I saw many solved examples about this topic but again I coudn't come up with any solutions about this question. How can I find the Fourier Series coefficients of the following signal ?

$x(t)=2 \cos(3\pi t) + \sin(100\pi t+\frac{\pi }{3})$

I know that;
$\cos(θ)=\frac12 (e^{jθ} + e^{−jθ})$ and,
$\sin(θ)=\frac{1}{2j} (e^{jθ} - e^{−jθ})$

I also know that I should use $$x(t) = \sum_{-\infty}^{\infty} a_{k} e^{jk(2\pi/T)t}$$

But I'm having trouble to define the fundamental period $T$ and the relation between sinusoidal terms and coefficients $a_k$, to sum all things together. Thanks to everyone...

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    $\begingroup$ do you know that $$ a_k = \frac1T \int\limits_{t_0}^{t_0 + T} x(t) e^{-jk(2\pi/T)t} \ dt $$ for any real $t_0$? and do you know what your period $T$ is? (but, of course, for this problem there is a much simpler way to go about it.) $\endgroup$ – robert bristow-johnson Jan 5 '16 at 4:33
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    $\begingroup$ Could you show us what you've done so far? That makes it easier to understand what your problem is. If you use those formulas for $\cos x$ and $\sin x$ then you're almost there. $\endgroup$ – Matt L. Jan 5 '16 at 7:58
  • $\begingroup$ This is just another example of homework or exercise based questions. There is nothing wrong as long as rules applied. Why on hold ? $\endgroup$ – Fat32 Jan 5 '16 at 14:40

First, recognise that the fundamental period of the signal x(t) (sum of two sinusoidals with periods 2/3 s and 1/50 s) is 2s. And the fundamental frequency, $f_0$, is 0.5 hz.

Then considering the CT Fourier Series expansion of the periodic signal x(t) wrt its fundamental frequency $f_0$ given as $\sum_{-\infty}^{\infty}{a_k e^{jk2\pi f_0t}}$ you can therefore find the coefficients corresponding to $2\cos(3\pi t)$ and $\sin(100\pi t + \pi/3)$ with k=3 and k=100 respectively.

Then, by expressing the sinusoidals in terms of complex exponentials using the Euler's identity this becomes $a_3=1 ,~ a_{-3}= 1 , ~~a_{100} = e^{j\pi/3}/2j, ~~a_{-100} = -e^{-j\pi/3}/2j $ which can be further manipulated if you want a rectengular representation of the complex coefficient instead of the above mixed-polar one.

Assuming no errors made during the arithmetic.

  • $\begingroup$ The coefficient corresponding to the sine cannot be real-valued. $\endgroup$ – Matt L. Jan 5 '16 at 12:01
  • $\begingroup$ yes sorry, thanks! I don't know how but during arithmetic I've mistaken the phase as $3\pi /2$ which is $\pi /3$ now I checked. On the other hand you should mean a "pure" sine, otherwise considering the phase for example, as mistaken above, to be $3\pi/2$ would happily produce a real coefficient, recognising already that it converts the sine into a cosine. It's dangerous therefore to make such implicit generalisations. $\endgroup$ – Fat32 Jan 5 '16 at 12:40
  • $\begingroup$ That's why I wrote "the" sine (as given in the example), not "a" sine. So no real danger here :) $\endgroup$ – Matt L. Jan 5 '16 at 12:43
  • $\begingroup$ then you should better said "is not" real instead of "cannot" which implies a generalisation :) $\endgroup$ – Fat32 Jan 5 '16 at 12:50
  • $\begingroup$ No, it not only "is not" real, but it also "cannot" be real in this case, as you've finally shown in your answer. Thanks for correcting! $\endgroup$ – Matt L. Jan 5 '16 at 12:55

Just wanted to add that one way to find the fundamental frequency is to calculate the greatest common divisor (gcd). In this case we have $\gcd(3,100) = 1$, as $100 = 2^2\cdot 5^2$ and $3$ don't share any prime factors.


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