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iFFT Scaling Problem

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cmd
cmd el 27 de Abr. de 2011
Comentada: Nina Nounou el 20 de Mzo. de 2017
Hi,
I want the transform a return loss versus frequency data to time domain. I have a .txt file from a cable analyzer within frequency return loss and phase information.
I wrote the ifft code. At time axis there is no problem , but the return loss values in time domain didn't match with the cable analyzer result. So i have to multiply it with a value. The code i wrote is as below:
fid = fopen('10.txt');
values= fscanf(fid,'%f %f %f \n',[3,inf]);
values = values';fclose(fid);
f = values(:,1);
rloss = values(:,2);
degree = values(:,3);
amp = 10.^(rloss/10);
radian = degree*pi/180;
complex = amp.*exp(i*radian);
GAMMA_F = complex;
%-------------IFFT-----------------------%
NFFT =517;
f1 = 700e6;
f2 = 1000e6;
Fstep = (f2-f1)/(NFFT-1);
dt = 1/(Fstep*(NFFT-1));
t = dt* [0:NFFT-1];
GAMMA_T = 30*ifft(GAMMA_F,NFFT);
RL = 10*log(abs(GAMMA_T));
In this code i have to multiply the ifft result with 30 in order to match with analyzer result. Why?
And if i change the source .txt file for a different experiment result. I have to multiply with a different factor?
I need to help to resolve this problem.
Thnaks in advance!
  1 comentario
Nina Nounou
Nina Nounou el 20 de Mzo. de 2017
hi i have the same problem as you ,can you help me please

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Respuestas (2)

Amit
Amit el 21 de Dic. de 2011
Hi, I too have the same problem. If you resolved it could you share your result with me. Thanks.

Wayne King
Wayne King el 21 de Dic. de 2011
It's hard to say anything definitive without knowing what is actually in this code:
rloss = values(:,2);
degree = values(:,3);
amp = 10.^(rloss/10);
I don't think this is your issue, but I doubt you want:
RL = 10*log(abs(GAMMA_T));
I suspect you minimally want
RL = 10*log10(abs(GAMMA_T));
or
RL = 20*log10(abs(Gamma_T));
  2 comentarios
Dr. Seis
Dr. Seis el 21 de Dic. de 2011
I would have suspected that the lack of symmetry (i.e., lack of negative frequencies) in the frequency domain data would have been the culprit.
Wayne King
Wayne King el 21 de Dic. de 2011
@Elige Grant that is also a very good point!

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