# Lag compensator

Lag compensator:-

A Lag compensator has a Transfer function of the form $G(s)&space;=&space;\frac{s+z_{c}}{s+p_{c}}------EQN(I)$

$G(s)&space;=&space;\frac{s+\frac{1}{\tau&space;}}{s+\frac{1}{\beta&space;\tau&space;}}$,    where $\beta&space;=\frac{z_{c}}{p_{c}}>&space;1$     and $\tau&space;>&space;0$

Pole-Zero Plot of Lag compensator:-

i.e, the pole is located to the right of the zero.

Realization of Lag compensator as Electrical Network:-

The lag compensator can be realized by an electrical Network.

Assume impedance of source is zero  $[Z_{s}&space;=0]$ and output load impedance to be infinite  .

The transfer function is $\frac{E_{o}(s)}{E_{i}(s)}&space;=&space;\frac{(R_{2}+\frac{1}{Cs})}{R_{1}+(R_{2}+&space;\frac{1}{Cs})}$

$\frac{E_{o}(s)}{E_{i}(s)}&space;=&space;\frac{R_{2}Cs+1}{(R_{1}+R_{2})Cs+1}$

after simplification

$\frac{E_{o}(s)}{E_{i}(s)}&space;=\frac{R_{2}}{(R_{1}+R_{2})}(\frac{s+\frac{1}{R_{2}C}}{s+\frac{1}{(R_{1}+R_{2})C}})$

after comparing the above equation with the transfer function of lag compensator has a zero at  $Z_{c}&space;=\frac{1}{R_{2}C}$  and has a pole at $p_{c}=\frac{1}{(R_{1}+R_{2})C}=\frac{1}{\beta&space;\tau&space;}$ .

from the pole $\beta&space;=\frac{(R_{1}+R_{2})}{R_{2}}$ and  $\tau&space;=R_{2}C$.

therefore  the transfer function  has a zero at $-\frac{1}{\tau&space;}$   and a pole at $-\frac{1}{\beta&space;\tau&space;}$.

$\frac{E_{o}(s)}{E_{i}(s)}&space;=\frac{1}{\beta&space;}\frac{s+\frac{1}{\tau&space;}}{s+\frac{1}{\beta&space;\tau&space;}}&space;=&space;(\frac{\tau&space;s+1}{\beta&space;\tau&space;s+1})--------EQN(II)$.

the values of the three parameters $R_{1}$ , $R_{2}$  and C are determined from the two compensator parameters $\tau$  and $\beta$.

using the EQN(II)

$\tau&space;=R_{1}C>&space;0$,    $\beta&space;=\frac{(R_{1}+R_{2})}{R_{2}}>&space;1$.

there is an additional degree of freedom in the choice of the values of the network components which is used to set the impedance level of the N/w.

the gain is $\left&space;|&space;G(j\omega&space;)&space;\right&space;|=\left&space;|&space;\frac{E_{o}(j\omega&space;)}{E_{i}(j\omega&space;)}&space;\right&space;|&space;=&space;\left&space;|&space;\beta(\frac{1+j\omega&space;\tau&space;}{1+\beta&space;\tau&space;j\omega&space;})&space;\right&space;|$

D.C gain at $\omega&space;=0$  is $\beta$  which is greater than 1.

Let the zero-frequency gain as unity, then the Transfer function is $G(j\omega&space;)&space;=&space;(\frac{1+j\omega&space;\tau&space;}{1+\beta&space;\tau&space;j\omega&space;})$.

Frequency-response of Lag compensator:-

Note:-“lag” refers to the property that the compensator adds positive phase to the system over some appropriate frequency range.

$G(j\omega&space;)&space;=&space;(\frac{1+j\omega&space;\tau&space;}{1+\beta&space;\tau&space;j\omega&space;})$,   let  $\beta&space;=1$.

the frequency response of lag compensator is $\left&space;|&space;G(j\omega&space;)&space;\right&space;|=&space;\sqrt{\frac{1+\omega&space;^{2}\tau&space;^{2}}{1+\omega&space;^{2}\beta&space;^{2}\tau&space;^{2}}}$

at $\omega&space;=\frac{1}{\tau&space;}$ $\Rightarrow\left&space;|&space;G(j\omega&space;)&space;\right&space;|=&space;\sqrt{\frac{2}{1+\beta&space;^{2}}}$.

$(1+j\omega&space;\tau&space;)\rightarrow$ has a slope +20 dB/decade with corner frequency $\frac{1}{\tau&space;}$.

$(1+\beta&space;\tau&space;j\omega)\rightarrow$ slope is -20 dB/decade with corner frequency $\frac{1}{\beta&space;\tau&space;}$.

$\Phi&space;=&space;\angle&space;G(j\omega&space;)=tan^{-1}\omega&space;\tau&space;-tan^{-1}\beta&space;\omega&space;\tau$

to find at which frequency the phase is minimum , differentiate $\Phi$ w.r to $\omega$ and equate it to zero.

$\Phi&space;=&space;tan^{-1}(\frac{\omega&space;\tau-\beta\omega&space;\tau}{1+\beta&space;\omega^{2}&space;\tau^{2}})$

$\frac{d\Phi&space;}{d\omega&space;}=0$

$\frac{1}{1+(\frac{\omega&space;\tau-\beta&space;\omega&space;\tau}{1+\alpha&space;\omega^{2}&space;\tau^{2}})^{2}}(\frac{((1+\beta&space;\omega^{2}&space;\tau^{2})\tau&space;(1-\beta&space;))-(\omega&space;\tau&space;(1-\beta&space;)2\omega&space;\beta&space;\tau&space;^{2})}{(1+\beta&space;\omega^{2}&space;\tau^{2})^{2}})=0$

${((1+\beta&space;\omega^{2}&space;\tau^{2})\tau&space;(1-\beta&space;))-(\omega&space;\tau&space;(1-\beta&space;)2\omega&space;\beta&space;\tau&space;^{2})}=0$

$\tau&space;(1-\beta&space;)(1+\beta&space;\omega^{2}&space;\tau^{2}-2\omega^{2}&space;\beta\tau&space;^{2})=0$

$\tau&space;(1-\beta)(1-\omega^{2}&space;\beta&space;\tau&space;^{2})=0$

$\because&space;\tau&space;\neq&space;0$    implies $(1-\beta)=0\Rightarrow&space;\beta&space;=1$   , which is invalid because $\beta&space;>&space;1$.

$(1-\omega^{2}&space;\beta&space;\tau&space;^{2})=0\Rightarrow&space;\omega^{2}&space;=\frac{1}{\beta&space;\tau&space;^{2}}$.

$\omega&space;=\frac{1}{\sqrt{\beta}&space;\tau&space;}$  , at this $\omega$  lead compensator has minimum phase given by

$\Phi&space;_{m}&space;=&space;tan^{-1}(\frac{1-\beta&space;}{2\sqrt{\beta}})$

$tan&space;\Phi&space;_{m}&space;=&space;\frac{1-\beta&space;}{2\sqrt{\beta}}$ implies $sin&space;\Phi&space;_{m}&space;=&space;\frac{1-\beta}{1+\beta&space;}$.

$\beta&space;=\frac{1-sin&space;\Phi&space;_{m}}{1+sin&space;\Phi&space;_{m}}$.

at $\omega&space;=\omega&space;_{m}$ ,    $\left&space;|&space;G(j\omega&space;)&space;\right&space;|&space;=&space;\frac{1}{\sqrt{\beta&space;}}$.

Choice of $\beta$ :-

Any phase lag at the gain cross over frequency of the compensated system is undesirable.

To prevent the effects of lag compensator , the corner frequency of the lag compensator must be located substantially lower than the $\omega&space;_{gc}$ of compensated system.

In the high frequency range , the lag compensator has an attenuation of $20&space;log(\beta&space;)$ dB, which is used to obtain required phase margin.

The addition of a lag compensator results in an improvement in the ratio of control signal to noise in the loop.

high frequency noise signals are attenuated by a factor $\beta&space;>&space;1$, while low-frequency control signals under go unit amplification (0 dB gain).

atypical value of $\beta&space;=10$.

Procedure for bode-plot of a lead compensator:-

Step 1:- Sketch the Bode-plot of the uncompensated system with the gain k. Set the value of k according to the steady-state error requirement.

Measure the gain cross over frequency and the phase margin of uncompensated system.

Step 2:-  find $\omega&space;_{gc}^{'}$ at which phase angle of uncompensated system is

$-180^{o}$ + given Phase Margin+ $\epsilon$.

$\epsilon&space;=5^{o}(or)15^{o}$   is a good assumption for phase-lag contribution.

Step 3:- find gain of the uncompensated system at $\omega&space;_{gc}^{'}$ and equate it to 20 log ($\beta$)  and then find $\beta$.

Step 4:- choose the upper corner frequency of the compensator to one octave to one decade  below $\omega&space;_{gc}^{'}$ and find $\tau$ value.

Step 5:- Calculate phase lag of compensator  at $\omega&space;_{gc}^{'}$, if it is less than $\epsilon$ go to next step.

Step 6:- Draw the Bode plot of compensated system  to meet the desired specifications.

(No Ratings Yet)