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Let us say we have the Heaviside unit step function $\Theta(t-t^\prime)$. I want to calculate its Fourier series $$ \Theta(t-t^\prime)=\frac{1}{T}\sum_{n,m}\Theta_{\omega_n,\omega_m}e^{-i\omega_n t}e^{-i\omega_m t^\prime}\, , $$ where $\omega_n=\frac{2\pi}{T}n$. So (according to my understanding) I have to execute the following integral $$ \Theta_{\omega_p,\omega_l}=\frac{1}{T}\int_0^Tdt\int_0^Tdt^\prime \Theta(t-t^\prime)e^{i\omega_p t}e^{i\omega_l t^\prime}\, . $$ The unit step function affects the integration boundaries, such that $$ \Theta_{\omega_p,\omega_l}=\frac{1}{T}\int_0^Tdt\int_0^tdt^\prime e^{i\omega_p t}e^{i\omega_l t^\prime}\, . $$ I calculated this integral for four different cases of $(p,l)\in \{(0,0),(p,0),(0,l),(p,-p)\}$ because all other cases are $0$. I get the following result $$ \Theta_{\omega_p,\omega_l}=\frac{1}{T}\int_0^Tdt\int_0^tdt^\prime e^{i\omega_p t}e^{i\omega_l t^\prime}= \begin{cases} &\frac{1}{2}\, , \quad &p=l=0\\ &-\frac{i}{\omega_p}\, , &l=0\neq p\\ &\frac{i}{\omega_l}\, , &p=0\neq l\\ &\frac{i}{\omega_p}\, , &p=-l\neq 0\, . \end{cases} $$ And then I plug it into the sum and get the following result $$ \Theta(t-t^\prime)=\frac{1}{2}+\frac{i}{T}\sum_{n\neq 0}\frac{1}{\omega_n}\left(e^{-i\omega_n t^\prime}-e^{-i\omega_n t}+e^{-i\omega_n (t-t^\prime)}\right)\, . $$ However, I am not sure if that is correct because the step function depends on the difference $t-t^\prime$, so I expected that the Fourier coefficients would also only depend on the frequency differences. Is there a flaw in the calculation?

The second question is: Since depends on the difference $t-t^\prime$, can one just write the Fourier series in the form of $$ \Theta(t-t^\prime)=\frac{1}{T}\sum_{n,m}\Theta_{\omega_n-\omega_m}e^{-i\omega_n (t-t^\prime)}e^{-i\omega_m (t-t^\prime)}\, , $$ or something similar, where can read out that it only depends on the time difference??

THX

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  • $\begingroup$ It is not a double Fourier series but just a single one. $\endgroup$ – Jon Nov 28 '18 at 15:00
  • $\begingroup$ Ok, so you mean it is $\endgroup$ – mr. curious Nov 28 '18 at 15:56
  • $\begingroup$ $$ \Theta(t-t^\prime)=\frac{1}{\sqrt{T}}\sum_n\Theta_{\omega_n}e^{-i\omega_n(t-t^{prime})} $$? I'm asking because I have two time variables, so I will have to integrate by $t$ and $t^\prime$? $\endgroup$ – mr. curious Nov 28 '18 at 15:57
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    $\begingroup$ The idea is that you just transform $\Theta(t)$ then, you can translate to whatever you want. The point is that, as far as I can say, just the Fourier transform has a meaning. $\Theta$ belongs to a distribution space that is meaningful only under integration sign.Check math.stackexchange.com/questions/73922/…. $\endgroup$ – Jon Nov 28 '18 at 17:17

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