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I feel problem in understanding the proof of Fourier series properties

  1. Time scaling
    \begin{align} b_k &= \frac{1}{T}\int_{T}x(t)e^{jk\omega_0t}dt\\ & = \frac{a}{T}\int_{T/a}x(at)e^{jk(\omega_0a)t}dt\tag{$\scriptstyle{a - \text{scaling factor}}$}\\ & = \frac{a}{T}\int_{T}x(L)e^{jk\omega_0L}\frac{dL}{a}\tag{$\scriptstyle{L=at}$}\\ &= \frac{1}{T}\int_{T}x(L)e^{jk\omega_0L}dL = a_k\\ \end{align} For this I want to ask that how $T/a$ term in the integral changes to $T$, and $dt$ changes to $dL/a$?

  2. Multiplication \begin{align} x(t)y(t)&= \sum^{+\infty}_{k\ =\ -\infty} a_k e^{jk\omega_0t}\sum^{+\infty}_{k\ =\ -\infty} b_k e^{jl\omega_0t}\\ &=\sum^{+\infty}_{k\ =\ -\infty} \sum^{+\infty}_{l\ =\ -\infty} a_k b_le^{j(k+l)\omega_0t}\\ &=\sum^{+\infty}_{m\ =\ -\infty} \left[\sum^{+\infty}_{l\ =\ -\infty} a_{m-l} b_l\right]e^{jm\omega_0t}\tag{$\scriptstyle{m=k+l}$} \end{align} I couldn't understand this last step, why the summation changes from $k$ to only $m$? Shouldn't it be $m-l$ and also is it done here to prove the multiplication property? or there are more steps?

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  • $\begingroup$ please write in comment the reason before giving down $\endgroup$
    – Aadnan Farooq A
    Commented Oct 12, 2015 at 8:07
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    $\begingroup$ There is a mistake in your last expression in time scaling. The integrating variable is dL and not dt. In your multiplication steps all the steps are correct. He is taking the substitution m=k+l. so now k becomes (m-l).But the limits of integration of k are expressed in terms of m. m=K+l. now if k=-infinity, m=-infinity and if k=+infinity,m=+infinity. All this is being done to express fourier coefficients of product of x(t)y(t) as discrete convolution of individual fourier series coefficients ak and bk. $\endgroup$
    – Karan Talasila
    Commented Oct 12, 2015 at 9:45
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    $\begingroup$ see when you are making a substitution of L=at in integral, here you are doing definite integrals , so you have to change the limits of integration also. So if L=at ,then you have to change everything with respect to L. so now the integration limits are t=0 and t=T/a before making substitution L=at. After L=at substitution, you have to change limits with respect to L. So when t=0 L=a.0=0 and when t=T/a, L=a.(T/a)=T.That's how limits are changing to T. with regard to dt/a, it is because L=at.differentiate b.s. you will get dL=dt.a. so dt=dL/a. change that dt/a to dL/a. That's a mistake. $\endgroup$
    – Karan Talasila
    Commented Oct 12, 2015 at 10:04
  • $\begingroup$ yes i have change it to dL, can you please help me to understand how $T/a$ term changes to$T$ in the integral changes and $dt$ to $dL/a$ $\endgroup$
    – Aadnan Farooq A
    Commented Oct 12, 2015 at 10:04
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    $\begingroup$ @AadnanFarooqA There shouldn't be those $dt$'s in your second question. Can you double-check the $x(t)y(t)$ equation ? $\endgroup$
    – Gilles
    Commented Oct 12, 2015 at 14:18

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  • Looking at the change of terms in the equality: $$\frac{a}{T}\int_{T/a}x(at)e^{jk(\omega_0a)t}dt = \frac{a}{T}\int_{T}x(L)e^{jk\omega_0L}\frac{dL}{a}$$ $$at = L\implies \begin{cases}\text{if }t=T/a\implies L=T\\adt=dL\implies dt=\frac{dL}{a}\end{cases}$$

  • Looking at the equality again: $$ \sum^{+\infty}_{k\ =\ -\infty} \sum^{+\infty}_{l\ =\ -\infty} a_k b_le^{j(k+l)\omega_0t}=\sum^{+\infty}_{m\ =\ -\infty} \left[\sum^{+\infty}_{l\ =\ -\infty} a_{m-l} b_l\right]e^{jm\omega_0t} $$ $$m=k+l\implies \begin{cases}k=m-l\quad({\scriptstyle{\rm indeed}})\\\text{and } m-l=\pm \infty \implies m=\pm\infty+l=\pm\infty\end{cases}$$

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