# Design of nth-Order Assisted Alignments

### Summary of Results So Far

When n > 4, the design of nth-order systems is restricted here to the Chebyshev, Butterworth and sub-Chebyshev alignments due to considerations previously discussed, . These three alignment types all share a common algorithm for the design of the loudspeaker portion of the system, consisting of the following steps.

1. Pick QT, QL, and the system order n.
2. For the given n, compute the list of Butterworth pole angles and pick two of them, denoted by θa and θb, to be assigned to the loudspeaker portion of the system.
3. Compute A through H using (76).
4. Use the given QT and QL, along with the expressions for a1 and a3 given in (81) and (83) respectively, to numerically solve (26) or its equivalent for k.
5. Using the computed k value, calculate a1, a2, a3 and T1 using (81), (82), (83) and (80) respectively.
6. Calculate h using (27).
7. Compute α using (28).
8. Compute T0 using (14).

• How are the transfer functions of the external filters calculated?
• What's the best way to pick θa and θb from the list of Butterworth pole angles for the given n, QT and QL?

The above two questions are answered in the next two sections.

### Calculating the Filter Transfer Functions

When designing the loudspeaker portion of the nth-order assisted system, the Butterworth pole angles are enumerated, and two of them, denoted by θa and θb, are associated with the low-pass prototype of the loudspeaker transfer function. When n > 5, each remaining Butterworth pole angle is associated with the low-pass prototype of a second-order external electrical filter which always has complex poles. Let the remaining Butterworth pole angles of the system, to be associated with the external electrical filters, be denoted by the collection θfi as follows.

 (96) ${\theta }_{fi}={\theta }_{f1}\text{,}{\theta }_{f2}\text{,}\phantom{\rule{0.5em}{0ex}}\text{...}$

In the notation of (96), f denotes "filter" and i is an integer index. For even n, i goes from 1 to (n - 4) / 2, and for odd n, i goes from 1 to (n - 5) / 2.

When n is odd, the system has a single real pole in addition to the other complex poles. For the case of the odd-order normalized Butterworth low-pass transfer function, the real pole is at s = -1. The poles of the normalized Chebyshev and sub-Chebyshev transfer functions are just the corresponding Butterworth poles with real parts multiplied by k. Therefore, the single real-valued pole of a normalized odd-order Chebyshev or sub-Chebyshev transfer function must be at s = -k. This can also be verified using (71).

As before, we denote the transfer function of the normalized low-pass prototype of the entire system and of the loudspeaker portion of the system as HLP(s) and GLP(s) respectively. GLP(s) is given by (2). We can then write HLP(s) as:

 (97) ${H}_{LP}\left(s\right)=\left\{\begin{array}{cc}{G}_{LP}\left(s\right)\prod _{i=1}^{\frac{n-4}{2}}\frac{{\omega }_{xi}^{2}}{{s}^{2}+\frac{{\omega }_{xi}}{{Q}_{fi}}s+{\omega }_{xi}^{2}}\text{,}\hfill & n\phantom{\rule{0.5em}{0ex}}\text{even}\hfill \\ {G}_{LP}\left(s\right)\frac{k}{s+k}\prod _{i=1}^{\frac{n-5}{2}}\frac{{\omega }_{xi}^{2}}{{s}^{2}+\frac{{\omega }_{xi}}{{Q}_{fi}}s+{\omega }_{xi}^{2}},\hfill & n\phantom{\rule{0.5em}{0ex}}\text{odd}\hfill \end{array}\right\$

For any value of i in (96), the second-quadrant pole sxi associated with the Butterworth pole angle θfi, and in turn the ith quadratic factor in the denominator of the product term in (97) is:

 (98) ${s}_{xi}=\text{-}k\mathrm{cos}\left({\theta }_{fi}\right)+j\mathrm{sin}\left({\theta }_{fi}\right)$

We need to find ωxi and Qfi in terms of θfi and k. This is done by forming a quadratic polynomial having sxi and its conjugate as roots, then equating coefficients. Following this approach, we obtain:

 (99) $\begin{array}{ccc}\hfill \left(s-{s}_{xi}\right)\left(s-{s}_{xi}^{*}\right)& \hfill =\hfill & {s}^{2}-2\mathrm{Re}\left({s}_{xi}\right)s+{\left|{s}_{xi}\right|}^{2}\hfill \\ \hfill & \hfill =\hfill & {s}^{2}+2k\mathrm{cos}\left({\theta }_{fi}\right)s+{k}^{2}{\mathrm{cos}}^{2}\left({\theta }_{fi}\right)+{\mathrm{sin}}^{2}\left({\theta }_{fi}\right)\hfill \\ \hfill & \hfill =\hfill & {s}^{2}+\frac{{\omega }_{xi}}{{Q}_{fi}}s+{\omega }_{xi}^{2}\hfill \end{array}$

Equating coefficients of like powers of s in the last two equations of (99), we get:

 (100) ${\omega }_{xi}=\sqrt{{k}^{2}{\mathrm{cos}}^{2}\left({\theta }_{fi}\right)+{\mathrm{sin}}^{2}\left({\theta }_{fi}\right)}$

and

 (101) ${Q}_{fi}=\frac{1}{2}\sqrt{1+\frac{1}{{k}^{2}}{\mathrm{tan}}^{2}\left({\theta }_{fi}\right)}$

Since each θfi is just picked from a list of known Butterworth pole angles for the given system order n, and k is known from the procedure of the previous section, it's easily seen that ωxi and Qfi are also known. In the odd-order case, the single pole of the normalized low-pass electrical filter at s = -k is also known. This leaves GLP(s), repeated below for clarity, to be considered.

 ${G}_{LP}\left(s\right)=\frac{1}{{\left(s{T}_{1}\right)}^{4}+{a}_{3}{\left(s{T}_{1}\right)}^{3}+{a}_{2}{\left(s{T}_{1}\right)}^{2}+{a}_{1}\left(s{T}_{1}\right)+1}$

By the procedure of the previous section, GLP(s) is known as well. We therefore know all the needed parameters to form the complete normalized low-pass prototype for the system transfer function HLP(s). All that remains is to find and apply the appropriate low-pass to high-pass transformation and frequency scaling that maps the normalized low-pass prototype to the physical realm. This is done by observing the physical transfer function G(s) of the loudspeaker portion of the system and comparing it to the normalized low-pass prototype GLP(s) of said transfer function. G(s) is repeated below for clarity.

 $G\left(s\right)=\frac{{\left(s{T}_{0}\right)}^{4}}{{\left(s{T}_{0}\right)}^{4}+{a}_{1}{\left(s{T}_{0}\right)}^{3}+{a}_{2}{\left(s{T}_{0}\right)}^{2}+{a}_{3}\left(s{T}_{0}\right)+1}$

It's straightforward to show by direct substitution that G(s) can be obtained by replacing sT1 in GLP(s) by 1 / (sT0). Therefore, by performing that transformation and frequency scaling to every term of HLP(s), we can obtain the denormalized transfer function H(s) of the entire system. The transformation is usually written as:

 $s{T}_{1}\to \frac{1}{s{T}_{0}}$

This can potentially cause confusion as there is no explicit T1 in the electrical filter portion of HLP(s) in (97). For this reason, it is better to write HLP(s) as a function of the dummy variable S, and let:

 (102) $\begin{array}{ccc}\hfill S{T}_{1}& \hfill =\hfill & \frac{1}{s{T}_{0}}\hfill \\ \hfill S& \hfill =\hfill & \frac{1}{s{T}_{0}{T}_{1}}\hfill \end{array}$

Performing this transformation on HLP(s) in (97) with ω0 = 1 / T0 and ω1 = 1 / T1 gives the overall system transfer function H(s) as follows:

 (103) $H\left(s\right)=\left\{\begin{array}{cc}G\left(s\right)\prod _{i=1}^{\frac{n-4}{2}}\frac{{s}^{2}}{{s}^{2}+\frac{{\omega }_{fi}}{{Q}_{fi}}s+{\omega }_{fi}^{2}}\text{,}\hfill & n\phantom{\rule{0.5em}{0ex}}\text{even}\hfill \\ G\left(s\right)\frac{s}{s+{\omega }_{p}}\prod _{i=1}^{\frac{n-5}{2}}\frac{{s}^{2}}{{s}^{2}+\frac{{\omega }_{fi}}{{Q}_{fi}}s+{\omega }_{fi}^{2}}\text{,}\hfill & n\phantom{\rule{0.5em}{0ex}}\text{odd}\hfill \end{array}\right\$

where

 (104) ${\omega }_{p}=\frac{{\omega }_{0}{\omega }_{1}}{k}$

and

 (105) ${\omega }_{fi}=\frac{{\omega }_{0}{\omega }_{1}}{{\omega }_{xi}}=\frac{{\omega }_{0}{\omega }_{1}}{\sqrt{{k}^{2}{\mathrm{cos}}^{2}\left({\theta }_{fi}\right)+{\mathrm{sin}}^{2}\left({\theta }_{fi}\right)}}$

The Qfi values in (97) are invariant under the low-pass to high-pass transformation, so the Qfi values of (103) are the same as those of (97), and are computed using (101). The ωp and ωfi values given above appear at first to have incorrect units. However, since k is a unitless scale factor, it can be seen from (80) and (76) that ω1 is a unitless normalized frequency. From (100), it can also be seen that ωxi is unitless. Therefore, both ωp and ωfi have units of radians/sec.

### Partitioning of System Transfer Function Between Loudspeaker and External Filters

#### An Extension of Thiele's Alignment Class Concept

When n > 5, there is more than one way to choose which system poles belong to the external filter. Thiele [5] resolved this potential ambiguity by originating the concept of alignment class for the specific case of n = 6. When n = 6, there are three second-quadrant complex poles, of which two get assigned to the loudspeaker plus box, with the remaining one assigned to the external filter. For each pole chosen in this way, its complex conjugate in the third quadrant is assumed to always be chosen as well, and associated with its corresponding second-quadrant pole to form a complex conjugate pair. It is easily seen that for n = 6, there are three ways to choose the filter. Thiele associated each choice with a class number as follows.

• Class 1: External filter has the highest Q
• Class 2: External filter has a middle Q value
• Class 3: External filter has the lowest Q

Consider the case of, say, n = 8. This case has four second-quadrant poles, of which two are assigned to the loudspeaker plus box. It's clear that the number of possible alignments is given by the number of combinations of 4 things taken 2 at a time, which is 6. It is useful to have a simple formula for the number of possible alignment classes for any system order n.

Let the number of second-quadrant complex poles for a given n be given by m. We have:

 (106) $m=\left\{\begin{array}{cc}\frac{n}{2}\text{,}& n\phantom{\rule{0.5em}{0ex}}\text{even}\hfill \\ \frac{n-1}{2}\text{,}& n\phantom{\rule{0.5em}{0ex}}\text{odd}\hfill \end{array}\right\$

In the general case, the number of different classes for a given system order n is the number of combinations of m things taken two at a time, where m is given by (106) above. This is computed as follows:

 (107) \text{Number of alignment classes}=\frac{m!}{2\left(m-2\right)!}

In the case of n = 6 considered by Thiele, there is only one complex conjugate pair of poles to be assigned to the electrical filter, so the alignment class can be a single integer. However, when n > 7, there are two or more such complex conjugate pole pairs to be assigned to the filter, so more than one integer is required to specify the filter poles. Two possible approaches might be taken. One might specify which two pole pairs are assigned to the loudspeaker, and always specify exactly two integers for the alignment class for all n > 5. However, I'd like the class designation to be compatible with Thiele's for n = 6, which requires that only a single integer be used in this situation. Compatibility with Thiele's approach implies that a collection of integers be used to specify which pole pairs are associated with the filters, not the speaker plus box. In this way, if the transfer function of the electrical filter contains only one complex pole pair as is true for the cases of n = 6 and n = 7, only a single integer is needed to specify the alignment class.

Since the procedure described herein for the Chebyshev, Butterworth and sub-Chebyshev alignments begins by specifying two Butterworth pole angles θa and θb to be associated with the poles of the speaker plus box, with the remaining Butterworth pole angles θfi associated with the electrical filters, we'd like to relate the θfi to the Q values of the electrical filter(s) to get identical class numbering to Thiele's for n = 6. For a given k value, the relationship between θfi and Qfi is specified in (101). Since 0 < θfi < π / 2, it can be seen from (101) that Qfi approaches 0.5 when θfi approaches zero, and approaches infinity as θfi approaches π / 2. Therefore, by considering the θfi as an array and sorting the array in decreasing order, the corresponding filter Qfi will be sorted in decreasing order as well. The filter class can then be specified as a collection of integer indices into this array from which the filter poles are picked. When n = 6, only one such index is needed, and when its value is 1, this gives the highest Qfi, consistent with Thiele's class specification. When the index is increased to 2, then 3, this gives ever smaller Qfi values, also consistent with Thiele's class specification.

An example should help clarify the class naming convention. Figure 1 below for a sixth-order Butterworth system shows how the Butterworth pole angles are partitioned between θa, θb and θf1 for class 1, 2 and 3 alignments. The filter Q, Qf1, computed using (101) with k = 1, is shown for each class.

Class θa θb θf1 Qf1
1 π / 12 3π / 12 5π / 12 1.9319
2 π / 12 5π / 12 3π / 12 0.7071
3 3π / 12 5π / 12 π / 12 0.5176

Table 1. Alignment Classes for Sixth-Order Assisted Butterworth Alignment

By first sorting the collection of sixth-order Butterworth pole angles in (44) in decreasing order of angle, a relationship between the alignment class and the position of θf1 in the sorted array can be seen in Table 1. Specifically, the index of the Butterworth pole angle θf1 into said sorted array is the same as the alignment class. To clarify this idea, Table 2 below shows the array of Butterworth pole angles from (44) for n = 6 sorted in decreasing order, along with their corresponding array indices. These should be compared with θf1 and the alignment class of Table 1 respectively.

Index θ
1 5π / 12
2 3π / 12
3 π / 12

Table 2. Sorted Butterworth Pole Angles and Their Indices for n = 6

When more than one second-order filter is needed, as is the case for n > 7, more than one index is also needed to specify the alignment class. In such cases, the alignment class consists of a collection of integers, each of which is the index of the corresponding filter's θfi value into the sorted array of Butterworth pole angles for a given n. By convention, the first index will be associated with θf1, the second with θf2, and so on. As before, the sorting is assumed to be in decreasing order of Butterworth pole angle. An example enumerating all possible alignment classes for an eighth-order assisted Butterworth alignment is given below in Table 3.

Class θa θb θf1 Qf1 θf2 Qf2
{1, 2} π / 16 3π / 16 7π / 16 2.5629 5π / 16 0.9000
{1, 3} π / 16 5π / 16 7π / 16 2.5629 3π / 16 0.6013
{1, 4} 3π / 16 5π / 16 7π / 16 2.5629 π / 16 0.5098
{2, 3} π / 16 7π / 16 5π / 16 0.9000 3π / 16 0.6013
{2, 4} 3π / 16 7π / 16 5π / 16 0.9000 π / 16 0.5098
{3, 4} 5π / 16 7π / 16 3π / 16 0.6013 π / 16 0.5098

Table 3. Alignment Classes for Eighth-Order Assisted Butterworth Alignment

Table 4 below shows the sorted list of Butterworth pole angles for n = 8 from (45) to help illustrate the alignment class convention.

Index θ
1 7π / 16
2 5π / 16
3 3π / 16
4 π / 16

Table 4. Sorted Butterworth Pole Angles and Their Indices for n = 8

#### Picking a Suitable Alignment Class for a Given n

Given the number of alignment classes possible, especially for large n, it would be useful to have a procedure for narrowing down the alignment class choices to those likely to result in reasonable physical characteristics of the system. We'd like to avoid such problems as excessive boost in the external filters and boxes that are too large. We'll look at this problem using the Dayton Audio RSS390HF-4 subwoofer driver from Parts Express as an example. This driver has a QTS value of 0.423. Equation (19) allows the computation of the QT value required for a Butterworth alignment for any n, provided θa and θb are known, and a suitable estimate of QL is chosen. The Dayton driver has a VAS of 9.95 ft3, so it's safe to assume the box will be large. For this reason, we'll pick an estimated QL value of 5. To make things interesting, we'll choose n = 8. Table 3 above gives all possible combinations of θa and θb for an eighth-order alignment, and together with the estimated QL value of 5, the QTB value for each alignment class can be computed for n = 8. The results are shown in Table 5.

Class θa θb QTB
{1, 2} π / 16 3π / 16 0.2920
{1, 3} π / 16 5π / 16 0.3481
{1, 4} 3π / 16 5π / 16 0.3885
{2, 3} π / 16 7π / 16 0.4647
{2, 4} 3π / 16 7π / 16 0.5396
{3, 4} 5π / 16 7π / 16 0.7684

Table 5. Values of QTB for n = 8 and QL = 5

Since this is a subwoofer application, it's desirable to take advantage of the properties of the Chebyshev alignment, allowing the system -3dB frequency to be lower than the driver's free-air resonant frequency fs, as Thiele [5] observed. To avoid the need for an impractically large box, a good first estimate for the design might consist of picking the alignment class having the largest QTB that's less than QT, that is, the Chebyshev alignment closest to the Butterworth condition. The alignment class meeting this criterion has the lowest frequency response ripple of any realizable Chebyshev alignment for the given QT, QL and n. This requirement is met by the class {1,4} alignment of Table 5, assuming the use of the Dayton Audio RSS390HF-4 driver.

It's been noted earlier that Thiele observed that the system -3dB frequency can be considerably lower than fs before the frequency response ripple becomes significant. But what happens for larger values of n? It appears from (84) and (85) that for a given k, the ripple becomes less as n increases. Let's calculate the ripple in dB for the hypothetical case of k = 0.5 with n = 4 and n = 8.

• For k = 0.5 and n = 4, the ripple is 0.2145 dB.
• For k = 0.5 and n = 8, the ripple is 0.0026 dB.

This is a fortunate state of affairs for the higher-order alignments, such as n = 8.

To determine the specifics of a proposed design, it becomes necessary to have software to compute all the necessary parameters both for the box and for the external filters, then simulate the proposed design to determine its strengths and weaknesses. The design software is the subject of the next section.