Second order all pass filter with variation in Q in numerator and denominator

Question

This is a problem from Microelectronic Circuits by Sedra and Smith.

Consider a second order all pass filter of the form:

$$T(s) = {a_2} \frac{s^2 - s\frac{ω_0}{Q} + {ω_0}^2 }{s^2 + s\frac{ω_0}{Q} + {ω_0}^2 } $$

Flat band gain \$a_2\$ = 1.

a) Errors in component values result in the Q factor of the zeroes being greater than Q factor of the poles. Roughly sketch the output of the transfer function |T|. Repeat for the case in which Q factor of the zeroes is lower than Q factor of the poles.

b) Errors in component values result in frequency of the zeroes being greater than frequency of the poles. Roughly sketch the output of the transfer function |T|. Repeat for the case in which frequency of the zeroes is lower than frequency of the poles.

Solution: All I know is that the all pass filter is just a phase shifter with constant transmission (gain.) Quality factor has the ability to bend a transfer function and make it have peaks or nearly flat response. High Q can make a band pass filter more selective by having a narrow bandwidth by having a narrow peak. One method is to draw a transfer function with just the numerator polynomial and make the denominator polynomial equal to 1. Next, draw the transfer function with numerator equal to 1 and make the denominator polynomial with higher Q. Find |T| for each and superimpose them. The shape of the transfer function of a bandpass filter would be just important near the center frequency \${ω_0}\$ which defines selectivity of the band pass filter. The flat band seems to be due to effects of transfer functions of numerator and denominator balancing each other and making the transmission constant. If Q is made high in numerator compared to denominator, the transfer function of numerator of the flat band would have a peak around \${ω_0}\$ compared to the denominator. The opposite case also seems analogous.

Is this a reasonable method for a)? What is the correct technique to determine the shape of |T(s)|?

If the question has an error, please state how and why.An answer to this question is appreciated. Thanks very much!

Reference: Microelectronic circuits, 7e, Oxford university press india, 2017, pages 1001, 1055.

Answer 1

This is a-- well, you'll know best from context, which has been omitted here, but it sounds like a Bode diagram problem.

That is, the task is to plot the transfer function (gain and phase), in straight line segments approximating the real (smooth polynomial) function; and to do so on inspection of the equation, not having to plot it numerically at all.

What do we know about the expression?

When we have pole-zero cancellation, we can flatten out the gain. If a single real pole and zero are nearby, there is a step change in gain, up or down, depending on which one comes first. If they're on top of each other, the step approaches zero and we say they're cancelled out.

Here, we have 2nd order poles and zeroes, but we also have zeroes in the RHP. The gain cancels out, but not the phase, so we get an all-pass filter.

What if the Qs differ?

$$ T(s) = {a_2} \frac{ s^2 - s\frac{\omega_0}{Q_z} + {\omega_0}^2 }{ s^2 + s\frac{\omega_0}{Q_p} + {\omega_0}^2 } $$

First off, what is Q? Q relates to the magnitude and width of a peaked response. Suppose we vary the Qs and examine gain at \$s = j / \omega_0\$. We get:

$$ T(j \omega_0) = {a_2} \frac{-\frac{j}{Q_z} + {\omega_0}^2 - {\omega_0}^{-2} }{\frac{j}{Q_p} + {\omega_0}^2 - {\omega_0}^{-2} } $$

With respect to the Qs, the right part is constant (we're not varying \$\omega_0\$), and we have an expression of the form

$$ \frac{ \frac{j}{a} + b }{ \frac{j}{c} + d } $$

If \$a\$ is small (approaching zero), the numerator blows up (and phase approaches 90°); vice versa for \$c\$ (and -90°). If either grows large, its contribution disappears; for both large, the result approaches \$\frac{b}{d}\$.

Of course since \$b = d\$ in the above expression, in this case overall gain simply trends towards \$a_2\$.

Note that we expect phase balanced at zero in the middle, so that at these extremes, we get an imbalance, what should be a canceled out phase, isn't.

Thus we can conclude, if \$Q_z < Q_p\$, phase and gain will be high, and vice versa. We can begin to plot this on the Bode diagram:

But it's not complete yet. Consider the meaning of Q with respect to each polynomial: it's a bandwidth and peaking effect. In the end, the asymptotes cancel out (we have the same left-side passband, and complementary right-side stop/boost asymptotes, at identical corner frequencies). All that's in question is the peaking. At high Q, the poles' gain has a strong peak in the transition band, or at low Q it's a slow gradual transition. The zero is opposite of course, having a valley rather than a peak. But since we know the strength of the peaks vary (as proven algebraically above), we know the widths of those peaks also have to vary.

So we'll actually get this: for \$Q_z < Q_p\$, the pole peak is stronger and narrower, and there's a "Mexican hat" sort of depression around the peak.

I'm not actually entirely sure how to draw phase here, but I think it has to take extra dips and peaks, to complement the \$\frac{d|T|}{d\omega}\$? The textbook may offer clues about this case. In any case, there's ripple in both gain and phase:

Or of course the upside-down of this for \$Q_z > Q_p\$.

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In lieu of context from the textbook, a check doesn't hurt. Wolfram Alpha turns out to be pretty good at this, actually:
https://www.wolframalpha.com/input?i=bode+plot+%28s%5E2-0.1s%2B1%29%2F%28s%5E2%2B0.11s%2B1%29

It seems the difference is not so exaggerated (or the order of the polynomials simply matches and that's all that matters) as to give the wobble I drew earlier, and in fact the "equalizer" response (a brief dip or peak depending on direction) is the correct result. The corresponding change in phase however is quite subtle, with the overall 360° shift dominating the plot.