Fourier transform
- Thread starter ecesoul
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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
I thought about initial value theorem, but don't know whether we can use it in Fourier domain
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.
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.
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.
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.
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.
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.
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
