FFT function in matlab

Hi

i just wanted to check out something regarding the fft function in matlab?

Does the spectra produced by the fft function give amplitudes or power?

Answer 1

[Rick Rosson]

Neither. It is a straight implementation of the definition of the DFT. Depending on how you choose to scale the result, you can compute one of a variety of different forms of the spectrum. It is in general proportional to the amplitude spectrum and not the power spectrum.

What are you trying to compute using fft?

Answer 2

[Image Analyst]

Maybe you'd rather use pwelch() in the Signal Processing Toolbox.

Answer 3

[Bran]

Hey everyone,

Well I am playing aound with fourier synthesis and using my spectrum as a guide. I understand that I need to sum cosine and sine waves to get back to my time signal. I think if I understand correctly that the coefficients of the sine waves are governed by the imaginary values of the FFT and cosine are governed by the real parts. I wanted to check that FFT gave amplitudes so I could use it as an accuaret guide because at the moment I am not recreating the signal I would expect

[Rick Rosson]

Please post your code.

[Bran]

b1 = -1.5153e+004; 
a1 = 4.4371e+004;

Total1 = abs(complex(b1,a1));
phaseangle1 = atan(a1/b1);
b2 = -6.5515e+001;
a2 = 2.7679e+004;

Total2 = abs(complex(b2,a2));
phaseangle2 = atan(a2/b2);

f1 = 0.8241;
f2 = 1.648;

t = linspace(0,1*60,60*60);

Yend = Total1*cos(2*pi*f1*t+phaseangle1)+Total2*cos(2*pi*f2*t+phaseangle2);

%made mistake of not keeping freq = freq1 looks alot more like what rowan
%would get

bb1 = 1.8949e+004;
aa1 = 1.0083e+004;

Total11 = abs(complex(bb1,aa1));
angle11 = atan(aa1/bb1);
bb2 = -7.5509e+003;
aa2 =9.4735e+003;
Total22 = abs(complex(bb2,aa2));
angle22 = atan(aa2/bb2);

bb3 = -2.9012e+003;
aa3 = 8.1011e+003;

Total33 = abs(complex(bb3,aa3));
angle33 = atan(aa3/bb3);
bb4 =-7.1940e+002;
aa4 = 7.3516e+002;
Total44 = abs(complex(bb4,aa4));
angle44=atan(aa4/bb4);
ff1 = 0.8927;
ff2 = 1.801;
ff3 = 2.718;
ff4 = 3.567;

yy1 = sin(2*pi*ff1*t);
yy2 = sin(4*pi*ff1*t);
yy3 = sin(6*pi*ff1*t);
yy4 = sin(8*pi*ff1*t);
YY11 = cos(2*pi*ff1*t);
YY22 = cos(4*pi*ff1*t);
YY33 = cos(6*pi*ff1*t);
YY44 = cos(8*pi*ff1*t);

Yend2 = Total11*cos(2*pi*ff1*t+angle11)+Total22*cos(2*pi*ff2*t+angle22)+Total33*cos(2*pi*ff3*t+angle33)+Total44*cos(2*pi*ff4*t+angle44);

plot(Yend,Yend2)

Please note that I got the complex number information of the peaks by looking at the values for FFT without taking abs()

[Bran]

Unfortunately this does not give the shape I would expect

[Rick Rosson]

Can you also post the code you have where you computed the complex values using the FFT?