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Next: 5 Optimizing the Sideband Up: Originally published in August Previous: 3 Analyzing the HA5WH

4 Analysis of the Ideal Cyclic Network

 The design of the HA5WH network, as shown in the handbook, has four identical resistors and four identical capacitors in each of the six network sections. This means the network is invariant under a cyclic interchange of the ordering of its ports. That is if we were to relabel the ports by letting 1 become 2, 2 become 3, 3 become 4, and 4 become 1, we would obtain exactly the same equations describing the network. Such invariances are treated generally using the mathematics of group theory[2], which greatly simplifies the study of system with symmetries. The ideal HA5WH network has what is known as cyclic 4 or C4 symmetry. The network equations can be analyzed using group theory. Analysis of the character of the matrix that represents the cyclic operator shows that each of the 4 irreducible representation of C4 appears once. These therefore correspond to the 4 eigenvectors of the $\tilde{A}$ matrices, which can then be written down immediately.

Most hams probably are unfamiliar with group theory, however, the results can be easily verified without using group theory. The right eigenvectors $\psi^{(m)}_{}$ and the eigenvalues $\lambda_{m}^{}$ of a matrix M are defined by finding the solutions to the equations,

 
M$\displaystyle\psi^{(m)}_{}$ = $\displaystyle\lambda_{m}^{}$$\displaystyle\psi^{(m)}_{}$. (18)

That is multiplying the eigenvector by the matrix gives the same eigenvector back as the result, simply multiplied by the eigenvalue. The effect of multiplying a matrix times one of its eigenvectors is to simply multiply the eigenvector by the eigenvalue.

The cyclic eigenvectors in our basis, are those that change by a constant phase between the elements, with the same phase change between the last and first elements. This gives,

 
$\displaystyle\left (
\begin{array}
{c}
1 \\ 1 \\ 1 \\ 1 \\ \end{array}\right),$$\displaystyle\left (
\begin{array}
{c}
1 \\ -1 \\ 1 \\ -1 \\ \end{array}\right),$$\displaystyle\left (
\begin{array}
{c}
1 \\ j \\ -1 \\ -j \\ \end{array}\right),$$\displaystyle\left (
\begin{array}
{c}
1 \\ -j \\ -1 \\ j \\ \end{array}\right).$ (19)

By direct matrix multiplication, it is easily verified that these are the eigenvectors of all the M matrices if all the R and C values are the same in a network section. This is a direct consequence of the cyclic 4 symmetry. Further, since the $\tilde{A}$ matrices are combinations of products of the M matrices, these same vectors are the eigenvectors of the $\tilde{A}$ matrices. Since Vout is given as a combination of $\tilde{A}$ matrices times Vin , if we express Vin as a linear combination of the four eigenvectors, Vout will be given by taking this same linear combination and multiplying each term by an appropriate eigenvalue. The network must then be designed to produce a 90 o relative phase shift.

The input to the HA5WH network contains only the last two eigenvectors written above. That is

 
Vin = $\displaystyle\left (
\begin{array}
{c}
1 \\ 1 \\ -1 \\ -1 \\ \end{array}\right)=$$\displaystyle{\frac{1-j}{2}}$$\displaystyle\left (
\begin{array}
{c}
1 \\ j \\ -1 \\ -j \\ \end{array}\right)+$$\displaystyle{\frac{1+j}{2}}$$\displaystyle\left (
\begin{array}
{c}
1 \\ j \\ -1 \\ -j \\ \end{array}\right)\equiv$$\displaystyle{\frac{1-j}{2}}$$\displaystyle\psi_{a}^{}$ + $\displaystyle{\frac{1+j}{2}}$$\displaystyle\psi_{b}^{}$, (20)

where the last line defines the relevant eigenvectors as $\psi_{a}^{}$ and $\psi_{b}^{}$ . Further, the output is also not sensitive to the first 2 eigenvectors if the output impedances are identical and the operational amplifiers have good common mode rejection. Having both of the conditions will be helpful if the cyclic symmetry is broken because of component tolerances.

With the input as in Eq. 23, the output will in general be

 
Vout = $\displaystyle{\frac{1-j}{2}}$ga$\displaystyle\psi_{a}^{}$ + $\displaystyle{\frac{1+j}{2}}$gb$\displaystyle\psi_{b}^{}$, (21)

and the two outputs to the balanced modulators will be
 
VA = (1 - j)ga + (1 + j)gb        
VB = (1 - j)jga - (1 + j)jgb      (22)
the suppression in dB is using Eq. 3,

 
20$\displaystyle\log_{10}^{}$$\displaystyle\left \vert \frac{g_a}{g_b} \right\vert.$ (23)

So to design a good network, we must eliminate one of these final two eigenvectors.

The analysis so far shows how the HA5WH network can be motivated. The C4 eigenvectors have equal amplitudes for the 4 voltages, and have a phase shift between adjacent ports of 0 o , + 90 o , 180 o , and 270 o , this last is equivalent to a phase shift of - 90 o . We want to choose the network drive, connections, and component values to select out one of the two 90 o phase shifted eigenvectors. As an aside, the same ideas could be used to design a 60 o relative phase shift by using a network invariant under the group C6 , or a 45 o shift from C8 , etc.

The first step in selecting the component values is to calculate the eigenvalues of the four M matrices. By direct multiplication, I get
 

$\displaystyle\lambda_{11}^{a}$ = $\displaystyle\lambda_{11}^{b}$ = $\displaystyle{\textstyle\frac{1}{R}}$ + j$\displaystyle\omega$C,        
$\displaystyle\lambda_{12}^{a}$ = - $\displaystyle{\textstyle\frac{1}{R}}$ - $\displaystyle\omega$C,$\displaystyle\lambda_{12}^{b}$ = - $\displaystyle{\textstyle\frac{1}{R}}$ + $\displaystyle\omega$C,        
$\displaystyle\lambda_{21}^{a}$ = $\displaystyle{\textstyle\frac{1}{R}}$ - $\displaystyle\omega$C,$\displaystyle\lambda_{21}^{b}$ = $\displaystyle{\textstyle\frac{1}{R}}$ + $\displaystyle\omega$C,        
$\displaystyle\lambda_{22}^{a}$ = $\displaystyle\lambda_{22}^{b}$ = - $\displaystyle{\textstyle\frac{1}{R}}$ - j$\displaystyle\omega$C,      (24)
where the superscript a or b indicates the eigenvalue corresponds to the eigenvector $\psi_{a}^{}$ or $\psi_{b}^{}$ respectively.

The effect of one of the A matrices, when a single eigenvector is input, is given by replacing the M matrices in Eq. 11 by their eigenvalues. After a little algebra, I get,

A a = $\displaystyle{\textstyle\frac{1}{1+\omega R C }}$$\displaystyle\left (
\begin{array}
{cc}
1+j \omega R C & -R \\ -2j \omega C & 1+j \omega R C\\ \end{array}\right)$   

 
A b = $\displaystyle{\textstyle\frac{1}{1-\omega R C }}$$\displaystyle\left (
\begin{array}
{cc}
1+j \omega R C & -R \\ -2j \omega C & 1+j \omega R C\\ \end{array}\right)$ (25)

The A b matrix is proportional to A a. If we feed the section of the network with a linear combination of $\psi_{a}^{}$ and $\psi_{b}^{}$ , the section suppresses $\psi_{a}^{}$ relative to $\psi_{b}^{}$ by a factor of

 
$\displaystyle{\frac{1- \omega R C }{1 + \omega R C }}$. (26)

The HA5WH network has the properties that the magnitude of the ratio given in Eq. 30 is always less than 1 for positive frequencies, and it is exactly zero for $\omega$ = 1/(RC) . The first property says that additional network sections can only improve the relative 90 o phase shift of the outputs. The second says that we can set the frequencies of exact 90 o phase shift by selecting the RC values of single network sections. These two properties greatly simplify the design and optimization of the network.

The sideband suppression at a single frequency is given for an n section network, with RC values in section i given by Ri and Ci , as

 
Suppression = 20$\displaystyle\sum_{i=1}^{n}$$\displaystyle\log_{10}^{}$$\displaystyle\left \vert \frac
{1 - \omega R_i C_i}
{1 + \omega R_i C_i} \right\vert.$ (27)

A simple method of picking the RC values for each section is to use a computer to plot the above result, and adjust n and RiCi to achieve the required suppression. This is in fact the obvious technique to use if you are trying to design with a set of parts already in your junk box. However, the form of the suppression makes it easy to select optimum values as seen in the next section.


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Next: 5 Optimizing the Sideband Up: Originally published in August Previous: 3 Analyzing the HA5WH