Fourier Series of a step function

[Oreith]

Homework Statement

[/B]
$$f(x)=\left\{\begin{array}{cc}0,&\mbox{ if }<br /> 0< x < 2\\1, & \mbox{ if } 2<x<4\end{array}\right.$$
Show that the Cosine Fourier Series of f(x) for the range [0,4] is given by:

$$A + B\sum^{\infty}_{n=0}\frac{(-1)^n}{(2m+1)}cos(\frac{(2m +1) \pi x}{2})$$

Homework Equations

$$a_n = \frac{2}{L}\int^{x_0 + L}_{x_0} f(x)cos(\frac{2\pi nx}{L})dx$$

The Attempt at a Solution

L = 4
$$a_n = \frac{1}{2}\int^{4}_{0} f(x)cos(\frac{\pi nx}{2})dx =\frac{1}{2}\int^{2}_{0} 0cos(\frac{\pi nx}{2})dx+\frac{1}{2}\int^{4}_{2} cos(\frac{\pi nx}{2})dx = \frac{1}{2}\int^{4}_{2} cos(\frac{\pi nx}{2})dx$$

$$= \frac{1}{2}[\frac{2 sin(2n\pi)}{\pi n} - \frac{2 sin(n\pi)}{\pi n} = \frac{1}{2}[0 - 0] = 0$$

I do get a non-zero a₀ term but it seems weird to me that B would be zero, is my L wrong?

 

[Fightfish]

When dealing with Fourier cosine and sine series, you are actually extending a non-periodic function onto a periodic even or odd domain. Hence the effective period is actually twice as large instead, that is to say, you are actually working with the interval -4 < x < 4 here as the basic unit.

 

[Oreith]

When dealing with Fourier cosine and sine series, you are actually extending a non-periodic function onto a periodic even or odd domain. Hence the effective period is actually twice as large instead, that is to say, you are actually working with the interval -4 < x < 4 here as the basic unit.

I see, is this symmetrisation of the interval around x = 0 always necessary or does it depend on f(x)?

 

[Fightfish]

Well, if you want to express a non-periodic function in terms of a Fourier series, then you will have to choose how to extend it to a periodic function - there are arbitrarily many different ways of doing so, but for convenience, usually we will choose the odd or even extensions, which lead respectively to the Fourier sine and cosine series.

 

[Oreith]

My function is only defined for 0 < x < 4. To make it periodic does my function become:

$$f(x)=\left\{\begin{array}{cc}<br /> 0,&\mbox{ if }<br /> -4< x < -2\\<br /> 1,&\mbox{ if }<br /> -2< x < 0\\<br /> 0,&\mbox{ if }<br /> 0< x < 2\\1, & \mbox{ if } 2<x<4\end{array}\right.$$

 

[vela]

No. Because you're asked to find the cosine series, you have to extend f(x) such that it's even about x=0.

 

[Oreith]

Thanks! Is the choice completely arbitrary for the complex exponential Fourier Series since it has both sine and cosine components?

 

[LCKurtz]

Homework Statement

[/B]
$$f(x)=\left\{\begin{array}{cc}0,&\mbox{ if }<br /> 0< x < 2\\1, & \mbox{ if } 2<x<4\end{array}\right.$$
Show that the Cosine Fourier Series of f(x) for the range [0,4] is given by:

$$A + B\sum^{\infty}_{n=0}\frac{(-1)^n}{(2m+1)}cos(\frac{(2m +1) \pi x}{2})$$

Homework Equations

$$a_n = \frac{2}{L}\int^{x_0 + L}_{x_0} f(x)cos(\frac{2\pi nx}{L})dx$$

Your problem is in that formula for $a_n$. In your problem with $L=4$ it should read
$$a_n = \frac 2 4 \int_0^4 f(x) \cos(\frac{n\pi x}{4})~dx$$

 

[Oreith]

I'm sorry but I don't really see that?

 

[LCKurtz]

I'm sorry but I don't really see that?

Look here, for example:
http://www.intmath.com/fourier-series/4-fourier-half-range-functions.php

 

[vela]

Thanks! Is the choice completely arbitrary for the complex exponential Fourier Series since it has both sine and cosine components?

That's right. There's no symmetry requirement for the complex exponential series.