# Laplace of third-order lead-lag filter

For Butterworth low-pass filters, the Laplace function of the third-order filter is $$H(s)= \frac {1}{\left(1+\frac{s}{\omega_c}\right)\left(1+\frac{s}{\omega_c}+\frac{s^2}{\omega_c^2}\right)}$$ which is not equal to just cascading the transfer functions of the first-order filters $$H(s)= \left(\frac {1}{1+\frac{s}{\omega_c}}\right)^3 = \frac {1}{\left(1+\frac{s}{\omega_c}\right)\left(1+2\frac{s}{\omega_c}+\frac{s^2}{\omega_c^2}\right)}$$ The Bode plot of the third-order Butterworth is "sharper" than the plot of three cascaded filters.

Now my question: Is there anything similar for a lead-lag filter with $$H(s)= \frac {1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}$$ ?

How can I design a third-order lead-lag filter without just taking this transfer function to the power of three?

• Are you aware that the 3rd-order function (your first function) is not a general one but the special case of cascading a 1st order with a 2nd-order stage with a quality factor of unity? In the second expression the second-order quality factor is 0.5 only.
– LvW
Commented Feb 4, 2022 at 7:57
• third order lead-lag filter - isn't that the same as trying to grow apples that taste of tomato? Commented Feb 4, 2022 at 8:03
• @LvW, I'm partly aware of this. Is there a way to raise the quality factors when cascading three lead-lag filters? Similarily how the first equation raised the quality factor of the second one? Commented Feb 4, 2022 at 8:21
• @Andyaka, well, if there are some apples which show a similar frequency response like three cascaded lead-lag filters, but with a higher quality factor, I don't mind if they taste like tomatoes or not. Commented Feb 4, 2022 at 8:22
• The point is this: a lead lag filter is a first order type for very sensible reasons just as an apple tastes like apple and not tomato. Commented Feb 4, 2022 at 9:12

If you need a lead-lag filter with the same order in the numerator as in the denominator, just use the same quality factor for the zeroes, but with a different frequency (it wouldn't make sense to have it the same as the denominator). So, if the Butterworth normalized 3rd order transfer function is (1), use the denominator in the numerator but with a different frequency:

\begin{align} H(s)&=\dfrac{1}{\left(\dfrac{s}{\omega_p}\right)^3+2\left(\dfrac{s}{\omega_p}\right)^2+2\left(\dfrac{s}{\omega_p}\right)+1} \tag{1} \\ G(s)&=\dfrac{\left(\dfrac{s}{\omega_z}\right)^3+2\left(\dfrac{s}{\omega_z}\right)^2+2\left(\dfrac{s}{\omega_z}\right)+1}{\left(\dfrac{s}{\omega_p}\right)^3+2\left(\dfrac{s}{\omega_p}\right)^2+2\left(\dfrac{s}{\omega_p}\right)+1} \tag{2} \end{align}

And here is a quick verification, showing that the poles have the same quality factor, while the zeroes have the same Q as the poles:

Note that the frequencies here are two decades apart, so the phase is noce and clean. If you'll make them closer than a decade (or so), the poles and the zeroes will influence each other -- this is normal.

• Well, I guess I didn't see the forest for the trees at that point... Looks perfect! Commented Feb 4, 2022 at 9:57

Anything similar ?

Using the same idea, I plot (EE&O) two functions.

The last is more "abrupt" (scale).