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While trying to do frequency domain filtering of audio signals using windowed overlap-add methodology, I came across some pre-calculated frequency domain coefficients. I noticed that the coefficients, complex numbers, are not conjugate symetric. Can that be? What does it tell me about the filter?

If the input filter is FIR, described by y(n)=b(1)x(n)+b(2)x(n−1)+⋯+b(nb+1)x(n−nb), then the discrete Fourier transform of this real filter coefficients should be conjugate symetric as for any other real input. Furthermore, after the input signal is transformed to frequency domain, multiplied with the filter coefficients, and transformed back to time domain using inverse DFT, the signal (audio_filtered) should again be real?

A = fft(audio);
H = fft(filter);
Afiltered = A .* H;
audio_filtered = ifft(Afiltered)

If the input filter is IIR, are things different with respect to conjugate symetry?

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  • $\begingroup$ " I noticed that the coefficients, complex numbers, are not conjugate symetric. " please provide the data or a link to it Without context it's impossible to answer this $\endgroup$
    – Hilmar
    May 12 at 16:12
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I noticed that the coefficients, complex numbers, are not conjugate symetric.

Check what is the frequency associated with each coefficient, usually the coefficients are defined such that $a_i$ corresponds to the frequency $f_s i / N$, where $f_s$ is the sampling frequency, and $N$ is the number of coefficients. So what you have is $a_i = conj(a_{N-i})$, in some cases they may be circularly shifted, and so on.

the signal (audio_filtered) should again be real?

Yes, if you think in the time domain, the convolution involves only real values, so the result is a real signal. If you think in the frequency domain, both the filter and the signal have Hermitian symmetry, so the product of them will have hermitian symmetry as well.

Proof, let the signal with FFT $X$, and the filter with FFT $H$, the FFT of the filtered signal $Y$ is

$Y_{i} = X_i H_i = conj(X_{N-i})conj(H_{N-i}) = conj(X_{N-i}H_{N-i}) = Y_{N-i}

If the input filter is IIR, are things different with respect to conjugate symetry?

No, if you look at the fourier transform of an infinite signal

$$X(\omega) = \sum_{k=0}^{\infty} u_k exp(-2j\pi k \omega)$$

Assume $u_k$ is real, and will will notice that since $e^{-j x}$ is the complex conjugate of $e^{+jx}$, each term will be conjugated and so, $X(\omega)$ is the complex conjugate of $X(-\omega)$

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  • $\begingroup$ Sorry, which questin do you anser with "No"... I asked many. :-) $\endgroup$
    – Danijel
    May 12 at 14:27
  • $\begingroup$ If the input filter is IIR, are things different with respect to conjugate symetry? $\endgroup$
    – Bob
    May 12 at 15:52
  • $\begingroup$ Have I addressed them now? $\endgroup$
    – Bob
    May 12 at 16:04

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