# Fourier transform

Discussion in 'Homework Help' started by ecesoul, Sep 9, 2012.

1. ### ecesoul Thread Starter New Member

Sep 8, 2012
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0
Given fourier transform H(jw)=(2cosw)(sin2w)/w.
what is h(0)?
please help, i think it is a tricky question

2. ### WBahn Moderator

Mar 31, 2012
18,085
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We are not mind readers. What do you think is tricky about it?

3. ### ecesoul Thread Starter New Member

Sep 8, 2012
8
0
sorry for the word 'tricky'. It was asked as a multiple choice question in an exam with limited time.So I thought there might be some easier way than conventional.

I thought about initial value theorem, but don't know whether we can use it in Fourier domain

4. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
Step 1 is to apply the double angle formula
sin2ω = 2sinωcosω

Step 2 is to replace the cosine terms with 1 as ω -> 0

Step 3 is to take the limit as ω -> 0 of sinω/ω which is the well known sinc function, which has a value at ω=0 of 1.

Step 4 is to write the final answer of 4.

That's the trick if you can call it a trick.

5. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
You can get it even more directly by noting that it can be written as:

f(w) = g(w) h(w)

Where

g(w) = 2cos(w)

h(w) = sin(2w)/w

The value of f(w) as w goes to 0 is simply the products of the limits of g(w) and h(w) as they individually go to 0 and, since g(w) is well behaved at w=0 (it's equal to 2), we can focus on h(w).

This goes to 0/0, which is one of the forms for which we can apply L'Hospital's rule by taking the ratio of the derivatives of the numerator and denominator. which gives us that h(w) as w goes to 0 is 2cos(2w)/1, which evaluates to 2. Taking the product gives us 4.

Another way is to recall that, for sufficiently small angles, sin(theta) is approximately (theta). Well, approaching zero certainly qualifies as 'sufficiently small', so sin(2w) is about 2w, giving us:

f(w) = 2cos(w)(2w/w) = 4cos(w) for small w. For w approaching zero, this approaches 4.

6. ### ecesoul Thread Starter New Member

Sep 8, 2012
8
0
Sorry I think you missed an important point
H(jw) is given ie in frequency domain
But we have to find h(0), which is Limit t->0 h(t) ,this is in time domain

So we have to find inverse transform, I dont know how that could be done in this case

7. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
You're right. I thought you were looking for H(0), not h(0). My vision is getting blurry enough that I have a hard time telling the difference between h and H. I got focused on the removable singularity and thought that was what wa giving you problems.

Now your comment regarding the initial value theorem makes much more sense. I'm not sure whether you can apply it or not. I thought you probably could, but haven't been able to confirm that and have seen some things that imply you can't. This makes sense because the Fourier transform is for periodic signals that exist for all time, so the very notion of an 'initial' or 'final' value is problematic.

Now, given a Laplace transform, you can convert it to a Fourier transform by setting s=jw. Depending on whether information was irretrievable lost in that process, you might be able to get the Laplace transform back by setting w = -js. But I'm skeptical that this is going to be valid.

Sorry I don't an answer or even firm guidance for you, but hopefully something I've said will get you to looking in a direction that eventually leads to someplace worthwhile.

8. ### Tesla23 Active Member

May 10, 2009
323
67
given

$h(t)=\frac{1}{2\pi}\int H(\omega)e^{i\omega t}d\omega$

then

$h(0)=\frac{1}{2\pi}\int H(\omega)d\omega$

you need to do some trigonometry on $H(\omega)$ to evaluate this using some standard integrals