I am trying to learn about lead compensator from above link

But i have bit confusion regarding two terms,"T" and "aT" here, as shown higlighted in attached snapshot? What do they denote?

They are related to some time parameters but it is not exactly clear, whose those parameters are?enter image description here

  • 1
    \$\begingroup\$ The 1st sentence from the screenshot? \$\endgroup\$ Dec 8 '20 at 10:33
  • 2
    \$\begingroup\$ T equals to "T" and aT equals to a*T that is all, not a big deal. \$\endgroup\$
    – emre iris
    Dec 8 '20 at 10:37

These are time constants and the formula would more advantageously be written as \$C(s) = K\frac{1+\tau s}{1+\alpha\tau s}\$ where \$\tau\$ is a time constant and has a dimension of time while \$\alpha\$ is unitless. \$K\$ represents the gain of this compensator for \$s=0\$.

In a first-order system, the pole is the inverse of the time constant and this expression can be reformulated as follows: \$C(s)=C_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$ with \$\omega_z=\frac{1}{\tau}\$ and \$\omega_p=\frac{1}{\alpha\tau}\$. I replaced \$K\$ by \$C_0\$ indicating that this is the gain of \$C\$ when \$s=0\$.

This type of compensator involves a zero \$f_z\$ and pole \$f_p\$ linked by the coefficient \$\alpha\$. If \$\alpha\$ is 1, then the pole and zero are coincident and cancel each other in magnitude and phase. If the zero is now lower than the pole (\$\alpha < 1\$), then you can determine the phase of the system and see that it increases: we say the compensator boosts the phase between the zero and the pole. As you spread them (the zero goes down the frequency axis while the pole runs opposite towards higher frequency), the phase boost increases and peaks to 90° as a maximum. You will actually organize the zero and pole so that it produces the expected phase boost at the point you need it the most which is usually the crossover frequency \$f_c\$. And if you do the maths ok, you'll find that the phase boost peaks at \$f_c=\sqrt{f_zf_p}\$. If you now add a pole at the origin (with an op-amp based circuit), you have created a type 2 compensator, extremely popular in switching converters.

  • \$\begingroup\$ What do you mean?"K" is overall gain or dc gain?? since you can see in ogata, the term appears as "Kc a" \$\endgroup\$
    – engr
    Dec 18 '20 at 7:55
  • \$\begingroup\$ In my expression, \$K\$ is unitless so it has the dimension of a gain here. I however prefer \$C_0\$ as a more appropriate notation. \$\endgroup\$ Dec 18 '20 at 19:48

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