FFT function in matlab

2014 10 6
3 个回答
更新时间：2014 10 7
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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?

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Rick Rosson
Rick Rosson 2014-10-6
编辑：Rick Rosson 2014-10-6

0 个投票

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?
Image Analyst
Image Analyst 2014-10-6

0 个投票

Maybe you'd rather use pwelch() in the Signal Processing Toolbox.
Bran
Bran 2014-10-6

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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

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Rick Rosson
Rick Rosson 2014-10-7
Please post your code.
Bran
Bran 2014-10-7
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
Bran 2014-10-7
Unfortunately this does not give the shape I would expect
Rick Rosson
Rick Rosson 2014-10-7
Can you also post the code you have where you computed the complex values using the FFT?

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Bran
Bran
2014-10-6

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Rick Rosson
Rick Rosson
2014-10-7

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