6 Effects of Amplitude Variations and Component Tolerances

  So far, I have only looked at the relative phase shift of the two outputs. To have a high quality audio signal, the network must have a flat output. Usually, this is handled by constructing an all pass network. Since the HA5WH network is not an all pass form, we must examine its attenuation as a function of frequency. In figs. 3, 4, and 5, I plot the sideband suppression, and the amplitude and phase variations of one of the output signals, respectively for the optimal 4, 6, and 8 section filters designed for the frequency range 300 Hz to 3000 Hz with equal value resistors. The network sections are ordered from largest RC value to smallest as in the original HA5WH design. As shown, the amplitude variations are less than $\pm$ 1dB , the phase variation is smooth, and the sideband suppression is of the equal ripple form as expected.

  

Figure 3: Ratio of the magnitude of the unwanted to wanted sideband for the 4, 6, and 8 optimal Chebychev networks for the frequency range 300 Hz to 3000Hz.

  

Figure 4: Relative amplitude of one output signal for the 4, 6, and 8 optimal Chebychev networks for the frequency range 300 Hz to 3000Hz.

  

Figure 5: The phase shift of one output signal for the 4, 6, and 8 optimal Chebychev networks for the frequency range 300 Hz to 3000Hz.

One of the main selling points of the handbook description of this network is the claim that low tolerance components can be used to obtain a high performance network. From the analysis of section 5, if cyclic symmetry is maintained, the network will perform perfectly at the n selected frequencies corresponding to f = 1/(2$\pi$RC) for each network section. Since matching components is generally easier than measuring their values accurately, I examine the effect of a change of these node frequencies caused by perfectly matched, but low tolerance components. Since both the resistors and capacitors can vary, using 10 percent components can vary the nodal frequency values by approximately 20 percent if both components change value in the same direction. A worst case condition would be for all the sections to have too high or too low of a frequency by 20 percent. This simply shifts the network center frequency by 20 percent. For the optimal 6 section filter from 300 Hz to 3000 Hz, this changes the sideband suppression from over 60 dB to about 42 dB. If a 10 percent variation of network node frequencies is assumed, that is 5 percent components, and again all the frequency changes are assumed to be in the same direction, the suppression is nearly 50 dB. This shows that relatively low tolerance but well matched components can give excellent results. Eq. 31 can be easily used to predict the effect of changing the RC values of each filter section due to component tolerances when the components are perfectly matched in each section.

The case where unmatched R and C values in each section are used is of course the one with the most practical interest. Here, we can get an idea of what the worst case possibilities are by looking at the cross terms between $\psi_{a}^{}$ and $\psi_{b}^{}$ when the M matrices are no longer cyclic. Typical terms give contributions like,

 

$$\left ( \frac{1}{R_1}+\frac{1}{R_3} \right)- \left ( \frac{1}{R_2}+\frac{1}{R_4} \right),$$ (29)

or

 

$$\omega \omega$$ (30)

where here the subscripts 1,2,3,4 indicate the position in the network section as in fig. 2. This indicates that a single section with a tolerance t (t = 0.1 would be 10 percent tolerance) can reduce the overall suppression to roughly 20$\log_{10}^{}$(t) dB. That is 10 percent components could give suppressions as low as 20 dB, and 1 percent components as low as 40 dB if the components in a network section are not matched. Notice that to be sure to obtain 60 dB opposite sideband attenuation, components with short and long term tolerances of 0.1 percent would need to be used.

As a concrete example of this sensitivity to unmatched components, I calculated the the suppression of the original HA5WH 6 section filter for the case where only the resistors in the last section have been changed by 10 percent. R₁ and R₃ have been raised by 10 percent, and R₂ and R₄ have been lowered by 10 percent. For ideal components, the suppression is greater than 57 dB. Changing just the resistors in the last section reduces the suppression to 26 dB, in rough agreement with the simple calculation above. Using these results to try to cook up a near worst case, I tried changing all the resistors in exactly the same manner in each section. In addition I changed all the capacitors by making raising the C₂ and C₄ values by 10 percent and lowering the C₁ and C₃ values by 10 percent. The result was to further lower the unwanted sideband suppression to about 17 dB. Clearly, 10 percent components and bad luck will produce an unacceptable sideband suppression.

One last comment on the handbook circuit is the design of operational amplifier circuit for the output. One section of this circuit is shown in fig. 6. All the resistors have the same value in the handbook circuit. This does not give a balanced output, and would be another source of phasing errors. If I assume that the operational amplifier input impedances are very large, the input impedance at point 2 is clearly 2R₂ . The voltage at the noninverting input is therefore V₂/2 . The current drawn from input 1 is therefore ${\frac{V_1 - V_2/2}{R_1}}$ , and since V₁ = - V₂ with perfect phasing, the impedance seen at input 1 is 1.5R₁ . So 2R₂ should be equal to 1.5R₁ , and in addition, dc balancing of the operational amplifiers may be required to compensate for input bias current. For the handbook circuit, the unbalanced output resistance reduces the sideband suppression even in the ideal component case to about 35 dB.

  

Figure 6: The schematic diagram of one operational amplifier section.

Note Added after original publication as mentioned in the errata, the above paragraph is incorrect. The original circuit will work fine. The circuit with the resistor ratio of 4/3 also works.