# 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.