I have an existing zero-delay feedback (ZDF) 2-pole state variable filter implementation (along the lines of the theory presented in VA Filter Design by V. Zavalishin), and I wish to determine the frequency response so that I can plot it. My usual approach with other filters (e.g. biquads) has been to determine the difference equation by inspection, apply the Z-transform, and then substitute $z=e^{j\omega}$ and plot phase and magnitude against $\omega$ from $0$ to $\pi$.

However I am having difficulty determining the difference equation of this filter.

Here is a diagrammatic representation of the implementation I have:ZDF SVF

Here's the implementation of an iteration, corresponding to a single input sample, in pseudo-code. Fs is the sample rate, F is the filter frequency of interest, R is the resonance, x is the input sample, and lp, bp, hp represent the low, band and high-pass outputs respectively:

# z1 and z2 contain state from previous iteration

k = tan(pi / Fs * F)
b = R + k
a = 1 / (1 + b * k)

hp = (x - (z2 + z1 * b)) * a
x1 = hp * k  # temporary
bp = x1 + z1
x2 = bp * k  # temporary
lp = x2 + z2

z1 = x1 + bp
z2 = x2 + lp

For analysis, I assign $v[n]$ to be the output of z1, and $w[n]$ to be the output of z2. From this I am able to determine by inspection the following equations:

$$hp[n] = x[n] - (b.v[n] + w[n])$$ $$v[n] = k.hp[n-1] + bp[n-1]$$ $$w[n] = k.bp[n-1] + lp[n-1]$$ $$bp[n] = k.hp[n] + v[n]$$ $$lp[n] = k.bp[n] + w[n]$$

I'm not sure how to handle such a system of equations algebraically. For example, if I wish to analyse the low-pass response, then I will need to determine a difference equation for $lp[n]$ in terms of $x[n]$. However I have not had success manipulating these equations by rearrangement and substitution even with the help of the Z transform. Is it possible to do this or are there too many unknown variables?

If an algebraic solution is impossible, is there an alternative method to determine the frequency response of this filter?


2 Answers 2


Solving for the input-output relations of such a structure may be a bit tedious, but not difficult. For the sake of generality I define the value of the first multiplier $k$ (closer to the input) as $c$, and the value of the second multiplier $k$ as $d$. In your case $c=d=k$. Furthermore I define an auxiliary signal $w[n]$ (with $\mathcal{Z}$-transform $W(z)$) at the output of the adder at the bottom of your figure, which is fed back to the input. The output signals are $y_h[n]$, $y_b[n]$, and $y_l[n]$ for the high-pass, band-pass and low-pass signals, respectively.

In the $\mathcal{Z}$-transform domain you get the following $4$ equations for the $4$ unknowns ($W(z)$, $Y_h(z)$, $Y_b(z)$, $Y_l(z)$):

$$\begin{align}W(z)&=bz^{-1}[cY_h(z)+Y_b(z)]+z^{-1}[dY_b(z)+Y_l(z)]\\ Y_h(z)&=a[X(z)-W(z)]\\ Y_b(z)&=c(1+z^{-1})Y_h(z)+Y_b(z)z^{-1}\\ Y_l(z)&=d(1+z^{-1})Y_b(z)+Y_l(z)z^{-1}\end{align}$$

This system of equations can be solved for the (transforms of the) three output signals in terms of the input $X(z)$. The solutions have the form


where the polynomial $P(z^{-1})$ has the form


and where all constants ($A$, $B$, $g_b$, $g_h$, $g_l$) of course depend on the multiplier values $a$, $b$, $c$, and $d$.

  • $\begingroup$ Thank you. My algebra was leading my into trouble as I ended up with [n-1], [n-2], [n-3], ... terms. I did not think to convert to the Z transform earlier. However, the bit that I would ask for an additional hint with please - how do you "solve" that system of equations? Does the standard linear algebra technique using matrices for solving four equations in four unknowns apply? I just need prompting in the right direction here as it's been a long time since I've tackled this sort of thing. It is reassuring to know that it is possible though - I just wasn't sure if I was wasting my time. $\endgroup$
    – davidA
    May 20, 2017 at 23:58
  • $\begingroup$ @meowsqueak: It's just like solving any other system of linear equations. In this case you could: plug Eq1 into Eq2, use Eq3 to express $Y_h$ in terms of $Y_b$ and plug that into Eq2. Then you can express $Y_b$ in terms of $X$. Now you can use Eqs 3 and 4 to express $Y_h$ and $Y_l$ in terms of $Y_b$. $\endgroup$
    – Matt L.
    May 21, 2017 at 8:59
  • $\begingroup$ I think $W(z)$ should be $W(z)=bz^{-1}[cY_h(z)+Y_b(z)] + z^{-1}(dY_b(z)+Y_l(z))$ otherwise only the feedback from the first integrator stage is used. $\endgroup$
    – davidA
    May 22, 2017 at 3:40
  • 1
    $\begingroup$ @meowsqueak: You're right, I forgot to type the second part of the equation. Corrected. $\endgroup$
    – Matt L.
    May 22, 2017 at 7:00

I do not see any reason why this could not be derived algebraically. But this would be a good application for Mason's gain formula; the references below may be helpful for that approach:

Rick Lyon's Tutorial

Mason's Gain Formula - Wikipedia

Youtube tutorial

Tutorial's Point tutorial

  • $\begingroup$ Thank you, I haven't encountered the Mason's Gain technique before. I will see if I can use it with this system. BTW, in a comment on a page related to one of those, someone mentioned using matrices to solve the system flow equations. I'd like to know more about this technique - do you know what it's called or where I can find more info on this please? $\endgroup$
    – davidA
    May 21, 2017 at 0:00
  • $\begingroup$ see "Solutions of Linear Systems" at this link: en.wikipedia.org/wiki/Linear_algebra $\endgroup$ May 21, 2017 at 0:57

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