Use the defining equation for the Fourier Series coefficients to evaluate the Fourier Series representation of the following signal: $$x(t)=\sum_{m=-\infty}^{+\infty}=(\delta(t-m/3)+\delta(t-2m/3))$$

I calculated $T=2/3$ and $w_0=3\pi$, however, I'm not sure whether $X[k]$ will be $$X[k]=3/2 * \int_{0}^{2/3}(2\delta(t)+\delta(t-1/3)+2\delta(t-2/3))*e^{-jk3\pi t}dt$$or$$X[k]=3/2 * \int_{0}^{2/3}(\delta(t)+\delta(t-1/3)+\delta(t-2/3))*e^{-jk3\pi t}dt$$ And if $X[k]=3/2 * \int_{0}^{2/3}(2\delta(t)+\delta(t-1/3)+2\delta(t-2/3))*e^{-jk3\pi t}dt$, is this following calculation correct (since I never did integral with $\delta$ before)? $$X[k]=3/2*(2e^{-jk3\pi 0} + e^{-jk3\pi 1/3} + 2e^{-jk3\pi 2/3})=3+3/2*e^{-jk\pi} + 3e^{-jk2\pi}$$


1 Answer 1


You're getting yourself into unnecessary trouble by choosing the integration limits exactly at those values of $t$ where you have Dirac impulses. Note that you can choose any integration limit as long as you integrate over one period of the given function.

E.g., if you choose some positive $\epsilon$ satisfying $0<\epsilon\le\frac16$ and you integrate from $-\epsilon$ to $\frac13+\epsilon$ then you have the relevant portion of the signal during one period inside the integral, and there's no question as to the scaling factor of the Dirac impulses. The result of the integration will be a real-valued constant for even $k$, and another real-valued constant for odd $k$. I'm sure you can derive the final result yourself.


Your Answer

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

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