Usually we have two versions of cepstrum: complex and real cepstrum which depend on whether we take logarithm of FFT or the absolute value of FFT. Suppose a real signal x(n) of length N and a real FIR filter h(n) of length M, N>>M.
Question 1: the complex cepstrum is also real, i.e., ifft(ln(fft(h(n),M))) is real. How to prove this?
Question 2: I have verified using Matlab that if I change the FFT point, say calculating ifft(ln(fft(h(n),2*M))), then the results become complex. How to explain this?
Question 3: According to the above two questions, we come to the classical application of separating signal and system convolution into addition. Let C denote cepstrum, the system output is y(n)=filter(h(n), 1, x(n)) which leads to Cy=Cx+Ch. In all literature I reviewed, it is implied that Cy is real, however, to obtain Cy we need to calculate fft(y(n), N)=fft(x(n),N)*fft(h(n),N). Note that ifft(ln(fft(h(n),N))) would never get real values because of Question 2.
How the above are actually calculated?
Hi Guang
Did you have a look at following:
http://www.xavieranguera.com/tdp_2011/8-Cepstral-Analysis.pdf
Hope it helps.
Hi Guang
Did you have a look at following:
http://www.xavieranguera.com/tdp_2011/8-Cepstral-Analysis.pdf
Hope it helps.
I have some referances however send it after I discuss this in our techie group . Plz excuse till such time
Dear Guang,
I will try to answer the question of when the inverse transform of FFT is real. For this, you should be familiar with Euler's formula (see link attached). Basically, any number on the complex plane can be expressed in polar coordinates using an angle and a distance from the origin. So suppose you have a complex number Z=r
I have tried several times but still haven't figured out how to reply an answer in this tricky system. Anyway, Parminder, Mandar, and especially Fernando, tthank you guys so much for your answers and concerns.
Fernando,
Yes. You are definitely correct. Now I am thinking that it may be the Matlab calculation precision problem that gave me imaginary parts when I run IFFT.
Best wishes,
Guang Hua
Yes, it sometimes happens that calculation errors yield non-zero imaginary parts, but they should still be very close to zero, like 1e-16. If they are significantly larger than that, then there may be something more running behind the scenes.
Best wishes,
Fernando
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