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

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

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

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

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

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

after simplification

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

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

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

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

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

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

using the EQN(II)

$\tau&space;=R_{1}C>&space;0$,    $\alpha&space;=\frac{R_{2}}{R_{1}+R_{2}}<&space;1$.

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

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

attenuation $\frac{1}{\alpha&space;}$ 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 $\frac{1}{\alpha&space;}$.

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.

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

the frequency response of lead compensator is

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

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

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

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

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

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

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

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

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

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

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

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

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

$({1+\alpha&space;})sin&space;\Phi&space;_{m}&space;=&space;{1-\alpha&space;}$

$sin&space;\Phi&space;_{m}+\alpha&space;sin&space;\Phi&space;_{m}&space;=&space;{1-\alpha&space;}$

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

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

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 $\frac{1}{\alpha&space;}$ 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 :-  $\epsilon&space;=5^{o}$ is a good assumption for -40 dB/decade.

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

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

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

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

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

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

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