Lead Compensator

Lead compensator:-

A Lead 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}{\alpha \tau }},    where \alpha =\frac{z_{c}}{p_{c}}< 1     and \tau > 0

Pole-zero plot of Lead compensator:-

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

  • A lead compensator speeds up the transient response and increases margin of stability of a system.
  • It also helps  to increase the system error constant through a limited range.

Realization of Lead compensator as an Electrical Network:-

The lead 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}}{R_{2}+(R_{1}|| \frac{1}{Cs})}

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

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

after simplification

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

\frac{E_{o}(s)}{E_{i}(s)} =\frac{s+\frac{1}{R_{1}C}}{s+\frac{1}{(\frac{R_{2}}{R_{1}+R_{2}})R_{1}C}}  by comparing this equation with the transfer function of lead compensator has a zero at  Z_{c} =\frac{1}{R_{1}C}  and the pole is p_{c}=\frac{1}{(\frac{R_{2}}{R_{1}+R_{2}})R_{1}C} .

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

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

\frac{E_{o}(s)}{E_{i}(s)} =\frac{s+\frac{1}{\tau }}{s+\frac{1}{\alpha \tau }} = \alpha (\frac{\tau s+1}{\alpha \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 \alpha.

using the EQN(II)   

 \tau =R_{1}C> 0,    \alpha =\frac{R_{2}}{R_{1}+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 | \alpha (\frac{1+j\omega \tau }{1+\alpha \tau j\omega }) \right |

D.C gain at \omega =0  is \alpha  which is less than 1.

attenuation \frac{1}{\alpha } is used to determine the steady state performance.

while using a lead N/w , it is important to increase the loop gain by an amount of \frac{1}{\alpha }.

A lead compensator is visualized as a combination of a N/w and an amplifier.

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

Frequency-response of a lead compensator:-

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

the frequency response of lead compensator is 

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

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

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

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

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

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

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

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

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

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

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

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

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

({1+\alpha })sin \Phi _{m} = {1-\alpha }

sin \Phi _{m}+\alpha sin \Phi _{m} = {1-\alpha }

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

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

when there is a need for phase leads of more than 60^{o}, two cascaded lead networks are used where each N/w provides half of the required phase.

for phase leads more than 60^{o}\alpha decreases sharply and if single N/w is used \alpha will be too low.

Choice of \alpha :-

In choosing parameters of compensator \tau depends on R_{1} and C . The \tau value may be anything but for \alpha there is a constraint. It depends on inherent noise in Control systems.

from the lead N/w , it’s been observed that the high frequency noise is amplified by \frac{1}{\alpha } while low frequencies by unity.

more (or) less \alpha should not be less than 0.07.

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:- using the relation 

Additional phase lead required = specified phase margin- Phase Margin of uncompensated system.

\epsilon  is a margin of safety required by the fact that the gain cross over frequency will increase due to compensation.

for example :-  \epsilon =5^{o} is a good assumption for -40 dB/decade.

\epsilon =15^{o}  (or) 20^{o} \Rightarrow -60 dB/decade.

Step 3:- Set the maximum phase of the lead compensator    \Phi _{m} = Additional phase lead required  and compute \alpha =\frac{1-sin \Phi _{m}}{1+sin \Phi _{m}}.

Step 4:- Find the frequency at which the uncompensated system has a gain of -20 log(\frac{1}{\sqrt{\alpha }}) dB, which gives the new gain cross over frequency.

with \omega _{gc} as the gain cross over frequency the system has a phase margin of \gamma _{1}

where as with \omega _{gc}^{'} as the gain cross over frequency the system has a phase margin of \gamma _{2}

Step 5:- Now \omega _{gc}^{'} = \omega _{m} = \frac{1}{\tau \sqrt{\alpha }}.    find the value of \tau and the transfer function of lead compensator  \frac{1+j\omega \tau }{1+\alpha j\omega \tau }.

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Author: vikramarka

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