Find the Fourier Integral Representation

[erok81]

Homework Statement

Find the Fourier integral representation of the given function.

$$f(x)= \left\{\begin{array}{ll}<br /> 1-\cos(x) & \mbox{ if } -\frac{\pi}{2} < x < \frac{\pi}{2} \\<br /> 0 & \mbox{ otherwise}<br /> \end{array}\right.$$

On a side note...how does one properly enter the above using LaTeX?

Homework Equations

$$f(x)=\int_{0}^{\infty} A(\omega)cos(\omega x)~+~B(\omega)sin(\omega x)$$

Where

$$A(\omega)=\frac{1}{\pi} \int_{- \infty}^{\infty}f(t)cos(\omega t) ~dt$$

$$B(\omega)=\frac{1}{\pi} \int_{- \infty}^{\infty}f(t)sin(\omega t) ~dt$$

The Attempt at a Solution

So first I start solving for A(ω)

$$A(\omega)=\frac{1}{\pi} \int_{- \infty}^{\infty}f(t)cos(\omega t) ~dt$$

$$A(\omega)=\frac{1}{\pi} \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}(1-cos(x))cos(\omega t) ~dt$$

I broke this up into two different integrals.

The first was easy...let's call this D where A(ω)=D-E. Since (1-cos(t))cos(ωt) = cos(ωt)-cos(t)cos(ωt)

$$D=\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}cos(\omega t)~dt~=~\frac{2}{\omega}sin( \frac{1}{2}\omega \pi)$$Next up is E...the one I am having trouble with.

$$E=\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}cos(\omega t)cos(t)~dt$$

I have three different scenarios with the above integral Or at least it looks that way. When ω=0,1,and ω>1

How do I represent these in my final answer for A(ω)?

 

[I like Serena]

See the following tex source for the proper representation:

$$f(x)= \left\{\begin{array}{ll}<br /> 1-\cos(x) & \mbox{ if } -\frac{\pi}{2} < x < \frac{\pi}{2} \\<br /> 0 & \mbox{ otherwise}<br /> \end{array}\right.$$

If you left-click it, it will pop up in a window that also has a hyperlink to a quick-reference for LaTeX.As for your problem, perhaps the following formula from wikipedia can be of use?

[edit]Oops, I forgot a minus sign in your original formula. I've just added it.[/edit]

 

[erok81]

Thanks for the LaTeX help. I'd never have figured that out.

As for the problem, I'll give that identity a try. The only problem I see is I'll still have an issue with the different values for omega. Since, depending on what omega is I'll be able to get three different integrals.

 

[I like Serena]

Thanks for the LaTeX help. I'd never have figured that out.

As for the problem, I'll give that identity a try. The only problem I see is I'll still have an issue with the different values for omega. Since, depending on what omega is I'll be able to get three different integrals.

How would the value of omega make a difference in the integral?

 

[erok81]

How would the value of omega make a difference in the integral?

$$E_{\omega}=\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}cos(\omega t)cos(t)~dt$$ This gives a rather large and messy answer.

$$E_{0}=\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}cos(t)~dt~=~0$$

$$E_{1}=\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}cos(t)cos(t)~dt~=~\frac{1}{2}\pi$$

So I guess it technically doesn't make a difference in the integral, but it will in the answer. Maybe I am just so used to Fourier series problems and orthogonality/n values that have to be accounted for I am looking to far into this problem and making it more than it is.

 

[I like Serena]

$$E_{\omega}=\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}cos(\omega t)cos(t)~dt$$ This gives a rather large and messy answer.

$$E_{0}=\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}cos(t)~dt~=~0$$

$$E_{1}=\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}}cos(t)cos(t)~dt~=~\frac{1}{2}\pi$$

So I guess it technically doesn't make a difference in the integral, but it will in the answer. Maybe I am just so used to Fourier series problems and orthogonality/n values that have to be accounted for I am looking to far into this problem and making it more than it is.

I'd say that if you ignore the values of omega and integrate the integral, you'll get an answer in which you don't really have to make distinctions for the values of omega.
However, the outcome will contain a fraction with (omega - 1) in the denominator.
But you can deal with that after the integration (the limit exists and is continuous if you let omega approach 1).

Btw, note that A(omega) is a function of omega, which means that omega should still be present in the answer.

 

[erok81]

I'd say that if you ignore the values of omega and integrate the integral, you'll get an answer in which you don't really have to make distinctions for the values of omega.
However, the outcome will contain a fraction with (omega - 1) in the denominator.
But you can deal with that after the integration (the limit exists and is continuous if you let omega approach 1).

Btw, note that A(omega) is a function of omega, which means that omega should still be present in the answer.

Ah! Good point on the omega comment. I think I am putting too much thought into it. I'll go with leaving it as is, try the identity that you posted, and see what I come up with.

 

[haselwhat?]

Remember omega is a continuous variable so considering its different values is more of a Fourier Series type of thinking... just integrate the way it is.

 

[erok81]

Thanks to both of you. You guys were right. I was in Fourier Series mode. I just proceeded as they were normal integrals. I used the above identity and my answer came out correct.

Thanks again for the help.