I am trying to find the Cosine/Sine Fourier series coefficients for the given equation:

$$x(t)=2\cos(4t) + 4\sin(10t)$$

$\cos(4t)$ has a period of $T=\frac{\pi}{2}$, and $\sin(10t)$ has a period of $T=\frac{\pi}{5}$. Therefore, the fundamental period of $x(t)$ is:


And the fundamental angular frequency is:


I know that the equations to find all the coefficients are: $$a_0=\frac{1}{T}\int_0^Tx(t)dt$$ $$a_n=\frac{2}{T}\int_0^Tx(t)\cos(\omega nt)dt$$ $$b_n=\frac{2}{T}\int_0^Tx(t)\sin(\omega nt)dt$$

However, after a long calculation showed that:

$$a_0=0$$ $$a_n=\frac{2}{\pi}\left(\frac{n\sin(2\pi n)}{n^2-4}-\frac{20\sin^2(\pi n)}{n^2-25}\right)$$ $$b_n=\frac{4}{\pi}\left(\frac{n\sin^2(\pi n)}{n^2-4}+\frac{5\sin(2\pi n)}{n^2-25}\right)$$ But since n is an integer, you can see that $a_n=0$ and $b_n=0$. But isn't that impossible because any function can be formed with a Fourier series? Did I make a mistake somewhere?

  • $\begingroup$ "But since n is an integer, you can see that $a_n=0$ and $b_n=0$." No, what I can see is that the formulas are absurd if $n^2=4$ or $n^2=25$ hence these $a_n$ and $b_n$ should be computed more rigorously. $\endgroup$ – Did Nov 26 '14 at 6:40
  • $\begingroup$ Oh wow, I didn't notice that. So I need to add a limit when $n^2$ approaches 4 or 25? $\endgroup$ – vxs8122 Nov 26 '14 at 6:58
  • $\begingroup$ No, n2 does not "approach 4 or 25", rather you have to redo the computation in this case. $\endgroup$ – Did Nov 26 '14 at 7:00
  • $\begingroup$ $a_0=0$ is correct since the function average is zero. In the formulas you gave the result is zero unless n=-2,2,-5,5. You have to either do the computation by scratch or take the limit as $n$ tends to one of the values above, that should work too. Hint: as n->0, Sin(n)/n->1 $\endgroup$ – Georgy Nov 26 '14 at 7:43

The formulae you have derived are invalid for $n=2,5$ but are correct otherwise. You would need to compute the coefficients individually in the cases where $n=2$ or $5$. However, you are making this more difficult than it needs to be.

Since your original function is of the form $$x(t)=a\cos(2nt)+b\sin(2mt),$$ you should be able to see that your original function is already a Fourier expansion with most coefficients equal to zero. By the uniqueness of Fourier expansions this going to be the Fourier expansion of $x$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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