I have a question regarding lead high pass filter and lag low pass filter.

I do not quite understand the reasoning behind these two special case (when z=0) ?

Poles and zeros of high/low pass filters


  • \$\begingroup\$ You haven't really asked a question but rather said what you don't understand and I cannot fathom out what it is you are having problems with. \$\endgroup\$
    – Andy aka
    Apr 1, 2016 at 9:08

1 Answer 1


Lag low-pass and lead high-pass are in fact the "standard" low-pass and high-pass filters, in the sense that an ideal low-pass filter should have a gain of zero for \$\omega\rightarrow\infty\$, and an ideal high-pass filter should have a gain of zero at \$\omega=0\$. These conditions are satisfied by the lag low-pass filter (with a zero at \$s\rightarrow\infty\$), and by the lead high-pass filter (with a zero at \$s=0\$).

The phase of the lead low-pass filter is greater (i.e., less negative) than the phase of the lag low-pass filter (\$\Rightarrow\$ "lead"), but the magnitude is worse because its gain only decays from \$z_1/p_1>1\$ to \$1\$ for \$\omega\rightarrow \infty\$. A similar thing is true for the lag high-pass filter. Its gain is not zero at \$\omega=0\$ but it equals \$z_1/p_1<1\$. Its phase is smaller (i.e., less positive) than the phase of the lead high-pass filter (\$\Rightarrow\$ "lag").

  • \$\begingroup\$ Thanks for the answer. (1) I would like to know why the compensator or filter \$\frac{s+z}{s+p}\$ with \$z < p\$ is labeled 'lag' while the one with \$p < z\$ is labeled 'lead'. It seems to contradict the definition of lead and lag compensation. (2) Is it accurate to say that in general, lead and lag filters have the behavior of high and low pass filters? \$\endgroup\$
    – kbakshi314
    Mar 24, 2021 at 2:15

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