---
date: '2024-03-01'
description: assignment on second-order systems with pid controller design for overshoot and settling time, routh-hurwitz stability criterion, and steady-state error analysis.
id: content
modified: 2026-06-05 15:08:42 GMT-04:00
tags:
  - sfwr3dx4
title: Second-order systems
created: '2024-03-01'
published: '2024-03-01'
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---
### problem 1.

Consider the following system:
![[thoughts/university/twenty-three-twenty-four/sfwr-3dx4/a2/a1-system.webp]]

> \[!question\] 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 $\omega_n$:

$$
\begin{align*}
\%{OS} &= e^{\frac{-\zeta\pi}{\sqrt{1-\zeta^2}}} \times 100 \% \\\
T_s &= \frac{4}{\zeta\omega_n}
\end{align*}
$$

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 $\omega_n$: $\omega_n = \frac{4}{\zeta T_s} \approx 6.76 \space rad \space s$.

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 $K_d$:

$$
\begin{align*}
7+K_pK_d &= 2\zeta\omega_n \\\
5+K_d &= \omega_n^2
\end{align*}
$$

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

The following is the [[thoughts/university/twenty-three-twenty-four/sfwr-3dx4/a2/p1.py|code]] snippet for generating the graphs and results:

![[thoughts/university/twenty-three-twenty-four/sfwr-3dx4/a2/p1.webp]]

```python title="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:
![[thoughts/university/twenty-three-twenty-four/sfwr-3dx4/a2/p2.webp]]

> \[!question\] 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.

```python
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.

> \[!question\] set b.
>
> Fix $K_i = 10$. Using the [[thoughts/Routh-Hurwitz criterion|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 $1 + G(s) = 0$:

$$
\begin{align*}
1 + \frac{K_d s^2+K_p s + 10}{s^3+3s^2+s} &= 0 \\\
s^3+3s^2+s + K_d s^2 + K_p s + 10 &= 0 \\\
s^3 + (3+K_d) s^2 + (K_p + 1)s + 10 &= 0
\end{align*}
$$

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

```python
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:

```prolog
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:

$$
\begin{align*}
K_d + 3 &> 0 \\\
-K_d + K_p - 2 &> 0
\end{align*}
$$

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

$$
\begin{align*}
K_d &> 0 \\\
K_p &> 2
\end{align*}
$$

> \[!question\] 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

$$
\lim_{s\to0}s\cdot R(s) \cdot (1-T(s))
$$

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

$$
SSE = 0.1 = \lim_{s\to0} s \cdot \frac{1}{s} \cdot (1 - \frac{K_p + K_d s}{s^2+7s +5 + K_p + K_d s})
$$

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