MHB Calculate the integral using the Fourier coefficients

AI Thread Summary
The discussion focuses on calculating the integral of the square of a periodic signal using its Fourier coefficients. The user confirms the relationship between the integral of the squared signal and the sum of the squares of its Fourier coefficients, questioning their calculation due to an unexpected result. Another participant suggests a correction in the summation limits of the Fourier coefficients and reinforces that the integral of the squared signal can be equated to the integral of the squared magnitude of the signal. The conversation highlights the importance of correctly applying Fourier series properties in signal analysis. The final conclusion emphasizes the relationship between the Fourier coefficients and the integral of the signal's square.
mathmari
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Hey! :o

A real periodic signal with period $T_0=2$ has the Fourier coefficients $$X_k=\left [2/3, \ 1/3e^{j\pi/4}, \ 1/3e^{-i\pi/3}, \ 1/4e^{j\pi/12}, \ e^{-j\pi/8}\right ]$$ for $k=0,1,2,3,4$.
I want to calculate $\int_0^{T_0}x^2(t)\, dt$.

I have done the following:

It holds that $$\frac{1}{T_0}\int_{T_0}|x(t)|^2\, dt=\sum_{k=-\infty}^{+\infty}|X_k|^2$$ right? (Wondering)

Then do we get $$\int_{T_0}|x(t)|^2\, dt=2\sum_{k=-\infty}^{+\infty}|X_k|^2=2\left [\left(\frac{2}{3}\right )^2+\left(\frac{1}{3}\right )^2+\left(\frac{1}{3}\right )^2+\left(\frac{1}{4}\right )^2+1\right ]$$ But the result that I get is not one of the choices. So have I done something wrong? (Wondering)
 
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Hey mathmari!

Shouldn't it be:
$$\frac{1}{T_0}\int_{T_0}|x(t)|^2\, dt
=\sum_{k=-N}^{+N}|X_k|^2 \\
\int_{T_0}|x(t)|^2\, dt
=T_0\sum_{k=-N}^{+N}|X_k|^2
=2\left\{\left(\frac{2}{3}\right )^2 + 2\left [\left(\frac{1}{3}\right )^2+\left(\frac{1}{3}\right )^2+\left(\frac{1}{4}\right )^2+1\right ]\right\}$$
(Wondering)

Oh, and since it's given that $x(t)$ is a real signal, we can write $\int_{T_0}|x(t)|^2\, dt = \int_{T_0}x(t)^2\, dt$, can't we? (Wondering)
 
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