# Homework Help: Need help understanding Fourier transform in Hz vs radians

1. Feb 10, 2010

### Rib5

Hi guys,

I'm having some issues understanding something about the Fourier transform. In my first signals and systems class we used the angular frequency omega. Doing it like that you end up with a weighing factor or 1/(2pi) when you take the transform. Now in the dsp class I am taking now we are using the frequency in Hz.

The thing I don't get is how can the amplitude in one frequency be different than in another for the same signal. I also read about another way of doing it where in both directions you multiply it by 1/sqrt(2pi), helping to preserve duality.

Is the frequency transform basically different based on how it is interpreted? Can someone help me out here, I don't know exactly what I am confused about but I don't see how it can just be arbitrarily defined and have different amplitudes for what is apparently the same thing just in a different frequency?

For example a sine wave has an amplitude of 1. So it would seem reasonable that in the Fourier transform it would have an impulse of 1 at the correct frequency. But if you use radians for the Fourier transform, it has a different amplitude!

Last edited: Feb 10, 2010
2. Feb 10, 2010

### marcusl

There shouldn't be any fundamental difference. When you take the FT in Hz, the argument of the exponential contains a factor of 2*pi, that is, exp(i*2*pi*f*t). Using angular freqency, the argument is exp(i*omega*t). You deal with real signals in your engineering classes, so the normalizing factor will be chosen to give the proper power spectral density (in Watts/Hz, e.g.) given a signal amplitude in volts.

3. Feb 11, 2010

### Rib5

So the Fourier Transform is still the Fourier Transform, regardless of what you scale it by? This seems weird because it would mean that if you tell someone, the FT of this is X, it could mean anything unless you give the scaling factor.

4. Feb 28, 2010

### marcusl

If you define the FT as

$$G(f)=\int_{-\infty}^{\infty} g(t)exp(-i2\pi ft) dt$$,

then the inverse transform is

$$g(t)=\int_{-\infty}^{\infty}G(f)exp(i2\pi ft) df$$.

If you change variables to

$$\omega=2\pi f$$

you necessarily get a factor of $$1/(2\pi)$$ in front of the inverse transform:

$$g(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)exp(i\omega t) d\omega$$.

In this case there is no difference in the amplitudes, both inverse expressions give the same thing. Notice that forward and reverse FT's are asymmetric in the latter case.

In mathematics the normalization constant is sometimes apportioned evenly between fwd and reverse transforms (using angular frequencies omega) by multiplying each by $$1/\sqrt{2\pi}$$. This restores symmetry but changes the normalization. Thus the first two conventions are preferred in physics and engineering.

EDIT: Don't know why Latex put primes and dots next to the differentials...