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I have come across this question that asks to find Fourier series coefficients of the following signal. $$1+\sin (\omega_0 t) + \cos (\omega_0 t) + \cos (2\omega_0 t + \pi / 4) $$

In my intuition, the signal is already in Fourier series form, and the question asks just to find the trigonometric Fourier coefficients.

I started breaking down the original signal and rearranging, which gives: $$1+\cos(\omega_0 t) + \frac{1}{\sqrt 2}\cos (2\omega_0 t) + \sin(\omega_0 t) - \frac{1}{\sqrt 2}\sin(2\omega_0 t)$$

While figuring out the pattern in the signal, the coefficient of cosine term is always 1, while the coefficient of sine term alternates in sign. So, in more general terms, $$1+\sum_{n=1}^{2}{(\frac{1}{\sqrt 2})^{n-1}.\cos(n\omega_0 t) + (-\frac{1}{\sqrt 2})^{n-1}\sin(n\omega_0 t)}$$

With an analogy to the trigonometric Fourier series, the coefficients are found out to be $$a_0 = 1, a_n = (\frac{1}{\sqrt 2})^{n-1}, b_n=(-\frac{1}{\sqrt 2})^{n-1}$$

Is this what am I required to do when asked to find the Fourier coefficients? Or should I take that signal and then follow the whole procedure of finding a0, an and bn using the Euler's coefficient formula?

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    \$\begingroup\$ The idea of this website is people ask questions and other people answer them. Then, everybody else who may have the same, or similar, question can see both the original question and the answer. But what you've done is re-written your question substantially, using the material provided in an answer. That benefits no one. I'll remove my answer. \$\endgroup\$ – Chu Sep 16 '18 at 11:31
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Because as you rightly point out it is already given in something very close to a Fourier series, there is no need to perform the whole procedure. Just rearrange to get it in exactly the right format.

I think you've over-cooked it with the summation pattern. There are just three frequency terms (DC, \$\omega_0\$ and \$2\omega_0\$) so you can just list them individually.

You can rearrange your second equation thusly:

$$ 1 + cos(\omega_{0}t) + sin(\omega_{0}t) + \frac{1}{\sqrt{2}}cos(2\omega_{0}t) - \frac{1}{\sqrt{2}}sin(2\omega_{0}t) $$

And then read the coefficients straight off:

$$ a_0 = 1, a_1 = 1, a_2 = \frac{1}{\sqrt{2}}, b_1 = 1, b_2 = -\frac{1}{\sqrt{2}} $$

Note some texts prefer \$f(x) = \frac{1}{2}a_0 + ...\$ as their Fourier series, in which case \$a_0 = 2\$.

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