Lag compensator

Lag compensator:-

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

G(s) = \frac{s+\frac{1}{\tau }}{s+\frac{1}{\beta \tau }},    where \beta =\frac{z_{c}}{p_{c}}> 1     and \tau > 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} =0] and output load impedance to be infinite .

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

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

after simplification

\frac{E_{o}(s)}{E_{i}(s)} =\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} =\frac{1}{R_{2}C}  and has a pole at p_{c}=\frac{1}{(R_{1}+R_{2})C}=\frac{1}{\beta \tau } .

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

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

\frac{E_{o}(s)}{E_{i}(s)} =\frac{1}{\beta }\frac{s+\frac{1}{\tau }}{s+\frac{1}{\beta \tau }} = (\frac{\tau s+1}{\beta \tau 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 =R_{1}C> 0,    \beta =\frac{(R_{1}+R_{2})}{R_{2}}> 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 | G(j\omega ) \right |=\left | \frac{E_{o}(j\omega )}{E_{i}(j\omega )} \right | = \left | \beta(\frac{1+j\omega \tau }{1+\beta \tau j\omega }) \right |

D.C gain at \omega =0  is \beta  which is greater than 1.

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

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 ) = (\frac{1+j\omega \tau }{1+\beta \tau j\omega }),   let  \beta =1.

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

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

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

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

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

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

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

\frac{d\Phi }{d\omega }=0

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

{((1+\beta \omega^{2} \tau^{2})\tau (1-\beta ))-(\omega \tau (1-\beta )2\omega \beta \tau ^{2})}=0

\tau (1-\beta )(1+\beta \omega^{2} \tau^{2}-2\omega^{2} \beta\tau ^{2})=0

\tau (1-\beta)(1-\omega^{2} \beta \tau ^{2})=0

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

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

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

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

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

\beta =\frac{1-sin \Phi _{m}}{1+sin \Phi _{m}}.

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

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 _{gc} of compensated system.

In the high frequency range , the lag compensator has an attenuation of 20 log(\beta ) 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 > 1, while low-frequency control signals under go unit amplification (0 dB gain).

atypical value of \beta =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 _{gc}^{'} at which phase angle of uncompensated system is 

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

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

Step 3:- find gain of the uncompensated system at \omega _{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 _{gc}^{'} and find \tau value.

Step 5:- Calculate phase lag of compensator  at \omega _{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.

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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.