3 Analyzing the HA5WH Network

  Fig. 1 gives the circuit diagram of the HA5WH network as shown in the ARRL Handbook.

  

Figure 1: The schematic diagram of the HA5WH wideband phase shift network.

Given this circuit, it is easy to analyze the network numerically using a mesh or nodal analysis. The disadvantage of this brute force approach is that it gives no insight into why the network works, or how changes in the network affect its performance. I will therefore describe a method that is both more efficient numerically, and by using the symmetry of the ideal network, leads to simple design equations.

Clearly, the network consists of 6 sections each with 4 input connections and 4 output connections. One of these sections is shown in fig. 2.

  

Figure 2: The schematic diagram of 1 section of an HA5WH network.

I have labeled the input voltages and currents V₁ , V₂ , V₃ , V₄ , I₁ , I₂ , I₃ , I₄ . The corresponding output voltages and currents are labeled V'₁ , V'₂ , V'₃ , V'₄ , I'₁ , I'₂ , I'₃ , I'₄ . A straightforward nodal analysis of this network gives the 8 linear equations represented by the matrix equation  

$$\left( \begin{array} {c} I \\ I' \\ \end{array}\right ) = \left... ...y}\right ) \left( \begin{array} {c} V \\ V' \\ \end{array}\right )$$ (5)

where V , V' , I , I' are length 4 vectors, and the M_ij are 4 by 4 matrices. Eq. 5 compactly represents the 8 equations that are the requirements of current conservation at each of the nodes of the network section. The M_ij matrices are

$$\left ( \begin{array} {cccc} \frac{1}{R_1}+j \omega C_1 & 0 & 0 & ... ... \omega C_3 & 0 \\ 0 & 0 & 0 &\frac{1}{R_4}+j \omega C_4 \\ \end{array}\right),$$

$$\left ( \begin{array} {cccc} -\frac{1}{R_1}& 0 & 0 & -j \omega C_1... ...ac{1}{R_3} & 0 \\ 0 & 0 & -j \omega C_4 & -\frac{1}{R_4} \\ \end{array}\right),$$

$$\left ( \begin{array} {cccc} \frac{1}{R_1} & j \omega C_2 & 0 & 0 ... ...} & j \omega C_4 \\ j \omega C_1 & 0 & 0 & \frac{1}{R_4} \\ \end{array}\right),$$

 

$$\left ( \begin{array} {cccc} -\frac{1}{R_1}-j \omega C_2 & 0 & 0 &... ...\omega C_4 & 0 \\ 0 & 0 & 0 &-\frac{1}{R_4}-j \omega C_1 \\ \end{array}\right).$$ (6)

In exact analogy with cascading two-port networks using ABCD matrices, to cascade these network sections, I define a new matrix equation,  

$$\left( \begin{array} {c} V' \\ I' \\ \end{array}\right ) = \lef... ...ay}\right ) \left( \begin{array} {c} V \\ I \\ \end{array}\right )$$ (7)

Solving for the A_ij matrices gives,
 

A₁₁ = - M ^{- 1}₁₂M₁₁,

A₁₂ = M ^{- 1}₁₂

A₂₁ = M₂₁ - M₂₂M ^{- 1}₁₂M₁₁

A₂₂ = M₂₂M ^{- 1}₁₂ (8)

where M₁₂^{- 1} is the inverse of the matrix M₁₂ .

Labeling the 8 by 8 matrices for each of the n sections of the network by A ⁽¹⁾, A ⁽²⁾, ... A ⁽ⁿ⁾, the matrix relating the input to the output of the full network is $\tilde{A}$ , made up of the four 4 by 4 matrices $\tilde{A}_{ij}^{}$ ,  

$$\left( \begin{array} {c} V_{out} \\ I_{out} \\ \end{array}\righ... ...) \left( \begin{array} {c} V_{in} \\ I_{in} \\ \end{array}\right )$$ (9)

where $\tilde{A}$ is the matrix product A ⁽¹⁾A ⁽²⁾A ⁽³⁾...A ⁽ⁿ⁾.

The handbook circuit drives 4 resistors on the 4 output connections. Labeling these as R ^(out)₁ , R ^(out)₂ , R ^(out)₃ , R ^(out)₄ , and defining a 4 by 4 load matrix L,

 

$$\left ( \begin{array} {cccc} \frac{1}{R^{(out)}_1} & 0 & 0 & 0 \\ ... ...{1}{R^{(out)}_3} & 0 \\ 0 & 0 & 0 & \frac{1}{R^{(out)}_4} \\ \end{array}\right)$$ (10)

I can write the relationship between the output voltage and current as,

 

(I_out) = L(V_out). (11)

Solving for I_out , and back substituting gives the final network matrix equation relating the 4 output voltages to the 4 input voltages,

 

$$\tilde{A}_{12}^{} \tilde{A}^{-1}_{22} \tilde{A}_{11}^{} \tilde{A}_{12}^{} \tilde{A}^{-1}_{22} \tilde{A}_{21}^{}$$ (12)

where 1 in Eq. 15 stands for the unit matrix

 

$$\left ( \begin{array} {cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array}\right).$$ (13)

If the output resistors are large compared to the other circuit impedances, L can be taken to be zero. In that case the equations simplify to,

 

$$\tilde{A}_{11}^{} \tilde{A}_{12}^{} \tilde{A}^{-1}_{22} \tilde{A}_{21}^{}$$ (14)

The handbook network has (V_in) proportional to

 

$$\propto \left ( \begin{array} {c} 1 \\ 1 \\ -1 \\ -1 \\ \end{array}\right).$$ (15)

After calculating the $\tilde{A}$ and L matrices from the circuit values, the output signals need to be combined as,
 

V_out,2 - V_out,4 = V_B (16)

and the sideband suppression is given by Eq. 3. The relative amplitude and phase of the signals can also be calculated. Most phase shift networks are based on all pass networks so that the amplitude of all signals are equally attenuated. The HA5WH network is not an all pass network. Ideally, we want both good sideband suppression and we want V_A and V_B to be constant in amplitude and phase across the passband of the audio circuit.

I have written a Fortran program to implement the analysis of this section. It is given in the Appendix. If the matrices that are inverted become singular, the above analysis breaks down at the singular points. For example, M₁₂ becomes singular when

 

$$\omega^{4}_{} {\textstyle\frac{1}{ R_1 R_2 R_3 R_4 C_1 C_2 C_3 C_4}}$$ (17)

Near these points, roundoff error in the calculations will be large. For the analysis done here, this is not a big problem. However, analysis on networks with many sections or near singular points will require more numerical care than I have taken in the program in the appendix, or the use of a the standard formulation where the full set of network equations are solved at once.