Sin and Cos frequency Question

1. Apr 30, 2010

frenzal_dude

Hi, if cosine and sine waves are exactly the same wave except cosine starts earlier in time, why do they have different Fourier Transforms? I know the cos would have an impulse at f=Fc with amplitude of 1, and sin would have an impulse at f=Fc and amplitude of 1/j. I don't understand why the amplitude of the sine goes complex in the amplitude spectrum just from having a different phase.

Hope you guys can explain it to me :)
Dave

2. May 1, 2010

MATLABdude

The MAGNITUDE of the Fourier Transform (|F(w)|) remains the same. The phase is what changes, according to the translation property:
http://en.wikipedia.org/wiki/Fourier_transform#Basic_properties

You can try this for yourself when you compare the Fourier transform of cos(x-pi/2) with the transform of sin(x), realizing that they have exactly the same graph.

When you say the amplitude changes (by a factor of j) realize that the magnitude hasn't changed, but phase has.

3. May 1, 2010

Phrak

From a heuristic point of view, a Fourier transform is invertible; there is no information lost. The Fourier transform of a Fourier transform returns the original function (or is it the negative of it?) Something has to encode phase information to distinguish cos from sin. As phi increases from 0 to 2pi the amplitude of cos(x+phi) rotates through the complex plane.

4. May 2, 2010

frenzal_dude

So basically if you take the FT of sin(2piFct) you'd get 2 impulses, one at f=-Fc with amplitude of -1/(2j), and one at f=Fc with amplitude +1/(2j).

So in the single sided spectrum you'd get an impulse with f=Fc and magnitude 1/j.

Does this mean that the amplitude is actually 1, but with a phase of pi/2 (because 1/j is on the +ve imaginary axis which corresponds to pi/2)?

But what if you multiply 1/j by j/j, you'd end up with -j which would then mean you have an amplitude of 1 with a phase of 3pi/2 because it's now on the -ve imaginary axis?

5. May 2, 2010

frenzal_dude

Wait I think I get it now. So if you take the FT of sin(2piFct) you can see from the frequency domain that your function is made up of one Cosine with a frequency of f=Fc, and with an amplitude of 1, and with a phase of pi/2 or 3pi/2 (since cos(pi/2)=cos(3pi/2)=0).

6. May 2, 2010

Phrak

Yes, 1/j = -j.

Rather than trust memory, I used Mathematica which came up with the following:

The transform of sin(2pi Fc t) has an amplitude of j sqrt(pi/2) DiracDelta at f=Fc,
The amplitude is -j sqrt(pi/2) DiracDelta at f=-Fc, where Fc is positive, real.

Essentially you're in agreement. Your phases agree. I don't what a unit impulse amplitude should be.

To establish a unit impulse you might use

FourierTransform[Exp(-j t omega)= Sqrt(2/pi) DiracDelta(omega-1),

or you may already using something different.

7. May 2, 2010

frenzal_dude

For your double sided spectrum, how did you get -j sqrt(pi/2) at f=-Fc. Isn't the amplitude just -1/(2j) (from Euler's Formula) ?

8. May 2, 2010

Phrak

A Fourier transform can be defined in several ways. The choice of defintion effects the amplitude. Mathematica uses

F(omega) = 1/(Sqrt[2pi]) Integral[f(t)ej omega t] dt

This is why I was questioning what a unit impulse was in your definition.

9. May 3, 2010

frenzal_dude

ahk, that's a bit different to what we got taught in Signal Theory.
Here's the formula we use:

$$G(f)=\int_{-\infty}^{\infty }g(t)e^{-j2\pi ft}dt$$

A unit impulse (from what my lecturer says) is just a straight line with amplitude of 1 at a certain frequency Fc.

10. May 3, 2010

Phrak

OK. I errored and left the negative sign out of the exponent by the way. omega = 2 pi F, so that is unchanged.

The unit amplitude should be a scaled Dirac delta function. A Dirac delta function has unit area and zero width. So a pure sine wave has "infinite amplitude". Normalizing this to one doesn't make a lot of sense to me unless you are only dealing with linear combinations of sine waves. Talk to your lecturer about this.

11. May 3, 2010

frenzal_dude

This is something my lecturer couldn't even explain. Apparantely a dirac delta function aka unit impulse, has infinite amplitude and 0 width, with unit area.

So why do they talk about sin and cos functions having a frequency content with 2 impulse functions with a height of 1/2 or 1/(2j) when the height has to be infinite?

12. May 3, 2010

Phrak

Let's look at the impulse areas instead. In your units each impulse has an area of one half. The total area is one.

|j/2| Integral[Dirac Delta] + |-j/2| Integral[Dirac Delta]

The integral is taken over all frequencies.