Designing a lead compensation for a third-order system

Question

Consider the control system shown in Fig. 1. Design a compensator such that the unit-step response curve will exhibit a maximum overshoot of 25% or less and a settling time of 5 sec or less.

What I tried:
I'm not sure how to approach the solution but what I did was indicate the \$\zeta\$ & \$\omega_n\$ from the equations of Mp & ts as in here & then approache the typical solution which yielded the following results:

Results I could get: \$\zeta = 0.4\$

\$\omega_n = 2\$

\$G_c =\frac{60.944(s+0.438)}{(s+8.041)}\$

\$G_cGH =\frac{60.944(s+0.438)}{s^2(s+4)(s+8.041)}\$

The question doesn't state any further information; it's a mid-term exam.

Answer 1

To verify the solution, you should obtain the closed-loop transfer function using a feedback of \$H(s) = 1\$, then check if the new system matches a maximum overshoot of 25% of less and that the settling time was reduced to at least 5 seconds.

Here is the code used in verification (works for both Matlab & octave):

num1 = [1]
den1 = [1 4 0 0]
num2 = [60.944 24.3776];
den2 = [1 12.041 32.164 0 0]

g1 = tf(num1, den1)
g2 = tf(num2, den2)
sys1 = 1 / (1 + g1)
sys2 = 1 / (1 + g2)

t = 0:0.05:20
c1 = step(sys2, t)
c2 = step(sys1, t)

plot(t,c1,'-',t,c2)
title('Exam 3. Question 1.')
xlabel('t Sec')
ylabel('Outputs c1, c2')
text(10, -0.25,'Compensated System')
text(6,1.2,'Uncompensated System')