is there any problem in my FFT?

6 views (last 30 days)
SUNHAK KIM on 28 Oct 2021
Commented: David Goodmanson on 28 Oct 2021
AB = C (time domain data)
A and C : known
B : unkown
length of A,B and C is 512
N1 = 512;
t = t1(1:N1-1);
A = A1(1:N1-1);
C =C1(1:N1-1);
T1 = tpp*N1;
fpp = 1/T1;
a1 = 0.3906;
nr = N1*a1;
fs = 1/tpp;
df1 = 0:fs/N1:(N1-1)*fs/N1;
dfp1 = df1(1:nr);
fft1 = fft(A);
fft_Mag_A(1) = (1/N1)*abs(fft1(1));
fft_Mag_A(2:nr) = (2/N1)*abs(fft1(2:nr));
fft_Phase_A = angle(fft1(1:nr));
fft2 = fft(C);
fft_Mag_C(1) = (1/N1)*abs(fft1(1));
fft_Mag_C(2:nr) = (2/N1)*abs(fft1(2:nr));
fft_Phase_C = angle(fft1(1:nr));
B = fft_Mag_C/fft_Mag_A*cos(w*t+fft_Phase_C - fft_Phase_A);
fft3 = fft(B);
fft_Mag_B(1) = (1/N1)*abs(fft3(1));
fft_Mag_B(2:nr) = (2/N1)*abs(fft3(2:nr));
fft_Phase_B(1:nr) = angle(fft3(1:nr));
C1 = fft_Mag_A*fft_Mag_B*cos(w*t+fft_Phase_A+fft_Phase_B)
here is a problem.
why C1 and C(data) doesn't match perfectly each other when i draw graph?
There's a slight error.
  1 Comment
David Goodmanson
David Goodmanson on 28 Oct 2021
Hi SK,
could you explain the purpose of doubling the magnitudes between 2 and nr, and also why the magnitude and phase of C are based on fft1 and not fft2?

Sign in to comment.

Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!