# Folding of a two signals

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?

• What is Y(e^jw) and X(e^ja) ? – nidhin Mar 15 '15 at 12:07
• Y(e^jw) is the output signal and X(e^ja) is the input signal (both in frequency domain) – m79 Mar 15 '15 at 15:10

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:

$$Y(e^{jw}) = \frac{1}{2\pi}\times (2\pi - |2w|) = 1-\frac{|w|}{\pi}$$

• Thank you for your detailed answer! I think I understand now why the result is 1-|w|/pi – m79 Mar 18 '15 at 8:35