Chapter 8: Raising the filter order

As the order of the filter grows, there are more and more different choices of the transfer function. Particularly, there is more than one way to introduce the resonance into a transfer function of order higher than 2. Some of the most interesting options were already discussed in the previous chapters.

We have also introduced the state-space form as a general representation for differential systems. However, being so general, the state-space form leaves lots of open questions in regards to the choice of topology and the user-facing parameters.

In this chapter we are going to discuss a number of standard topologies which can be used to construct a system of any given order and also a number of ways to map commonly used user-facing parameters, such as cutoff and resonsance, to the internal parameters of such systems. Note, however, that these structures and techniques are useful only occasionally, for rather specific purposes.

8.1 Generalized SVF

We have seen that the idea of the ladder filter can be generalized from a 4-pole to other numbers of poles, even though there are problems arising at pole counts other than 4. Could we somehow attempt to generalize the SVF?

The most natural way to generalize the SVF is probably to treat it as the so-called controllable canonical form, (Fig. 8.1) which is the analog counterpart of direct form II (Fig. 3.33). Apparently, the main difference between Fig. 3.33 and Fig. 8.1 is simply that all unit delays are replaced by integrators. The other differerence, namely the inverted feedback is merely a matter of convention, resulting in opposite signs of the coefficients $a_n$ compared to what they would have been in the absence of the feedback inversion. We chose the convention with the inverted feedback mainly because it’s more similar to the 2-pole SVF structure in Fig. 4.1.

The controllable canonical form allows to implement an arbitrary transfer function of $N$-th order (the requirement that the transfer function is a non-strictly proper rational function being implicitly understood). Indeed, it’s not difficult to figure out that the tranfer function of the system in Fig. 8.1 is

Figure 8.1: Generalized SVF (controllable canonical form).

$$H(s)=\frac{\sum _{n=0}^N{b}_n{s}^{-n}}{1+\sum _{n=1}^N{a}_n{s}^{-n}}=\frac{\sum _{n=0}^N{b}_n{s}^{N-n}}{1+\sum _{n=1}^N{a}_n{s}^{N-n}}=\frac{\sum _{n=0}^N{b}_{N-n}{s}^n}{s^N+\sum _{n=0}^{N-1}{a}_{N-n}{s}^n}$$

Thus $a_n$ and $b_n$ are simply the denominator and numerator coefficients of the transfer function. Notice that $b_n$ are essentially modal pickups and we can share the feedback part of the structure (consisting of integrators and $a_n$ gains) among several different sets of pickup coefficients $b_n$ to simultaneously implement a number of filters sharing a common denominator.

Normally Fig. 8.1 assumes unit-cutoff integrators, because the $a_n$ and $b_n$ coefficients provide enough freedom to implement any transfer function of the given order. However, in music DSP applications cutoff control is a common feature, therefore we could also allow the integrators to take identical non-unit cutoffs. Further, letting $N=2$, $a_2=1$ and $a_1=2R$ we obtain an SVF with $b_n$ serving as modal mixing coefficients for HP, BP and LP outputs. On the other hand, at $N=1$, $a_1=1$ we obtain the 1-pole filter we discussed in the beginning of this book.

Generally, letting $a_N$ have a fixed value is a good way to remove the redundancy introduced into the system control by the embedded cutoffs of the integrators. It is not difficult to realize that

$$a_N=\prod (-p_n)$$

where $p_n$ are the positions of the system poles when ${\omega }_c=1$. Notably, although it is mostly academic, this also can support the case of real poles of opposite signs, which cannot be implemented by a classical 2-pole SVF due to $a_N$ being fixed to 1.

Unfortunately, there is no clear answer to what the coefficients $a_n$ should be for $N>2$. The simplicity of the 2-pole case was due to the fact that the denominator of a 2-pole transfer function essentially has only 2 degrees of freedom (corresponding to $a_1$ and $a_2$), one degree being taken by the cutoff, and we are being left with the remaining degree which just happens to correspond to the resonance. With the 1-pole there was only one freedom degree, being taken by the cutoff. At $N>2$ there are too many different options of how to map the freedom degrees to filter control parameters and there is no definite answer to that, although some of the options will be discussed later in this chapter.

With the numerator coefficients $b_n$ there is a bit more clarity, as there are certain general considerations applying more or less for any choice of $a_n$. E.g. if the numerator is equal to $a_N$, we get some kind of an $N$-th order lowpass, since $H(0)=1$ and $H(s)\sim a_N/s^N$ for $s\to \infty$. For the $s^N$ numerator we have $H(\infty )=1$ and $H(s)\sim s^N/a_N$ for $s\to 0$, corresponding to some kind of an $N$-th order highpass. For an even $N$ and an $a_N^{1/2}{s}^{N/2}$ numerator we get $H(s)\sim s^{N/2}/a_N^{1/2}$ for $s\to 0$ and $H(s)\sim a_N^{1/2}/s^{N/2}$ for $s\to \infty$, corresponding to some kind of a bandpass. This however defines only the asymptotic behavior at 0 and $\infty$, the amplitude response shape in the middle can be pretty much arbitrary, being defined by the denominator.

By transposing the controllable canonical form one obtains the so-called observable canonical form. We are not going to address it in detail, as most of the discussion of the controllable canonical form above applies to the observable canonical form as well.

8.2 Serial cascade representation

Another structure which allows implementing arbitrary transfer functions is the serial cascade. It is probably the one most commonly used. Compared to the generalized SVF, in the serial cascade representation we are using only 1- and 2-pole filters and we can choose commonly known and well-studied structures to implement those.¹ The benefit compared to the parallel implementation (discussed later in this chapter) is that the serial cascade form doesn’t get ill-conditioned when system poles get close to each other.

Cascade decomposition

Given an arbitrary $N$-th order real transfer function, let’s write it in the multiplicative form:

$$H(s)=g\cdot \frac{\prod _{n=1}^{N_z}(s-z_n)}{\prod _{n=1}^{N_p}(s-p_n)}\text{\hspace{1em}(8.1)}$$

where $N_z\le N_p$, since $H(s)$ must be nonstrictly proper. Since $H(s)$ has real coefficients, all complex poles of $H(s)$ will come in conjugate pairs, and the same can be said about the zeros.

Now we are going to write each pair of conjugate poles as a purely real 2nd-order factor in the denominator:

$$(s-p)(s-p^{\ast })=s^2-s\cdot 2\mathrm{Re}p+|p{|}^2$$

and we are going to write each pair of conjugate zeros as a purely real 2nd-order factor in the numerator:

$$(s-z)(s-z^{\ast })=s^2-s\cdot 2\mathrm{Re}z+|z{|}^2$$

Further, if necessary, we can combine any two real poles into a 2nd-order factor in the denominator:

$$(s-p_1)(s-p_2)=s^2-(p_1+p_2)\cdot s+p_1{p}_2$$

and we can combine any two real zeros into a 2nd-order factor in the numerator:

$$(s-z_1)(s-z_2)=s^2-(z_1+z_2)\cdot s+z_1{z}_2$$

Thus we can distribute all conjugate pair of poles and zeros into 2nd-order real rational factors of the form

$$\frac{s^2+as+b}{s^2+cs+d}$$

unless we do not have enough zeros, in which case there will be one or more 2nd-order real rational factors of the form

$$\frac{s+b}{s^2+cs+d}\text{and/or}\frac{1}{s^2+cs+d}$$

The remaining pairs of real poles and zeros can be combined into 1st-order real rational factors of the form

$$\frac{s+a}{s+b}\text{and/or}\frac{1}{s+b}$$

or they can be also combined into 2nd-order real rational factors, e.g.:

$$\frac{s+a_1}{s+b_1}\cdot \frac{s+a_2}{s+b_2}=\frac{s^2+(a_1+a_2)s+a_1{a}_2}{s^2+(b_1+b_2)s+b_1{b}_2}$$

Thus the entire transfer function is represented as a product of purely real 2nd- and 1st-order factors:

$$H(s)=g\cdot \prod _{n=1}^{N_2}{H}_{2n}(s)\cdot \prod _{n=1}^{N_1}{H}_{1n}(s)\text{\hspace{1em}(8.2)}$$

where $H_{2n}(s)$ and $H_{1n}(s)$ are the 2nd- and 1st-order factors respectively. The gain coefficient $g$, if desired, can be factored into the numerator of one or several of the factors $H_{2n}(s)$ and $H_{1n}(s)$, so that the product expression gets a simpler form:

$$H(s)=\prod _{n=1}^{N_2}{H}_{2n}(s)\cdot \prod _{n=1}^{N_1}{H}_{1n}(s)\text{\hspace{1em}(8.3)}$$

Now recall that 1-pole multimode can implement any stable real 1st-order transfer function and SVF can implement any stable real 2nd-order transfer function. This means that we can implement pretty much any $H(s)$ as a serial chain of SVFs² and 1st-order multimodes.³ We will refer to the process of representing $H(s)$ is a cascade form as cascade decomposition of $H(s)$.

Cutoff control

The denominator $1+s/{\omega }_c$ of a 1-pole filter is controlled by a single parameter, which is the filter cutoff. The denominator $1+2Rs/{\omega }_c+(s/{\omega }_c{)}^2$ of a 2-pole filter is controlled by cutoff and damping. Thus each of the 2- and 1-poles in (8.3) has a cutoff, defined by the positions of the respective poles. Writing explicitly these cutoff parameters in (8.3) we obtain

$$H(s)=\prod _{n=1}^{N_2}{\bar{H}}_{2n}(s/{\omega }_{2n})\cdot \prod _{n=1}^{N_1}{\bar{H}}_{1n}(s/{\omega }_{1n})$$

where ${\bar{H}}_{2n}$ and ${\bar{H}}_{1n}$ are unit-cutoff versions of the same 2- and 1-poles and ${\omega }_{2n}$ and ${\omega }_{1n}$ are the respective cutoffs.

Suppose the above $H(s)$ defines a unit-cutoff filter. Then non-unit cutoff for $H(s)$ is achieved by

$$H(s/{\omega }_c)=\prod _{n=1}^{N_2}{\bar{H}}_{2n}(s/{\omega }_c{\omega }_{2n})\cdot \prod _{n=1}^{N_1}{\bar{H}}_{1n}(s/{\omega }_c{\omega }_{1n})\text{\hspace{1em}(8.4)}$$

which means that the cutoffs of the underlying 2- and 1-poles are simply multiplied by ${\omega }_c$ and we have ${\omega }_c{\omega }_{2n}$ and ${\omega }_c{\omega }_{1n}$ as the 2- and 1-pole cutoffs.

One should remember, that it is important to apply one and the same prewarping for all filters in the cascade, as discussed in Section 3.8. E.g. we could choose to prewarp (8.4) at $\omega ={\omega }_c$, which means that we prewarp only ${\omega }_c$ (rather than individually prewarping the 2- and 1-pole cutoffs ${\omega }_c{\omega }_{2n}$ and ${\omega }_c{\omega }_{1n}$), thereby obtaining its prewarped version ${\tilde{\omega }}_c$, and then simply substitute ${\tilde{\omega }}_c$ for ${\omega }_c$ in (8.4):

$$H(s/{\tilde{\omega }}_c)=\prod _{n=1}^{N_2}{\bar{H}}_{2n}(s/{\tilde{\omega }}_c{\omega }_{2n})\cdot \prod _{n=1}^{N_1}{\bar{H}}_{1n}(s/{\tilde{\omega }}_c{\omega }_{1n})$$

Thus, the 2- and 1-pole cutoffs become ${\tilde{\omega }}_c{\omega }_{2n}$ and ${\tilde{\omega }}_c{\omega }_{1n}$ respectively.

Cascaded model of a ladder filter

As an example of the just introduced technique we are going to implement the transfer function of a 4-pole lowpass ladder filter by a serial chain of two SVFs. A 4-pole lowpass ladder filter has no zeros and two conjugate pairs of poles for $k>0$. By considering two coinciding poles on a real axis also as mutually conjugate, we can assume $k\ge 0$.

Since there are no zeros, we simply need a 2-pole lowpass SVF for each conjugate pair of poles. Let $p_1$, $p_1^{\ast }$, $p_2$, $p_2^{\ast }$ be the poles of the ladder filter. According to (5.2)

$$p_{1,2}=-1+\frac{\pm 1+j}{\sqrt{2}}{k}^{1/4}\text{\hspace{1em}(8.5)}$$

By (4.13), the cutoffs of the 2-pole lowpasses ${\omega }_{1,2}=|p_{1,2}|$ and $R=-\mathrm{Re}{p}_{1,2}/|p_{1,2}|$. Respectively the transfer function of the ladder filter can be represented as

$$H(s)=g\frac{1}{{\left(\frac{s}{{\omega }_1}\right)}^2+2R_1\frac{s}{{\omega }_1}+1}\cdot \frac{1}{{\left(\frac{s}{{\omega }_2}\right)}^2+2R_2\frac{s}{{\omega }_2}+1}\text{\hspace{1em}(8.6)}$$

The unknown gain coefficient $g$ can be found by evaluating (5.1) at $s=0$, obtaining the condition $H(0)=1/(1+k)$. Evaluating (8.6) at $s=0$ yields $H(0)=g$. Therefore

$$g=\frac{1}{1+k}$$

This gives us a cascade of 2-poles implementing a unit-cutoff ladder filter. Extending (8.6) to arbitrary cutoffs is respectively done by

$$H(s)=\frac{1}{1+k}\cdot \frac{1}{{\left(\frac{s}{{\omega }_c{\omega }_1}\right)}^2+2R_1\frac{s}{{\omega }_c{\omega }_1}+1}\cdot \frac{1}{{\left(\frac{s}{{\omega }_c{\omega }_2}\right)}^2+2R_2\frac{s}{{\omega }_c{\omega }_2}+1}$$

Cascaded multimode

The cascade decomposition can be also used to provide modal outputs, sharing the same transfer function denominator. In order to demonstrate this we will consider a serial connection of two SVFs.⁴

The transfer function of such structure can have almost any desired 4th order stable denominator.⁵ We would like to construct modal outputs for such connection, so that by mixing those modal signals we should be able to obtain arbitrary numerators. This should allow us to share this chain of SVFs for generation of two or more signals which share the same transfer function’s denominator.

We have several options of connecting two SVFs in series, depending on which of the modal outputs of the first SVF is connected to the second SVF’s input. The most symmetric option seems to be picking up the bandpass output (Fig. 8.2).

Now let

$$\begin{array}{rl}{D}_1(s)& =s^2+2R_1{\omega }_{1}s+{\omega }_1^2\\ D_2(s)& =s^2+2R_2{\omega }_{2}s+{\omega }_2^2\end{array}$$

be the denominators of the transfer functions of the two SVFs and let $D(s)=D_1(s)D_2(s)$ be their product. Writing out the transfer functions for the signals at the SVF outputs (in respect to the input signal $x(t)$ in Fig. 8.2) we obtain

Figure 8.2: A multimode cascade of two SVFs.

$$\begin{array}{rl}{H}_{LP1}(s)& =\frac{{\omega }_1^2}{D_1(s)}=\frac{{\omega }_1^2{D}_2(s)}{D(s)}\\ H_{BP1}(s)& =\frac{{\omega }_{1}s}{D_1(s)}=\frac{{\omega }_{1}s{D}_2(s)}{D(s)}\\ H_{HP1}(s)& =\frac{s^2}{D_1(s)}=\frac{s^2{D}_2(s)}{D(s)}\\ H_{LP2}(s)& =\frac{{\omega }_2^2}{D_2(s)}\cdot H_{BP1}(s)=\frac{{\omega }_2^2{\omega }_{1}s}{D(s)}\\ H_{BP2}(s)& =\frac{{\omega }_{2}s}{D_2(s)}\cdot H_{BP1}(s)=\frac{{\omega }_2{\omega }_1{s}^2}{D(s)}\\ H_{HP2}(s)& =\frac{s^2}{D_2(s)}\cdot H_{BP1}(s)=\frac{{\omega }_1{s}^3}{D(s)}\end{array}$$

Or, since we have the common denominator $D(s)$ everywhere, we could concentrate just on the numerators:

$$\begin{array}{rl}{N}_{LP1}(s)& ={\omega }_1^2{D}_2(s)\\ N_{BP1}(s)& ={\omega }_{1}s{D}_2(s)\\ N_{HP1}(s)& =s^2{D}_2(s)\\ N_{LP2}(s)& ={\omega }_2^2{\omega }_{1}s\\ N_{BP2}(s)& ={\omega }_2{\omega }_1{s}^2\\ N_{HP2}(s)& ={\omega }_1{s}^3\end{array}$$

Noticing from Fig. 8.2 that BP1 can be obtained as LP2 + $2R_2$BP2 + HP2 anyway, we can drop the respective numerator from the list and try to arrange the remaining ones in the order of the descending polynomial order:

$$\begin{array}{rl}{N}_{HP1}(s)& =s^2{D}_2(s)=s^4+2R_2{\omega }_2{s}^3+{\omega }_2^2{s}^2\\ N_{HP2}(s)& ={\omega }_1{s}^3\\ N_{BP2}(s)& ={\omega }_2{\omega }_1{s}^2\\ N_{LP2}(s)& ={\omega }_2^2{\omega }_{1}s\\ N_{LP1}(s)& ={\omega }_1^2{D}_2(s)={\omega }_1^2{s}^2+2R_2{\omega }_1^2{\omega }_{2}s+{\omega }_1^2{\omega }_2^2\end{array}$$

The last line doesn’t really fit, and the first one looks more complicated than the next three, but we can fix that by replacing the first and the last lines by linear combinations:

$$\begin{array}{rl}{N}_{HP1}(s)-2R_2\frac{{\omega }_2}{{\omega }_1}{N}_{HP2}(s)-\frac{{\omega }_2}{{\omega }_1}{N}_{BP2}(s)& =s^4\\ N_{HP2}(s)& ={\omega }_1{s}^3\\ N_{BP2}(s)& ={\omega }_2{\omega }_1{s}^2\\ N_{LP2}(s)& ={\omega }_2^2{\omega }_{1}s\\ N_{LP1}(s)-\frac{{\omega }_1}{{\omega }_2}{N}_{BP2}(s)-2R_2\frac{{\omega }_1}{{\omega }_2}{N}_{LP2}(s)& ={\omega }_1^2{\omega }_2^2\end{array}$$

Thus we can obtain all powers of $s$ from linear combinations of LP1, HP1, LP2, BP2 and HP2, thereby being able to construct arbitrary polynomials of orders up to 4 for the numerator.

Notably, instead of connecting the bandpass output of the first SVF to the input of the second SVF, as it has been shown in Fig. 8.2, we could have connected the lowpass or the highpass output. This would have resulted in somewhat different math, but essentially gives the same modal mixture options.

8.3 Parallel representation

Real poles

Given a transfer function which has only real poles which are all distinct, we could expand it into a sum of 1st-order partial fractions. Each such 1st-order fraction corresponds to a 1-pole and we could implement the transfer function as a sum of 1-poles. Essentially this is identical to the diagonal state-space form, which, provided all system poles are real and sufficiently distinct (so that no ill-conditioning occurs), is just a set of parallel Jordan 1-poles.

In the case of a single-input single-output system, which we are currently considering, the transfer function of such diagonal system, given by (7.18), has the form

$$H(s)=\sum _{n=1}^N\frac{c_n{b}_n}{s-p_n}+d\text{\hspace{1em}(8.7)}$$

where $b_n$ and $c_n$ are the input and output gains respectively. Given a particular nonstrictly rational $H(s)$, the partial fraction expansion (8.7) uniquely defines $d$ and the products $c_n{b}_n$. The respective freedom of choice of $c_n$ and $b_n$ can be resolved by letting $b_n=1\forall n$ and thus we control the numerator of the transfer function by the output mixing coefficients $c_n$ (Fig. 8.3).⁶

Figure 8.3: Implementation by parallel Jordan 1-poles.

We could also replace Jordan 1-poles by ordinary 1-pole lowpasses, where we need to divide the mixing coefficients by the respective cutoffs ${\omega }_{cn}$ (Fig. 8.4).

Figure 8.4: Implementation by parallel 1-pole lowpasses.

The global cutoff control of the entire filter in Fig. 8.3 or Fig. 8.4 is achieved in the same way as with serial cascades. Obviously, the usual consideration of common prewarping of the 1-pole components applies here as well.

Complex poles

If system poles are complex we need to use the real diagonal form, which replaces the complex Jordan 1-poles with Jordan 2-poles. For a single-input single- output system, equation (7.28) takes the form

$$H(s)=\sum _{\mathrm{Im}{p}_n>0}\frac{{\alpha }_{n}s+{\beta }_n}{s^2-2\mathrm{Re}{p}_n\cdot s+|p_n{|}^2}+\sum _{\mathrm{Im}{p}_n=0}\frac{c_n{b}_n}{s-p_n}+d\text{\hspace{1em}(8.8)}$$

We could obtain the explicit expressions for ${\alpha }_n$ and ${\beta }_n$ from the derivation of (7.28), but it would be more practical to simply obtain their values from the partial fraction expansion of $H(s)$. That is, given $H(s)$, we find ${\alpha }_n$ and ${\beta }_n$ (as well as, of course, $c_n{b}_n$ and $d$) from (8.8). We also should remember that, according to the freedom of choice of the state space basis vectors lengths, we could choose any non-zero input gains vector, e.g. $(10{)}^{\mathsf{T}}$ which means that we are using only the “real part” input of the Jordan 2-pole.⁷ According to (7.26), the contribution of such Jordan 2-pole to $H(s)$ will be

$$\frac{1}{s^2-2\mathrm{Re}{p}_n\cdot s+|p_n{|}^2}\left(\begin{array}{cc}{c}_n& c_{n+1}\end{array}\right)\left(\begin{array}{cc}s-\mathrm{Re}{p}_n& -\mathrm{Im}{p}_n\\ \mathrm{Im}{p}_n& s-\mathrm{Re}{p}_n\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=$$

$$=\frac{c_n(s-\mathrm{Re}{p}_n)+c_{n+1}\mathrm{Im}{p}_n}{s^2-2\mathrm{Re}{p}_n\cdot s+|p_n{|}^2}=\frac{c_{n}s+(c_{n+1}\mathrm{Im}{p}_n-c_n\mathrm{Re}{p}_n)}{s^2-2\mathrm{Re}{p}_n\cdot s+|p_n{|}^2}$$

Thus

$$\begin{array}{rl}{\alpha }_n& =c_n\\ {\beta }_n& =c_{n+1}\mathrm{Im}{p}_n-c_n\mathrm{Re}{p}_n\end{array}$$

from where

$$\begin{array}{rl}{c}_n& ={\alpha }_n\\ c_{n+1}& =\frac{{\beta }_n+{\alpha }_n\mathrm{Re}{p}_n}{\mathrm{Im}{p}_n}\end{array}$$

Thus, having found ${\alpha }_n$ and ${\beta }_n$, we can find $c_n$ and $c_{n+1}$. The respective structure is shown in Fig. 8.5. Notice that as $\mathrm{Im}{p}_n$ becomes smaller, $c_{n+1}$ becomes larger. This is the ill-conditioning effect of the diagonal form discussed in Section 7.11.

Figure 8.5: Implementation by parallel Jordan 2- and 1-poles. Disconnected imaginary part inputs are receiving zero signals.

Similarly to how we could replace Jordan 1-poles with ordinary 1-pole lowpasses, we could replace Jordan 2-poles by some other 2-poles, e.g. by SVFs. Finding the output mixing coefficients becomes simpler, since, apparently, the coefficients ${\alpha }_n$ and ${\beta }_n$ in (8.8) now simply correspond to SVF bandpass and lowpass output gains (properly scaled by the cutoff). Fig. 8.6 illustrates.

Figure 8.6: Implementation by parallel SVFs and 1-pole lowpasses.

Another benefit of an SVF is that it doesn’t have a problem at the point where its poles coincide and also can support the case of real poles, meaning that we could convert arbitrary pairs of parallel 1-poles into an SVF. The same apparently could be done by an SKF/TSK. There would still be a problem though, if poles of different parallel 2-poles coincide, resulting in the already known ill-conditioning effect.

Regarding the cutoff control of the entire system, there is no difference from the parallel 1-poles case.

Coinciding poles

Generally, anything with repeated or close to each other poles cannot be implemented in a parallel form and needs some non-parallel implementation (SVF, a chain of SVFs, Jordan chain, etc.) However the implementation could still be partially parallel, where the poles may be repeated within each block, but different parallel blocks shouldn’t have poles at the same locations.

8.4 Cascading of identical filters

So we have learned a number of different ways to implement higher-order transfer functions, of which cascaded form is said to be usually the best option, however, how do we construct these transfer functions in the first place? E.g. how do we generalize a resonating 2-pole transfer function to a 4-th or 8-th order? Or how do we generalize a 1-st order lowpass to a 5-th or 8-th order?

One possible way which could immediately occur to us is to stack several identical filters together. Note that, given a filter with the transfer function $G(s)$ and another one with the transfer function $H(s)=G^N(s)$ and looking at their decibel-scale amplitude responses, we notice that the latter is simply the former multiplied by $N$, that is the amplitude response becomes scaled $N$ times vertically (obviously, the same scaling is happening to the phase response). Particularly this means that the rolloff slope of the filter becomes $N$ times steeper.

Therefore in order to generalize a 1-st order lowpass $1/(1+s)$ to the $N$-th order we could simply connect $N$ such lowpasses in series:

$$H(s)={\left(\frac{1}{1+s}\right)}^N$$

resulting in the amplitude response curve in Fig. 8.7. It looks as if the cutoff of $H(s)=G^N(s)$ is is too low. In principle we could address this by shifting the filter cutoff, so that $|G^N(j)|=1/\sqrt{2}$. In order to do so we solve the equation

$$\frac{1}{|1+j\omega {|}^N}=\frac{1}{\sqrt{2}}$$

obtaining the frequency which should be treated as the cutoff point of each of the chain’s elements:

$$\omega =\sqrt{2^{1/N}-1}$$

so the transfer function becomes

$$H(s)={\left(\frac{1}{1+s\sqrt{2^{1/N}-1}}\right)}^N\text{\hspace{1em}(8.9)}$$

This looks a bit better (Fig. 8.8) and can be taken as a possible option.

Figure 8.7: Amplitude response of a 1-pole lowpass filter (dashed) vs. amplitude response of a serial chain of 4 identical 1-pole lowpass filters (solid).

In the same way we could generalize a resonating 2-nd order lowpass $1/(1+2Rs+s^2)$ to the $2N$-th order by connecting $N$ of such lowpasses together

$$H(s)={\left(\frac{1}{1+2Rs+s^2}\right)}^N$$

However in this case the situation is somewhat worse than with 1-poles. First we notice that the resonance peak becomes much higher at the same damping (Fig. 8.9). At first sight it doesn’t look like a big problem, we could simply use smaller values of the damping. However if we compare the amplitude response curves of a 2-pole vs. $N$ stacked 2-poles with the damping adjusted to produce the same peak height,⁸ we notice that due to the now smaller damping value the resonance peak of the 2-pole chain is much wider than the peak of a single 2-pole (Fig. 8.10), all in all not a very desirable scenario.

Figure 8.8: Amplitude response of a 1-pole lowpass filter (dashed) vs. amplitude response of a serial chain of 4 identical 1-pole lowpass filters with adjusted cutoff (solid).

8.5 Butterworth transformation

We have seen that cascading $N$ identical filters is one possible way to obtain higher-order filters, which effectively scales the decibel-scale amplitude response and the phase response of the filter $N$ times vertically, respectively making the filter rolloff $N$ times steeper.

Another way to make the rolloff $N$ times steeper would be finding a transformation which shrinks the amplitude response in the logarithmic frequency scale $N$ times:

$$\log \omega \leftarrow N\log \omega \omega \ge 0N=2,3,4,\dots$$

(where we don’t care about $\omega <0$ because for real filters $|H(j\omega )|=|H(-j\omega )|$, and where $\log 0=-\infty$). Or equivalently

$$\omega \leftarrow {\omega }^N\omega \ge 0\text{\hspace{1em}(8.10)}$$

The readers may recall the LP to HP transformation $s\leftarrow 1/s$ which flips the responses in the logarithmic frequency axis. One could try to draw an analogy and attempt substitutions of the form $s\leftarrow s^N$ or $s\leftarrow a{s}^N$ ($a\in \mathbb{C}$, $|a|=1$), however it’s not difficult to convince oneself that such substitutions do not work. Nevertheless, the basic direction is mostly right. Just instead of of performing an argument substitution on the transfer function, we will directly apply (8.10) to the amplitude response $|H(j\omega )|$. That is we will be looking for such $H^{\prime }(s)$ that

Figure 8.9: Amplitude response of a 2-pole filter (dashed) vs. amplitude response of a serial chain of 4 identical 2-pole filters (solid).

Figure 8.10: Amplitude response of a 2-pole filter (dashed) vs. amplitude response of a serial chain of 4 identical 2-pole filters with adjusted damping (solid).

$$|H^{\prime }(j\omega )|=|H(j{\omega }^N)|\omega \ge 0\text{\hspace{1em}(8.11)}$$

We will refer to the transformation of $H(s)$ into $H^{\prime }(s)$ defined by (8.11) as Butterworth transformation.⁹ The integer $N$ will be respectively referred to as the order of the Butterworth transformation. We will denote Butterworth transformation as

$$H^{\prime }(s)=\mathcal{B}[H(s)]$$

or, if we want to explicitly specify the order

$$H^{\prime }(s)={\mathcal{B}}_N[H(s)]$$

where $H^{\prime }(s)$ denotes the new transfer function obtained as the result of the transformation.¹⁰

Without having developed the transformation details yet, we can already establish several properties of this transformation, which follow from (8.11):

• the transformation doesn’t change a constant function:

$$\mathcal{B}[a]=a\text{\hspace{1em}(8.12a)}$$

• a constant gain can be simply factored out of the transformation:

$$\mathcal{B}[g\cdot H(s)]=g\cdot \mathcal{B}[H(s)]\text{\hspace{1em}(8.12b)}$$

• a change of the cutoff is shrunk $N$ times in the logarithmic scale after the transformation:

$${\mathcal{B}}_N[H(s/a)]={\mathcal{B}}_N[H(s)]{|}_{s\leftarrow s/a^{1/N}}\text{\hspace{1em}(8.12c)}$$

• the transformation commutes with LP to HP substitution

$${\mathcal{B}}_N[H(1/s)]={\mathcal{B}}_N[H(s)]{|}_{s\leftarrow 1/s}\text{\hspace{1em}(8.12d)}$$

• the transformation distributes over multiplication:

$$\mathcal{B}[H_1(s)H_2(s)]=\mathcal{B}[H_1(s)]\cdot \mathcal{B}[H_2(s)]\text{\hspace{1em}(8.12e)}$$

• the transformation distributes over division:

$$\mathcal{B}[H_1(s)/H_2(s)]=\mathcal{B}[H_1(s)]/\mathcal{B}[H_2(s)]\text{\hspace{1em}(8.12f)}$$

• Butterworth transformations can be chained:

$${\mathcal{B}}_N[{\mathcal{B}}_M[H(s)]]={\mathcal{B}}_{N\cdot M}[H(s)]\text{\hspace{1em}(8.12g)}$$

Since (8.11) doesn’t uniquely define the transformation result, the above properties have to be understood in the sense that the right-hand side can be taken as one possible result of the transformation in the left-hand side. However the amplitude responses of the transformation results are uniquely defined and in those terms the above properties can be understood as usual equalities. E.g. the property (8.12g) can be understood as

$$|{\mathcal{B}}_N[{\mathcal{B}}_M[H(s)]]|=|{\mathcal{B}}_{N\cdot M}[H(s)]|\forall s=j\omega ,\omega \in \mathbb{R}$$

Instead of developing Butterworth transformation immediately for arbitrary order filters we are going to first find a way to apply it to 1-pole filters and then to 2-pole filters. At that point we will be able to simply use the property (8.12e) to apply Butterworth transformation to arbitrary-order filters by representing these arbitrary order filters as cascades of 1-st and 2-nd order filters.

8.6 Butterworth filters of the 1st kind

As we just mentioned, first we will develop a way to apply Butterworth transformation to 1-pole filters, in which case we will more specifically refer to this transformation as Butterworth transformation of the 1st kind. The results of Butterworth transformation of the 1st kind coincide with filters commonly known as Butterworth filters. However in this book later we will generalize the idea of Butterworth filters to include the results of Butterworth transformation of filters of orders higher than 1. In order to be able to tell between different kinds of Butterworth filters, we are going to more specifically refer to the filters obtained by Butterworth transformation of 1-pole filters as Butterworth filters of the 1st kind.

Considering that a 1-pole transfer function is essentially a ratio of two 1st-order polynomials

$$H(s)=\frac{P_1(s)}{P_2(s)}$$

and that the amplitude response of $H(s)$ can be written as a ratio of formal amplitude responses of these polynomials:

$$|H(j\omega )|=\frac{|P_1(j\omega )|}{|P_2(j\omega )|}$$

it is sufficient to develop the transformation for 1st-order polynomials. The transformation of $H(s)$ can be then trivially obtained as:

$$H^{\prime }(s)=\mathcal{B}[H(s)]=\frac{\mathcal{B}[P_1(s)]}{\mathcal{B}[P_2(s)]}=\frac{P_1^{\prime }(s)}{P_2^{\prime }(s)}$$

where $P_1^{\prime }(s)$ and $P_2^{\prime }(s)$ are transformed polynomials $P_1(s)$ and $P_2(s)$.

Transformation of polynomial $P(s)=s+1$

We begin by obtaining the Butterworth transformation of the polynomial $P(s)=s+1$. Its formal amplitude response is

$$|P(j\omega )|=\sqrt{1+{\omega }^2}$$

and we wish to find $P^{\prime }(s)=\mathcal{B}[P(s)]$ such that

$$|P^{\prime }(j\omega )|=|P(j{\omega }^N)|=\sqrt{1+{\omega }^{2N}}$$

In order to get rid of the square root we can deal with squared amplitude response instead

$$\begin{array}{rl}|P(j\omega ){|}^2& =1+{\omega }^2\\ |P^{\prime }(j\omega ){|}^2& =1+{\omega }^{2N}\end{array}$$

Now we would like to somehow obtain $P^{\prime }(s)$ from the latter equation.

In order to do so, let’s notice that

$$|P(j\omega ){|}^2=1+{\omega }^2=1-(j\omega {)}^2=(1+j\omega )(1-j\omega )=P(j\omega )P(-j\omega )=Q(j\omega )$$

where $Q(s)=P(s)P(-s)$, so the roots of $Q(s)$ consist of the root of $P(s)$ at $s=-1$ and of its origin-symmetric image at $s=1$, the latter being the root of $P(-s)$. This motivates to introduce $Q^{\prime }(s)$ such that

$$Q^{\prime }(j\omega )=1+{\omega }^{2N}$$

and then try to factor it into $P^{\prime }(s)P^{\prime }(-s)$ in such a way that

$$|P^{\prime }(j\omega ){|}^2=P^{\prime }(j\omega )P^{\prime }(-j\omega )=Q^{\prime }(j\omega )$$

In order to find the possible ways to factor $Q(s)$ into $P^{\prime }(s)P^{\prime }(-s)$ let us find the roots of $Q(s)$. Instead of solving $Q^{\prime }(s)=0$ for $s$ let’s solve $Q^{\prime }(j\omega )=0$ for $\omega$, where we formally let $\omega$ take complex values. The solutions in terms of $s$ are related to the solutions in terms of $\omega$ through $s=j\omega$.

Solving $Q^{\prime }(j\omega )=1+{\omega }^{2N}=0$ for $\omega$ we obtain

$$\omega =(-1{)}^{1/2N}=e^{j\alpha }\alpha =\pi \frac{2n+1}{2N}=\pi \frac{\frac{1}{2}+n}{N}n=0,\dots ,2N-1\text{\hspace{1em}(8.13)}$$

The solutions are illustrated in Figs. 8.11 and 8.12 where the complex plane can be alternatively interpreted in terms of $s$ or in terms of $\omega$ (note the labelling of the axes), thus these figures simultaneously illustrate the solutions in terms of $\omega$ or in terms of $s$. Thus the $2N$ roots of $Q^{\prime }(s)$ are equally spaced on a unit circle with an angular step of $\pi /N$. If $N$ is odd there will be roots at $s=\pm 1$ otherwise there are no real roots.

Another possible way to look at the solutions of $Q^{\prime }(j\omega )=0$ is to rewrite the equation $1+{\omega }^{2N}=0$ as

$$1+{\omega }^{2N}={\omega }^{2N}-j^2=({\omega }^N+j)({\omega }^N-j)=0$$

In this case the roots obtained from the equation ${\omega }^N-j=0$ will be interleaved with the roots obtained from the equation ${\omega }^N+j=0$ (Figs. 8.11 and 8.12 illustrate). Sometimes therefore such roots are referred to as even and odd roots respectively, since they occur respectively at even and odd $n$ in (8.13). This distinction usually can be ignored, but occasionally becomes important.

Figure 8.11: Roots of $Q^{\prime }(s)$ for the Butterworth transformation of the 1st kind of an even order ($N=6$). White and black dots correspond to even and odd roots.

Figure 8.12: Roots of $Q^{\prime }(s)$ for the Butterworth transformation of the 1st kind of an odd order ($N=5$). White and black dots correspond to even and odd roots.

Having found the roots of $Q^{\prime }(s)$ how do we split them into the roots of $P^{\prime }(s)$ and the roots of $P^{\prime }(-s)$? Obviously we cannot do this splitting in an arbitrary way, since there are several special properties which need to be satisfied.

• For any possible polynomial $P^{\prime }(s)$ its roots are origin-symmetric to the roots of $P^{\prime }(-s)$, so our root splitting must respect this property.

• $P^{\prime }(s)$ must be a real polynomial. This requires its roots to be either real or coming in complex conjugate pairs.

• If $P^{\prime }(s)$ is the denominator of the filter’s transfer function, then its roots must be located in the left complex semiplane (in order for the filter to be stable).

• The requrement $|P^{\prime }(j\omega ){|}^2=P^{\prime }(j\omega )P^{\prime }(-j\omega )$ implies $|P^{\prime }(j\omega )|=|P^{\prime }(-j\omega )|$. In order to satisfy the latter, the roots of $P^{\prime }(s)$ must be symmetric to the roots of $P^{\prime }(-s)$ with respect to the imaginary axis (essentially it is the same reasoning which we had in the discussion of minimum phase and maximum phase zero positioning).

Looking at Figs. 8.11 and 8.12 it’s not difficult to notice that all of the above requirements will be satisfied if we choose the roots in the left complex semiplane to be the roots of $P^{\prime }(s)$ and the roots in the right complex semiplane as the roots of $P^{\prime }(-s)$ respectively.¹¹

Having found the roots $p_n^{\prime }$ of $P^{\prime }(s)$ we still need to find the leading coefficient $g^{\prime }$ of $P^{\prime }(s)$:

$$P^{\prime }(s)=g^{\prime }\cdot \prod _n(s-p_n^{\prime })$$

In order to do so, notice that (8.11) implies $|P^{\prime }(0)|=|P(0)|$. Since $P(0)=1$ and $|P(0)|=1$ we should have $|P^{\prime }(0)|=1$. Actually, if we let $g^{\prime }=1$ we will obtain $P^{\prime }(0)=1$. Indeed,

$$P^{\prime }(0)=\prod _{n=1}^N(0-p_n^{\prime })=\prod _{n=1}^N(-p_n^{\prime })$$

That is $P^{\prime }(0)$ is equal to the product of all roots of $P^{\prime }(-s)$. Looking at Figs. 8.11 and 8.12 we notice that the product of all roots of $P^{\prime }(-s)$ is equal to 1 and thus $P^{\prime }(0)=1$.¹²

Thus, by finding the roots and the leading coefficient of $P^{\prime }(s)$ we have obtained a real polynomial $P^{\prime }(s)=\mathcal{B}[P(s)]$ in the multiplicative form. In practical filter implementations the complex conjugate pairs of factors of $P^{\prime }(s)$ will be represented by 2nd-order filter sections, the purely real factor of $P^{\prime }(s)$ appearing for odd $N$ will be represented by a 1st-order filter section:

$$P^{\prime }(s)=(s+1{)}^{N\wedge 1}\cdot \prod _n(s^2+2R_{n}s+1)$$

where

$$N\wedge 1=\{\begin{array}{ll}1& \text{if }N\text{ is odd}\\ 0& \text{if }N\text{ is even}\end{array}$$

stands for bitwise conjunction.

Arbitrary 1st-order polynomials

Considering $P(s)$ of a more generic form $P(s)=s+a$ ($a>0$) we notice that essentially the procedure is the same as for $P(s)=s+a$ except that instead of the equation ${\omega }^{2N}+1=0$ we obtain the equation

$${\omega }^{2N}+a^2=1$$

This means that the roots of $Q^{\prime }(s)$ are no longer located on the unit circle but on a circle of radius $a^{1/N}$. It is not difficult to see that the leading coefficient of $P^{\prime }(s)$ is still equal to 1.

The above result also could have been obtained by rewriting $P(s)$ as $P(s)=a\cdot (s/a+1)$ and applying properties (8.12b) and (8.12c), which on one hand gives a more intuitive understanding of why the circle of roots is scaled by $a^{1/N}$, on the other hand can serve as an explicit proof of (8.12c) for the case of Butterworth transformation of the 1st kind.

The case of $a=0$ ($P(s)=s$) can be obtained as a limiting case¹³ $a\to +0$ resulting in $P^{\prime }(s)=s^N$.

If $a<0$ then, noticing that the amplitude responses of $P(s)=s+a$ and $P(s)=s-a$ are identical (for $a\in \mathbb{R}$), we could obtain $P^{\prime }(s)$ as Butterworth transformation of $P(s)=s-a$. However, since the root of $P(s)$ is in the right semiplane, it would be logical to also pick the right semiplane roots of $Q^{\prime }(s)$ as the roots of $P^{\prime }(s)$. Particularly, if $P(s)$ is the numerator of a maximum phase filter, the transformation result will retain the maximum phase property.

The 1st-order polynomials of the most general form $P(s)=a_{1}s+a_0$ can be treated by rewriting them as $P(s)=a_1\cdot (s+a_0/a_1)$, if $a_1\ne 0$. The case of $a_1=0$ can be simply treated as a limiting case $a_1\to 0$, where we drop the vanishing higher-order terms of $P^{\prime }(s)$, resulting in $P^{\prime }(s)=a_0$.

Lowpass Butterworth filter of the 1st kind

Given

$$H(s)=\frac{1}{s+1}\text{\hspace{1em}(8.14)}$$

we transform the denominator $P(s)=s+1$ according to the previous discussion of the Butterworth transformation of a 1st order polynomial. The roots of the transformed polynomial (located on the unit circle) become the poles of $H^{\prime }(s)$. The numerator of $H^{\prime }(s)$ is obviously unchanged by the transformation. Thus we obtain

$$H^{\prime }(s)={\left(\frac{1}{s+1}\right)}^{N\wedge 1}\cdot \prod _n\frac{1}{s^2+2R_{n}s+1}$$

where the $1/(s+1)$ term occurs in case of an odd $N$ (“$N\wedge 1$” standing for bitwise conjunction). Therefore $H^{\prime }(s)$ can be implemented as a series of 1-pole and 2-pole lowpass filters, where the 1-pole appears in case of an odd $N$.

Fig. 8.13 compares the amplitude response of a Butterworth lowpass filter of the 1st kind ($N=2$) against the prototype 1-pole lowpass filter. One can observe the increased steepness of the cutoff slope resulting from the shrinking along the logarithmic frequency axis. Fig. 8.14 compares the same Butterworth lowpass filter against cascading of identical 1st order lowpasses, that is comparing the shrinking along the logarithmic frequency axis vs. stretching along the logarithmic amplitude axis. One can see that the Butterworth lowpass filter has the sharpest cutoff corner among different filters in Figs. 8.13 and 8.14.

Figure 8.13: 2nd order lowpass Butterworth filter of the 1st kind (solid line) vs. 1st order lowpass filter (dashed line).

Figure 8.14: 2nd order lowpass Butterworth filter of the 1st kind (solid line) vs. duplicated 1st order lowpass filter without and with cutoff adjustment (dashed lines).

It is useful to know and recognize the expression for the squared amplitude response of a Butterworth lowpass filter of the 1st kind. Since the squared amplitude response of (8.14) is

$$|H(j\omega ){|}^2=\frac{1}{1+{\omega }^2}$$

after the substitution $\omega \leftarrow {\omega }^N$ we obtain

$$|H^{\prime }(j\omega ){|}^2=\frac{1}{1+{\omega }^{2N}}\text{\hspace{1em}(8.15)}$$

This expression is used in traditional derivation of Butterworth filters. Essentially the $N$-th order lowpass Butterworth filter is traditionally defined as a filter whose the amplitude response satisfies (8.15). Note that by (8.15) the 1-pole lowpass is the Butterworth filter of order 1. We can formally treat it as a 1st-order Butterworth transformation of itself

$$\frac{1}{1+s}={\mathcal{B}}_1\left[\frac{1}{1+s}\right]$$

It is also useful to explicitly know the transfer function of the Butterworth lowpass filter of the 1st kind of order $N=2$. It’s not difficult to realize that for $P(s)=s+1$ the roots of $P^{\prime }(s)$ are located $45^{\circ }$ away from the negative real semiaxis. Thus the respective damping is $R=\arccos 45^{\circ }=1/\sqrt{2}$ and

$$H^{\prime }(s)=\frac{1}{s^2+\sqrt{2}s+1}$$

This damping value and the 2nd-order term $s^2+\sqrt{2}s+1$ appears in all Butterworth filters of the 1st kind of order $N=2$ (highpass, bandpass, etc.) The readers may also recall the appearance of the damping value $R=1/\sqrt{2}$ in the discussion of 2-pole filters, where it was mentioned that at $R=1/\sqrt{2}$ the 2-pole filter turns into a Butterworth filter. This also corresponds to the fact that among all non-resonating (in the sense of the missing resonance peak) 2nd-order filters the Butterworth filter is the one with the sharpest possible cutoff corner in the amplitude response.

Highpass Butterworth filter of the 1st kind

For

$$H(s)=\frac{s}{1+s}$$

we have the same denominator as for the respective lowpass. Thus the result of the denominator transformation is the same as for the lowpass. The result of the numerator transformation is $s^N$ and thus

$$H^{\prime }(s)={\left(\frac{s}{s+1}\right)}^{N\wedge 1}\cdot \prod _n\frac{s^2}{s^2+2R_{n}s+1}$$

That is we obtain the same result as for the 1-pole lowpass, except that instead of a series of lowpasses we should take a series of highpasses. Fig. 8.15 illustrates the respective amplitude response.

It is not difficult to verify that the highpass Butterworth filter obtained in the described above way is identical to the result of LP to HP substitution applied to the lowpass Butterworth filter of the same order, which is in agreement with (8.12d).

Figure 8.15: 2nd order highpass Butterworth filter of the 1st kind vs. 1st order highpass filter (dashed line).

Bandpass Butterworth filter of the 1st kind

For an even $N$, by formally putting a numerator $s^{N/2}$ over the Butterworth transformation of a polynomial $P(s)=1+s$ we obtain a kind of a bandpass filter:

$$H^{\prime }(s)=\prod _n\frac{s^{N/2}}{s^2+2R_{n}s+1}$$

(Fig. 8.16), which can be also formally seen as a Butterworth transformation of $H(s)=s^{1/2}/(s+1)$.

Figure 8.16: 2nd order bandpass Butterworth filter of the 1st kind.

Note that thereby this bandpass filter doesn’t have any parameters to control, except the cutoff. As we will see a bit later in the discussion of Butterworth filters of the 2nd kind, this filter also can be obtained by an order $N/2$ Butterworth transformation of the 2-pole bandpass $H(s)=s/(s^2+\sqrt{2}s+1)$. Therefore there is not much point in specifically using Butterworth bandpass filters of the 1st kind, one can simply use Butterworth bandpass filters of the 2nd kind instead, achieving exactly the same response at a particular resonance setting.

A bandpass filter which has controllable bandwidth can be obtained by applying the LP to BP substitution to a Butterworth lowpass filter of the 1st kind. Apparently this produces a normalized bandpass (Fig. 8.17). This filter does not coincide with the result of the Butterworth transformation of the normalized 2-pole bandpass $H(s)=\sqrt{2}s/(s^2+\sqrt{2}s+1)$. The reason is that in the first case we have a Butterworth transformation of a 1-pole lowpass $1/(1+s)$ followed by the LP to BP substitution, while in the second case we first have the LP to BP substitution (with an appropriately chosen bandwidth) applied to $1/(1+s)$ yielding $H(s)=\sqrt{2}s/(s^2+\sqrt{2}s+1)$, which is then followed by the Butterworth transformation. So it’s the opposite order of the application of LP to BP substitution and the Butterworth transformation.

Figure 8.17: A bandwidth-tuned LP to BP substitution of a lowpass Butterworth filter of the 1st kind vs. Butterworth transformation of $H(s)=\sqrt{2}s/(s^2+\sqrt{2}s+1)$ (dashed line).

The LP to BP substitution can be performed algebraically on the transfer function of the Butterworth lowpass. In order to simplify things, the substitution can be applied in turn to the poles of each of the underlying 1- and 2-pole filters of the cascaded implementation of the Butterworth lowpass. After organizing the transformed poles into mutually conjugate pairs, we can simply construct the result as a series of normalized 2nd order bandpasses, defined by those pole pairs. Alternatively the LP to BP substitution can be implemented using the integrator substitution technique (Fig. 4.19).

8.7 Butterworth filters of the 2nd kind

Now we are going to apply the Butterworth transformation to 2nd order polynomials and respectively 2nd order filters. Such transformation will be referred to as Butterworth transformation of the 2nd kind and the filters obtained as the results of the tranformation will be referred to Butterworth filters of the 2nd kind.

Transformation of polynomial $P(s)=s^2+2Rs+1$

We will first consider the following 2nd order polynomial

$$P(s)=s^2+2Rs+1$$

corresponding to the denominator of a unit-cutoff 2-pole filter.

It will be most illustrative to obtain the Butterworth transformation of the 2nd kind as a combination of two opposite perturbations of two Butterworth transformations of the 1st kind. Factoring $P(s)$ we obtain

$$P(s)=(s+a_1)(s+a_2)=P_1(s)P_2(s)$$

At $R=1$ we have $a_1=a_2=1$ and $P(s)$ is a product of two 1st-order polynomials $P_1(s)=P_2(s)=s+1$. Applying the Butterworth transformation of the 1st kind to each of the polynomials $P_1^{\prime }(s)$ and $P_2^{\prime }(s)$ we obtain two identical sets of the roots of $P_1^{\prime }(s)$ and $P_2^{\prime }(s)$ respectively. We can also consider the respective (also identical) extended polynomials

$$\begin{array}{rl}{Q}_1(s)& =P_1(s)P_1(-s)\\ Q_2(s)& =P_2(s)P_2(-s)\\ Q(s)& =P(s)P(-s)=Q_1(s)Q_2(s)\\ Q_1^{\prime }(s)& =P_1^{\prime }(s)P_1^{\prime }(-s)\\ Q_2^{\prime }(s)& =P_2^{\prime }(s)P_2^{\prime }(-s)\\ Q^{\prime }(s)& =P^{\prime }(s)P(-s)=Q_1^{\prime }(s)Q_2^{\prime }(s)\end{array}$$

which additionally contain the right-semiplane roots. As we should remember from the discussion of the Butterworth transformation of the 1st kind, the roots in each of the two sets corresponding to $Q_1^{\prime }(s)$ and $Q_2^{\prime }(s)$ are equally spaced on the unit circle.¹⁴

Now suppose we initially have $R=1$ and then increase $R$ to a value $R>1$, resulting in $a_1$ growing and $a_2$ decreasing, staying reciprocal to each other:

$$a_1=R+\sqrt{R^2-1}{a}_2=R-\sqrt{R^2-1}(a_1{a}_2=1)$$

(Fig. 8.18). Since $a_1$ and $a_2$ are the “cutoffs” of the 1st-order polynomials $s+a_1$ and $s+a_2$, from the properties of the Butterworth transformation of the 1st kind we obtain that the radii of the circles, on which the roots of $Q_1^{\prime }(s)$ and $Q_2^{\prime }(s)$ are located, become equal to

$$r_1^{\prime }=(R+\sqrt{R^2-1}{)}^{1/N}{r}_2^{\prime }=(R-\sqrt{R^2-1}{)}^{1/N}(r_1^{\prime }{r}_2^{\prime }=1)$$

Thus, one circle grows and the other circle shrinks, while their radii are staying reciprocal to each other (Fig. 8.19).

Now let’s decrease $R$ from 1 to a value $0<R<1$. This makes $a_1$ and $a_2$ complex:

$$a_1=e^{j\alpha }{a}_2=e^{-j\alpha }(\cos \alpha =R,a_1{a}_2=1)$$

Figure 8.18: Roots of $Q(s)$ for $R>1$ (black dots are roots of $Q_1(s)$, white dots are roots of $Q_2(s)$) and their positions at $R=1$ (indicated by circled dots, where each such dot denotes a root of $Q_1(s)$ coinciding with a root of $Q_2(s)$).

Figure 8.19: Roots of $Q^{\prime }(s)$ for $R>1$ (black dots are roots of $Q_1^{\prime }(s)$, white dots are roots of $Q_2^{\prime }(s)$) and their positions at $R=1$ (indicated by circled dots, where each such dot denotes a root of $Q_1^{\prime }(s)$ coinciding with a root of $Q_2^{\prime }(s)$). Butterworth transformation order $N=2$.

Writing out the “amplitude response” we have

$$\begin{array}{rl}|P(j\omega ){|}^2& =P(j\omega )P(-j\omega )=P_1(j\omega )P_2(j\omega )\cdot P_1(-j\omega )P_2(-j\omega )=\\ & =P_1(j\omega )P_1(-j\omega )\cdot P_2(j\omega )P_2(-j\omega )=\\ & =Q_1(j\omega )Q_2(j\omega )=({\omega }^2+a_1^2)\cdot ({\omega }^2+a_2^2)\end{array}$$

Respectively, our goal is to have

$$|P^{\prime }(j\omega ){|}^2=Q_1^{\prime }(j\omega )Q_2^{\prime }(j\omega )=Q_1(j{\omega }^N)Q_2(j{\omega }^N)=({\omega }^{2N}+a_1^2)\cdot ({\omega }^{2N}+a_2^2)$$

So how do we find the roots of $Q_1^{\prime }(s)$ and $Q_2^{\prime }(s)$? If $a_1=1$ ($\alpha =0$, $R=1$) then, as we just discussed, $Q_1^{\prime }(s)$ simply generates a set of the Butterworth roots of the 1st kind on the unit circle. Now if we replace $\alpha =0$ with $\alpha >0$ (corresponding to replacing $R=1$ with $R<1$) this means a rotation of $a_1$ by the angle $\alpha$ (Fig. 8.20). This rotates all roots of $Q_1^{\prime }(j\omega )={\omega }^{2N}+a_1^2$ by $\alpha /N$ (Fig. 8.21). At the same time $a_2$ will be rotated by $-\alpha$ and respectively all roots of $Q_2^{\prime }(j\omega )={\omega }^{2N}+a_2^2$ by $-\alpha /N$.