1
$\begingroup$

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...

$\endgroup$
3
  • 1
    $\begingroup$ 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.) $\endgroup$ Jan 5, 2016 at 4:33
  • 1
    $\begingroup$ 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, 2016 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, 2016 at 14:40

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ The coefficient corresponding to the sine cannot be real-valued. $\endgroup$
    – Matt L.
    Jan 5, 2016 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, 2016 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, 2016 at 12:43
  • $\begingroup$ then you should better said "is not" real instead of "cannot" which implies a generalisation :) $\endgroup$
    – Fat32
    Jan 5, 2016 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, 2016 at 12:55
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.