Fourier series of a bandwidth limited periodic function

Thread starter hotvette
Start date Oct 31, 2017
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Bandwidth Fourier Fourier series Function Periodic Series

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SUMMARY

The discussion focuses on finding the Fourier coefficients of a periodic function x(t) with a Fourier Transform X(f) defined as (1-f^2)^2 for |f|<1 and a period T_0=4. The participant notes that the standard formula for Fourier coefficients, FC=\hat x_T(k,T_0)=\sum_{k=-\infty}^\infty\frac{1}{T_0}X\left(k/T_0\right), is not applicable due to the bandwidth limitation of the Fourier Transform. They reference a sampling theorem related to waveform sampling, indicating a need for further guidance on applying these concepts to derive the Fourier series in this specific context.

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Understanding of Fourier Transform and Fourier Series
Knowledge of bandwidth limitations in signal processing
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Basic principles of waveform sampling and reconstruction

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Oct 31, 2017

hotvette

Homework Helper

Homework Statement

Find Fourier coefficients of the periodic function whose template is x(t) where the Fourier Transform of x(t) is X(f) = (1-f^2)^2 where \left|f\right|<1 and period T_0= 4.

Homework Equations

FC=\hat x_T(k,T_0)=\sum_{k=-\infty}^\infty\frac{1}{T_0}X\left(k/T_0\right)

The Attempt at a Solution

I know the expression for FC only applies with the Fourier transform X(f) has no limit on bandwidth, so something different needs to be done. We've had nothing in lectures about finding the Fourier series for this situation (i.e. where FT is zero outside of a specific frequency range), so I'm not sure where to go from here.

There is a short section in the book about waveform sampling stating that for T=1/(2f_c):
x(t) = T\sum_{k=-\infty}^\infty x(kT) \frac{sin 2 \pi f_c (t-kT)}{\pi(t-kT)} which I think is just saying:
T\sum_{k=-\infty}^\infty x(kT) \frac{sin 2 \pi f_c (t-kT)}{\pi(t-kT)} <==> (1-f^2)^2 where |f|<1 and f_c=1, but I don't immediately see how that's useful. Any tips or suggestions?