(Questions)
(Questions)
Line 48: Line 48:
 
&x(n) = \sum_{n=-\infty}^\infty sin(pi*n)\\
 
&x(n) = \sum_{n=-\infty}^\infty sin(pi*n)\\
 
&a_k = sinc(k) \\
 
&a_k = sinc(k) \\
&Therefore, x(n) = sinc(0)*sin(pi*n) + sinc(1)*sin(pi*n) + ... \\
+
&Therefore, x(n) = sinc(0)sin(pi*n) + sinc(1)sin(pi*n) + ... \\
 
\end{align}
 
\end{align}
 
</math>
 
</math>

Revision as of 11:40, 16 September 2010

This pages contains exercises to practice computing the Fourier series of a CT signal

Note: This is a collective study page. You are expected to participate by adding content/comment/questions/exercises etc. This is a wiki after all!


Recall the basic formulas

  • Fourier series of a continuous-time signal x(t) periodic with period T
$ x(t)=\sum_{n=-\infty}^\infty a_n e^{j \frac{2\pi}{T}nt} $
  • Fourier series coefficients of a continuous-time signal x(t) periodic with period T
$ a_n=\frac{1}{T} \int_{0}^T x(t) e^{-j \frac{2\pi}{T}nt}dt $

Case 1: For some periodic functions, the Fourier series coefficients must be obtained by integration



Case 2: Some periodic functions (e.g. sine and cosine) can be directly expanded into a linear combination of complex exponentials

The following pages contain a periodic signal along with a computation of the Fourier series coefficients of that signal. These were contributed by your peers in ECE301. Check whether the answers are correct. Are all the steps explained clearly and logically? Do you have questions? Feel free to comment directly on the pages!


Questions

  • What is the difference between the Fourier series of a signal, and the Fourier series coefficients for a signal?

The Fourier series coefficients are in terms of k, and are the coefficients of each nth progression of the signal. The Fourier series is the sum of the Fourier coefficient and nth term of the signal product.

For example: $ \begin{align} &x(n) = \sum_{n=-\infty}^\infty sin(pi*n)\\ &a_k = sinc(k) \\ &Therefore, x(n) = sinc(0)sin(pi*n) + sinc(1)sin(pi*n) + ... \\ \end{align} $

At least that's how I understand it now.

~ksoong


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