Chapter 9: Classical signal processing filters

In Chapter 8 we have introduced, among other ideas, Butterworth filters of the 1st kind. In this chapter we are going to construct further similar filter types by allowing the amplitude response to have ripples of equal amplitude (a.k.a. equiripples) in the pass- or stop-band, or in both. These filters as well as Butterworth filters of the 1st kind are the filter types used in classical signal processing. They have somewhat less prominent role in music DSP, therefore we first concentrated on other filter types. Still, they are occasionally useful.

9.1 Riemann sphere

Before we begin discussing equiripple filters we need to go into some detail of complex algebra, concerning the Riemann sphere and some derived concepts. The key feature of the Riemann sphere is that infinity is treated like any other point, and this (among with some other possibilities arising out of using the Riemann sphere) will be quite helpful in our discussions. It seems there are a number of slightly different conventions regarding the Riemann sphere. We are going to introduce now one particular convention which will be most useful for our purposes.¹

Given a complex plane $w=u+jv$ we introduce the third dimension, thereby embedding the plane into the 3-dimensional space $(x,y,z)$. The $x$ and $y$ axes coincide with $u$ and $v$ axes, the $z$ axis is directed upwards. The Riemann sphere will be the sphere of a unit radius $x^2+y^2+z^2=1$ (Fig. 9.1). Thus, the intersection of the Riemann sphere with the complex plane is at the “equator” which in terms of $w$ is simply the complex unit circle $|w|=1$. The center of projection will be at the “north pole” $(0,0,1)$ of the Riemann sphere, which thereby is the image of $w=\infty$. Respectively, the complex unit circle $|w|=1$ coincides with its own projection image on the Riemann sphere.

We will denote and refer to the points on the Riemann sphere by the complex values that they represent. E.g. the “north pole” will be simply denoted as $\infty$, the “south pole” as $0$, the point on the “zero meridian” as $1$, the point on the $90^{\circ }$ meridian as $j$ etc. Some of these points are shown on Fig. 9.1.

Figure 9.1: Riemann sphere.

Real Riemann circle

The 2-dimensional subspace $(x,z)$ of the $(x,y,z)$ space in Fig. 9.1 contains just the real axis $u$ of the complex plane and the real axis’s image on the Riemann sphere which is a circle of unit radius $x^2+z^2=1$ (Fig. 9.2). It will be intuitive to refer to this circle as the real Riemann circle.

Figure 9.2: Real Riemann circle. The 0, 1, $\infty$, -1 labels denote the points on the circle which correspond to these values.

We can use the polar angle $\phi$ (defined as shown in Fig. 9.2) as the coordinate on the Riemann circle. One of the reasons for this choice of definition of $\phi$ are the following convenient mappings between $u$ and $\phi$:

$$\begin{array}{rl}u=0& ⟺\phi =2\pi n\\ u=1& ⟺\phi =\frac{\pi }{2}+2\pi n\\ u=-1& ⟺\phi =-\frac{\pi }{2}+2\pi n\\ u=\infty & ⟺\phi =\pi +2\pi n\end{array}$$

Also, if we restrict $\phi$ to $(-\pi ,\pi )$, then

$$\begin{array}{rl}u=0& ⟺\phi =0\\ u>0& ⟺\phi >0\\ u<0& ⟺\phi <0\end{array}$$

From Fig. 9.2, using some basic geometry it’s not difficult to find that $u$ and $\phi$ are related as

$$u=\tan \frac{\phi }{2}\text{\hspace{1em}(9.1)}$$

and that

$$u=\frac{x}{1-z}\text{\hspace{1em}(9.2a)}$$

By introducing the “homogeneous” coordinate $\bar{z}=1-z$ the equation (9.2a) can be rewritten in a more intuitive form:

$$u=x/\bar{z}\text{\hspace{1em}(9.2b)}$$

Conversely, using Fig. 9.2 and equation (9.1) we have

$$x=\sin \phi =\frac{2u}{u^2+1}\text{\hspace{1em}(9.3a)}$$

$$z=-\cos \phi =\frac{u^2-1}{u^2+1}\text{\hspace{1em}(9.3b)}$$

$$\bar{z}=\frac{2}{u^2+1}\text{\hspace{1em}(9.3c)}$$

Symmetries on the real Riemann circle

Certain symmetries between a pair of points on the real axis correspond to symmetries on the real Riemann circle. Specifically, from (9.1) we obtain:

$$u_1+u_2=0⟺{\phi }_1+{\phi }_2=2\pi n\text{\hspace{1em}(9.4a)}$$

$$u_1{u}_2=1⟺{\phi }_1+{\phi }_2=\pi +2\pi n\text{\hspace{1em}(9.4b)}$$

$$u_1{u}_2=-1⟺{\phi }_1-{\phi }_2=\pi +2\pi n\text{\hspace{1em}(9.4c)}$$

or, by restricting ${\phi }_1$ and ${\phi }_2$ to $[-\pi ,\pi ]$

$$u_1+u_2=0⟺{\phi }_1+{\phi }_2=0\text{\hspace{1em}(9.5a)}$$

$$u_1{u}_2=1⟺\frac{{\phi }_1+{\phi }_2}{2}=\pm \frac{\pi }{2}\text{\hspace{1em}(9.5b)}$$

$$u_1{u}_2=-1⟺{\phi }_1-{\phi }_2=\pm \pi \text{\hspace{1em}(9.5c)}$$

Fig. 9.3 illustrates.

Figure 9.3: Symmetries of the values on the real Riemann circle.

Coordinate relationships for the Riemann sphere

The equations (9.2) and (9.3) generalize to the 3-dimensional space $(x,y,z)$ containing the complex plane $w=u+jv$ and the Riemann sphere $x^2+y^2+z^2=1$ in an obvious way as:

$$u=\frac{x}{1-z}=x/\bar{z}\text{\hspace{1em}(9.6a)}$$

$$v=\frac{y}{1-z}=y/\bar{z}\text{\hspace{1em}(9.6b)}$$

$$w=u+jv=\frac{x+jy}{\bar{z}}\text{\hspace{1em}(9.6c)}$$

and

$$x=\frac{2u}{|w{|}^2+1}\text{\hspace{1em}(9.7a)}$$

$$y=\frac{2v}{|w{|}^2+1}\text{\hspace{1em}(9.7b)}$$

$$z=\frac{|w{|}^2-1}{|w{|}^2+1}\text{\hspace{1em}(9.7c)}$$

$$\bar{z}=\frac{2}{|w{|}^2+1}\text{\hspace{1em}(9.7d)}$$

$$\text{\hspace{1em}(9.7e)}$$

The equation (9.1) can be generalized if we restrict $\phi$ to $[0,\pi ]$, in which case

$$|w|=\tan \frac{\phi }{2}\text{\hspace{1em}(9.8)}$$

In principle we could also introduce the spherical azimuth angle, which is simply equal to $\arg w$, but we won’t do it in this book.

Imaginary Riemann circle

The $(y,z)$ subspace of the $(x,y,z)$ space in Fig. 9.1 contains just the imaginary axis $v$ of the complex plane and the imaginary axis’s image on the Riemann sphere which is a circle of unit radius $x^2+y^2=1$. We will refer to this circle as the imaginary Riemann circle.

There are no essential differences to the real Riemann circle. The same illustrations and formulas hold, except that we should use $v$ in place of $u$. Fig. 9.4 provides a simple illustration of the circla and its symmetries. Note that the reciprocal symmetries change sign if expressed in terms of the complex variable $w$, since $1/jv=-j/v$.

Figure 9.4: Symmetries of the values on the imaginary Riemann circle, where $w=jv$, $v=\mathrm{Im}w$.

9.2 Arctangent scale

From the Riemann circle one can derive a special scale which will be useful for plotting function graphs with interesting behavior around infinity. One commonly known special scale is a logarithmic scale $x^{\prime }=\log x$ which maps the logical values $x$ to the geometric positions $x^{\prime }$ on the plot. In a similar fashion, we introduce the arctangent scale

$$x^{\prime }=2\arctan x\text{\hspace{1em}(9.9)}$$

which is using the polar angle $\phi$ of the real Riemann circle as the geometric position $x^{\prime }$. It is easy to notice that (9.9) is equivalent to (9.1), where we have $x$ in place of $u$ and $x^{\prime }$ in place of $\phi$.

The actangent scale warps the entire real axis $(-\infty ,+\infty )$ into the range $(-\pi ,\pi )$. Due to the periodic nature of the Riemann circle’s polar angle it is not unreasonable to require the scale $x^{\prime }$ to be periodic as well, in which case we can also support the value $x=\infty$ which will map to $\pi +2\pi n$.

Treating the infinity like any other point, the arctangent scale provides a convenient means for plotting the functions where the range of the values of interest includes infinity. E.g. we could plot the graph of the cosecant function $y=\csc x=1/\sin x$ using the arctangent scale for the function’s value axis, as illustrated by Fig. 9.5

Symmetries in the arctangent scale

The symmetries of a graph plotted in the arctangent scale are occurring in agreement with Riemann circle symmetries (9.4) and (9.5) (illustrated in Fig. 9.3). Specifically:

1. Mutually opposite values are symmetric with respect to points $0$ and $\infty$.

Figure 9.5: Graphs of $y=\sin x$ (dashed) and $y=\csc x=1/\sin x$ (solid) using arctangent scale for the ordinate.

2. Mutually reciprocal values are symmetric with respect to points $1$ and $-1$.

3. Values whose product is $-1$ map are spaced by the distance equal to the half of the arctangent scale’s period, that is e.g. to the distance between $-1$ and $1$ or between $0$ and $\infty$.

One can observe all of these properties if Fig. 9.5, where the third property can be observed between the $\csc x$ and the half-period-shifted $\sin x$.

9.3 Rotations of Riemann sphere

We are going to introduce two special transformations of the complex plane:

$${\rho }_{+1}(w)=\frac{1+w}{1-w}\text{\hspace{1em}(9.10a)}$$

$${\rho }_{-1}(w)=\frac{w-1}{w+1}\text{\hspace{1em}(9.10b)}$$

where $w\in \mathbb{C}\cup \infty$. It is easy to check by direct substitution that ${\rho }_{-1}({\rho }_{+1}(w))={\rho }_{+1}({\rho }_{-1}(w))=w$, that is the transformations ${\rho }_1$ and ${\rho }_{-1}$ are each other’s inverses.² As we shall see, ${\rho }_{\pm 1}$ are simply rotations of the Riemann sphere by $90^{\circ }$ in two opposite directions.

Letting $w=u+jv$ where $u$ and $v$ are the real and imaginary parts of $w$, we have

$$\begin{array}{rl}{w}^{\prime }=u^{\prime }+j{v}^{\prime }={\rho }_{+1}(w)& =\frac{1+w}{1-w}=\frac{1+(u+jv)}{1-(u+jv)}=\frac{(1+u)+jv)}{(1-u)+jv}=\\ & =\frac{((1+u)+jv)((1-u)+jv)}{(1-u{)}^2+v^2}=\frac{(1-u^2-v^2)+2jv}{1+u^2+v^2-2u}=\frac{(1-|w{|}^2)+2jv}{1+|w{|}^2-2u}\end{array}$$

That is

$$u^{\prime }=\frac{1-|w{|}^2}{1+|w{|}^2-2u}$$

$$v^{\prime }=\frac{2v}{1+|w{|}^2-2u}$$

On the other hand,

$$|w^{\prime }{|}^2={\left|\frac{1+w}{1-w}\right|}^2=\frac{|1+w{|}^2}{|1-w{|}^2}=\frac{(1+u{)}^2+v^2}{(1-u{)}^2+v^2}=\frac{1+|w{|}^2+2u}{1+|w{|}^2-2u}$$

and

$$|w^{\prime }{|}^2+1=\frac{(1+|w{|}^2+2u)+(1+|w{|}^2-2u)}{1+|w{|}^2-2u}=2\frac{1+|w{|}^2}{1+|w{|}^2-2u}$$

from where by (9.7d)

$${\bar{z}}^{\prime }=\frac{2}{|w^{\prime }{|}^2+1}=\frac{1+|w{|}^2-2u}{1+|w{|}^2}$$

Then, using (9.7) we obtain

$$\begin{array}{rl}{x}^{\prime }& ={\bar{z}}^{\prime }{u}^{\prime }=\frac{1-|w{|}^2}{1+|w{|}^2}=-z\\ y^{\prime }& ={\bar{z}}^{\prime }{v}^{\prime }=\frac{2v}{1+|w{|}^2}=y\\ z^{\prime }& =1-{\bar{z}}^{\prime }=\frac{(1+|w{|}^2)-(1+|w{|}^2-2u)}{1+|w{|}^2}=\frac{2u}{1+|w{|}^2}=x\end{array}$$

That is $x^{\prime }=-z$, $y^{\prime }=y$, $z^{\prime }=x$ which is simply a rotation by $90^{\circ }$ around the $y$ axis in the direction from the (positive) $x$ axis towards the (positive) $z$ axis. Thus ${\rho }_{+1}$ simply rotates the Riemann sphere around the imaginary axis of the complex plane $w$ by $90^{\circ }$ in the direction from $1$ to $\infty$, or, which is the same, in the direction from $0$ to $1$ (where by $0$, $1$ and $\infty$ we mean the points on the Riemann sphere which are the projection images of $w=0$, $w=1$ and $w=\infty$ respectively). The points $\pm j$ are thereby untouched, which can be also seen by directly evaluating ${\rho }_{+1}(\pm j)=\pm j$. The transformation ${\rho }_{-1}$ being the inverse of ${\rho }_{+1}$ simply rotates in the opposite direction.

In terms of the real Riemann circle ${\rho }_{\pm 1}$ clearly correspond to a counterclockwise (for ${\rho }_{+1}$) or clockwise (for ${\rho }_{-1}$) rotation by $90^{\circ }$ (Fig. 9.6).³ Respectively, in the arctangent scale they correspond to shifts by a quarter of the arctangent scale’s period.

The imaginary Riemann circle is rotated into the unit circle $|w^{\prime }|=1$, which is its own image on the Riemann sphere. Therefore for ${\rho }_{+1}$ the polar angle $\phi$ from Fig. 9.2 becomes the polar angle in the complex plane (since $\phi =0⟺w=0$ and since $\arg {\rho }_{+1}(0)=\arg 1=0$), therefore $\phi =\arg {\rho }_{+1}(w)$ and by (9.1) we have

$${\rho }_{+1}\left(j\tan \frac{\phi }{2}\right)=e^{j\phi }$$

This can also be verified by direct substitution, where it’s easier to use the inverse transformation:

$${\rho }_{-1}(e^{j\phi })=\frac{e^{j\phi }-1}{e^{j\phi }+1}=\frac{e^{j\phi /2}-e^{-j\phi /2}}{e^{j\phi /2}+e^{-j\phi /2}}=$$

Figure 9.6: Transformation of the real Riemann circle by ${\rho }_{+1}$.

$$=j\frac{e^{j\phi /2}-e^{-j\phi /2}}{2j}\cdot \frac{2}{e^{j\phi /2}+e^{-j\phi /2}}=j\tan \frac{\phi }{2}$$

For ${\rho }_{-1}$ the polar angle $\phi$ gets mapped into $\arg {\rho }_{-1}(w)=\pi -\phi$. Conversely a polar angle equal to $\pi -\phi$ will be mapped into $\arg {\rho }_{-1}(w)=\phi$, thus

$$e^{j\phi }={\rho }_{-1}\left(j\tan \frac{\pi -\phi }{2}\right)={\rho }_{-1}\left(j\tan \left(\frac{\pi }{2}-\frac{\phi }{2}\right)\right)={\rho }_{-1}\left(\frac{j}{\tan \frac{\phi }{2}}\right)$$

Summing up:

$${\rho }_{+1}\left(j\tan \frac{\phi }{2}\right)=e^{j\phi }\text{\hspace{1em}(9.11a)}$$

$${\rho }_{-1}\left(\frac{j}{\tan \frac{\phi }{2}}\right)=e^{j\phi }\text{\hspace{1em}(9.11b)}$$

$${\rho }_{-1}(e^{j\phi })=j\tan \frac{\phi }{2}\text{\hspace{1em}(9.12a)}$$

$${\rho }_{+1}(e^{j\phi })=\frac{j}{\tan \frac{\phi }{2}}\text{\hspace{1em}(9.12b)}$$

Note that (9.12) do not simply give the inverse versions of (9.11), but also reflect the fact the complex unit circle gets rotated into the imaginary Riemann circle.

Symmetries

The transformations of symmetries by ${\rho }_{\pm 1}$ can be derived in an intuitive way from the symmetries on the real Riemann circle and the arctangent scale and from the fact that these transformations are $\pm 90^{\circ }$ rotations of the real Riemann circle or (equivalently) shifts of the arctangent scale by the scale’s quarter period, if we assume $w\in \mathbb{R}\cup \infty$.

Particularly, a $\pm 90^{\circ }$ rotation of the real Riemann circle maps $0$ and $\infty$ to $\pm 1$ and vice versa. Respectively, points on the Riemann circle which are symmetric relatively to $0$ and $\infty$ map to points symmetric relatively to $\pm 1$ and vice versa. Thus, mutually opposite values map to mutually reciprocal values and vice versa:

$${\rho }_{+1}(w){\rho }_{+1}(-w)=1\text{\hspace{1em}(9.13a)}$$

$${\rho }_{-1}(w){\rho }_{-1}(-w)=1\text{\hspace{1em}(9.13b)}$$

$${\rho }_{+1}(w)+{\rho }_{+1}(1/w)=0\text{\hspace{1em}(9.13c)}$$

$${\rho }_{-1}(w)+{\rho }_{-1}(1/w)=0\text{\hspace{1em}(9.13d)}$$

Fig. 9.7 illustrates, where more properties are immediately visible. E.g. one could notice that ${\rho }_{\pm 1}(w)$ are obtained by rotating $w$ by $\pm 90^{\circ }$, therefore the results are $180^{\circ }$ apart, which means:

$${\rho }_{+1}(w){\rho }_{-1}(w)=-1\text{\hspace{1em}(9.14)}$$

or that rotating the opposite points $\pm w$ by $90^{\circ }$ in opposite directions produces opposite points:

$${\rho }_{\pm 1}(-w)=-{\rho }_{\mp 1}(w)\text{\hspace{1em}(9.15)}$$

etc.

Figure 9.7: Symmetries of the transformations ${\rho }_{\pm 1}$ on the real Riemann circle, where they represent $90^{\circ }$ rotations.

The formulas (9.13), (9.14) and other similarly obtained symmetry-related properties of ${\rho }_{\pm 1}$ work not only for $w\in \mathbb{R}\cup \infty$ but actually for any $w\in \mathbb{C}\cup \infty$ which can be verified algebraically. However, the interpretation of these symmetries in terms of the real Riemann circle obviously works only for $w\in \mathbb{R}\cup \infty$.

The imaginary Riemann circle gets rotated by ${\rho }_{\pm 1}$ into the complex unit circle, where the results of the two transformations are thereby symmetric relatively to the imaginary axis. Fig. 9.8 illustrates some of the symmetries arising out of this rotation. More illustrations of this kind can be created, particularly for the transformation of the values lying on the complex unit circle, but the ones we are already having should be sufficient for our purposes in this book.

Unit circle rotations

The Riemann sphere rotations ${\rho }_{\pm 1}$ can be described as rotations of the Riemann sphere around the imaginary axis or as rotations of the real Riemann circle. Similarly the Riemann sphere rotations around the vertical axis $z$ can be thought of as rotations of the unit circle.

Apparently, such rotations are simply achived by a multiplications of complex values by a unit-magnitude complex constant, where we are not restricted to rotations by multiples of $90^{\circ }$. We won’t need a special notation for this transformation and will simply write

Figure 9.8: Symmetries of the results of the transformation ${\rho }_{+1}$ of the imaginary axis (lying on the complex unit circle).

$$w^{\prime }=aw(|a|=1)$$

Obviously

$$|w^{\prime }|=|w|$$

$$\arg w^{\prime }=\arg w+\arg a$$

The rotations by $+90^{\circ }$ and $-90^{\circ }$ are simply multiplications by $j$ and $-j$ respectively.

There is not much more to say in this respect as this rotation is pretty trivial.

Imaginary rotations

We could also wish to rotate the imaginary Riemann circle. We will denote the respective transformations as ${\rho }_{\pm j}$, where the subscript, as with ${\rho }_{\pm 1}$, denotes the image of the zero after the transformation.

We could rotate the imaginary circle by doing the three steps in succession:

1. First, we rotate the Riemann sphere by $90^{\circ }$ around its “vertical” axis, thereby turning the imaginary Riemann circle into the real Riemann circle, where $w=\pm j$ is transformed into $w=\pm 1$ respectively, while $0$ and $\infty$ stay in place. Such rotation is simply achieved by multiplying $w$ by $-j$.

2. Now we apply ${\rho }_{\pm 1}$ to rotate the real circle.

3. We convert the real circle back to the imaginary one by multiplying the rotation result by $j$.

Therefore we simply have

$${\rho }_{\pm j}(w)=j{\rho }_{\pm 1}(-jw)=-j{\rho }_{\mp 1}(jw)\text{\hspace{1em}(9.16)}$$

(where the second expression is obtained in the same way as the first one, except that we rotate the Riemann sphere in the other direction, both vertically and horizontally).

In order to distinguish between ${\rho }_{\pm 1}$ and ${\rho }_{\pm j}$ we could refer to the former as real rotations of the Riemann sphere and to the latter as imaginary rotations of the Riemann sphere.

9.4 Butterworth filter revisited

In Chapter 8 we have developed the lowpass Butterworth filters of the 1st kind (also simply known as Butterworth filters) as a Butterworth transformation of the 1-pole lowpass filter, where we also mentioned that the tranditional definition of the Butterworth filter simply defines the (lowpass) Butterworth filter as a (stable) filter whose amplitude response is

$$|H(j\omega ){|}^2=\frac{1}{1+{\omega }^{2N}}(\omega \in \mathbb{R})\text{\hspace{1em}(9.17)}$$

Apparently both definitions are equivalent.

We could generalize this idea by replacing ${\omega }^N$ in (9.17) by some other polynomial function $f(\omega )$:

$$|H(j\omega ){|}^2=\frac{1}{1+f^2(\omega )}(\omega \in \mathbb{R})\text{\hspace{1em}(9.18)}$$

The practical application of (9.18) essentially follows the steps of the Butterworth transformation of the 1st kind, which includes solving $1+f^2=0$ to obtain the poles of $|H(s){|}^2$ and then discarding the right-semiplane poles, thereby effectievely obtaining the desired transfer function $H(s)$ expressed in the cascade form. Note that there are two implicit conditions which need to be fulfilled in order for this procedure to work:

• There is the symmetry of the poles of $|H(s){|}^2$ with respect to the real axis: if $s$ is a pole then so is $s^{\ast }$ (where $s^{\ast }$ may be the same pole as $s$ if $s$ is real). This is necessary in order for $H(s)$ to be a real function of $s$.

• There is the pairwise symmetry of the poles $|H(s){|}^2$ with respect to the imaginary axis: if $s$ is a pole then $-s^{\ast }$ is another pole (it must be another pole even if $s=-s^{\ast }$, in which case it simply means that the pole is duplicated). This guarantees that we can split the poles into the left- and right-semiplane halves with identical contributions to the amplitude response. Therefore by discarding the right-semiplane half of the poles, we effectively go from $|H(j\omega ){|}^2$ to $|H(j\omega )|$.

We could expect that these properties will not be fulfilled for an arbitrary $f(\omega )$. However, let’s require that

• The function $f(\omega )$ is a real function of $\omega$.

• The function $f(\omega )$ is odd or even.

The readers can convince themselves that under these restrictions the poles of $|H(s){|}^2$ defined by (9.18) will have the necessary symmetries with respect to the real and imaginary axes.

Rational $f(\omega )$

We could allow $f(\omega )$ to be not just a polynomial but a rational function. In this case $H(s)$ has not only poles, but also zeros at locations where $f(\omega )$ has poles and respectively the denominator of $|H(j\omega ){|}^2$ turns to $\infty$, which means that both $|H(j\omega ){|}^2$ and $|H(j\omega )|$ turn to zero. In order to see that the multiplicities of the zeros of $H(s)$ are equal to the multiplicities of the respective poles of of $f(\omega )$ simply consider

$$|H(j\omega ){|}^2=\frac{1}{1+\frac{P_1^2(\omega )}{P_2^2(\omega )}}=\frac{P_2^2(\omega )}{P_1^2(\omega )+P_2^2(\omega )}$$

Thus the set of zeros of $|H(j\omega ){|}^2$ is the duplicated set of zeros of $P_2(\omega )$, however after switching to $H(j\omega )$ we should drop the duplicates, being left only with a single set of zeros of $P_2(\omega )$, which are the poles of $f(\omega )$. Note that $H(s)$ shouldn’t have more zeros than poles, which means that the order of the denominator of $f(\omega )$ should not exceed the order of the numerator of $f(\omega )$.

In order for $H(s)$ to be real, its zeros must be conjugate symmetric, which will be ensured if $f(\omega )$ is real odd or even function. Indeed, in this case the poles of $f(\omega )$ are both conjugate symmetric and symmetric with respect to the origin, which implies that they are also symmetric with respect to the imaginary $\omega$ axis, which is the same as the symmetry with respect to the real $s$ axis.

Representations of linear scaling

We are now going to develop another way of looking at the Butterworth filter generating function $f(x)=x^N$. It will be highly useful with other functions $f(x)$ occurring in place of $f(x)=x^N$ in (9.18).

Let $f(x)=x^N$. Since $x^N=\exp (N\ln x)$, we can write

$$f(x)=\exp (N\ln x)\text{\hspace{1em}(9.19)}$$

Introducing auxiliary variables $u$ and $v$ we can rewrite (9.19) as a set of equations:

$$\begin{array}{rl}x& =\exp u\\ v& =Nu\\ f(x)& =\exp v\end{array}$$

which also allows to define $f(x)$ implicitly as a function satisfying the equation

$$f(\exp u)=\exp (Nu)\text{\hspace{1em}(9.20)}$$

We could consider $x$ and $f(x)$ as representations of $u$ and $v$, where the connection between the preimages $u$ and $v$ and their respective representations $x$ and $f(x)$ is achieved via the exponential mapping $x=\exp u$ (Fig. 9.9). In terms of preimage domain the function $f(x)=x^N$ is simply a multiplication by $N$.

Figure 9.9: The preimage and representations domains.

Now consider that the function $\exp u$ is periodic in the imaginary direction:

$$\exp u=\exp (u+2\pi j)$$

Therefore preimages are $2\pi j$-periodic, that is if $u$ is a representation of $x$, then so is $u+2\pi jn\forall n\in \mathbb{Z}$ (Fig. 9.10).

Figure 9.10: Periods of the preimage of the representation $x=\exp u$. Each strip (where there is no difference between gray and white strips) denotes a preimage of the entire complex plane with the exception of zero. The preimages denoted by the dots are preimages of one and the same value.

A multiplication by an integer in the preimage domain $v=Nu$ expands one period $\mathrm{Im}u\in [0,2\pi ]$ to $N$ periods $\mathrm{Im}v\in [0,2\pi N]$. Respectively the function $f(x)$ takes each value $N$ times as $x$ goes across all possible values in $\mathbb{C}$. Conversely, a division by an integer $u=v/N$ shrinks $N$ periods to one period and a previously single value of $f(x)$ turns into $N$ different values of $x$. This is another possible way to explain the fact that the equation $x^N=a$ has $N$ different solutions. Fig. 9.11 demonstrates the result of transformation of all preimages in Fig. 9.10 by a division by 2 corresponding to the equation $u=v/2$. Notice that the preimages in Fig. 9.11 correspond to two different representation values, while the peimages in Fig. 9.10 were corresponding to one and the same representation value.

Figure 9.11: Transformation of the preimage points in Fig. 9.10 by division by 2.

The preimage of the real axis $x\in \mathbb{R}$ consists of two “horizontal” lines $\mathrm{Im}u=0$ and $\mathrm{Im}u=\pi$, or, more precisely, of their periodic repetitions $\mathrm{Im}u=2\pi n$ and $\mathrm{Im}u=\pi +2\pi n$, where $n\in \mathbb{Z}$. The line $\mathrm{Im}u=0$ (and its repetitions) corresponds to the positive real numbers, the line $\mathrm{Im}u=\pi$ (and its repetitions) corresponds to the negative real numbers. The preimage of the zero $x=0$ exists only in the limiting sense $\mathrm{Re}u\to -\infty$. Thus the multiplication of $u$ by $N$ is mapping the preimage of the real axis onto itself, therefore this multiplication’s representation $x^N$ maps the real axis onto itself.

The “vertical” lines $\mathrm{Re}u=a$ are preimages of circles of radius $e^a$, where moving upwards along such lines corresponds to the counterclockwise movement along the respective circle (Fig. 9.12), Particularly the imaginary axis is the preimage of the unit circle (note that a section of such line extending over a single imaginary period $\mathrm{Im}u\in [b,b+2\pi )$ is sufficient to generate all possible values on such circle). Multiplication by $N$ maps the imaginary axis onto itself, thereby its representation $x^N$ maps the unit circle onto itself. A single preimage period $[0j,2\pi j]$ is thereby mapped to $N$ periods $[0,2\pi jN]$, which corresponds to the unit circle being mapped to itself “$N$ times”. We will shortly see that the mapping of the unit circle onto itself is the reason for the Butterworth poles being located on the unit circle.

Figure 9.12: A circular trajectory and its preimage.

Even/odd poles

The poles of (9.18) are given by $1+f^2=0$, which can be rewritten as $f=\pm j$. In the Butterworth case the solutions of $f=\pm j$ were interleaved on the unit circle and also corresponded to even and odd values of $n$ in the solution formula for $1+f^2=0$, where $f=j$ was defining the even poles and $f=-j$ was defining the odd poles.

We will take effort to keep the same correspondence between the equations $f=\pm j$ and the even/odd values of $n$ for other functions $f(\omega )$. In that regard it is instructive to first review the Butterworth case, but now using the just introduced linear scaling representation form, as it will then nicely generalize to other $f(\omega )$ that we are going to use.

Let $\omega$ move along the unit circle in the counterclockwise direction. Its preimage $u$ defined by $\omega =\exp u$ will respectively move upwards along the imaginary axis and so will $v=Nu$. Respectively $f(\omega )=\exp v$ moves along the unit circle in the counterclockwise direction, just $N$ times faster, so while $f(\omega )$ completes one circle, $\omega$ will complete only $1/N$-th of a circle. The value of $f(\omega )$ will be passing through the points $j$ and $-j$, since they are lying on the unit circle. At these moments the value of $\omega$ will be the solution of the equations $f(\omega )=j$ and $f(\omega )=-j$ respectively (Figs. 9.13 and 9.14). There will be no other solutions since if $\omega$ moves in a circle of any other radius, this circle will map to a circle of a non-unit radius and $f(\omega )$ will not go through the points $\pm j$. Thus Fig. 9.14 contains the full set of Butterworth poles, where the interleaving of the white/black dots on the circle in Fig. 9.14 arises from the interleaving of the white/black dots on the circular tajectory in Fig. 9.13.

Figure 9.13: $f(\omega )$ moving in a unit-radius circular trajectory, the points $f(\omega )=\pm j$ and their preimages.

Figure 9.14: Transformation of Fig. 9.13 by $u=v/N$ (for $N=2$). The white and black dots on the circle are even/odd Butterworth poles in terms of $\omega$.

Apparently, the passing of $f(\omega )$ through $\pm j$ corresponds to

$$v=j\pi \left(\frac{1}{2}+n\right)$$

where $f(\omega )=j$ occurs at even $n$ and $f(\omega )=-j$ occurs at odd $n$. The value of $u$ at these moments is

$$u=j\pi \frac{\frac{1}{2}+n}{N}$$

and the value of $\omega$ is

$$\omega =\exp \left(j\pi \frac{\frac{1}{2}+n}{N}\right)$$

which is pretty much the same as the expression (8.13) we have developed before. The even/odd values of $n$ still correspond to the solutions of $f=j$ and $f=-j$ respectively and thus we have a constency in referring to the solutions of $f=j$ as even poles and to the solutions of $f=-j$ as odd poles.

Lowpass, bandpass and highpass filters

If we want $H(s)$ in (9.18) to be a (unit cutoff) lowpass filter, then we should impose some additional requirements on $f(\omega )$:

$$\begin{array}{rlrl}f(\omega )& \approx 0& & \text{for }\omega \ll 1\\ f(\omega )& \approx \infty & & \text{for }\omega \gg 1\end{array}\text{\hspace{1em}(9.21)}$$

where around $\omega =1$ the absolute magnitude of $f(\omega )$ should smoothly grow from 0 to $\infty$. Apparently the Butterworth filter’s function $f(\omega )={\omega }^N$ satisfies these requirements.

Similarly to Butterworth filter, with other filter types arising from (9.18) we will not be constructing $f(x)$ which give a highpass or bandpass response. Instead, highpass and bandpass filters can be simply obtained by LP to HP and LP to BP substitutions respectively.

9.5 Trigonometric functions on complex plane

The trigonometric functions, such as $\sin x$, $\cos x$, $\tan x$ and so on can be evaluated for complex argument values. In that regard they occur to be closely related to the hyperbolic functions $\sinh x$, $\cosh x$, $\tanh x$ and so on. The extention to $x\in \mathbb{C}$ can be obtained by simply evaluating the formulas

$$\begin{array}{rl}\cosh x& =\frac{e^x+e^{-x}}{2}\end{array}\text{\hspace{1em}(9.22)}$$

$$\sinh x=\frac{e^x-e^{-x}}{2}\text{\hspace{1em}(9.23)}$$

$$\tanh x=\frac{\sinh x}{\cosh x}\text{\hspace{1em}(9.24)}$$

$$\cos x=\frac{e^{jx}+e^{-jx}}{2}=\cosh jx\text{\hspace{1em}(9.25)}$$

$$\sin x=\frac{e^{jx}-e^{-jx}}{2j}=-j\sinh jx\text{\hspace{1em}(9.26)}$$

$$\tan x=\frac{\sin x}{\cos x}=\frac{1}{j}\cdot \frac{e^{jx}-e^{-jx}}{e^{jx}+e^{-jx}}=-j\tanh jx\text{\hspace{1em}(9.27)}$$

etc., where the function $e^x$ is allowed to take complex argument values. Notice that thereby we immediately obtain the “imaginary argument properties”:

$$\sin jx=j\sinh x\text{\hspace{1em}(9.28a)}$$

$$\cos jx=\cosh x\text{\hspace{1em}(9.28b)}$$

$$\sinh jx=j\sin x\text{\hspace{1em}(9.28c)}$$

$$\cosh jx=\cos x\text{\hspace{1em}(9.28d)}$$

etc., where intuitively we assume $x\in \mathbb{R}$, but the formulas also work for $x\in \mathbb{C}$.

By direct evaluation one could verify that all basic properties and fundamental trigonometric and hyperbolic identities continue to hold in complex domain. Particularly

$$\begin{array}{rl}\sin (-x)& =-\sin x\\ \cos (-x)& =\cos x\\ \cos (x+2\pi )& =\cos x\\ \cos (x-\pi /2)& =\sin x\\ {\sin }^{2}x+{\cos }^{2}x& =1\\ \sinh (-x)& =-\sinh x\\ \cosh (-x)& =\cosh x\\ {\cosh }^{2}x-{\sinh }^{2}x& =1\end{array}$$

etc. Also, apparently, conjugation commutes with the respective functions:

$$\begin{array}{rl}\sin x^{\ast }& =(\sin x{)}^{\ast }\\ \cos x^{\ast }& =(\cos x{)}^{\ast }\\ \sinh x^{\ast }& =(\sinh x{)}^{\ast }\\ \cosh x^{\ast }& =(\cosh x{)}^{\ast }\end{array}$$

etc.

A direct corollary of (9.28) and the trigonometric formulas for the sum of arguments are the formulas allowing to express a trigonometric function a complex argument via the real and imaginary parts of the argument. E.g.

$$\cos (u+jv)=\cos u\cos jv-\sin u\sin jv=\cosh v\cos u-j\sinh v\sin u\text{\hspace{1em}(9.29a)}$$

$$\sin (u+jv)=\sin u\cos jv+\cos u\sin jv=\cosh v\sin u+j\sinh v\cos u\text{\hspace{1em}(9.29b)}$$

etc.

Periodicity

Since the periodicity property is retained in the complex domain, the former real periods of the trigonometric functions turn into strips on the complex plane. E.g. the $2\pi$ periods of $\cos x$ are shown in Fig. 9.15.

Figure 9.15: Periods of $\cos x$ in the complex plane. All dots are preimages of one and the same value.

Due to the even symmetry of the cosine, almost every value occurs twice on a period (as illustrated by the dots in Fig. 9.15). That is if the value $y$ occurs at $x$ (that is $y=\cos x$), then $y$ also occurs at $-x$. The exceptions are being $\cos x=1$ and $\cos x=-1$, which are mapped to themselves by $x\leftarrow -x$ if the periodicity of $\cos x$ is taken into account.

Inverse functions

Inverting (9.22) and (9.23) we obtain

$$\begin{array}{rl}{\cosh }^{-1}x& =\ln \left(x\pm \sqrt{x^2-1}\right)\\ {\sinh }^{-1}x& =\ln \left(x\pm \sqrt{x^2+1}\right)\end{array}$$

The principal value of the complex square root is defined as

$$\sqrt{x}=\exp \frac{\ln x}{2}=\sqrt{|x|}\cdot \exp j\frac{\arg x}{2}\text{\hspace{1em}(9.30)}$$

where $\arg x\in [-\pi ,\pi ]$,⁴ in which case $\mathrm{Re}\sqrt{x}\ge 0\forall x\in \mathbb{C}$ thereby (9.30) is a generalization of arithmetric square root of real argument to complex domain. Notice that (9.30) gives the values on the upper imaginary semiaxis for real $x<0$ (provided $\arg x=\pi$ for $x<0$).

The principal value of $\ln x$ is defined in the usual way:

$$\ln x=\ln |x|+j\arg x$$

Respectively we can introduce the principal values of ${\cosh }^{-1}$ and ${\sinh }^{-1}$ as:

$${\cosh }^{-1}x=\ln \left(x+\sqrt{x^2-1}\right)\text{\hspace{1em}(9.31a)}$$

$${\sinh }^{-1}x=\ln \left(x+\sqrt{x^2+1}\right)\text{\hspace{1em}(9.31b)}$$

where we chose the signs in front of the square roots in such as way as to ensure that ${\cosh }^{-1}x\ge 0\forall x\ge 1$ and ${\sinh }^{-1}x\in \mathbb{R}\forall x\in \mathbb{R}$.

The formulas (9.31a), (9.31b) raise concerns of numerical robustness in cases where the two terms under the logarithm sign are nearly opposite. By reciprocating the values under the logarithm signs we can rewrite them equivalently as

$${\cosh }^{-1}x=-\ln \left(x-\sqrt{x^2-1}\right)\text{\hspace{1em}(9.31c)}$$

$${\sinh }^{-1}x=-\ln \left(\sqrt{x^2+1}-x\right)\text{\hspace{1em}(9.31d)}$$

where the choice between (9.31a), (9.31b) and (9.31c), (9.31d) should be made based on comparing the complex arguments of the two terms under the logarithm sign. We should choose the formulas where we are adding two numbers whose complex arguments are not further than $90^{\circ }$ apart. Particularly, for real $x$ we may write

$${\sinh }^{-1}x=\mathrm{sgn}x\cdot \ln \left(|x|+\sqrt{x^2+1}\right)(x\in \mathbb{R})\text{\hspace{1em}(9.31e)}$$

whereas the formula (9.31a) already works well for real $x\ge 1$.

Using (9.28) we can construct the principal values:

$$\begin{array}{rl}\arccos x& =-j{\cosh }^{-1}x\\ & =-j\ln \left(x+\sqrt{x^2-1}\right)=j\ln \left(x-\sqrt{x^2-1}\right)\end{array}\text{\hspace{1em}(9.32a)}$$

$$\begin{array}{rl}\arcsin x& =-j{\sinh }^{-1}jx\\ & =-j\ln \left(jx+\sqrt{1-x^2}\right)=j\ln \left(\sqrt{1-x^2}-jx\right)\end{array}\text{\hspace{1em}(9.32b)}$$

However besides the precision issues there are issues related to the principal values of $\arg x$ switching leaf on the negative real axis. Technically this means that there is a discontinuity in the principal values of x​ and $\ln$ on the negative real axis. With (9.31) this was generally tolerable, as the discontinuities weren’t arising for the “usual” values of the argument, which are $x\ge 1$ for ${\cosh }^{-1}x$ and $x\in \mathbb{R}$ for ${\sinh }^{-1}x$, since neither the argument of the square root nor the argument of the logarithm become real negative in such cases. With (9.32a), on the other hand, we do have a negative expression under the square root for real $x\in (-1,1)$, which is the most important argument range.

We could therefore adjust (9.32a) to

$$\arccos x=-j\ln \left(x+j\sqrt{1-x^2}\right)=j\ln \left(x-j\sqrt{1-x^2}\right)\text{\hspace{1em}(9.32c)}$$

The formulas (9.32b) and (9.32c) work well for $x\in [-1,1]$, particularly precision-wise it doesn’t matter which of the two options in (9.32b) and (9.32c) are taken for $x\in [-1,1]$, however they exibit a discontinuity for real $x:|x|>1$. On the other hand, for purely imaginary argument values the formula (9.32b) can be rewritten essentially as (9.31e) to automatically choose the best option precision-wise:

$$\arcsin jx=j{\sinh }^{-1}x=j\mathrm{sgn}x\cdot \ln \left(|x|+\sqrt{x^2+1}\right)(x\in \mathbb{R})\text{\hspace{1em}(9.32d)}$$

Preimages of the real line with respect to $\cos x$

By (9.29a) $\cos x$ attains purely real values iff $x\in \mathbb{R}$ or $\mathrm{Re}x=\pi n$ where $n\in \mathbb{Z}$. However, due to periodicity and evenness properties, each value is attained infinitely many times. We would like to choose a principal preimage of the real line with respect to $\cos x$. That is, we are interested in a (preferably continuous) minimal set of points, whose image under transformation $y=\cos x$ is $\mathbb{R}$.

Note, that we do not really have to choose this principal preimage, as the discussions where we are going to refer to it should lead to exactly the same results no matter which of the preimages of the real line is taken. However, for the sake of clarity of discussion it is convenient to have an unambiguous reference preimage.

Apparently there are infinitely many choice possibilities, among which there are at least several “reasonable” ones. For the purposes of this text we will choose the principal preimage as shown in Fig. 9.16. The same figure also shows the periodic repetitions of the principal preimage.⁵

This principal preimage of the real axis thereby consists of three parts:

$$\begin{array}{rl}x\in [0,\pi ]& ⟺y\in [-1,1]\\ x\in [0,+j\infty )& ⟺y\in [1,+\infty )\\ x\in [\pi ,\pi +j\infty )& ⟺y\in (-\infty ,-1]\end{array}$$

Apparently the principal preimage alone doesn’t cover all preimage points of the real line. Neither does it if we add its periodic repetitions in Fig. 9.16, since the lower-semiplane points of lines $\mathrm{Re}x=\pi n$ are still not included. We can include them by simply rotating all preimages in Fig. 9.16 around the origin, which corresponds to multiplication of all points $x$ by $-1$. Notice that by adding periodic repetitions we addressed the periodicity of $\cos x$, while by adding the preimages multiplied by $-1$ we addressed the evenness property of $\cos x$.

Figure 9.16: The principal preimage (solid line) of the real axis, with respect to $y=\cos x$, and its periodic repetitions (dashed lines).

Representations of horizontal preimage lines by $\cos x$

Equation (9.29a) implies that if the imaginary part $v$ is fixed and the real part $u$ is varying, that is the argument of the cosine is moving in a line parallel to the real axis, then the value of $\cos (u+jv)$ is moving along an ellipse in the complex plane, the real semiaxis of the ellipse being equal to $\cosh v$ and the imaginary semiaxis being equal to $\sinh v$. Fig. 9.17 illustrates. At $v=0$ the real semiaxis is 1 and the imaginary semiaxis is zero. As $|v|$ grows both semiaxes grow, the imaginary semiaxis staying smaller than the real one in absolute magnitude (Fig. 9.18). Both semiaxes become equal in the limit $v\to \infty$ where the ellipse turns into a circle.

Figure 9.17: An elliptic trajectory and two its preimages. The picture is qualitative. In reality, for this kind of ellipse proportions the preimages would need to be located much closer to the real line.

Given $v>0$ and increasing $u$, the movement of the point $\cos (u+jv)$ along the ellipse will be in the negative (clockwise) direction, due to the $-$ sign in front of the imaginary part in (9.29a). Respectively the positive (counterclockwise) direction movement will occur either for a decreasing $u$ or for a negative $v$, where the latter is illustrated in Fig. 9.17.

The even symmetry of the cosine ($\cos (-x)=\cos x$) implies that for each horizontal trajectory $u+jv$ of the cosine’s argument, there is a symmetric trajectory $-(u+jv)$ which produces exactly the same cosine trajectory. This other trajectory is shown in Fig. 9.17 by the dashed line. Notice how this is related to the fact that flipping the sign of $v$ flips the direction of the movement along the ellipse: flipping the sign of $v$ is the same as flipping the sign of the entire cosine’s argument (that is flipping the signs of both $u$ and $v$), which leaves the elliptic trajectory unaffected, and then flipping the sign of $u$, which reverts the direction of movement of both $u+jv$ and $\cos (u+jv)$.

From the fact that the semiaxes of the ellipse are $\cosh v$ and $\sinh v$ and therefore their absolute magnitudes are monotonically growing with $|v|$ (as one can see in Fig. 9.18) we can deduce that ellipses corresponding to different $v$ do not overlap, except for a switch from $v$ to $-v$, which simply changes the direction of the movement along the ellipse. That is for a given ellipse with $\cosh v$ and $\sinh v$ semiaxes there are only two preimages, as shown in Fig. 9.17. An exception occurs when the imaginary semiaxis of the ellipse is zero, in which case there is only one preimage, which is the real line.

Figure 9.18: A family of elliptic trajectories generated from horizontal preimages $v=\text{const}<0$.

Representations of horizontal preimage lines by $\sec x$

The secant function $\sec x=1/\cos x$ is obviously $2\pi$-periodic. In fact it bears quite a few further similarities to $\cos x$, which are easier to see if we write it in the polar form:

$$|\sec x|=\frac{1}{|\cos x|}\text{\hspace{1em}(9.33a)}$$

$$\arg \sec x=-\arg \cos x\text{\hspace{1em}(9.33b)}$$

Consider a horizontal line $u+jv$ (where $u$ is varying and $v=\text{const}$) and its respective representation $y=\sec (u+jv)$. According to (9.33), $y$ moves in an ellipse-like curve around the origin, as shown in Fig. 9.19. At smaller magnitudes of $v$ the curve begins to look more like a figure of 8 (Fig. 9.20). The curve is not an ellipse anymore due to the reciprocation in (9.33a). On the other hand, by (9.33b) the “angular velocity” is the same as with $y=\cos x$, except that is has the opposite sign, therefore the rotation is happening in the opposite direction. Therefore for an increasing $u$ we get counterclockwise rotation iff $v>0$ rather than iff $v<0$.

Figure 9.19: A quasielliptic trajectory and two its preimages (qualitatively).

Figure 9.20: A family of quasielliptic trajectories generated from horizontal preimages $v=\text{const}>0$.

9.6 Chebyshev polynomials

An $N$-th order Chebyshev polynomial is defined as:

$$T_N(x)=\cos (N\arccos x)\text{\hspace{1em}(9.34)}$$

Fig. 9.21 illustrates. Notice the bipolar oscillations of equal amplitude (referred to as equiripples) on the range $[-1,1]$. As one can see in Fig. 9.21, the equiripple amplitude is unity.

Somewhat surprisingly, (9.34) can be equivalently written as an $N$-th order real polynomial of $x$ at any $N\in \mathbb{N}$, e.g. for $N=4$ we have $T_4(x)=8x^4-8x^2+1$, which is why they are called polynomials.

Figure 9.21: Chebyshev polynomials of even (solid) and odd (dashed) orders.

Note that $\arccos x$ takes complex values for $x>1$ and $x<-1$. We will also often assume $x$ taking complex values, therefore $\arccos x$ and $T_N(x)$ will be complex as well. For $|x|>1$ even though $\arccos x$ becomes complex, the value $\cos (N\arccos x)$ is still real, and so is the polynomial itself.

A proper discussion of Chebyshev polynomials falls outside the scope of the book, as the respective information can be easily found elsewhere. Here we shall concentrate on the details which will be important for our purposes.

Chebyshev polynomials as representations of linear scaling

Introducing auxiliary variables $u$ and $v$ we rewrite (9.34) as

$$\begin{array}{rl}x& =\cos u\\ v& =Nu\\ T_N(x)& =\cos v\end{array}$$

or, in the implicit form:

$$T_N(\cos u)=\cos (Nu)\text{\hspace{1em}(9.35)}$$

Thus the function $T_N(x)$ is a representation of the linear scaling $v=Nu$, the mapping function being $x=\cos u$. Note that by multiplying the whole preimage domain by $j$ we obtain a different (but equivalent) representation:

$$\begin{array}{rl}x& =\cosh u\\ v& =Nu\\ T_N(x)& =\cosh v\end{array}$$

which gives us another equivalent expression for (9.34)

$$T_N(x)=\cosh \left(N{\cosh }^{-1}x\right)\text{\hspace{1em}(9.36)}$$

and its respective implicit form

$$T_N(\cosh u)=\cosh (Nu)$$

In our discussion we will stick to using the cosine-based representation. The readers may draw parallels to the hyperbolic cosine-based representation if they wish.

In case of Butterworth filter-generating functions $x^N$ represented via $\exp u$ mapping, the preimages were $2\pi j$-periodic. This time they are $2\pi$-periodic. Additionally there is an even symmetry: $u$ and $-u$ are preimages of the same $x$.

Considering the effect the linear scaling $v=Nu$ on the principal preimage of the real line shown in Fig. 9.16, we obtain the following:

• The principal real half-period $[0,\pi ]$ is expanded to to $[0,\pi N]$, which is responsible for the occurrence of the equiripples on the segment $x\in [-1,1]$. Since $N$ is integer, other real half-periods expand similarly, without generating yet more representation values of $f(x)$. Thus $f(x)$ is a single-valued function.
• The principal preimage $[0,+j\infty )$ of $x\in [1,+\infty )$ maps onto itself. The non-principal preimages $[2\pi n+0j,2\pi n+j\infty )$ of $x\in [1,+\infty )$ map onto some other preimages $[2\pi Nn+0j,2\pi Nn+j\infty )$ of $x\in [1,+\infty )$. Similar mappings occur for the preimages of $x\in [1,+\infty )$ located in the lower complex semiplane. Thus $x\in [1,+\infty )$ is mapped by $T_N(x)$ onto itself, corresponding to the monotonically growing behavior of $T_N(x)$ for $x\ge 1$.
• The principal preimage $[\pi +0j,\pi +j\infty )$ of $x\in (-\infty ,-1]$ is mapped onto $[\pi N+0j,\pi N+j\infty )$, which is a preimage of $x\in [1,+\infty )$ is $N$ is even and of $x\in (-\infty ,-1]$ if $N$ is odd. The non-principal preimages of $x\in (-\infty ,-1]$ (both those in the upper complex semiplane and in the lower complex semiplane) are mapped similarly. Thus $x\in (-\infty ,-1]$ is mapped by $T_N(x)$ onto itself if $N$ is odd, or onto $x\in [1,+\infty )$ if $N$ is even, corresponding to the monotonic behavior of $T_N(x)$ for $x\le -1$.

Notice that these results correspond to the graphs in Fig. 9.21.

Now consider a line $\mathrm{Im}u=\beta$ (where $\beta$ is some real constant value) parallel to the real axis in the preimage domain. Such lines are are, as we know from the previous discussion of the cosine of complex argument, the preimages of ellipses of various sizes, where the size grows with $|\beta |$ (Fig. 9.18). These ellipses are also not overlapping each other, except that the ellipses corresponding to $\beta$ and $-\beta$ are identical (but have opposite orientations). Therefore $T_N(x)$ maps any ellipse from this family onto another ellipse from this family and vice versa, similarly to how $x^N$ mapped the unit circle onto itself and mapped circles onto other circles. This time, however, the ellipse which is mapped onto itself, is the one with a zero imaginary semiaxis.

Notice that since the line $\mathrm{Im}u=\beta$ is mapped to $\mathrm{Im}v=N\beta$, the line stays in the same (upper or lower) semiplane after such mapping and goes in the same (to the right or to the left) direction. Thus, if $x$ is moving in an ellipse in a counterclockwise or respectively clockwise direction, then $T_N(x)$ moves in the same direction.

Even/odd property

Since $\cos (u\pm \pi )=-\cos u$, a negation of $x$ corresponds to a shift of its preimage $u$ by $\pi$. Respectively $v$ is shifted by $N\pi$, which will result in a negation of $T_N(x)$ if $N$ is odd and will not change $T_N(x)$ is $N$ is even. Therefore $T_N(x)$ is even/odd if $N$ is even/odd:

$$T_N(-x)=(-1{)}^N{T}_N(x)\text{\hspace{1em}(9.37)}$$

Values at special points

The principal preimage of $x=1$ is $u=0$. Therefore $v=0$ and $T_N(x)=1$. Therefore

$$T_N(1)=1$$

By (9.37)

$$T_N(-1)=(-1{)}^N$$

The principal preimage of $x=0$ is $u=\pi /2$. Respectively $v=N\pi /2$ and

$$T_N(0)=\mathrm{Re}{j}^N=\{\begin{array}{ll}0& \text{if }N\text{ is odd}\\ (-1{)}^{N/2}& \text{if }N\text{ is even}\end{array}$$

where $\mathrm{Re}{j}^N$ is a way of writing the sequence $1,0,-1,0,\dots$ in the same way how $(-1{)}^N$ is a way of writing the sequence $1,-1,1,-1,\dots$.

Leading coefficient

Knowing that $T_N(x)$ is a real polynomial of order $N$, we can obtain its leading coefficient $a_N$ by evaluating the limit

$$\begin{array}{rl}{a}_N& =\underset{x\to +\infty }{\lim }\frac{T_N(x)}{x^N}=\underset{x\to +\infty }{\lim }\frac{\cosh (N{\cosh }^{-1}x)}{x^N}=\underset{x\to +\infty }{\lim }\frac{\exp (N{\cosh }^{-1}x)}{2x^N}=\\ & =\underset{x\to +\infty }{\lim }\frac{\exp (N\ln 2x)}{2x^N}=\underset{x\to +\infty }{\lim }\frac{(2x{)}^N}{2x^N}=\underset{x\to +\infty }{\lim }\frac{2^N{x}^N}{2x^N}=2^{N-1}\end{array}$$

For the purposes of this book’s material, we won’t need to be able to explicitly find the other coefficients and will therefore skip this topic.

Zeros

Rather than being interested in the values of the coefficients of Chebyshev polynomials, for our purposes it will be more practical to know the locations of their zeros. Letting $T_N(x)=0$ we have

$$v=\pi \left(\frac{1}{2}+n\right)$$

Respectively

$$u=\pi \frac{\frac{1}{2}+n}{N}$$

and

$$x=\cos \left(\pi \frac{\frac{1}{2}+n}{N}\right)$$

which means that the zeros are

$$z_n=\cos \left(\pi \frac{\frac{1}{2}+n}{N}\right)\text{\hspace{1em}(9.38)}$$

where there are $N$ distinct values corresponding to $0<u<\pi$. Notice that the zeros are all real and lie within $(-1,1)$. Also notice that $z_n=-z_{N-1-n}$, therefore the zeros are positioned symmetrically around the origin. Consequently, if $N$ is odd, one of $z_n$ will be at the origin.

Using (9.38) we can write $T_N(x)$ in the factored form:

$$T_N(x)=x^{N\wedge 1}\cdot \prod _{z_n>0}\frac{x^2-z_n^2}{1-z_n^2}\text{\hspace{1em}(9.39)}$$

where we are taking the product only over the positive zeros using the symmetry of the zeros relatively to the origin, and the odd factor $x^{N\wedge 1}$ (where $N\wedge 1$ denotes bitwise conjunction) appears only for odd $N$ where one of the zeros is at the origin. The normalizations by $(1-z_n^2)$ are simply appearing from the requirement that each factor must be equal to 1 at $x=1$, so that $T_N(x)=1$.

Renormalized Chebyshev polynomials

The factored form (9.39) offers some nice insights into the comparison of the behavior of $T_N(x)$ and $x^N$. Writing $x^N$ is a comparable factored form we have

$$x^N=x^{N\wedge 1}\cdot \prod _{z_n>0}{x}^2\text{\hspace{1em}(9.40)}$$

where “taking the product over $z_n>0$” means that we are having as many factors as there are positive zeros in the Chebyshev polynomial $T_N(x)$. Thus the difference between $x^N$ and $T_N(x)$ is that the factors $x^2$ are replaced by $(x^2-z_n^2)/(1-z_n^2)$.

Apparently, if $z_n\to 0\forall n$ then $(x^2-z_n^2)/(1-z_n^2)\to x^2$ and respectively (9.39) is approaching $x^N$. Unfortunately we cannot express this as simply $T_N(x)\to x^N$, since the zeros of $T_N(x)$ are fixed.

To mathematically express this variation of zeros, we can notice that the value of (9.39) at $x=1$ is always unity. Therefore we can introduce the polynomials

$${\hat{T}}_N(x,\lambda )=\frac{T_N(x/\lambda )}{T_N(1/\lambda )}\text{\hspace{1em}(9.41)}$$

to which we refer as renormalized Chebyshev polynomials. By construction ${\hat{T}}_N(1,\lambda )=1\forall \lambda$, while the zeros of ${\hat{T}}_N(x,\lambda )$ are ${\hat{z}}_n=\lambda z_n$. Therefore

$${\hat{T}}_N(x,\lambda )=x^{N\wedge 1}\cdot \prod _{z_n>0}\frac{x^2-(\lambda z_n{)}^2}{1-(\lambda z_n{)}^2}\text{\hspace{1em}(9.42)}$$

and

$$\underset{\lambda \to 0}{\lim }{\hat{T}}_N(x,\lambda )=x^N$$

The formula (9.41) cannot be evaluated for $\lambda =0$, however we apparently can take the limit at $\lambda \to 0$ as the value of ${\hat{T}}_N(x,0)$ and thus

$${\hat{T}}_N(x,0)=x^N$$

$${\hat{T}}_N(x,1)=T_N(x)$$

Notice that the formula (9.42) perfectly works at $\lambda =0$.

Since the equiripple amplitude of $T_N(x)$ is unity, by (9.41) the equiripple amplitude of ${\hat{T}}_N(x)$ is $1/T_N(1/\lambda )$. The equiripple range $x\in [-1,1]$ of $T_N(x)$ is respectively transformed into the equiripple range $x\in [-\lambda ,-\lambda ]$ of ${\hat{T}}_N(x,\lambda )$. Thus $\lambda$ simultaneously controls the equiripple amplitude and the equiripple range of ${\hat{T}}_N(x,\lambda )$ (Fig. 9.22).

Figure 9.22: Renormalized Chebyshev polynomial ${\hat{T}}_7(x,\lambda )$ for $\lambda =1$ (solid), $\lambda =0.93$ (dashed) and $\lambda =0$ (thin dashed).

As we won’t need $\lambda <0$, we won’t consider that option. As for the large values of $\lambda$, it will be practical to restrict the value of $\lambda$ so that $|\lambda z_n|<1\forall n$. Apparently this means ${\lambda }_{\max }=1/\max \{z_n\}$ where $0\le \lambda <{\lambda }_{\max }$. Notice that since $|z_n|<1\forall n$, it follows that ${\lambda }_{\max }>1$.

Often it will be even more practical to restrict $\lambda$ to $0\le \lambda \le 1$. At $\lambda =1$ the equiripple amplitude of ${\hat{T}}_N(x,\lambda )$ is already unity. As $\lambda$ grows further the equiripple amplitude quickly grows, reaching $\infty$ at $\lambda ={\lambda }_{\max }$. Also the equiripple range exceeds $[-1,1]$, which could become inconvenient for our purposes.

In order to simplify the notation, often we will omit the $\lambda$ parameter, understanding it implicitly, and simply write ${\hat{T}}_N(x)$ instead of ${\hat{T}}_N(x,\lambda )$.

Slope at $|x|\ge 1$

Let’s compare the factors of (9.42) and (9.40). Computing the differences:

$$\frac{x^2-(\lambda z_n{)}^2}{1-(\lambda z_n{)}^2}-x^2=\frac{x^2-(\lambda z_n{)}^2-x^2+(\lambda z_n{)}^2{x}^2}{1-(\lambda z_n{)}^2}=\frac{(\lambda z_n{)}^2(x^2-1)}{1-(\lambda z_n{)}^2}\text{\hspace{1em}(9.43)}$$