Stability Analysis of a First-Order System with Proportional Control – Bode Plot

Let’s analyze the stability of a first-order system with a proportional controller using Bode Plot method. The process has a time constant of 10 seconds, and the proportional controller gain is $K_p$ .

The transfer function of the process under control is:

$P(s) = \frac{1}{10s + 1}$

The transfer function of the open-loop system is:

$G(s) = C(s) \cdot P(s) = \frac{K_p}{10s + 1}$

The closed-loop transfer function is:

$T(s) = \frac{G(s)}{1 + G(s)} = \frac{\frac{K_p}{10s + 1}}{1 + \frac{K_p}{10s + 1}} = \frac{K_p}{10s + 1 + K_p}$

Bode Plot and Stability Analysis for $C(s)$ and $P(s)$

The Bode plot is a powerful tool used to analyze the frequency response of a control system. Even though it is drawn based on the open-loop transfer function, $G(s) = C(s)P(s)$ (where $C(s)$ is the controller and $P(s)$ is the process), it provides valuable insights into the stability of the closed-loop system.

The Role of the Open-Loop Transfer Function in Closed-Loop Stability

For a system with unit feedback, the closed-loop transfer function is:

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

The closed-loop system’s stability is determined by the characteristic equation:

$1 + G(s) = 1 + C(s)P(s) = 0$

The roots of this equation (poles of the closed-loop system) dictate stability. By analyzing $G(s) = C(s)P(s)$ , we can predict whether the closed-loop poles lie in the left-half $s$ -plane (stable) or the right-half $s$ -plane (unstable).

Key Concepts for Stability Analysis Using the Bode Plot

Open-Loop Gain ():
- The magnitude of $G(j\omega)$ tells how much the system amplifies signals at different frequencies.
Phase Shift:
- The phase of $G(j\omega)$ shows how much delay is introduced at each frequency.
Stability Margins:
- Gain Margin (GM): Indicates how much the gain of $G(j\omega)$ can increase before the closed-loop system becomes unstable.
- Phase Margin (PM): Indicates how much additional phase lag can be tolerated before instability occurs.

Steps for Stability Analysis Using the Bode Plot

Construct the Bode Plot:
- Plot the magnitude ( $|G(j\omega)|$ ) and phase ( $\angle G(j\omega)$ ) of $G(s) = C(s)P(s)$ across a range of frequencies ( $\omega$ ).
Identify Critical Frequencies:
- Gain Crossover Frequency ( $\omega_g$ ): The frequency where $|G(j\omega)| = 1$ (or $0 \, \text{dB}$ ).
- Phase Crossover Frequency ( $\omega_p$ ): The frequency where $\angle G(j\omega) = -180^\circ$ for $K_p > 0$ , or $-360^\circ$ for $K_p < 0$ .
Determine Stability Margins:
- Phase Margin (PM): At $\omega_g$ , measure the difference between the phase and the critical phase ( $-180^\circ$ for $K_p > 0$ , $-360^\circ$ for $K_p < 0$ ). A positive PM indicates stability.
- Gain Margin (GM): At $\omega_p$ , calculate the gain increase (in dB) required to make $|G(j\omega)| = 1$ . A positive GM indicates stability.

Why Are There Different Critical Phases for Positive and Negative Gains?

Stability analysis using the Bode plot is based on the Nyquist Stability Theorem, which states that the stability of a closed-loop system depends on the encirclement of the critical point $-1 + j0$ in the Nyquist plot.

For $K_p > 0$ : The critical phase is $-180^\circ$ . The Nyquist plot must avoid encircling $-1 + j0$ to ensure stability.
For $K_p < 0$ : The critical phase becomes $-360^\circ$ . This accounts for the additional $-180^\circ$ phase shift introduced by the negative gain.

Explanation of Critical Phase Thresholds

When $K_p > 0$ :

The phase plot starts at $0^\circ$ and decreases with increasing frequency.
Instability occurs if the total phase shift reaches $-180^\circ$ , at which point the Nyquist plot risks encircling $-1 + j0$ .

When $K_p < 0$ :

The negative gain introduces an additional $-180^\circ$ phase shift.
As a result, the phase plot starts at $-180^\circ$ at low frequencies and decreases further.
Instability now occurs if the total phase shift reaches $-360^\circ$ , which includes the initial $-180^\circ$ from the negative gain.

Thus, for $K_p < 0$ , the critical phase threshold shifts from $-180^\circ$ to $-360^\circ$ to account for the added phase shift. This ensures consistency with the Nyquist criterion, which examines encirclement of the $-1$ point in the complex plane.

Visualizing with Nyquist Plots

Consider the Nyquist plots for $K_p = 1$ , $K_p = -0.5$ , and $K_p = -2$ :

For $K_p = 1$ : The plot starts at $0^\circ$ and avoids encircling $-1 + j0$ , ensuring stability.
For $K_p = -0.5$ : The plot starts at $-180^\circ$ . Stability is maintained as it does not encircle $-1 + j0$ by reaching $-360^\circ$ .
For $K_p = -2$ : The plot starts at $-180^\circ$ and crosses $-360^\circ$ , leading to instability due to encirclement of $-1 + j0$ .

This shift in the critical phase aligns with the Nyquist stability criterion and reflects the impact of the sign of $K_p$ on the system’s stability.

Stability Conditions for the Closed-Loop System

The closed-loop system is stable if:
- Gain Margin $> 0 \, \text{dB}$
- Phase Margin $> 0^\circ$
The system is marginally stable or integrator-driven unstable if:
- Gain Margin $= 0 \, \text{dB}$
- Phase Margin $= 0^\circ$
The system is unstable if:
- Gain Margin $< 0 \, \text{dB}$ or Phase Margin $< 0^\circ$

Why the Bode Plot of $G(s)$ Predicts Stability of the Closed-Loop System

Although the Bode plot is based on the open-loop transfer function $G(s) = C(s)P(s)$ , it predicts closed-loop stability because the characteristic equation $1 + G(s) = 0$ involves $G(s)$ . When $G(s)$ has a magnitude of $|G(j\omega)| = 1$ and a phase of $-180^\circ$ , the system is on the verge of instability because it satisfies $1 + G(j\omega) = 0$ .

By ensuring sufficient gain and phase margins in the Bode plot of $G(s)$ , you can guarantee the closed-loop poles remain in the left-half $s$ -plane.

Bode Plot for the System

For the transfer function $P(s) = \frac{1}{10s + 1}$ , the Bode plot can be derived as follows:

Magnitude Plot:
- The magnitude of $P(j\omega)$ is:
  $|P(j\omega)| = \frac{1}{\sqrt{(10\omega)^2 + 1}}$
- In decibels (dB):
  $20 \log_{10} |P(j\omega)| = -20 \log_{10} \sqrt{(10\omega)^2 + 1}$
Phase Plot:
- The phase of $P(j\omega)$ is:
  $\angle P(j\omega) = -\tan^{-1}(10\omega)$
Key Frequencies:
- At low frequencies ():
  - Magnitude: $|P(j\omega)| \approx 0 \, \text{dB}$
  - Phase: $\angle P(j\omega) \approx 0^\circ$
- At high frequencies ():
  - Magnitude: $|P(j\omega)| \to -\infty \, \text{dB}$
  - Phase: $\angle P(j\omega) \to -90^\circ$

Python Code to Generate the Bode Plot

The following Python code uses the control library to generate the Bode plot for the system:

import numpy as np
import matplotlib.pyplot as plt
from control import tf, bode

# Define the transfer function
num = [1]  # Numerator of G(s)
den = [10, 1]  # Denominator of G(s)
G = tf(num, den)

# Generate Bode plot
plt.figure()
mag, phase, omega = bode(G, dB=True, omega=np.logspace(-2, 2, 500), plot=True)

# Customize the plot
plt.suptitle("Bode Plot of the System", fontsize=14)
plt.show()

Bode plot for $P(s) = \frac{1}{10s + 1}$ is depicted below:

Wait! What Happened to $C(s)$ ?

You’re absolutely right—what happened to $C(s)$ ? We can’t forget it! In the open-loop transfer function $G(s)$ , we must include $C(s)$ . Luckily, in this example, $C(s)$ is a pure gain: $C(s) = K_p$ . Let’s explore how this gain impacts the Bode plot.

Magnitude Plot

The magnitude of $G(j\omega) = C(j\omega)P(j\omega)$ is given by:

$|G(j\omega)| = |C(j\omega)||P(j\omega)| = \frac{K_p}{\sqrt{(10\omega)^2 + 1}}$

In decibels (dB), this becomes:

$20 \log_{10} |G(j\omega)| = 20 \log_{10} K_p - 20 \log_{10} \sqrt{(10\omega)^2 + 1}$

This means the magnitude plot shifts vertically when $K_p$ changes. Increasing $K_p$ raises the plot, while decreasing $K_p$ lowers it.

Phase Plot

For $K_p > 0$ :

The phase of $G(j\omega)$ is:

$\angle G(j\omega) = \angle C(j\omega) - \angle P(j\omega) = 0^\circ - \tan^{-1}(10\omega) = -\tan^{-1}(10\omega)$

The phase plot for $K_p > 0$ remains unchanged, as $K_p$ contributes no phase shift.

For $K_p < 0$ :

When $K_p < 0$ , the gain introduces a $-180^\circ$ phase shift:

$\angle G(j\omega) = \angle C(j\omega) - \angle P(j\omega) = -180^\circ - \tan^{-1}(10\omega)$

In this case, the phase plot shifts downward by $-180^\circ$ .

Bode Plot Analysis for $K_p > 0$

Let’s consider $G(s) = \frac{K_p}{10s + 1}$ . For $K_p > 0$ , the magnitude plot shifts vertically with $K_p$ , but the phase plot remains the same.

Gain Margin:

The gain margin is $\infty$ , meaning no amount of gain can make the system unstable. For instability to occur in a closed-loop system, the phase plot must cross $-180^\circ$ , but this doesn’t happen here.

Phase Margin:

The phase margin measures how much additional phase lag can occur before instability. It is determined at the frequency where the magnitude crosses 0 dB.

For $K_p = 1$

At low frequencies ( $\omega \approx 0$ ), the phase is already $0^\circ$ , and the magnitude is 0 dB. This means the phase margin is:

$\text{Phase Margin} = 180^\circ -0^\circ = 180^\circ$

For $K_p = 10$

For $K_p = 10$ , the magnitude crosses 0 dB at a frequency $\omega_c$ . Solving for $\omega_c$ when $|G(j\omega)| = 1$ :

$\frac{|10|}{\sqrt{(10\omega_c)^2 + 1}} = 1$

$\sqrt{(10\omega_c)^2 + 1} = 10$

$(10\omega_c)^2 = 100 - 1 = 99$

$\omega_c = \frac{\sqrt{99}}{10}\approx0.99 \, \text{rad/s}$

The magnitude crosses 0 dB at $\omega \approx 0.99 \, \text{rad/s}$ . At this frequency, the phase is:

$\angle G(j0.99) = -\tan^{-1}(10 \cdot 0.99) \approx -84.2^\circ$

The phase margin is:

$\text{Phase Margin} = 180^\circ - 84.2^\circ = 95.8^\circ$

Stability Range

The closed-loop system is stable for all $K_p > 0$ because the phase plot never reaches $-180^\circ$ .

What Does the Phase Margin Tell Us?

The phase margin gives us an idea of the system’s robustness. For $K_p = 10$ , a phase margin of $95.8^\circ$ means the system can tolerate some delay before becoming unstable.

Time Delay Interpretation

Let’s add a time delay $T_d$ to the loop:

A delay introduces a phase shift proportional to frequency:

$\phi = -\omega T_d \quad \text{(in radians)} \quad \text{or} \quad \phi = -\frac{180}{\pi} \cdot \omega T_d \quad \text{(in degrees)}.$

To reduce the phase margin by $95.8^\circ$ at $\omega = 0.99 , \text{rad/s}$ , the time delay is:

$T_d = \frac{\phi}{-\frac{180}{\pi} \cdot \omega}$

$T_d = \frac{-95.8}{-\frac{180}{\pi} \cdot 0.99} \approx 1.69 , \text{seconds}.$

This shows that the system can tolerate a delay of approximately 1.69 seconds before losing stability when $K_p = 10$ . Below, step response of the closed loop system is depicted for $K_p = 10$ with various time delays.

Calculate how much delay can be added if $K_p=2$ . You will see that the Phase margin is approximately $120.2^\circ$ that means the closed loop can tolerate 12.13 seconds of delay before becoming unstable.