Electronic – Sampling Theorem

For any periodic signal you get

$$\int_{-\infty}^{\infty}|x(t)|^2dt=\infty$$

i.e. the integral does not converge, and, consequently, the energy is infinite. The power is finite and can be computed from the following formula (which differs from yours by a factor of \$\frac12\$):

$$\overline{x^2(t)}=\frac{1}{2T}\int_{-T}^T|x(t)|^2dt$$

Note that \$\cos^2 x=\frac12 (1+\cos(2x))\$ and \$\sin^2 x=\frac12 (1-\cos(2x))\$. So after integrating over one period, the terms with double frequency and also the cross-term cancel out. So you finally get for the power

$$\overline{x^2(t)}=10^2\cdot \frac12 + 5^2\cdot\frac12=62.5$$

Note that the actual value of \$T\$ is irrelevant. This is a good thing because the value you got is wrong.

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:

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