# Fourier Transform of sinusoid long-hand

Discussion in 'Math' started by blah2222, Jun 12, 2013.

1. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
581
38
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

Last edited: Jun 13, 2013
2. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
581
38
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

Last edited: Jun 13, 2013
3. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
581
38
$
= \ \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]
$

Last edited: Jun 13, 2013
4. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
581
38
$
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...

Last edited: Jun 12, 2013
5. ### WBahn Moderator

Mar 31, 2012
23,396
7,100
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.

blah2222 likes this.
6. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
581
38
Good catch, I meant cos(t).

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...

7. ### WBahn Moderator

Mar 31, 2012
23,396
7,100
No, the cos(ω_o t) is fine, it's the dx that is the problem. It should be dt.

You need the ω_o because it's what parks the detla functions.

8. ### mikeleeson New Member

Aug 22, 2012
26
4
I cannot see where 'n' is used in your integrand. Shouldn't ω be written with some dependence upon n?

9. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
581
38
Again, good catch. The math should be clear of variable typos now.

10. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
581
38
No, w = 2∏f. N is just used as a placeholder for the integration limits because they are set to ±∞.

11. ### WBahn Moderator

Mar 31, 2012
23,396
7,100
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).

12. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
581
38
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.

13. ### WBahn Moderator

Mar 31, 2012
23,396
7,100
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.

blah2222 likes this.

May 3, 2010
581
38
15. ### GopherT AAC Fanatic!

Nov 23, 2012
8,025
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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.