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

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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

Periodic square wave with period two
Periodic $|\text{sin}(\pi t)|$ with period one
write a page with another example here.

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:

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

At least that's how I understand it now.

~ksoong

Ok, that's a good start. (And great job with the latex, by the way!) Now let's do the "variable test" on the above equations. (This is an excellent trick to use when doing transforms!). First, let's look at the statement regarding the coefficients:

a_{\color{OliveGreen}k} = sinc({\color{OliveGreen}k})

Notice that both the left side and the right side of the equations only depend on the variable k. That's good! This is what you would expect since, as you meant above, the coefficients depend on the variable k, and you do indeed have an expression in terms of k on the right-hand side.

Now compare this with the Fourier series formula stated:

x( {\color{blue}n} ) = \sum_{{\color{blue}n}=-\infty}^\infty sin(\pi*{\color{blue}n})

Notice that the left-hand side depends on the variable n. However, the right-hand side actually does NOT depend on n. This is because n is the variable used to index all the terms of the summation: it is not a meaningful variable, because at the end when all the terms are summed together (so to get the left-hand-side value x[n]), there is no need for the variable n anymore. The variable used to index the terms of the summation is an example of what is called a "dummy variable", just like the integration variable in an integral. Now, the fact that the left hand side depends on n, and the right-hand side does not is problematic: you would expect both sides to depend on the same variable n. Can you/somebody propose a way to solve this problem? --Mboutin 19:39, 16 September 2010 (UTC)

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Recommended exercise Fourier series computation - Rhea

Contents

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

Recall the basic formulas

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

Questions

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