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Given Fourier series co-efficient of $x(t)$ is $X$; how do I go about solving fourier series of $x(at)$ when $a = 0$ ? I can easily get to the step where I insert $x(0)$ into the Fourier coefficient equation, but how do I simply from there?

Similarly how do I determine fourier series of $x(at)$ when $a>0, a<0$?

My main objective is to see the effect of shrinking/expanding of the signal x(t) on the fourier series coefficienct

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To clarify, $x(t)$ is a function, and you know that it's Fourier series is $X(t) = \sum_{n=-\infty}^\infty c_n e^{i n\pi t / L}$ (or maybe in terms of sines and cosines?) Is that correct? If so, are you asking how to write the series for the (constant) function $x(0)$ and then for the functions $x(at)$ (using the series $X$, i.e., without re-computing the coefficients from scratch)? –  BaronVT Sep 9 '12 at 21:48
    
yes, using exponential representation(not sines and cosines as we don't know if x(t) will be real) –  rrazd Sep 9 '12 at 21:52
    
@rrazd you dont need actually to know if x(t) is real or not to have Fourier series in terms of cosines and sines as you will still have complex terms. –  Seyhmus Güngören Sep 9 '12 at 21:53
    
@SeyhmusGüngören I meant to say that the trigonometric Fourier series expansion is valid only for real periodic x(t) –  rrazd Sep 9 '12 at 21:57
    
@rrazd - assuming the original function $x(t)$ is periodic, with period $2L$, do you want the new functions to have the same period, or (perhaps more naturally) have a period of $2L/|a|$? –  BaronVT Sep 9 '12 at 22:53

1 Answer 1

up vote 2 down vote accepted

For the Fourier transform on $\mathbb{R}$, the answer is simple: $$ \hat{f}(\xi)=\int_{-\infty}^\infty f(x)\,e^{-2\pi i x\cdot\xi}\,\mathrm{d}x $$ thus, we have with the change of variables $y=ax$ $$ \begin{align} \int_{-\infty}^\infty f(ax)e^{-2\pi i x\cdot\xi}\,\mathrm{d}x &=\frac1{|a|}\int_{-\infty}^\infty f(y)\,e^{-2\pi i y\cdot\xi/a}\,\mathrm{d}y\\ &=\frac1{|a|}\hat{f}(\xi/a) \end{align} $$ However, for Fourier series on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, scaling the function only gives a nice answer if $a\in\mathbb{Z}$.

First, suppose $(a,n)=b$, then $$ \begin{align} \color{#C00000}{\sum_{k=0}^{a-1}e^{-2\pi ink/a}} &=b\sum_{k=0}^{a/b-1}e^{-2\pi i(n/b)k/(a/b)}\\ &=\color{#00A000}{b\frac{e^{-2\pi in/b}-1}{e^{-2\pi in/a}-1}}\\ &=\left\{\begin{array}{cl}\color{#C00000}{a}&\color{#C00000}{\text{if }a\;|\;n}\\\color{#00A000}{0}&\color{#00A000}{\text{if }a\!\not|\;n}\end{array}\right. \end{align} $$ By definition, $$ c(n)=\int_0^1f(x)\,e^{-2\pi inx}\,\mathrm{d}x $$ therefore, for positive $a$, $$ \begin{align} \int_0^1f(ax)\,e^{-2\pi inx}\,\mathrm{d}x &=\frac1{a}\int_0^af(y)\,e^{-2\pi iny/a}\,\mathrm{d}y\\ &=\frac1{a}\sum_{k=0}^{a-1}\int_0^1f(y)\,e^{-2\pi in(y+k)/a}\,\mathrm{d}y\\ &=\frac1{a}\sum_{k=0}^{a-1}e^{-2\pi ink/a}\int_0^1f(y)\,e^{-2\pi iny/a}\,\mathrm{d}y\\ &=\left\{\begin{array}{cl}c(n/a)&\text{if }a\;|\;n\\0&\text{if }a\!\not|\;n\end{array}\right. \end{align} $$ which, upon checking signs, works for all $a\not=0$.

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