Second-order systems

problem 1.

Consider the following system:

Question

Using the properties of second-order systems, determine $K_p$ and $K_d$ such that the overshoot is 10 percent and the settling time is 1 second. Confirm that your design meets the requirements by plotting the step response.

Given the percent overshoot $\%OS$ and settling time based on the damping ratio $\zeta$ and natural frequency ${\omega }_n$:

$$\begin{array}{rl}\%OS& =e^{\frac{-\zeta \pi }{\sqrt{1-{\zeta }^2}}}\times 100\%\\ T_s& =\frac{4}{\zeta {\omega }_n}\end{array}$$

For 10% overshoot, we can solve for $\zeta$: $\zeta =\frac{-\ln (\%OS/100)}{\sqrt{{\pi }^2+{\ln }^2(\%OS/100)}}\approx 5.916\times e-1$. For 1 second settling time, we can solve for ${\omega }_n$: ${\omega }_n=\frac{4}{\zeta T_s}\approx 6.76rads$.

Given second-order systems’ transfer function:

$$G(s)=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$$

and the transfer function of the PID controller in the given system is given by:

$$G_c(s)=K_p+K_{d}s$$

The transfer function is then followed by:

$$T(s)=G(s)G_c(s)=\frac{K_p+K_{d}s}{s^2+7s+5}$$

We then have ${\omega }_n$ and $\zeta$ to solve for $K_p$ and $K_d$:

$$\begin{array}{rl}7+K_p{K}_d& =2\zeta {\omega }_n\\ 5+K_d& ={\omega }_n^2\end{array}$$

Thus, $K_p=40.784365358764106$ and $K_d=0.9999999999999991$.

The following is the code snippet for generating the graphs and results:

p1.py

from scipy.optimize import fsolve
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import TransferFunction, step
 
OS, Ts = 0.10, 1.0
zeta = fsolve(lambda z: np.exp(-z*np.pi/np.sqrt(1-z**2)) - OS, 0.5)[0]
wn = 4 / (zeta * Ts)
 
# Coefficients from the standard second-order system
a1 = 2 * zeta * wn  # coefficient of s
a0 = wn**2          # constant coefficient
 
# Equating the coefficients to solve for Kp and Kd
# 7 + Kd = a1 and 5 + Kp = a0
Kd = a1 - 7
Kp = a0 - 5
 
# Confirm the design by plotting the step response
# First, define the transfer function of the closed-loop system with the calculated Kp and Kd
G = TransferFunction([Kd, Kp], [1, 7+Kd, 5+Kp])
 
# Now, generate the step response of the system
time = np.linspace(0, 5, 500)
time, response = step(G, T=time)
 
print(Kp, Kd, zeta, wn)
# Plot the step response
plt.figure(figsize=(10, 6))
plt.plot(time, response)
plt.title('Step Response of the Designed PD Controlled System')
plt.xlabel('Time (seconds)')
plt.ylabel('Output')
plt.grid(True)
plt.show()

---

problem 2.

Consider the following system:

set a.

If $K_d=K_p=K_i=1$, is the system stable? (Please determine this by explicitly finding the poles of the closed-loop system and reasoning about stability based on the pole locations.)

Given that $K_d=K_p=K_i=1$, The PID controller transfer function is:

$$C(s)=K_p+\frac{K_i}{s}{K}_{d}s=1+\frac{1}{s}+s$$

The open-loop transfer function is given by: $G(s)=C(s)P(s)=(1+s+\frac{1}{s})\frac{1}{s^2+3s+1}$.

Thus the closed-loop transfer function is given by $T(s)=\frac{G(s)}{1+G(s)}=\frac{s^3+s^2+1}{s^3+s^2+4s+2}$.

We need to solve $s^3+s^2+4s+2=0$ to find the poles of the closed-loop system.

import numpy as np
print(np.roots([1,1,4,2]))

which yields [-0.23341158+1.92265955j -0.23341158-1.92265955j -0.53317683+0.j] as poles. Since all the poles have negative real parts, the system is stable.

set b.

Fix $K_i=10$. Using the Routh-Hurwitz criterion, determine the ranges of $K_p$ and $K_d$ that result in a stable system.

The open-loop transfer function is given by

$$G(s)=C(s)P(s)=(K_p+\frac{K_i}{s}+K_{d}s)\frac{1}{s^2+3s+1}=\frac{K_d{s}^2+K_{p}s+10}{s^3+3s^2+s}$$

The characteristic equation of the closed-loop system is given by $1+G(s)=0$:

$$\begin{array}{rl}1+\frac{K_d{s}^2+K_{p}s+10}{s^3+3s^2+s}& =0\\ s^3+3s^2+s+K_d{s}^2+K_{p}s+10& =0\\ s^3+(3+K_d)s^2+(K_p+1)s+10& =0\end{array}$$

Applying the Routh-Hurwitz criterion, we have the following table:

from sympy import symbols, Matrix
Kd, Kp = symbols('Kd Kp')
a0 = 10
a1 = Kp + 1
a2 = 3 + Kd
a3 = 1
 
routh = Matrix([
  [a3, a1],
  [a2, a0],
  [a1 - (a2*a3)/a3, 0],
  [a0, 0]
])
 
print(routh)

which results in the following table:

Matrix([[1, Kp + 1], [Kd + 3, 10], [-Kd + Kp - 2, 0], [10, 0]])

The conditions for stability from the Routh-Hurwitz criterion states that all the elements in the first column of the Routh array must be positive. Thus, we have the following inequalities:

$$\begin{array}{rl}{K}_d+3& >0\\ -K_d+K_p-2& >0\end{array}$$

Solving for $K_d$ and $K_p$ yields the following ranges:

$$\begin{array}{rl}{K}_d& >0\\ K_p& >2\end{array}$$

set c.

For the system in the first question, suppose that you want the steady-state error to be $10\%$. What should the values of $K_p$ and $K_d$ be? (Hint: the system is not in the unity gain form that we discussed in detail in lecture, so be careful.)

The open-loop transfer function is given by:

$$G(s)H(s)=(K_p+K_{d}s)\frac{1}{s^2+7s+5}$$

The transfer function for closed-loop is given by:

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

From final value theorem, the steady-state error is given by

$$\underset{s\to 0}{\lim }s\cdot R(s)\cdot (1-T(s))$$

For step input $R(s)=\frac{1}{s}$ we got

$$SSE=0.1=\underset{s\to 0}{\lim }s\cdot \frac{1}{s}\cdot (1-\frac{K_p+K_{d}s}{s^2+7s+5+K_p+K_{d}s})$$

$K_p=\frac{5}{8}$