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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    – Chu
    Commented Sep 16, 2018 at 11:31

1 Answer 1


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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