Fourier series signal

[username]

Derive the expression for coefficients of Fourier series in exponential form for the sequence of rectangular pulses (with amplitude A, period T and duration ?) shown in this image:



Derive the expression for signal power depending on the coefficients of Fourier series.

Attempt:



For the case when k is equal to zero, the integral becomes



Average power is given by:




Average power is not dependent on Fourier series coefficients. Is this a mistake in evaluation? Is the signal power depending on the coefficients of Fourier series equal to the average power? How to evaluate the signal power depending on the coefficients of Fourier series?

[T3sl4co1l]

You found the power of the signal s(t), do it for a_k(t) = sin(n w t) thingys instead.  Then you get a series p_k.

Tim

[egonotto]

Hi,

your doubts are justified. But in this case there is no mistake.
It is a consequence of Parseval's identity.

In your case it states, that the sum of the average power of the elements of the Fourier series is the average power of s.

Fourier series are not easy at all.
Books:
Walter Rudin: Principles of Mathematical Analysis
Otto Forster: Analysis 1 (in german)


Best regards
egonotto

[orolo]

If you want to integrate the Fourier components, due having an orthonormal basis, you will be adding the series \$\sum a_k^2 = \sum_{k}\frac{1}{k^2}\, \sin^2(k\cdot t/2) \$, where in your case \$ t = w_0\cdot \theta\$.

Summing that series is easy, but a bit of a hassle. Use \$\sin^2 x = \frac{1- cos(2x)}{2}\$ to remove the square from the sine, and you get a classical series \$ \sum\frac{1}{2k^2} = \pi^2/12\$ and another Fourier series \$\sum\frac{1}{2k^2}\cos(k\cdot t)\$ which is just the integral of a sawtooh wave (a parabolic wave) evaluated at 't'. From there, computing the series is easy, and you arrive at the same value than you got using the simpler way.

Or, as said above, you can trust Parseval and save yourself the trouble.