2
\$\begingroup\$

Consider the compensation transfer function \$\frac{s+z}{s+p}\$ so that the inequalities \$0<z<p\$ and \$0<p<z\$ correspond to lead and lag compensation respectively.

At very low frequencies of input signals \$|s|\approx0\$, and at higher frequencies of the input signal \$|s|\approx+\infty\$. Therefore, the gain due the lead compensation at very low frequencies is \$\approx \frac{|z|}{|p|}<1\$ and at very high frequencies is \$\approx 1\$, while the gain due the lead compensation at very low frequencies is \$\approx \frac{|z|}{|p|}>1\$ and at very high frequencies is \$\approx 1\$.

This indicates that lead compensation attenuates lower frequency signals while lag compensation amplifies high frequency signals. In fact, the classical high pass filter (\$\frac{s}{s+p}\$) is recovered from the lead compensator in the limit \$|z|\rightarrow 0\$, while the classical low pass filter is mathematically identical to the the lag compensator for high values of \$|z|\rightarrow +\infty\$.

In general, lead and lag filters have the behavior of high and low pass filters. Is it accurate to say that this similarity with respect to the objective of filters holds true even if (either or both of) the zero and pole used are complex?

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

It's possible you're overthinking this. \$s\$ is the Laplace notation for the complex frequency, \$j2\pi f\$, or \$j\omega\$, and it spans the entire frequency range. When you're saying that \$|s|\rightarrow0\$ or \$|s|\rightarrow\infty\$, what you're probably referring to is the frequency, itself, as it approaches DC or as it goes to infinity, not \$s\$.

The transfer function you're showing is that of a 1st order system, and it can only have simple, real roots. Therefore using the absolute notation, \$|p|\$, or \$|z|\$, is unnecessary.

The lead or lag compensators are also called shelf filters, because of the way their magnitude looks like when plotted. Since you, yourself, said just before that the gains converge towards \$z/p\$ or \$p/z\$, then you know that there is no infinite attenuation towards that frequency. You can't talk about "recovering" some "underlying" filter from it. The topologies for lowpass and highpass are ones, and for lead and lag filters are anothers. You can't mix the two. The only similarity between a lead compensator and a highpass is the part of the spectrum situated at the geometrical mean between the zero and the pole, and until the very end; similar for lowpass. Note: similar, not identical. See below a comparison between \$\dfrac{s+0.1}{s+1}\$ and \$\dfrac{s}{s+1}\$, and notice how the magnitudes have similar slopes until around \$\sqrt{0.1\cdot 1}\$:

comp

If you want to talk about how the limit of the pole or the zero approaches zero or infinity, then chances are you are no longer talking about lead/lag compensators, but of low-/highpass filters, instead. Otherwise, the same conclusion could be drawn about Cauer/elliptic filters that converge towards a Butterworth when the ripples are set to zero, and the zeroes to infinity: you may end up with a Butterworth, but it's not one. Fun fact: the same Cauer/elliptic also converges towards both a Chebyshev, or an inverse Chebyshev, with the proper settings; it's still not either of them.

Therefore, lead compensators do not have the behaviour of highpass filters, in the same way the same lead compensators do not have the behaviour of lowpass filters -- they have them both and clearly defined. It's what makes them shelf filters. Similar for lag compensator. Similar for xomplex poles/zeroes.


I'll add here the responses to your comments, since they might be a bit longer. For your first comment, the table in the linked OP is awfully confusing. As noted in the answer, those are plain high-/lowpass filters, so yes, they are (very much) misnomers.

For the second comment, w.r.t. OP's awfully confusing table, you are referring to the case where \$0<z<p\$, which is labeled "lag high-pass", but the answerer is referring to the "lead low-pass", which is the 3rd entry, or \$0<p<z\$ (blue is the compensator, orange the lowpass):

phase

So, no, the answerer is right, and I'll venture to say it's most probable that he is not wrong (as his profile on dsp.ee might make it more clear). And since I fully agree that the table is horrible, I'll put this minor mishap on the account of that utterly repulsive monstrosity of a table (did I mention it's awfully confusing?). Unfortunately, you will see in the literature various people trying to impose their names through the usage of "new and improved" ways of re-defining already defined definitions, seemingly not counting the effect that it may have on the readers.

As for the topologies, there is no "standard" way to build these things, though some schematics may make more sense than others (i.e. using RC instead or RL), but if the name is not clear, think of the topology as the transfer function, and then all things should become clear.


To avoid the information going in the comments, I'll add this to the answer, but diverting too much from the topic means a new question is needed.

A pole in the right-half plane would make the system unstable (the real axis is positive). The zero could be there without affecting stability, but you're showing \$s+z\$, not \$s-z\$. In either case, when you say \$z<p\$, I take it to understand the absolute value (what I said earlier still applies, unless comparing strictly values). But even with a RHP zero, the transfer function becomes:

$$\dfrac{s-1}{s+2}\;\rightarrow\;\dfrac{j\omega-1}{j\omega+2}\cdot\dfrac{-j\omega+2}{-j\omega+2}\;\rightarrow\;\dfrac{(\omega^2-2)+j3\omega}{\omega^2+4}$$

For completeness, for a LHP zero:

$$\dfrac{s+1}{s+2}\;\rightarrow\;\dfrac{j\omega+1}{j\omega+2}\cdot\dfrac{-j\omega+2}{-j\omega+2}\;\rightarrow\;\dfrac{(\omega^2+2)+j\omega}{\omega^2+4}$$

For even more completeness, the phases of the LHP zero (blue), RHP zero (orange), plus a lowpass with a pole at \$-2\$ (red), showing a more positive phase than the lowpass:

3

\$\endgroup\$
9
  • \$\begingroup\$ Thanks for the comprehensive and intuitive answer. I am not well versed with filter topologies, so thanks for reminding me to study that. Please allow me to press ahead on some minor issues. In the text cited (screenshot) by the OP in this question the expression \$\frac{s+z_1}{s+p_1}\$ with \$z_1<p_1\$ is referred to as a 'lag high pass' filter (and similar corresponding reference to lead filter). Is it accurate to say that this (these) are misnomers, since, for instance \$\frac{s+z_1}{s+p_1}\$, \$z_1<p_1\$ is clearly lead compensation? \$\endgroup\$
    – kbakshi314
    Mar 24, 2021 at 15:33
  • \$\begingroup\$ Further, in the question referenced in the comment above, the answer mentions that the phase of \$\frac{s+z_1}{s+p_1}\$, \$0<z_1<p_1\$ (dropping the name since that is a quibble I have) is 'less negative than...'. However, for any \$s=\omega j, \; 0<\omega\$, _the complex plane location of \${s+z_1}\$ is to the left of_ \${s+p_1}\$, thus leading to a positive argument or phase of \$\frac{s+z_1}{s+p_1}\$, \$0<z_1<p_1\$. Is it accurate to say that this is an error in the answer? \$\endgroup\$
    – kbakshi314
    Mar 24, 2021 at 15:40
  • 1
    \$\begingroup\$ I've updated my answer. \$\endgroup\$ Mar 24, 2021 at 16:25
  • \$\begingroup\$ Your updated answer addressing the extended questions in the comments is much appreciated. \$\endgroup\$
    – kbakshi314
    Mar 24, 2021 at 16:28
  • 1
    \$\begingroup\$ I've updated the answer, but please avoid too much content added to the OP, it's better to ask a new question, instead (what will future searches be finding?). \$\endgroup\$ Mar 24, 2021 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.