Power from a Fourier transform

[neil.thompson]

So I have been away from education for a little while now and I'm going through some refresher stuff - in particular I have been playing around with FFTs.

If i take (with MATLAB notation):

time = 0:0.01:10
y = fft(sin(2*pi*f*time))

with f = 5
then the maximum amplitude of the fft output is about 498.

with f = 10
the maximum amplitude of fft output is 492.

I understand the amplitude is 'halved' in both cases because this fft is ambiguous so the energy is spread over two peaks. But why is the energy less when the frequency increases? I have more cycles in the case with more frequency, but I suppose this means I have less samples. Also, is it usual to normalise this in some way? It seems like this is something you wouldn't want if you were dealing were plotting energy return from doppler shifts.

 

[DrClaude]

But why is the energy less when the frequency increases?

How is the energy related to the FT? Have you checked the width of the peaks?

Also, is it usual to normalise this in some way? It seems like this is something you wouldn't want if you were dealing were plotting energy return from doppler shifts.

There are different ways of normalizing the FFT, but it is customary that doing FFT^-1(FFT(f)) will return N × f, although MATLAB returns the original result.

 

[neil.thompson]

How is the energy related to the FT? Have you checked the width of the peaks?
.

Good point - my language was imprecise. What I meant was peak power, or peak amplitude -- I guess what I'm asking is: if I have a frequency resolution on the FFT such that all of the signal should be confined to a signal frequency bin (e.g., each bin spans is 2kHz, my input signal is a single tone at 4.5kHz so all of the signal should fold into the third bin) so why does the height of the amplitude peak depend on the frequency of the signal?

edit: and into the bargain, related to your point, if I have enough data to get a very fine resolution FT then I guess this will reveal that there will be some spread across bins so the peak isn't the same because the curve is wider/narrower.

There are different ways of normalizing the FFT, but it is customary that doing FFT^-1(FFT(f)) will return N × f, although MATLAB returns the original result.

Right, thanks.

 

[DrClaude]

Have you tried to take a finer grid an see what you get then?

 

[CellCoree]

The Fourier transform is a powerful tool in signal processing that allows us to decompose a signal into its individual frequency components. In your example, you are using the Fast Fourier Transform (FFT) to analyze a sinusoidal signal with a frequency of either 5 or 10 Hz.

The FFT works by breaking down a signal into smaller segments, known as bins, and analyzing the frequency content of each bin. The maximum amplitude of the FFT output represents the energy at that particular frequency. In your case, you are seeing a decrease in the maximum amplitude as the frequency increases from 5 to 10 Hz.

This decrease in energy is due to the fact that as the frequency increases, the signal becomes more "compressed" in time. This means that there are fewer samples available to capture the signal, resulting in a decrease in energy at that particular frequency. This is also why you see two peaks in the FFT output - the energy is spread over two bins.

In terms of normalization, it is common to normalize the FFT output to account for the number of samples used. This allows for a more accurate comparison of the energy at different frequencies. However, it is important to note that the normalization factor may vary depending on the specific application and signal being analyzed.

In conclusion, the decrease in energy at higher frequencies in your FFT output is a result of the signal becoming more compressed in time, leading to fewer samples available to capture the signal. Normalization may be necessary for accurate comparisons between different frequencies.