# Fourier Transform, Delta Function

1. Aug 1, 2008

### darkSun

Hey everybody.

I was studying Fourier transforms today, and I thought, what if you took the transform of an ordinary sine or cosine? Well, since they only have one frequency, shouldn't the transform have only one value? That is, a delta function centered at the angular frequency of the wave.

But when I was reading a random pdf I googled, it said that the transform of a cosine wave was a pair of delta functions, centered at plus/minus the frequency. I don't get it.

It also said that the transform of a constant, a straight horizontal line, is a delta function. I understand the opposite, that the transform of a delta function is a straight line, but why is the inverse true?

This kind of made me realize there is something fundamental I am not understanding. In the frequency space of the transform, are those frequencies of sines or cosines? That is, to get the original function would you add up all sine waves with those amplitudes (of the transform), or cosine waves? I think even this question is not correct (or clear).

I appreciate any clarification!

2. Aug 1, 2008

### mathman

cos is an even function. Therefore transform of cos(fx) is same as transform of cos(-fx)
In general, Fourier transform of a Fourier transform will lead to a similar function, that is transform of f(x) to F(t), then transform of F(x) will get to f(-t). Since f is even in the situation described, you get what is stated.
One way to look at it is that any function can be split up into an odd function and an even function. The even part is transformed by the cos, while the odd part is transformed by the sin.

3. Aug 1, 2008

### darkSun

Thanks mathman.

I might be oversimplifying, but if I had a transform of a function, I would add all all the sine and cosine waves with those amplitudes.
If the function was even, then the transform would be symmetrical with respect to the y axis, and all the sines would cancel, and for odd functions, transform would be symmetrical to the y axis but negative on one side so the cosines would cancel. Or something like that.

4. Aug 2, 2008

### mathman

That is the general idea.

5. Aug 4, 2008

Got it.

6. Oct 4, 2008

### darkSun

Now I have more questions about Fourier transforms... I thought I'd just revisit this thread.

Basically this stems from page 67 of Griffith's Intro to Quantum Mechanics, where the problem requires you to find the Fourier transform of exp[-a|x|], that is, e to negative of the absolute value of x times a constant(how can I use that fancy Latex font everyone else has?).

If I did the evaluation correctly, I obtained F(w)= 2iw/(a^2 + w^2). Now this confuses me. I know (or thought) that Fourier transforms can be interpreted, very roughly, (in the frequency domain) as the graph of the spread of the frequencies that make up the original function. But how can this function be graphed? It's not real!

I have read that sometimes the Fourier transform yields complex-valued functions, but why does this happen if what it can mean is distribution of frequencies? In fact, this whole imaginary business seems to come from the exp[-iwx] in the integral, which seems artificial to me. Why not just use cos(wx) + sin (wx), instead of the exponential? I know usually exponentials are used in place of trigonometric functions to simplify the analysis, but in this case it looks like it is making it more convoluted.

I appreciate the assistance.

7. Oct 4, 2008

### Crosson

To get from a Fourier transform to a power spectrum in the frequency domain you have to square the absolute value (in the sense of complex numbers) of the Fourier transform. In other words, multiply your function by its complex conjugate to get the power spectrum.

8. Oct 4, 2008

### Redbelly98

Staff Emeritus
See https://www.physicsforums.com/misc/howtolatex.pdf

People graph either the absolute value of the transform, or as Crosson said the square of the absolute value.

Using complex numbers is one way to keep track of the sine and cosine terms in the transform. Real part = cosine terms, Imaginary part = sine terms.

Some waveform operations are more easily done using complex numbers. For example, applying a filter whose frequency characteristics are known.