Chapter 5: Ladder filter

In this chapter we are going to discuss the most classical analog filter model: the transistor ladder filter. The main idea of this structure, which is to create resonance by means of a feedback loop, is encountered in many other filter designs, some of which we are also going to discuss. We will be referring to the class of such filters as simply ladder filters.¹

5.1 Analog model

The most classical example of a ladder filter is transistor ladder filter, which implements a 4-pole lowpass structure shown in Fig. 5.1.² The structure in Fig. 5.1 is not limited to transistor-based analog implementations. Particularly, there are many implementations of the same structure based on OTAs (operational transconductance amplifiers). The difference between transistor- and OTA-based ladders is, however, lying in the nonlinear behavior, which we are not touching at this point yet. The linear aspects of both are identical.

The LP${}_1$ blocks denote four identical (same cutoff) 1-pole lowpass filters (Fig. 2.2). The $k$ coefficient controls the amount of negative feedback, which creates resonance in the filter. Typically $k\ge 0$, although $k<0$ is also sometimes used.

Figure 5.1: Transistor (4-pole lowpass) ladder filter.

Let $$H_1(s)=\frac{1}{1+s}$$ be the 1-pole lowpass transfer function. Assuming complex exponential $x$ and $y$ we write $$y=H_1^4(s)\cdot (x-ky)$$ from where $$y(1+k{H}_1^4(s))=H_1^4(s)\cdot x$$ and the transfer function of the ladder filter is $$H(s)=\frac{y}{x}=\frac{H_1^4(s)}{1+k{H}_1^4(s)}=\frac{\frac{1}{(1+s{)}^4}}{1+k\frac{1}{(1+s{)}^4}}=\frac{1}{k+(1+s{)}^4}\text{\hspace{1em}(5.1)}$$

At $k=0$ the filter behaves as 4 serially connected 1-pole lowpass filters.

The poles of the filter are respectively found from $$k+(1+s{)}^4=0$$ giving $$s=-1+(-k{)}^{1/4}$$ where the raising to the 1/4th power is understood in the complex sense, therefore giving 4 different values: $$s=-1+\frac{\pm 1\pm j}{\sqrt{2}}{k}^{1/4}(k\ge 0)\text{\hspace{1em}(5.2)}$$ (this time $k^{1/4}$ is understood in the real sense). Thus there are 4-poles and we can also refer to this filter as a 4-pole lowpass ladder filter.

At $k=0$ all poles are located at $s=-1$, as $k$ grows they move apart in 4 straight lines, all going at “$45^{\circ }$ angles” (Fig. 5.2). As $k$ grows from 0 to 4 the two of the poles (at $s=-1+\frac{1\pm j}{\sqrt{2}}{k}^{1/4}$) are moving towards the imaginary axis, producing a resonance peak in the amplitude response (Fig. 5.3). At $k=4$ they hit the imaginary axis: $$\mathrm{Re}\left(-1+\frac{1\pm j}{\sqrt{2}}{4}^{1/4}\right)=0$$ and the filter becomes unstable.³

In Fig. 5.3 one could notice that, as the resonance increases, the filter gain at low frequencies begins to drop. Indeed, substituting $s=0$ into (5.1) we obtain $$H(0)=\frac{1}{1+k}$$ This is a general issue with ladder filter designs.

Figure 5.2: Poles of the 4-pole lowpass ladder filter.

Figure 5.3: Amplitude response of the 4-pole lowpass ladder filter for various $k$.

5.2 Feedback and resonance

Before we continue with discussing more practical aspects of the ladder filter, we’d like to make one important observation considering the resonance peaks created by the ladder filter feedback.

In Fig. 5.3 we can see that, similarly to the 2-pole case, the resonance frequency is approaching the filter cutoff frequency as the filter approaches selfoscillation at $k=4$. This is a manifestation of a more general principle concerning ladder filters as such. Consider a general ladder filter in Fig. 5.4, where $G(s)$ denotes a more or less arbitrary structure, whose transfer function is $G(s)$. Notice that the feedback in Fig. 5.4 is not inverted.

Figure 5.4: Structure of a generic ladder filter.

The transfer function of the entire structure is therefore $$H(s)=\frac{G(s)}{1-kG(s)}=\frac{1}{G^{-1}(s)-k}\text{\hspace{1em}(5.3)}$$ and the poles are defined by the equation $$G^{-1}(s)=k\text{\hspace{1em}(5.4)}$$ That is at $k=0$ the poles of $H(s)$ are the zeros of $G^{-1}(s)$ (the latter obviously simply being the poles of $G(s)$). As $k$ begins to deviate from zero, the solutions of (5.4) will move in the $s$-plane, usually in a continuous fashion. E.g. for the 4-pole lowpass ladder (Fig. 5.1) we had $G^{-1}(s)=(s+1{)}^4$ and (5.4) takes the form $(s+1{)}^4=-k$, where we take $-k$ instead of $k$ because of the inverted feedback in Fig. 5.1.

The value of $k$ at which the filter starts to selfoscillate should correspond to some of the poles being located on the imaginary axis. At this moment the infinitely high resonance peak in the amplitude response is occurring exactly at these pole positions. Denoting a purely imaginary pole position as $j\omega$, we rewrite (5.4) for such poles as $$G^{-1}(j\omega )=k$$ or $$kG(j\omega )=1\text{\hspace{1em}(5.5)}$$ We can refer to (5.5) as the selfoscillation equation for a feedback loop. This equation implies that selfoscillation appears at the moment when the total frequency response across the feedback loop $kG(j\omega )$ exactly equals 1 at some frequency $\omega$. That is the total amplitude gain must be 1, and the total phase shift must be $0^{\circ }$.

This is actually a pretty remarkable result. Of course it is quite intuitive that selfoscillation tends to occur at frequencies where the feedback signal doesn’t cancel the input signal, but rather boosts it. And such boosting tends to be strongest at frequencies where we have a $0^{\circ }$ total phase shift across the feedback loop. However, what is quite counterintuitive, is that selfoscillation can appear (as $k$ is reaching the respective threshold value) only at such frequencies.⁴

Therefore for $k>0$ the selfoscillation appears at frequencies where the phase response of $G(s)$ is $0^{\circ }$. For $k<0$ the selfoscillation appears at frequencies where the phase response of $G(s)$ is $180^{\circ }$. The respective value of $k$ can be found from (5.5) giving $$k=\frac{1}{G(j\omega )}\text{\hspace{1em}(5.6)}$$ or, rewriting (5.6) in terms of the amplitude response of $G(s)$: $$k=\pm \frac{1}{|G(j\omega )|}\text{\hspace{1em}(5.7)}$$ where we take the plus sign if the phase response of $G(s)$ at $\omega$ is $0^{\circ }$ and the minus sign if the phase response of $G(s)$ at $\omega$ is $180^{\circ }$.

The just discussed effects are the reason that we used negative feedback in the 4-pole lowpass ladder filter. We want the resonance to occur at the filter’s cutoff. The phase response of a single 1-pole lowpass at the cutoff frequency is $-45^{\circ }$, respectively the phase response of a chain of four 1-poles is $-180^{\circ }$, exactly what we need for the resonance peak, if we use negative feedback.

At the same time, the amplitude response of a 1-pole lowpass at the cutoff is $|1/(1+j)|=1/\sqrt{2}$, respectively the amplitude response of a chain of four 1-poles is $(1/\sqrt{2}{)}^4=1/4$. According to (5.7), the infinite resonance is attained at $k=1/(1/4)=4$.

At $\omega =0$ a chain of four 1-pole lowpasses will have a phase shift of $0^{\circ }$, while the amplitude response at $\omega =0$ is 1. Therefore, in Fig. 5.1 the “selfoscillation” at $\omega =0$ will occur at $k=-1$. However the amplitude response peak at $\omega =0$ hardly can count as resonance.

5.3 Digital model

A naive digital implementation of the ladder filter shouldn’t pose any problems. We will therefore immediately skip to the TPT approach.

Recalling the instantaneous response of a single 1-pole lowpass filter (3.29), we can construct the instantaneous response of a serial connection of four of such filters. Indeed, let’s denote the instantaneous responses of the respective 1-poles as $f_n(\xi )=g\xi +s_n$ (obviously, the coefficient $g$ is identical for all four, whereas $s_n$ depends on the filter state and therefore cannot be assumed identical). Combining two such filters in series we have $$f_2(f_1(\xi ))=g(g\xi +s_1)+s_2=g^2\xi +g{s}_1+s_2$$ Adding the third one: $$f_3(f_2(f_1(\xi )))=g(g^2\xi +g{s}_1+s_2)+s_3=g^3\xi +g^2{s}_1+g{s}_2+s_3$$ and the fourth one: $$\begin{array}{rl}{f}_4(f_3(f_2(f_1(\xi ))))& =g(g^3\xi +g^2{s}_1+g{s}_2+s_3)=\\ & =g^4\xi +g^3{s}_1+g^2{s}_2+g{s}_3+s_4=G\xi +S\end{array}$$ where $$G=g^4$$ $$S=g^3{s}_1+g^2{s}_2+g{s}_3+s_4$$ Using the obtained instantaneous response $G\xi +S$ of the series of 4 1-poles, we can redraw the ladder filter structure as in Fig. 5.5.

Figure 5.5: TPT 4-pole ladder filter in the instantaneous response form.

Rather than solving for $y$, let’s solve for the signal $u$ at the feedback point. From Fig. 5.5 we obtain $$u=x-ky=x-k(Gu+S)$$ from where $$u=\frac{x-kS}{1+kG}\text{\hspace{1em}(5.8)}$$ We can then use the obtained value of $u$ to process the 1-pole lowpasses one after the other, updating their state, and computing $y[n]$ as the output of the fourth lowpass.

Apparently the total instantaneous gain of the zero-delay feedback loop in Fig. 5.5 and in (5.8) is $-kG$. As we should recall from the discussion of 1-pole lowpass filters, $0<g<1$ for positive cutoff settings. Respectively $0<G<1$ and the filter doesn’t become instantaneously unstable provided $k\ge -1$.

5.4 Feedback shaping

We have observed that at high resonance settings the amplitude gain of the filter at low frequencies drops (Fig. 5.3). An obvious way to fix this problem would be e.g. to boost the input signal by the $(1+k)$ factor.⁵ However there’s another way to address the same issue. We could “kill” the feedback for the low frequencies only by introducing a highpass filter into the feedback path (Fig. 5.6). In the simplest case this could be a 1-pole highpass.

The cutoff of the highpass filter can be static or vary along with the cutoff of the lowpasses. The static version has a nice feature that it kills the resonance effect at low frequencies regardless of the master cutoff setting, which may be desirable if the resonance at low frequencies is considered rather unpleasant (Fig. 5.7).

Figure 5.6: Transistor ladder filter with a highpass in the feedback.

Figure 5.7: Amplitude response of the ladder filter with a static-cutoff highpass in the feedback for various lowpass cutoffs.

In principle one can also use other filter types in the feedback shaping. One has to be careful though, since this changes the total phase and amplitude responses of the feedback path, thus the frequency of the resonance peak and the value of $k$ at which selfoscillation is reached may be changed. E.g., quite counterintuitively, inserting a 1-pole lowpass into the feedback path can destabilize an otherwise stable filter.

In order to establish and analyse the latter fact mathematically, we’d need to find the total amplitude response across the feedback loop at the point where the total phase shift is $180^{\circ }$. Let $H_1(s)=1/(1+s)$ be the underlying 1-pole lowpass of the ladder filter and let $H_f(s)=1/(1+s/{\omega }_{cf})$ be the lowpass in the feedback, with a generally speaking different cutoff ${\omega }_{cf}$. The $180^{\circ }$ point is found from the equation $$\begin{array}{rl}4\arg H_1(j\omega )+\arg H_f(j\omega )& =4\arg \frac{1}{1+j\omega }+\arg \frac{1}{1+j\omega /{\omega }_{cf}}=\\ & =-4\arctan \omega -\arctan \frac{\omega }{{\omega }_{cf}}=-\pi \end{array}\text{\hspace{1em}(5.9)}$$ where we have used (2.8). The equation (5.9) looks a bit daunting, if having an analytic solution at all. Fortunately, we don’t actually need to know the frequency of the $180^{\circ }$ point, it would suffice to know the respective amplitude responses.

Let ${\phi }_1(\omega )$ be the negated phase response of $H_1(s)$: $${\phi }_1(\omega )=-\arg H_1(j\omega )=\arctan \omega >0\forall \omega$$ Expressing $\omega$ as a function of ${\phi }_1$ we have $\omega =\tan {\phi }_1$. Respectively, expressing the amplitude response as a function of the (negated) phase response we have $$A_1=|H_1(j\omega )|=\frac{1}{\sqrt{1+{\omega }^2}}=\frac{1}{\sqrt{1+{\tan }^2{\phi }_1}}=\cos {\phi }_1\text{\hspace{1em}(5.10)}$$ Thus, the total amplitude response of the four 1-poles in the feedforward path of the ladder filter is $$A_1^4(\omega )=\frac{1}{(1+{\phi }_1^2{)}^2}$$ and the total phase response of the feedforward path is $4{\phi }_1$.

Since (5.10) is cutoff-independent, it also holds for $H_f(s)$: $$A_f=\cos {\phi }_f$$ where $A_f=|H_f(j\omega )|$, ${\phi }_f=-\arg H_f(j\omega )$. Now let ${\omega }_0$ be the (unknown to us) solution of (5.9), that is the total phase shift at ${\omega }_0$ is $180^{\circ }$. In terms of the just introduced functions ${\phi }_1(\omega )$ and ${\phi }_f(\omega )$ equation (5.9) can be rewritten as $$4{\phi }_1({\omega }_0)+{\phi }_f({\omega }_0)=\pi \text{\hspace{1em}(5.11)}$$ Since ${\phi }_f(\omega )>0\forall \omega$, the $180^{\circ }$ phase shift is achieved earlier than without the feedback filter, that is ${\omega }_0<1$ (whatever the value of ${\omega }_{cf}$ is).

Computing the total amplitude response of all five 1-pole lowpasses at ${\omega }_0$ we have $$A_1^4({\omega }_0)\cdot A_f({\omega }_0)={\cos }^4{\phi }_1({\omega }_0)\cdot \cos {\phi }_f({\omega }_0)={\cos }^4\left(\frac{\pi }{4}-\frac{{\phi }_f({\omega }_0)}{4}\right)\cdot \cos {\phi }_f({\omega }_0)$$ Considering only the first factor we have $${\cos }^4\left(\frac{\pi }{4}-\frac{{\phi }_f}{4}\right)={\left(\frac{1+\cos \left(\frac{\pi }{2}-\frac{{\phi }_f}{2}\right)}{2}\right)}^2={\left(\frac{1+\sin \frac{{\phi }_f}{2}}{2}\right)}^2$$ (where we dropped the argument ${\omega }_0$, understanding it implicitly). Respectively $$A_1^4\cdot A_f=\frac{1}{4}\cdot {\left(1+\sin \frac{{\phi }_f}{2}\right)}^2\cdot \cos {\phi }_f(\omega ={\omega }_0)\text{\hspace{1em}(5.12)}$$ Fig. 5.8 contains the graph of (5.12). The interpretation of this graph is like follows. Suppose the feedback lowpass’s cutoff ${\omega }_{cf}$ is very large (${\omega }_{cf}\to +\infty$). In the limit the feedback lowpass has no effct and $${\omega }_0=1{\phi }_f({\omega }_0)=0A_f({\omega }_0)=1A_1^4({\omega }_0)A_f({\omega }_0)=\frac{1}{4}(\text{for }{\omega }_{cf}=+\infty )$$ As we begin to lower ${\omega }_{cf}$ back from the infinity, the value of ${\phi }_f({\omega }_0)$ grows from zero into the positive value range. The graph in Fig. 5.8 plots the total amplitude response of the five 1-pole lowpasses in the feedback loop against the growing ${\phi }_f({\omega }_0)$. We see that the amplitude response grows for quite a while. As long as it is above 1/4, the filter will explode at $k=4$. The zero amplitude response at ${\phi }_f=\pi /2$ corresponds to ${\omega }_{cf}=0$, where the extra lowpass is fully closed, thus the entire feedback loop is muted.

Figure 5.8: Total amplitude response of the four feedforward lowpass 1-poles plus the feedback lowpass 1-pole at the $180^{\circ }$ phase shift point, plotted against the phase shift by the feedback 1-pole.

At ${\omega }_{cf}=1$ (equal cutoffs of all 1-poles) from (5.11) we have ${\phi }_f({\omega }_0)={\phi }_1({\omega }_0)=\pi /5$. In Fig. 5.8 one can see that this is the “most unstable” situation among all possible ${\omega }_{cf}$.

In comparison, if we had a 1-pole highpass in the feedback, then we would have $\arg H_f(j\omega )>0$ and respectively ${\phi }_f(\omega )<0\forall \omega$. Therefore the $$180^{\circ }$$ point would be shifted to the right: ${\omega }_0>1$. Therefore $A_1^4({\omega }_0)<A_1^4(1)<1/4$, while $A_f(\omega )<1\forall \omega$, thus the total amplitude response $A_1^4{A}_f$ at the $180^{\circ }$ point would decrease and the filter won’t become “more unstable” than it was before the introduction of the extra highpass filter.

5.5 Multimode ladder filter

Warning! The multimode functionality of the ladder filter is a somewhat special feature. There are more straightforward ways to build bandpass and highpass ladders, discussed later in this chapter.

By picking up intermediate signals of the ladder filter as in Fig. 5.9 we obtain the multimode version of this filter.

Figure 5.9: Multimode ladder filter.

We then can use linear combinations of signals $y_n$ to produce various kinds of filtered signal.⁶

Suppose $k=0$. Apparently, in this case, the respective transfer functions associated with each of the $y_n$ outputs are $$H_n(s)=\frac{1}{(1+s{)}^n}(n=0,\dots ,4)\text{\hspace{1em}(5.13)}$$ If $k\ne 0$ then from $$H_4(s)=\frac{1}{k+(1+s{)}^4}$$ using the obvious relationship $H_{n+1}(s)=H_n(s)/(s+1)$ we obtain $$H_n(s)=\frac{(1+s{)}^{4-n}}{k+(1+s{)}^4}\text{\hspace{1em}(5.14)}$$

4-pole highpass mode

Considering that the 4th order lowpass transfer function (under the assumption $k=0$) is built as a product of four 1st order lowpass transfer functions $1/(1+s)$ $$H_{\mathrm{LP}}(s)=\frac{1}{(1+s{)}^4}$$ we might decide to build the 4th order highpass transfer function as a product of four 1st order highpass transfer functions $s/(1+s)$: $$H_{\mathrm{HP}}(s)=\frac{s^4}{(1+s{)}^4}$$ Let’s attempt to build $H_{\mathrm{HP}}(s)$ as a linear combination of $H_n(s)$. Apparently, a linear combination of $H_n(s)$ must have the denominator $k+(1+s{)}^4$, so let’s instead construct $$H_{\mathrm{HP}}(s)=\frac{s^4}{k+(1+s{)}^4}\text{\hspace{1em}(5.15)}$$ which at $k=0$ will turn into $s^4/(1+s{)}^4$. We also have $H_{\mathrm{HP}}(\infty )=1$ while the four zeros at $s=0$ provide a 24dB/oct rolloff at $\omega \to 0$, thus we are still having a more or less reasonable highpass. In order to express $H_{\mathrm{HP}}(s)$ as a sum of the modes we write $$\frac{s^4}{k+(1+s{)}^4}=\frac{a_0(1+s{)}^4+a_1(1+s{)}^3+a_2(1+s{)}^2+a_3(1+s)+a_4}{k+(1+s{)}^4}$$

that is

$$s^4=a_0(1+s{)}^4+a_1(1+s{)}^3+a_2(1+s{)}^2+a_3(1+s)+a_4$$

We need to find $a_n$ from the above equation, which generally can be done by equating the coefficients at equal powers of $s$ in the left- and right-hand sides. However, for the specific equation that we’re having here we could do a shortcut by simply formally replacing $s+1$ by $s$ (and respectively $s$ by $s-1$):

$$(s-1{)}^4=a_0{s}^4+a_1{s}^3+a_2{s}^2+a_{3}s+a_4$$

from where immediately

$$a_0=1,a_1=-4,a_2=6,a_3=-4,a_4=1$$

The amplitude response corresponding to (5.15) is plotted in Fig. 5.10.

Figure 5.10: Amplitude response of the highpass mode of the ladder filter for various k.

4-pole bandpass mode

A bandpass filter can be built as

$$H_{BP}(s)=\frac{s^2}{k+(1+s{)}^4}\text{\hspace{1em}(5.16)}$$

The two zeros at $s=0$ will provide for a $-12$dB/oct rolloff at low frequencies and will reduce the $-24$dB/oct rolloff at high frequencies to the same $-12$dB/oct. Notice that the phase response at the cutoff is zero:

$$H_{BP}(j)=\frac{-1}{k+(1+j{)}^4}=\frac{1}{4-k}$$

The coefficients are found from

$$\begin{array}{rl}{s}^2& =a_0(1+s{)}^4+a_1(1+s{)}^3+a_2(1+s{)}^2+a_3(1+s)+a_4\\ (s-1{)}^2& =a_0{s}^4+a_1{s}^3+a_2{s}^2+a_{3}s+a_4\end{array}$$

The amplitude response corresponding to (5.16) is plotted in Fig. 5.11.

Figure 5.11: Amplitude response of the bandpass mode of the ladder filter for various k.

Lower-order modes

Recalling the transfer functions of the modal outputs $y_n$ in the absence of the resonance (5.13), we can consider the modal signals $y_n$ and their respective transfer functions (5.14) as a kind of “$n$-pole lowpass filters with 4-pole resonance”.

“Lower-order” highpasses can be build by considering the zero-resonance transfer functions

$$H_{HP}(s)=\frac{s^N}{(s+1{)}^N}=\frac{(s+1{)}^{4-N}{s}^N}{(s+1{)}^4}$$

which for $k\ne 0$ turn into

$$H_{HP}(s)=\frac{(s+1{)}^{4-N}{s}^N}{k+(s+1{)}^4}$$

In a similar way we can build a “2-pole” bandpass

$$H_{BP}(s)=\frac{s}{(s+1{)}^2}=\frac{(s+1{)}^{2}s}{(s+1{)}^4}(k=0)$$

$$H_{BP}(s)=\frac{(s+1{)}^{2}s}{k+(s+1{)}^4}(k\ne 0)$$

Other modes

Continuing in the same fashion we can build further modes (the transfer functions are given for $k=0$):

etc. The principles are more or less similar. We are trying to attain a desired asymptotic behavior at $\omega \to 0$ and $\omega \to +\infty$ by having the necessary orders and coefficients of the lowest-order and highest-order terms in the numerator. E.g. by having $s^2$ as the lowest-order term of the numerator we ensure a 12dB/oct rolloff at $\omega \to 0$, or by having $s^4$ as the highest-order term we ensure $H(\infty )=1$. The notch at $\omega =1$ is generated by placing a zero at $s=\pm j$. The 2-notch version is obtained by explicitly writing out the transfer function of a 4-pole multinotch described in Section 11.3.

5.6 HP ladder

Performing an LP to HP transformation on the lowpass ladder filter we effectively perform it on each of the underlying 1-pole lowpasses, thus turning them into 1-pole highpasses. Thereby we obtain a “true” highpass ladder filter (Fig. 5.12). Obviously, the amplitude response of the ladder highpass is symmetric to the amplitude response of the ladder lowpass (Fig. 5.13).

Figure 5.12: A “true” highpass ladder filter.

The instantaneous gain of a 1-pole highpass is complementary to the instantaneous gain of the 1-pole lowpass:

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

Figure 5.13: Amplitude response of the 4-pole highpass ladder filter for various k.

where $g={\omega }_{c}T/2$. Thus the instantaneous gain of a single 1-pole highpass is varying within the range $(0,1)$ and so does the gain of the chain of four highpasses: $0<G<1$. Therefore, 4-pole highpass ladder doesn’t get instantaneously unstable for $k>-1$.

5.7 BP ladder

In order to build a “true” 4-pole bandpass ladder, we replace only half of the lowpasses with highpasses (it doesn’t matter which two of the four 1-pole lowpasses are replaced). The total transfer function of the feedforward path is thereby

$$\frac{s^2}{(1+s{)}^4}=\frac{s}{(1+s{)}^2}\cdot \frac{s}{(1+s{)}^2}$$

where each of the $s/(1+s{)}^2$ factors is built from a serial combination of a 1-pole lowpass and a 1-pole highpass:

$$\frac{s}{(1+s{)}^2}=\frac{s}{1+s}\cdot \frac{1}{1+s}$$

Apparently $s/(1+s{)}^2=s/(1+2s+s^2)$ is a 2-pole bandpass with damping $R=1$ and a serial combination of two of them makes a 4-pole bandpass. The frequency response of $s/(1+s{)}^2$ at $\omega =1$ is $1/2$, that is there is no phase-shift. Respectively the frequency response of $s^2/(1+s{)}^4$ at $\omega =1$ is $1/4$, also without a phase shift. Therefore we need to use positive rather than negative feedback (Fig. 5.14), the selfoscillation still occuring at $k=4$, the same as with lowpass and highpass ladders.

Figure 5.14: A “true” bandpass ladder filter.

Noticing that the filter structure is invariant relative to the LP to HP transformation, we conclude that its amplitude response must be symmetric (around $\omega =1$) in the logarithmic frequency scale (Fig. 5.15).

The question of instantaneous instability is more critical for the bandpass ladder, since the feedback is positive. The instantaneous gain of a lowpass-highpass pair is a product of the instantaneous gains of a 1-pole lowpass and a

Figure 5.15: Amplitude response of the 4-pole bandpass ladder filter for various k.

1-pole highpass:

$$\frac{g}{1+g}\cdot \frac{1}{1+g}$$

(where $g={\omega }_{c}T/2$). It’s not difficult to verify that the maximum gain of this pair is attained at $g=1$ and is equal to $1/4$. The maximum instantaneous gain of two of these pairs is therefore $1/16$, and thus the instantaneously unstable case doesn’t occur provided $k<16$.

Bandwidth control

Using (5.3) and the fact that the frequency response of $s^2/(1+s{)}^4$ at $\omega =1$ is $1/4$ we obtain the frequency response of the 4-pole bandpass ladder at the cutoff

$$H(j)=\frac{1}{4-k}$$

Therefore, by multiplying the output (or the input signal) of the 4-pole bandpass ladder by $4-k$ we can turn it into a normalized bandpass, where the bandwidth is controlled by varying $k$.

There is another way, however. Recall that the normalized 2-pole bandpass (4.15) is an LP to BP transformation of the 1-pole lowpass $1/(1+s)$. At the same time,

$$\frac{1}{1+s}\cdot \frac{s}{1+s}=\frac{s}{(1+s{)}^2}=\frac{s}{1+2s+s^2}=\frac{1}{2}\cdot \frac{2s}{1+2s+s^2}$$

is simply a halved version of (4.15) taken at $R=1$ and therefore is an LP to BP transformation fo the halved 1-pole lowpass $1/2(1+s)$. This means that Fig. 5.14 can be replaced by Fig. 5.16 which in turn is an LP to BP transformation of Fig. 5.17.

Figure 5.16: 4-pole bandpass ladder filter expressed in terms of normalized 2-pole bandpasses.

Figure 5.17: LP to BP transformation applied to this structure produces the 4-pole bandpass ladder in Fig. 5.16.

We don’t even specifically care to analyse the structure Fig. 5.17. What is important is that the damping parameter of the LP to BP transformation controls the transformation bandwidth and thereby the bandwidth of the bandpass ladder in Fig. 5.16. Thus, introducing the damping control into the normalized 2-pole bandpasses in Fig. 5.16 we can control the bandpass ladder’s bandwidth by simply varying the damping parameter of the underlying 2-pole bandpasses.

At the same time we still have the $k$ parameter available, which we still can use to control the bandwidth of the normalized bandpass (Fig. 5.18). Thus, $k$ and $R$ provide two different ways of bandwidth control, resulting in somewhat different amplitude response shapes (Fig. 5.19).⁷

Obviously, normalized 2-pole bandpasses with damping control could be implemented using an SVF. If nonlinearities are involved, however, using TSK/SKF 2-pole bandpasses might be a better option. Since we didn’t introduce the latter yet, we need to postpone the respective discussion. We will return to this question, however, in the discussion of 8-pole bandpass ladder in Section 5.9, where the bandwidth control via the 2-pole bandpass damping will be a particularly desired feature compared to being somewhat academic in the case of a 4-pole bandpass.

5.8 Sallen-Key filters

In this section we are going to introduce two special kinds of 2-pole bandpass ladder filters, the Sallen-Key filter and its transpose.⁸ They are important because of their nonlinear versions, since, as linear digital 2-pole filters go, the SVF filter could be sufficient for most applications, and it also provides probably the best performance among different TPT 2-poles.

Figure 5.18: 4-pole normalized bandpass ladder filter expressed in terms of normalized 2-pole bandpasses.

Figure 5.19: Amplitude response of the 4-pole normalized bandpass ladder filter in Fig. 5.18 for two different combinations of $k$ and $R$ resulting in comparable bandwidths.

For now we shall develop the linear versions of these filters. The Sallen-Key filter is more famous than its transpose, but we’ll start with the transpose, for the sake of a more systematic presentation of the material.

Transposed Sallen-Key (TSK) filters

Attempting to build a 2-pole lowpass ladder filter (Fig. 5.20) we don’t end up with a useful filter.

Figure 5.20: 2-pole lowpass ladder filter (not very useful).

Indeed, the transfer function of this filter is

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

and the poles are respectively at

$$s=-1\pm \sqrt{-k}=-1\pm j\sqrt{k}(k\ge 0)$$

Interpreting these pole positions in terms of 2-pole cutoff and damping (which we can do using (4.13)), we obtain

$$\{\begin{array}{l}{\omega }_c=\left|-1\pm j\sqrt{k}\right|=\sqrt{1+k}\\ R=\frac{-\mathrm{Re}\left(-1\pm j\sqrt{k}\right)}{\left|-1\pm j\sqrt{k}\right|}=\frac{1}{\sqrt{1+k}}\end{array}$$

Thus, firstly, there is coupling between the feedback amount and the effective cutoff of the filter. Secondly, as $k$ grows, $R$ stays strictly positive, thus the filter poles never go into the right semiplane (and, as with the 4-pole ladder filter, this would be quite desired once we make the filter nonlinear). So, all in all, not a very useful structure.

A similar situation occurs in an attempt to use two 1-pole highpasses instead of two 1-pole lowpass in the same structure (the readers may wish verify this on their own as an exercise).

This result is no wonder, considering that the transfer function of a chain of two 1-pole lowpasses is $1/(1+s{)}^2$, with the phase response being $0^{\circ }$ only at $\omega =0$ and being $180^{\circ }$ only at $\omega =\infty$ (for the highpasses the situation is opposite, we have $180^{\circ }$ only at $\omega =0$ and $0^{\circ }$ only at $\omega =\infty$, which doesn’t make a big difference for our purposes). Thus we don’t get a good resonance peak at any finite location. This however hints at the idea that we might still try to build a 2-pole bandpass ladder filter from a chain of a 1-pole lowpass and a 1-pole highpass, as the total phase shift at the cutoff would be $0^{\circ }$ in this case:

$$\left(\frac{1}{1+s}\cdot \frac{s}{1+s}\right){|}_{s=j}={\frac{s}{(1+s{)}^2}|}_{s=j}=\frac{j}{(1+j{)}^2}=\frac{1}{2}$$

The respective structure is shown in Fig. 5.21. Notice that we don’t invert the feedback.

Computing the transfer function of this filter we have

$$H(s)=\frac{\frac{s}{(1+s{)}^2}}{1-k\frac{s}{(1+s{)}^2}}=\frac{s}{(1+s{)}^2-ks}=\frac{s}{s^2+(2-k)s+1}$$

Figure 5.21: 2-pole bandpass ladder filter.

The obtained expression is identical to the transfer function of a 2-pole bandpass filter with a damping gain $2R=2-k$. That is, the filter in Fig. 5.21 is pretty much the same as a linear 2-pole SVF bandpass, at least from the frequency response perspective. Notice that $k=0$ corresponds to the resonance-neutral setting ($R=1$) while $k=2$ is the self-oscillation point ($R=0$). As we should remember from the 4-pole bandpass ladder discussion, the maximum possible instantaneous gain of the lowpass-highpass pair is $1/4$, therefore under the condition $k<4$ the TPT implementation of Fig. 5.21 doesn’t become instantaneously unstable.

It might seem that we have failed to construct a 2-pole lowpass filter using the above approach, but in fact with a slight modification we can obtain one from the bandpass filter in Fig. 5.21. Let’s replace the 1-pole highpass with a 1-pole multimode with highpass and lowpass outputs (Fig. 5.22).

Figure 5.22: 2-pole bandpass ladder filter with an extra output mode.

Obviously, the signal $y(t)$ is not affected by this replacement. Let’s find out what kind of signal is $y_1(t)$. In order to simplify the computation of the transfer function of the entire structure at $y_1$, consider first the transfer functions of the 1-pole multimode filter used in isolation:

$$H_{LP}(s)=\frac{1}{1+s}{H}_{HP}(s)=\frac{s}{1+s}$$

or, for complex sinusoidal signals of the form $e^{st}$

$$Y_{LP}(s)=\frac{1}{1+s}X(s)Y_{HP}(s)=\frac{s}{1+s}X(s)$$

where $X(s)e^{st}$ is the input signal of the multimode 1-pole and $Y_{LP}(s)e^{st}$ and $Y_{HP}(s)e^{st}$ are the respective output signals. This means that

$$Y_{LP}(s)=\frac{Y_{HP}(s)}{s}$$

Therefore a similar relationship exists between the outputs $y_1(t)$ and $y(t)$ of the filter in Fig. 5.22:

$$Y_1(s)=\frac{Y(s)}{s}$$

and there is the same relationship between their respective transfer functions

$$H_1(s)=\frac{H(s)}{s}=\frac{1}{s}\cdot \frac{s}{s^2+(2-k)s+1}=\frac{1}{s^2+(2-k)s+1}$$

where $H_1(s)$ the the transfer function for the signal $y_1(t)$ in respect to the input signal $x(t)$. Therefore $y_1(t)$ is an ordinary 2-pole lowpass signal with damping gain $2R=2-k$.

Thus we have obtained a multimode 2-pole ladder filter with the lowpass and bandpass outputs. We redraw the structure in Fig. 5.22 once again as Fig. 5.23 to reflect what we have just found out about this structure.

Figure 5.23: Transposed Sallen-Key (TSK) filter.

The structure in Fig. 5.23 happens to be a transpose of the Sallen-Key filter, therefore we will refer to it as the transposed Sallen-Key (TSK) filter.⁹ The transfer functions of the TSK filter are, as we have found out:

$$H_{LP}(s)=\frac{1}{s^2+(2-k)s+1}$$

$$H_{BP}(s)=\frac{s}{s^2+(2-k)s+1}$$

A 2-pole highpass output mode cannot be picked up in a straightforward way, but can be obtained with some extra effort. Let’s also turn the first lowpass into a multimode (Fig. 5.24). It is not difficult to realize that the transfer function for the signal at the LP output of ${\mathrm{MM}}_{1a}$, which is simultaneously the input signal of ${\mathrm{MM}}_{1b}$, is

$$H_{MM1aLP}(s)=H_{LP}(s)\cdot {\left(\frac{1}{s+1}\right)}^{-1}=\frac{s+1}{s^2+(2-k)s+1}$$

respectively for the signal at the HP output of ${\mathrm{MM}}_{1a}$ we have

$$H_{MM1aHP}(s)=s\cdot H_{MM1aLP}(s)=\frac{(s+1)s}{s^2+(2-k)s+1}$$

Thus we obtain

$$\begin{array}{rl}{H}_{\mathrm{HP}}(s)& =H_{MM1aHP}(s)-H_{\mathrm{BP}}(s)=\frac{(s+1)s}{s^2+(2-k)s+1}-\frac{s}{s^2+(2-k)s+1}=\\ & =\frac{s^2}{s^2+(2-k)s+1}\end{array}$$

Figure 5.24: Fully multimode TSK filter.

Alternative representations

Recall that 1-pole highpass signal can be obtained as the difference of the 1-pole lowpass filter’s input and output signals:

$$\frac{s}{1+s}=1-\frac{1}{1+s}$$

Then we can replace the multimode 1-pole in Fig. 5.23 by a 1-pole lowpass, constructing the highpass signal “manually” by subtracting the lowpass output from the lowpass input (Fig. 5.25). A further modification of Fig. 5.25 is formally using negative feedback (Fig. 5.26)

Figure 5.25: TSK filter (alternative representation).

Highpass TSK filter

Let’s take the filter in Fig. 5.21 and switch the order of lowpass and highpass 1-pole filters (Fig. 5.27). Since this doesn’t change the transfer function of the entire chain of 1-poles, the filter output stays the same, it is still a 2-pole bandpass.

Figure 5.26: TSK filter (alternative representation, negative feedback form).

Figure 5.27: 2-pole bandpass ladder filter with a different order of 1-pole lowpass and highpass filters.

Turning the 1-pole lowpass into a multimode we obtain the structure in Fig. 5.28. It’s not difficult to see that the signal at the other output of the multimode is a 2-pole highpass one. Therefore, in order to distinguish between the filters in Figs. 5.23 and 5.28 we will refer to the former more specifically as a lowpass TSK filter and to the latter as a highpass TSK filter. If necessary, we can add the lowpass output, using a way similar to Fig. 5.24.

Figure 5.28: Highpass TSK filter.

The highpass versions of Fig. 5.25 and Fig. 5.26 could have been built by performing transformations of Fig. 5.28 similarly to how we did with Fig. 5.23. However it’s easier just to apply the LP to HP substitution ($s\leftarrow 1/s$) to Figs. 5.25 and 5.26.

Sallen-Key filter (SKF)

We could take the structure in Fig. 5.27 and convert the 1-pole highpass filter into a tranposed multimode 1-pole (Fig. 5.29). By doing this one obtains a transpose of Fig. 5.23 which is (apparently) called Sallen-Key filter or shortly SKF. If necessary, the highpass input can be added, turning Fig. 5.23 into a transpose of Fig. 5.24.

Figure 5.29: Sallen-Key filter.

If instead we take the structure in Fig. 5.21 and convert the lowpass into a transposed multimode 1-pole, we can obtain the structure in Fig. 5.30. In order to distinguish between Fig. 5.29 and Fig. 5.30, we will, as we did with their transposes, refer to the structure in Fig. 5.29 more specifically as a lowpass Sallen-Key filter and to the structure in Fig. 5.30 as a highpass Sallen-Key filter. The lowpass input can be added to the highpass SKF using the transposed version of the idea of Fig. 5.24.

Figure 5.30: Highpass SKF.

The transposed versions of Fig. 5.25 and Fig. 5.26 make alternative representations of the lowpass SKF. E.g. by transposing the structure in Fig. 5.25 we obtain the one in Fig. 5.31.

Figure 5.31: Sallen-Key filter (alternative representation).

MIMO Sallen-Key filters

By turning both 1-poles in Fig. 5.27 into multimodes we’ll obtain a MIMO (multiple input multiple output) Sallen-Key filter, as illustrated in Fig. 5.32.

Note that the labelling of the inputs and outputs $x_{\mathrm{LP}}$, $x_{\mathrm{HP}}$, $y_{\mathrm{LP}}$, $y_{\mathrm{HP}}$ is thereby formal. The actual transfer functions are defined for signal paths from a given input to a given output:

Figure 5.32: MIMO Sallen-Key filter (HP-LP).

By putting the feedback path around lowpass-highpass chain rather than lowpass-highpass, Fig. 5.32 is turned into Fig. 5.33.

Figure 5.33: MIMO Sallen-Key filter (LP-HP).

Allpass TSK/SKF

Consider again the 2-pole bandpass ladder filter structure in Fig. 5.21. Suppose that we use 1-pole allpasses $(1-s)/(1+s)$ instead of low- and highpass filters. We also use negative, rather than positive feedback, although this is more a matter of convention. The result is shown in Fig. 5.34, where we also prepared the modal outputs.

Figure 5.34: 2-pole ladder filter based on allpasses (not so useful).

The transfer function of the main output is

$$\begin{array}{rl}{H}_2(s)& =\frac{{\left(\frac{1-s}{1+s}\right)}^2}{1+k{\left(\frac{1-s}{1+s}\right)}^2}=\frac{(1-s{)}^2}{(1+s{)}^2+k(1-s{)}^2}=\\ & =\frac{(1-s{)}^2}{(1+k)s^2+2(1-k)s+(1+k)}=\frac{1}{1+k}\cdot \frac{(1-s{)}^2}{s^2+2\frac{1-k}{1+k}s+1}\end{array}$$

which is not exactly a 2-pole allpass transfer function. The denominator of $H(s)$ however looks pretty usable, it’s a classical 2-pole transfer function denominator with damping $R=(1-k)/(1+k)$.

The transfer functions at the other two outputs can be obtained by “reverse application” of the transfer functions of the 1-pole allpasses to $H_2(s)$:

$$H_1(s)={\left(\frac{1-s}{1+s}\right)}^{-1}\cdot H_2(s)=\frac{1}{1+k}\cdot \frac{(1+s)(1-s)}{s^2+2\frac{1-k}{1+k}s+1}$$

$$H_0(s)={\left(\frac{1-s}{1+s}\right)}^{-1}\cdot H_1(s)=\frac{1}{1+k}\cdot \frac{(1+s{)}^2}{s^2+2\frac{1-k}{1+k}s+1}$$

We can try building the desired transfer function

$$H(s)=\frac{s^2-2Rs+1}{s^2+2Rs+1}=\frac{s^2-2\frac{1-k}{1+k}s+1}{s^2+2\frac{1-k}{1+k}s+1}$$

as a linear combination of $H_0(s)$, $H_1(s)$ and $H_2(s)$:

$$a_0{H}_0(s)+a_1{H}_1(s)+a_2{H}_2(s)=H(s)$$

Noticing that the denominators of $H_0(s)$, $H_1(s)$, $H_2(s)$ are all identical to the desired denominator already, we can discard the common denominator from the equation and simply write:

$$a_0\frac{(1+s{)}^2}{1+k}+a_1\frac{(1+s)(1-s)}{1+k}+a_2\frac{(1-s{)}^2}{1+k}=s^2-2\frac{1-k}{1+k}s+1$$

or

$$a_0(1+2s+s^2)+a_1(1-s^2)+a_2(1-2s+s^2)=(1+k)s^2-2(1-k)s+(1+k)$$

From where $a_0=k$, $a_1=0$, $a_2=1$. Thus

$$H(s)=H_0(s)+k{H}_2(s)=\frac{s^2-2\frac{1-k}{1+k}s+1}{s^2+2\frac{1-k}{1+k}s+1}$$

and the corresponding structure is shown in Fig. 5.35.^{10 11} The main idea of this structure is very similar to the one of a TSK filter with some “embedded” modal mixture. For that reason we can refer to the filter Fig. 5.35 as a allpass TSK filter, or we could call it a 2-pole allpass ladder filter.

Figure 5.35: Allpass TSK filter.

The 2-pole damping parameter $R$ is related to $k$ via

$$\begin{array}{rl}R& =(1-k)/(1+k)\\ k& =(1-R)/(1+R)\end{array}$$

so that for $k=-1\dots +\infty$ the damping varies from $+\infty$ to $-1$. The stable range $R=+\infty \dots 0$ corresponds to $k=-1\dots 1$.

Transposing the structure in Fig. 5.35 we obtain the structure Fig. 5.36 which for obvious reasons we will refer to as an allpass SKF.

Figure 5.36: Allpass SKF.

5.9 8-pole ladder

Connecting eight 1-pole lowpass filters in series instead of four we can build an 8-pole lowpass ladder filter (Fig. 5.37).

The transfer function of the 8-pole lowpass ladder is obviously

$$H(s)=\frac{1}{k+(1+s{)}^8}$$

Figure 5.37: 8-pole lowpass ladder filter.

and the pole positions are defined by

$$k+(1+s{)}^8=0$$

giving

$$s=-1+(-k{)}^{1/8}$$

where $(-k{)}^{1/8}$ is understood in the multivalued complex root sense:

$$(-k{)}^{1/8}=|k{|}^{1/8}{e}^{j\alpha }$$

where

$$\alpha =\frac{\pi +2\pi n}{8}$$

The main difference from the 4-pole ladder lowpass, besides the steeper cutoff slope, is that the $180^{\circ }$ phase shift by the chain of 1-pole lowpasses is no longer occurring at the cutoff. Instead, the phase response of the lowpass chain at the cutoff is $360^{\circ }$. In order to find the frequency at which $180^{\circ }$ phase shift is occurring we need to solve

$$\arg {\left(\frac{1}{1+j\omega }\right)}^8=-\pi$$

that is

$$\arg (1+j\omega )=\pi /8\text{or}\arg (1+j\omega )=3\pi /8$$

(apparently the values $5\pi /8$ an larger cannot be attained by $\arg (1+j\omega )$). This gives

$$\omega =\tan \pi /8\text{or}\omega =\tan 3\pi /8$$

The value of $\tan \pi /8$ can be easily found using the formula for the tangent of double angle:

$$\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$$

where letting $\alpha =\pi /8$ we obtain

$$\frac{2\tan \pi /8}{1-{\tan }^2\pi /8}=1$$

$$2\tan \pi /8=1-{\tan }^2\pi /8$$

$${\tan }^2\pi /8+2\tan \pi /8-1=0$$

$$\omega =\tan \pi /8=\sqrt{2}-1\approx 0.4142$$

For $\tan 3\pi /8$ we can use the formula for the tangent of the complementary angle:

$$\omega =\tan 3\pi /8=\tan (\pi /2-\pi /8)=\frac{1}{\tan \pi /8}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\approx 2.4142$$

Thus the resonance peak can occur at $\omega =\tan (\pi /4\pm \pi /8)=\sqrt{2}\pm 1$. Let’s find the values of $k$ at which the respective poles hit the imaginary axis. According to (5.7), $k$ is the reciprocal of the amplitude amplitude response of the chain of eight 1-pole lowpasses at the respective frequencies:

$$\begin{array}{rl}k& ={\left({\left|\frac{1}{1+j\omega }\right|}^8\right)}^{-1}={\left(\frac{1}{\sqrt{1+{\omega }^2}}\right)}^{-8}={\left(\sqrt{1+{\omega }^2}\right)}^{-8}=\\ & ={\left(\sqrt{1+{\tan }^2(\pi /4\pm \pi /8)}\right)}^8={\cos }^{-8}(\pi /4\pm \pi /8)\end{array}$$

finally giving

$$\begin{array}{rlrl}{\omega }_1& \approx 0.4142& k_1& \approx 1.884\\ {\omega }_2& \approx 2.4142& k_2& \approx 2174\end{array}$$

Thus the selfoscillation at ${\omega }_1$ is occurring way much earlier than the one at ${\omega }_2$. It is very unlikely that even in a nonlinear version of this filter, which allows going into unstable range of $k$, we will use $k$ as large as 2174. It also hints to the fact that the second resonance is way much weaker than the first one. Therefore, for practical purposes we will simply ignore the second resonance and say that the infinite resonance is occuring at $\omega =\sqrt{2}-1\approx 0.4142$ at $k\approx 1.884$. Fig. 5.38 illustrates the amplitude response behavior for various $k$.

Figure 5.38: Amplitude response of the 8-pole lowpass ladder filter for various $k$.

Considering that at $k=0$ the amplitude response of a chain of eight 1-poles at the cutoff is $(1/\sqrt{2}{)}^8=1/16$, which is ca. $-24$dB, we could treat the resonance frequency $\omega =\sqrt{2}-1$ as the “user-facing” cutoff frequency instead, and in practical implementations of the filter let the cutoff of the underlying 1-poles equal the “user-facing” cutoff multiplied by $1/(\sqrt{2}-1)=\sqrt{2}+1\approx 2.4142$.

One could ask the following question: the phase response of the chain of eight 1-poles at $\omega =1$ is $0^{\circ }$, therefore why don’t we simply use positive feedback to create the resonance peak at $\omega =1$? The problem is that the phase response at $\omega =0$ is also $0^{\circ }$. Since the amplitude response at $\omega =0$ is 1, the selfoscillation will occur already at $k=1$, whereas at $\omega =1$ it will occur only at $k=1/(1/\sqrt{2}{)}^8=16$.

The instantaneously unstable range of $k$ is found similarly to the 4-pole lowpass ladder and is $k<-1$.

Various modal mixtures for the 8-pole lowpass ladder filter can be built in a similar way to the 4-pole ladder filter. However the fact that the resonance frequency is noticeably lower than the cutoff frequency of the underlying 1-poles will affect the shapes of the resulting modal mixtures. Some smart playing around with the modal mixture coefficients can sometimes reduce the effect of this discrepancy.

8-pole highpass ladder

Replacing the 1-pole lowpasses with highpasses we obtain an 8-pole highpass ladder filter (Fig. 5.39). As we already know from the discussion of the 4-pole highpass, it essentially the same as lowpass except for the $s\leftarrow 1/s$ substitution. The instantaneously unstable range of $k$ is found similarly to the 4-pole highpass ladder and is $k<-1$.

Figure 5.39: 8-pole highpass ladder filter.

8-pole bandpass ladder

Replacing half of the lowpasses with highpasses in Fig. 5.37 we obtain the 8-pole bandpass ladder filter, where we shouldn’t forget that in a bandpass ladder the feedback shouldn’t be inverted (Fig. 5.40).

Figure 5.40: 8-pole bandpass ladder filter.

The total gain at the cutoff of the 1-pole chain is

$${\left(\frac{s}{(1+s{)}^2}\right)}^4{|}_{s=j}={\left(\frac{1}{2}\right)}^4=\frac{1}{16}$$

therefore selfoscillation occurs at $k=16$. Fig. 5.41 illustrates the amplitude response behavior at various $k$. Note that the amplitude response is pretty low (particularly, for $k=0$ it peaks at $-24$dB), therefore additional boosting of the output signal may be necessary in practical usage.

Figure 5.41: Amplitude response of the 8-pole bandpass ladder filter for various $k\ge 0$.

The instantaneously unstable range of $k$ is found similarly to the 4-pole bandpass ladder and is $k\ge 2^8=256$.

An interesting feature of the 8-pole bandpass ladder is that at negative $k$ the filter obtains two resonance peaks (Fig. 5.42).¹² Indeed, notice that the phase response of the 8-pole lowpass-highpass chain is the same as the one of the 8-pole lowpass chain:

$$\arg \frac{s^4}{(1+s{)}^8}=\arg \frac{1}{(1+s{)}^8}s=j\omega ,\omega \in \mathbb{R}$$

Thus we still have a $180^{\circ }$ phase shift at $\omega =\sqrt{2}\pm 1$.

The amplitude response of a single lowpass-highpass pair at $\omega =\sqrt{2}\pm 1$ is

$${\left|\left(\frac{s}{(1+s{)}^2}\right)\right|}_{s=j(\sqrt{2}\pm 1)}=\frac{\sqrt{2}\pm 1}{1+(\sqrt{2}\pm 1{)}^2}=\frac{1}{(\sqrt{2}\mp 1)+(\sqrt{2}\pm 1)}=\frac{1}{2\sqrt{2}}$$

therefore selfoscillation occurs at $k=-(2\sqrt{2}{)}^4=-64$.

8-pole bandpass ladder with bandwidth control

The occurence of two resonance peaks in an 8-pole bandpass ladder at $k<0$ motivates the introduction of the possibility to control the distance between these two peaks. In Section 5.7 we have introduced two different approaches to control the 4-pole bandpass ladder’s bandwidth. Apparently, the approach using the $k$ parameter is not good for our goal here, since we don’t want to affect the amplitude response shape in the vertical direction. Also, from Fig. 5.42 it seems that the variation of k in the negative range has little effect on the actual bandwidth. On the other hand, the approach using the damping of the underlying 2-pole bandpasses looks much more promising.

Figure 5.42: Amplitude response of the 8-pole bandpass ladder filter for various k < 0.

Representing the 8-pole bandpass ladder in terms of normalized 2-pole bandpasses (Fig. 5.43) we notice that it is an LP to BP transformation of the filter in Fig. 5.44. The filter in Fig. 5.44 is essentially the same as the ordinary 4-pole lowpass ladder (Fig. 5.1), except that

• the feedback is positive, so that selfoscillation at $\omega =1$ occurs at some negative value of $k$
• the output signal amplitude and the feedback amount are 16 times lower, thus selfoscillation at $\omega =1$ doesn’t occur at $k=-4$ but at $k=-64$ (which matches the already established fact of selfoscillation of Fig. 5.40 and equivalently Fig. 5.43 at $k=-64$).

Therefore by controlling the bandwidth of the LP to BP transformation, we will control the distance between the resonance peaks in Fig. 5.42.

Figure 5.43: 8-pole bandpass ladder filter expressed in terms of normalized 2-pole bandpasses.

Figure 5.44: 4-pole lowpass ladder filter with positive feedback and additional gains of 1/2. The LP to BP substitution applied to this filter produces the filter in Fig. 5.43.

Since the resonance peak in Fig. 5.44 is occurring at $\omega =1$, the formula (4.20) expresses $R$ in terms of the distance between the two images of this peak after the LP to BP transformation. Therefore we can directly use the formula (4.20) to control the distance between the resonance peaks in Fig. 5.43. The prewarping techniques described in Section 4.6 also apply, thereby allowing us to achieve the exact positioning of the resonance peaks (in the limit $k\to -64$).

There is an important question concerning the choice of the specific topology for the normalized bandpasses BPn. Of course, the most obvious choice would be to use an SVF. This should work completely fine in the linear case. In a nonlinear case, however, we might want to use a different topology. Particularly, we might want that at $R=1$ our controlled-bandwidth topology becomes fully identical to Fig. 5.40 (therefore obtaining the sound, which is identical to the one of the structure in Fig. 5.40 even in the presence of nonlinear effects).

Assuming that Fig. 5.40 implies interleaved 1-pole low- and highpasses (as shown in Fig. 5.45), a good solution is provided by the TSK/SKF filters. E.g. considering the structure in Fig. 5.21 (which is essentially the TSK filter from Fig. 5.23), we can notice that at $k=0$ it becomes fully equivalent to a single lowpass-highpass pair. This suggests that we could use this structure to construct a halved normalized bandpass (Fig. 5.46), where expressing the TSK feedback $k$ in terms of damping $R$ we have $k=2(1-R)$. Note that at $R=1$ not only the feedback path in Fig. 5.46 is disabled, but also the output gain element $R$ is becoming transparent. Using the halved normalized bandpass in Fig. 5.46, we could reimplement Fig. 5.45 as Fig. 5.47.

Figure 5.45: Fig. 5.40 implemented by interleaved 1-pole low- and high-passes.

Figure 5.46: Halved normalized TSK bandpass.

Figure 5.47: 8-pole bandpass ladder filter expressed in terms of halved normalized TSK bandpasses.

5.10 Diode ladder

In the diode ladder filter the serial connection of four 1-pole lowpass filters (implemented by the transistor ladder) is replaced by a more complicated structure of 1-pole filters (implemented by the diode ladder). The block diagram of the diode ladder is shown in Fig. 5.48, while the diode ladder filter adds the feedback loop around that structure, feeding the fourth output of the diode ladder into the diode ladder’s input (Fig. 5.49).

Figure 5.48: Diode ladder.

Figure 5.49: Diode ladder filter.

It is instructive to write out the 1-pole equations implied by Fig. 5.48:

$$\begin{array}{rl}{\dot{y}}_1& ={\omega }_c((x+y_2)-y_1)\\ {\dot{y}}_2& ={\omega }_c((y_1+y_3)/2-y_2)\\ {\dot{y}}_3& ={\omega }_c((y_2+y_4)/2-y_3)\\ {\dot{y}}_4& ={\omega }_c(y_3/2-y_4)\end{array}\text{\hspace{1em}(5.18)}$$

In this form it’s easier to guess the reason for the gain elements 1/2 used in Fig. 5.48, they perform the averaging between the feedforward and feedback signals. However this averaging in (5.18) and Fig. 5.48 is not done fully consistently. It would have been more consistent to have no 1/2 gain element at the input of the fourth lowpass, rather than of the first one:

$$\begin{array}{rl}{\dot{y}}_1& ={\omega }_c((x+y_2)/2-y_1)\\ {\dot{y}}_2& ={\omega }_c((y_1+y_3)/2-y_2)\\ {\dot{y}}_3& ={\omega }_c((y_2+y_4)/2-y_3)\\ {\dot{y}}_4& ={\omega }_c(y_3-y_4)\end{array}\text{\hspace{1em}(5.19)}$$

in which case the first lowpass would take $(x+y_2)/2$ as its input, the second lowpass would take $(y_1+y_3)/2$ as its input, the third lowpass would take $(y_2+y_4)/2$ as its input, and the fourth lowpass would take $y_3$ as its input. However, (5.18) is a more traditional way to implement a diode ladder filter. Anyway, the difference between (5.18) and (5.19) is actually not that large, since (as we are going to show below) they result in one and the same transfer function,

The more complicated connections between the 1-pole lowpasses present in the diode ladder “destroy” the frequency response of the ladder in a remarkable form, which, is responsible for the characteristic diode ladder filter sound.¹³ Generally, the behavior of the diode ladder filter is less “straightforward” than the one of the transistor ladder filter.

Transfer function

We are going to develop the transfer function for the diode ladder in a generalized form (Fig. 5.50), where $H_n(s)$ denote blocks with respective transfer functions. In the case of Fig. 5.48 and (5.18) we would have

$$H_1(s)=G(s)H_2(s)=H_3(s)=H_4(s)=\frac{G(s)}{2}\text{\hspace{1em}(5.20)}$$

while in the case of (5.19) we would respectively have

$$H_1(s)=H_2(s)=H_3(s)=\frac{G(s)}{2}{H}_4(s)=G(s)\text{\hspace{1em}(5.21)}$$

where

$$G(s)=\frac{1}{1+s}\text{\hspace{1em}(5.22)}$$

Figure 5.50: Generalized diode ladder in transfer function form.

Assuming complex exponential signals $e^{st}$, for the $H_4(s)$ block we have

$$y_4=H_4{y}_3$$

(where $H_4$ is short for $H_4(s)$), therefore

$$\frac{1}{H_4}{y}_4=y_3\text{\hspace{1em}(5.23)}$$

For the $H_3(s)$ block we have

$$y_3=H_3(y_2+y_4)$$

Substituting (5.23) we have

$$\frac{1}{H_4}{y}_4=H_3(y_2+y_4)$$

$$\frac{1}{H_{34}}{y}_4=y_2+y_4$$

$$\frac{1-H_{34}}{H_{34}}{y}_4=y_2\text{\hspace{1em}(5.24)}$$

where $H_{34}$ is a short notation for $H_3{H}_4$.

For the $H_2(s)$ block we have

$$y_2=H_2(y_1+y_3)$$