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I try to understand the implementation of the low-pass comb filter of the Freeverb reverberation algorithm:

https://ccrma.stanford.edu/~jos/pasp/Lowpass_Feedback_Comb_Filter.html

The original implementation is as follows:

inline float comb::process(float input)
{
    float output;
    output = buffer[bufidx];
    filterstore = (output*damp2) + (filterstore*damp1);
    buffer[bufidx] = input + (filterstore*feedback);
    if(++bufidx>=bufsize) bufidx = 0;
    return output;
}

I drew the closed loop of this algorithm:

enter image description here

In the link, it says:

Inspection of comb.h in the Freeverb source shows that Freeverb's ``comb'' filter is more specifically a lowpass-feedback-comb filter (LBCF4.11--§2.6.5). It is constructed using a delay line whose output is lowpass-filtered and summed with the delay-line's input. The particular lowpass used in Freeverb is a unity-gain one-pole lowpass having the transfer function

$\displaystyle H(z) = \frac{1-d}{1-d\,z^{-1}}. $

When $ d=0$ , the LBCF reduces to the feedback comb filter (FBCF) of §2.6.2 in which the feedback was not filtered. The overall LBCF transfer function is then

$ \displaystyle \hbox{LBCF}_{N}^{\,f,\,d} \;= \; \frac{z^{-N}}{1 -f\frac{1-d}{1-d\,z^{-1}}\,z^{-N}}. $

Apparently, this transfer function is implemented here. But can someone tell me, how to derive this implementation out of this transfer function?

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Your diagram looks correct. Let's call the transfer function in the feedback loop $G(z)$. Consider the signal $w[n]$ at the input to the delay line. Its $\mathcal{Z}$-transform satisfies

$$W(z)=X(z)+G(z)Y(z)\tag{1}$$

where $X(z)$ and $Y(z)$ are the $\mathcal{Z}$-transforms of the input and output sequences, respectively. The output is just a delayed version of $w[n]$, i.e.,

$$Y(z)=W(z)z^{-M}\tag{2}$$

From $(1)$ and $(2)$, the transfer function of the complete system can be derived as

$$H(z)=\frac{z^{-M}}{1-G(z)z^{-M}}\tag{3}$$

Now it remains to find $G(z)$. It is quite straightforward to show that

$$G(z)=\frac{d_2f}{1-d_1z^{-1}}\tag{4}$$

Combining $(3)$ and $(4)$ gives the desired result with $d_1=d$ and assuming that $d_2=1-d_1$.

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