# Calculate the integral using the Fourier coefficients

• MHB
• mathmari
Gold Member
MHB
Hey! 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)

$$\frac{1}{T_0}\int_{T_0}|x(t)|^2\, dt =\sum_{k=-N}^{+N}|X_k|^2 \\ \int_{T_0}|x(t)|^2\, dt =T_0\sum_{k=-N}^{+N}|X_k|^2 =2\left\{\left(\frac{2}{3}\right )^2 + 2\left [\left(\frac{1}{3}\right )^2+\left(\frac{1}{3}\right )^2+\left(\frac{1}{4}\right )^2+1\right ]\right\}$$
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)