Question

Design of Lead Compensator With Matlab...

G(s) = 9/(s^2+0.5s) and Gc(s) = 1

Transfer Function, maximum overshoot...

DESIGN of a LEAD COMPENSATOR with MATLAB

For the figure below, G(s)=9 / s(s+0.5)

a) For the compensator Gc(s)=1 Obtain

- Transfer function,

- Maximum overshoot and settling time for unit-step input

- Draw

i. unit step-response curve in MATLAB.

ii. unit ramp-response curve in MATLAB.

iii. Root- locus curve in MATLAB

- Obtain steady state error for unit-ramp input

b) Design a lead compensator Gc(s) to shift the poles at new locations of s₁=-4+j4 and s₂=-4-j4

- Obtain new transfer function by calculating Kc, α and Gc (s)

- Draw

i. unit step-response curve in MATLAB.

ii. unit ramp-response curve in MATLAB.

iii. Root- locus curve in MATLAB

- Obtain steady state error for unit-ramp input

Answer 1

The closed loop transfer function for the given figure is:

$$ \begin{gathered} T(S)=\frac{G_{C}(S) G(S)}{1+G_{C}(S) G(S)} \\ G(S)=\frac{9}{s(s+0.5)} \end{gathered} $$

a) For \(G_{C}(S)=1\),

$$ T(S)=\frac{G(S)}{1+G(S)} $$

Substitute \(\mathrm{G}(\mathrm{S})\) in above equation,

$$ \begin{aligned} &T(S)=\frac{\frac{9}{s(s+0.5)}}{1+\frac{9}{s(s+0.5)}} \\ &T(S)=\frac{9}{s^{2}+0.5 s+9} \end{aligned} $$

Compare the above equation with standard second order system,

\(T_{S}(S)=\frac{\omega_{n}{ }^{2}}{s^{2}+2 \varepsilon \omega_{n} s+\omega_{n}{ }^{2}}\) \(\omega_{n}{ }^{2}=9 \rightarrow \omega_{n}=3 \mathrm{rad} / \mathrm{s}\) \(2 \varepsilon \omega_{n}=0.5 \rightarrow \varepsilon=0.083\) Peak overshoot \(\left(\% M_{p}\right)=e^{-\frac{\varepsilon \pi}{\sqrt{1-\varepsilon^{2}}}} \times 100 \%\) Settling time( \(\left.t_{s}\right)=\frac{4}{\varepsilon \omega_{n}}=\frac{4}{0.083 \times 3}=16.06 \mathrm{~s}\)

Steady state error for unit ramp input \(=e_{g s}=\frac{1}{k_{\eta}}\)

$$ \begin{gathered} \underline{\underline{\text { Where }}}, K_{v}=\lim _{s \rightarrow 0} s G(S)=\lim _{s \rightarrow 0} s \frac{9}{s(s+0.5)} \\ \qquad K_{v}=\lim _{s \rightarrow 0} \frac{9}{s+0.5}=18 \\ e_{s s}=\frac{1}{k_{v}}=0.055 \end{gathered} $$

step response:

Ramp response:

Root locus: