpha})*X(e^{j\alpha+\omega})d\alpha} = (\frac{\pi}{2}-w)\times 2 + \frac{\pi}{2}\times 2 = 2\pi - 2w$$
If you do this for a right shift of signal, which corresponds to negative \$w\$, you will get the same result.
So we can write:
$$Y(e^{jw}) = \frac{1}{2\pi}\times (2\pi - |2w|) = 1-\frac{|w|}{\pi}$$