# Fourier series

1. Jun 28, 2013

### DUET

The Fourier series can be used to represent an arbitrary function within the interval from - π to + π even though function does not continue or repeat outside this interval. Outside this interval the Fourier series expression will repeat faithfully from period to period irrespective of whether the given function continues. The same remarks apply to any arbitrary function which is specified over any finite range, say from t=0 to t=t0. An infinite number of Fourier series expansion with fundamental periods T≥t0 can be found such that they all reproduce f(t) within the given range. Outside this range, different expansion may have entirely different values, depending upon the choice of T as compared with t0 and of the waveform in the interval from t= t0 to T, which is entirely arbitrary except that the Dirichlet conditions must be satisfied.

I was reading a book, Analysis of linear systems-by David K. Cheng, to learn Fourier series. I have understood every think except the above writing. Could someone please explain me the purpose of the above writing?

2. Jun 28, 2013

### Marioeden

Here's a simple example:

Suppose you have a function f(t) defined on the range 0<t<a

Then two possible fourier series for f(t) are the fourier sine series and the fourier cosine series.

You can obtain the sine series by defining f(t) to be an odd function i.e. define it on -a<t<a with f(-t) = -f(t) and finding the fourier series for this function which will just have sine terms.

Similarlly you could obtain a cosine series by defining f(t) even.

Basically, there's more than one way to represent a function as a fourier series in the given range, the only difference is what the fourier series does outside that range.

3. Jun 28, 2013

### DUET

The Fourier series can be used to represent an arbitrary function within the interval from - π to + π even though function does not continue or repeat outside this interval. Outside this interval the Fourier series expression will repeat faithfully from period to period irrespective of whether the given function continues.

What does the above writing suggest?

4. Jun 28, 2013

### DUET

Could you explain the underlined part a little bit more?

5. Jun 28, 2013

### LCKurtz

Here's a picture where $f(x) = x^2$ on the interval $(-\pi,\pi)$ is approximated by a few terms of a cosine FS. Notice that outside the interval, the FS does't represent the parabola.

6. Jun 29, 2013

### DUET

Fourier function

The Fourier series can be used to represent an arbitrary function within the interval from - π to + π even though function does not continue or repeat outside this interval. Outside this interval the Fourier series expression will repeat faithfully from period to period irrespective of whether the given function continues.

What the above two sentence mean?

7. Jun 29, 2013

### HallsofIvy

Staff Emeritus
Given a function defined on $[-\pi, \pi]$ or, more generally, [a, b] for any a< b, we can "extend" the function "by periociity". That is, define $f(x+ 2n\pi)= f(x)$ for any x outside that interval. Taking the Fourier series of a function from $-\pi$ to $\pi$ automatically makes that assumption.

Last edited by a moderator: Jun 29, 2013
8. Jun 29, 2013

### Marioeden

Well, your Fourier series is periodic. If you look at the sketch posted above you can see that the fourier series replicates the function in the period given and then just repeats itself as a periodic function outside that interval (in the same way the simple trig functions do).

Moreover, if you can have two different fourier series representations then they'll look the same for the period given but outside of that they could look very different.

As an example, try it for y=x^2 defines on the interval (0,1).

The cosine series will look as given in the sketch above, but the sine series will only look the same for half of the intervals and will behave differently elsewhere. Particularly in the interval (-1,0), where the cosine series will look like y=x^2 but the sine series will look like y=-x^2

9. Jun 29, 2013

### Redbelly98

Staff Emeritus
Moderator's note: related thread merged with this one (post #s 6 & 7)

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