0
$\begingroup$

I determined the complex Fourier series of a sinusoidal signal and arrived at the following expression: $$\sum_{n=\infty}^{\infty} \left[\frac{4e^{-j \frac{\pi}{2}n}}{\pi(1-n^2)}(e^{-j\pi n}+1)\right]e^{-j \frac{\pi}{2}tn} $$

However, we have a division by zero at $n = 1$ and $n = -1 $. Is there any way to determine $D_{-1}$ and $D_1$?

Improve this question
$\endgroup$
1
  • $\begingroup$ You don't have a division by $0$ at $n=\pm 1$, you have the indeterminate form $\dfrac 00$ because $e^{-jn\pi}+1 = 0$ when $n=\pm 1$. You could use L'Hopital's rule (treating $n$ as a continuous variable instead of being integer-valued), but it is better to follow MattL's suggestion and evaluate separately for $n=\pm 1$. $\endgroup$
    – Dilip Sarwate
    Mar 26, 2021 at 14:26

1 Answer 1

Reset to default
1
$\begingroup$

HINT:

You need to go back to the original definition of the Fourier coefficients via the integral, and treat $|n|=1$ as a special case. Your mistake happened there already (you divided by zero).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Can I just calculate the integral for $n = 1$ and $n = -1$ to determine these coefficients? $\endgroup$
    – carraro
    Mar 26, 2021 at 12:40
  • $\begingroup$ @JulyH.: Yes, sure, why not? The definition is valid for any value of $n$. $\endgroup$
    – Matt L.
    Mar 26, 2021 at 13:47

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.