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Guilherme, The relationship of your sampling rate and FFT size is one that will determine
the frequency resolution. I have used the TI RFFT at various resolutions and sampling rates and
have validated performance against spectrum analyzer outputs. In general your frequency
resolution will be SAMPLING_RATE/FFT_SIZE .

If this is all not news to you, then also performing the sum of squares on the rfft result
vector, or just the sum to experiment.

From TI's 54x DSP Lib Document:
"
On output x will contain the FFT(x) = y in the following format:
y(0)Re y(nx/2)im ? DC and Nyquist
y(1)Re y(1)Im
y(2)Re y(2)Im
"

I think I did one of the following,
Ignoring the DC y(0) and y(1),
FFTout(0) = y(2)^2 + y(3)^2
FFTout(1) = y(4)^2 + y(5)^2
etc..

Also try,
FFTout(0) = y(1)^2 + y(2)^2
FFTout(1) = y(3)^2 + y(4)^2
etc..

Ten

Spectrum from the RFFT 
Guilherme Martinhon - Feb 7 16:05:00 2006
	Hello people! I´pm working with the RFFT function and I´m testing my program with a sinusoidal
signal with some number of samples and different FFT length. But I noticed that the frequency
resulting of the DSP is different from the nominal frequency, from the signal generator (seeing
on the osciloscope, I can affirm that the generator is not the problem). In some cases, the
difference reaches 12%. Someone of you had reach some similar result?
 
Thank you
 
Guilherme
_____________________
Guilherme Martinhon
Engenheiro Eletricista
Mestrando em Engenharia Elétrica
Área: Instrumentação Eletrônica
Laboratório de Ultra-Som
UNESP - Campus de Ilha Solteira
	

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