Fourier Transform of sinusoid long-hand

Thread Starter

blah2222

Joined May 3, 2010
582
Hi all,

I have been trying to go through the calculation of the Fourier Transform of cos(w0t) and was wondering how they go from Equation 2 to the final dirac delta function on this webpage.

Any help would be much appreciated.
Thanks
 
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Thread Starter

blah2222

Joined May 3, 2010
582
This is what I have:

\(
f(t)\ = \ cos(w_{0}t) = \frac{e^{iw_{0}t} + e^{-iw_{0}t}}{2}
\)

\(
F(w)\ =\ \lim_{n \to \infty}\ \int_{-n}^n cos(w_{0}t)e^{-iwt}dt = \ \frac{1}{2}\ \lim_{n \to \infty}\ \int_{-n}^n (e^{iw_{0}t} + e^{-iw_{0}t})e^{-iwt}dt
\)

\(
= \ \frac{1}{2}\ \lim_{n \to \infty}\ \int_{-n}^n e^{it(w_{0}-w)} + e^{-it(w_{0}+w)}dt
\)


I have a feeling that this involves some algebraic maneuvering but I can't get it to work out so that:


\(
F(w) = \frac{1}{2}[\delta(w - w_{0}) + \delta(w + w_{0})]
\)

Where:

F(w0) = infinity
F(-w0) = infinity
F(other) = 0
 
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Thread Starter

blah2222

Joined May 3, 2010
582
\(
= \ \frac{1}{2}\ \lim_{n \to \infty}\ \left[\int_{-n}^n e^{it(w_{0}-w)}dt + \int_{-n}^n\ e^{-it(w_{0}+w)}dt\right]
\)


\(
= \ \frac{1}{2}\ \lim_{n \to \infty}\ \left[\int_{-n}^n e^{-it(w-w_{0})}dt + \int_{-n}^n\ e^{-it(w+w_{0})}dt\right]
\)


\(
= \ \frac{1}{2}\ \lim_{n \to \infty}\ \left[\frac{e^{-it(w-w_{0})}}{-i(w-w_{0})}|_{-n}^n\ +\ \frac{e^{-it(w+w_{0})}}{-i(w+w_{0})}|_{-n}^n\right]
\)


\(
= \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{e^{-in(w-w_{0})}}{w-w_{0}}\ - \ \frac{e^{in(w-w_{0})}}{w-w_{0}}\ +\ \frac{e^{-in(w+w_{0})}}{w+w_{0}}\ - \ \frac{e^{in(w+w_{0})}}{w+w_{0}}\right]
\)


\(
= \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{e^{-in(w-w_{0})}}{w-w_{0}}\ - \ \frac{e^{in(w-w_{0})}}{w-w_{0}}\ +\ \frac{e^{-in(w+w_{0})}}{w+w_{0}}\ - \ \frac{e^{in(w+w_{0})}}{w+w_{0}}\right]
\)


\(
= \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{(w+w_{0})e^{-in(w-w_{0})}\ - \ (w+w_{0})e^{in(w-w_{0})}\ +\ (w-w_{0})e^{-in(w+w_{0})}\ - \ (w-w_{0})e^{in(w+w_{0})}}{w^{2}-w_{0}^{2}}\right]
\)
 
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Thread Starter

blah2222

Joined May 3, 2010
582
\(
F(w0)\ = \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{(w_{0}+w_{0})e^{-in(w_{0}-w_{0})}\ - \ (w_{0}+w_{0})e^{in(w_{0}-w_{0})}\ +\ (w_{0}-w_{0})e^{-in(w_{0}+w_{0})}\ - \ (w_{0}-w_{0})e^{in(w_{0}+w_{0})}}{w_{0}^{2}-w_{0}^{2}}\right]
\)


\(
= \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{(2w_{0})e^{-in(0)}\ - \ (2w_{0})e^{in(0)}\ +\ (0)e^{-in(2w_{0})}\ - \ (0)e^{in(2w_{0})}}{0}\right]
\)


\(
= \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{2w_{0}\ - \ 2w_{0}\ +\ 0\ - \ 0}{0}\right]\ = \ \frac{0}{0}
\)





\(
F(-w0)\ = \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{(-w_{0}+w_{0})e^{-in(-w_{0}-w_{0})}\ - \ (-w_{0}+w_{0})e^{in(-w_{0}-w_{0})}\ +\ (-w_{0}-w_{0})e^{-in(-w_{0}+w_{0})}\ - \ (-w_{0}-w_{0})e^{in(-w_{0}+w_{0})}}{(-w_{0})^{2}-w_{0}^{2}}\right]
\)


\(
= \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{(0)e^{-in(-2w_{0})}\ - \ (0)e^{in(-2w_{0})}\ +\ (-2w_{0})e^{-in(0)}\ - \ (-2w_{0})e^{in(0)}}{0}\right]
\)


\(
= \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{0\ - \ 0\ -\ 2w_{0}\ + \ 2w_{0}}{0}\right]\ = \ \frac{0}{0}
\)





\(
A\ =\ w - w_{0}
B\ =\ w + w_{0}

if\ w\gt w_{0}:\ A = A;\ B = B
if\ w\lt w_{0}:\ A = -A;\ B = B
\)





\(
F(w\gt w_{0})\ = \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{(B)e^{-in(A)}\ - \ (B)e^{in(A)}\ +\ (A)e^{-in(B)}\ - \ (A)e^{in(B)}}{AB}\right]
\)


\(
=\ -\frac{1}{2i}\ \left[\frac{(B)e^{-\infty}\ - \ (B)e^{\infty}\ +\ (A)e^{-\infty}\ - \ (A)e^{\infty}}{AB}\right]\ =\ -\frac{1}{2i}\ \left[\frac{0\ - \ (B)e^{\infty}\ +\ 0\ - \ (A)e^{\infty}}{AB}\right]\ =\ \infty
\)






\(
F(w\lt w_{0})\ = \ -\frac{1}{2i}\ \lim_{n \to \infty}\ \left[\frac{(B)e^{-in(-A)}\ - \ (B)e^{in(-A)}\ +\ (-A)e^{-in(B)}\ - \ (-A)e^{in(B)}}{-AB}\right]
\)


\(
= \ -\frac{1}{2i}\ \left[\frac{(B)e^{\infty}\ - \ (B)e^{-\infty}\ -\ (A)e^{-\infty}\ + \ (A)e^{\infty}}{-AB}\right]\ = \ -\frac{1}{2i}\ \left[\frac{(B)e^{\infty}\ - \ 0\ -\ 0\ + \ (A)e^{\infty}}{-AB}\right]\ =\ \infty
\)




Not exactly what I expected...
 
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WBahn

Joined Mar 31, 2012
29,976
The first thing I notice is that you have t in your integrand but are integrating over x.

Why?

I think if you split things out sufficiently, you will eventually get to the point of having a pair of coefficients each multiplying the Fourier transform of 1, which is what gives you the delta functions. Now, proving this last part is tricky. I think on of the standard ways is to do the opposite and show either that the Inverse Fourier Transform of a delta function is a constant, namely 1. Another way would be to show that the Fourier Transform of a delta function is equal to 1 and then invoke duality.
 

Thread Starter

blah2222

Joined May 3, 2010
582
The first thing I notice is that you have t in your integrand but are integrating over x.

Why?
Good catch, I meant cos(t).

I think if you split things out sufficiently, you will eventually get to the point of having a pair of coefficients each multiplying the Fourier transform of 1, which is what gives you the delta functions. Now, proving this last part is tricky. I think on of the standard ways is to do the opposite and show either that the Inverse Fourier Transform of a delta function is a constant, namely 1. Another way would be to show that the Fourier Transform of a delta function is equal to 1 and then invoke duality.
Yeah, the duality is how it seems to be done often, but I thought I'd take a stab at the long route. Definitely showing to be tricky...
 

WBahn

Joined Mar 31, 2012
29,976
That's not how you do something like that.

Your integrand has a variable over which you are integrating. You originally had this as x since your integrand ended with 'dx'. After integration, you end up with a result that still has the variable x in it. You then take the difference between this expression evaluated with x equal to the upper limit and this expression evaluated with x at the lower limit. That is how the limits enter into the picture.

You are being very, very sloppy with the notation and this is going to cause you all kinds of problems (probably already are).
 

Thread Starter

blah2222

Joined May 3, 2010
582
That's not how you do something like that.

Your integrand has a variable over which you are integrating. You originally had this as x since your integrand ended with 'dx'. After integration, you end up with a result that still has the variable x in it. You then take the difference between this expression evaluated with x equal to the upper limit and this expression evaluated with x at the lower limit. That is how the limits enter into the picture.

You are being very, very sloppy with the notation and this is going to cause you all kinds of problems (probably already are).
My apologies for being "sloppy" (sarcasm). Writing this out in LaTeX makes it a lot harder to keep track of small typos. Whether the integration variable is stated as 'x' or 't', I figured it would be clear enough for those reading along. I have fixed all the cases of 'x' and 'dx' to their 't' counterparts. My apologies that this wasn't perfectly coded throughout this thread.

As for the 'n' limits. They make perfect sense and taking a limit of an integral with n going to infinity is logical and makes it easier to manipulate the equations later on without having to have infinity subbed in everywhere.

It's nice to know that people are picky for the right reasons... I honestly do appreciate the help though, so thank you for that.
 

WBahn

Joined Mar 31, 2012
29,976
I went back and looked quickly and how you brought 'n' into your equations in Post #3 is just fine. I misremembered how you had introduced them after reading a later reply. Sorry about that.
 

GopherT

Joined Nov 23, 2012
8,009
I tuned my radio to a negative frequency recently. I new it was negative because (insert punch line here).

It was a setup for a great joke, I am just drawing a blank.
 
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