Second-order system simulation in Simulink - Mini Project: Build & Apply

Second-order system simulation

📖 Scenario: You are working as a control engineer. You need to simulate a second-order system to understand how it behaves over time when given a step input. This helps in designing controllers for machines and processes.

🎯 Goal: Build a simple Simulink model that simulates a second-order system with given parameters and plots its response to a step input.

📋 What You'll Learn

Create a Simulink model with a second-order system block using the transfer function representation.

Set the system parameters: natural frequency and damping ratio.

Add a step input block to the model.

Simulate the model and plot the output response.

💡 Why This Matters

🌍 Real World

Second-order systems model many real-world devices like springs, circuits, and mechanical systems. Simulating them helps engineers design stable and efficient controllers.

💼 Career

Control engineers, system analysts, and automation specialists use these simulations daily to predict system behavior and improve designs.

1

Create the second-order system transfer function

Create a variable called wn and set it to 5 for the natural frequency. Create a variable called zeta and set it to 0.7 for the damping ratio. Then create a transfer function variable called sys representing the second-order system with numerator [wn^2] and denominator [1, 2*zeta*wn, wn^2].

Simulink

# Create variables wn and zeta
# Create transfer function sys

Hint

Use tf to create the transfer function. The denominator coefficients are 1, 2*zeta*wn, and wn^2.

2

Add a step input and simulation time

Create a variable called t as a time vector from 0 to 5 seconds with 0.01 second intervals. Create a variable called u as a step input vector of ones with the same length as t.

Simulink

wn = 5;
zeta = 0.7;
sys = tf([wn^2], [1, 2*zeta*wn, wn^2]);
# Create time vector t and step input u

Hint

Use 0:0.01:5 to create the time vector and ones(size(t)) for the step input.

3

Simulate the system response

Use the lsim function to simulate the response of sys to input u over time t. Store the output in a variable called y.

Simulink

wn = 5;
zeta = 0.7;
sys = tf([wn^2], [1, 2*zeta*wn, wn^2]);
t = 0:0.01:5;
u = ones(size(t));
# Simulate system response y

Hint

Use lsim(sys, u, t) to simulate the system output.

4

Plot the system output

Plot the output y versus time t using the plot function. Label the x-axis as 'Time (seconds)' and the y-axis as 'Output'. Add a title 'Second-order System Step Response'.

Simulink

wn = 5;
zeta = 0.7;
sys = tf([wn^2], [1, 2*zeta*wn, wn^2]);
t = 0:0.01:5;
u = ones(size(t));
y = lsim(sys, u, t);
# Plot y vs t with labels and title

Hint

Use plot, xlabel, ylabel, and title functions to create the graph.

Practice

1. What does the damping ratio (ζ) affect in a second-order system simulation in Simulink?

easy

A. The color of the simulation plot

B. The speed and overshoot of the system's response

C. The size of the Simulink blocks

D. The input signal amplitude

Solution

1. Step 1: Understand damping ratio role

   The damping ratio controls how oscillations decay in the system response.

2. Step 2: Connect damping ratio to response behavior

   A higher damping ratio reduces overshoot and speeds settling time, affecting speed and overshoot.

3. Final Answer:

   The speed and overshoot of the system's response -> Option B

4. Quick Check:

   Damping ratio = speed and overshoot control [OK]

Hint: Damping ratio controls oscillation and overshoot [OK]

Common Mistakes:

• Confusing damping ratio with input amplitude
• Thinking damping ratio changes block size
• Assuming damping ratio affects plot color

2. Which of the following is the correct way to represent a second-order transfer function in Simulink's Transfer Function block?

easy

A. Numerator: [1 0 0], Denominator: [1 2*zeta*wn wn]

B. Numerator: [1 2*zeta*wn wn^2], Denominator: [1]

C. Numerator: [1], Denominator: [1 2*zeta*wn wn^2]

D. Numerator: [wn^2], Denominator: [1 2*zeta*wn 1]

Solution

1. Step 1: Recall second-order transfer function form

   The standard form is numerator = [1], denominator = [1 2*zeta*wn wn^2].

2. Step 2: Match options to standard form

   Numerator: [1], Denominator: [1 2*zeta*wn wn^2] matches the correct numerator and denominator coefficients for a second-order system.

3. Final Answer:

   Numerator: [1], Denominator: [1 2*zeta*wn wn^2] -> Option C

4. Quick Check:

   Standard form = Numerator: [1], Denominator: [1 2*zeta*wn wn^2] [OK]

Hint: Denominator has 1, 2*zeta*wn, wn squared [OK]

Common Mistakes:

• Swapping numerator and denominator
• Using wrong powers for wn
• Missing damping term 2*zeta*wn

3. Given the Simulink transfer function block with numerator [1] and denominator [1 4 4], what is the natural frequency (wn) and damping ratio (ζ) of the system?

medium

A. wn=2, ζ=1

B. wn=4, ζ=2

C. wn=1, ζ=4

D. wn=2, ζ=0.5

Solution

1. Step 1: Identify denominator coefficients

   Denominator is [1 4 4], which matches [1 2*zeta*wn wn^2].

2. Step 2: Calculate wn and ζ from coefficients

   wn^2 = 4 -> wn = 2; 2*zeta*wn = 4 -> ζ = 4/(2*2) = 1.

3. Final Answer:

   wn=2, ζ=1 -> Option A

4. Quick Check:

   wn=2, ζ=1 from denominator [OK]

Hint: wn is sqrt of last denominator term [OK]

Common Mistakes:

• Confusing wn^2 with wn
• Dividing damping term incorrectly
• Ignoring coefficient order

4. You created a second-order system in Simulink with numerator [1] and denominator [1 0 4], but the simulation shows no oscillations. What is the likely error?

medium

A. The damping ratio is zero, causing no oscillations

B. The input signal is missing

C. The numerator should be [1 0 4]

D. The denominator coefficients are incorrect for oscillations

Solution

1. Step 1: Analyze denominator coefficients

   Denominator [1 0 4] means 2*zeta*wn = 0, so damping ratio ζ = 0.

2. Step 2: Understand effect on oscillations

   Zero damping ratio means undamped oscillations, but the system denominator form is incorrect for natural frequency 2 (since wn^2=4).

3. Step 3: Identify correct denominator form

   Correct denominator should be [1 0 4] only if system is undamped, but zero middle term can cause simulation issues.

4. Final Answer:

   The denominator coefficients are incorrect for oscillations -> Option D

5. Quick Check:

   Incorrect denominator = no oscillations [OK]

Hint: Check denominator matches 1, 2*zeta*wn, wn^2 [OK]

Common Mistakes:

• Assuming zero damping means no oscillations
• Swapping numerator and denominator
• Ignoring input signal presence

5. You want to simulate a second-order system with a natural frequency of 5 rad/s and a damping ratio of 0.3 in Simulink. What should be the denominator coefficients for the Transfer Function block?

hard

A. [1 3 25]

B. [1 3 5]

C. [1 3 15]

D. [1 3 30]

Solution

1. Step 1: Calculate denominator coefficients

   Given wn=5, ζ=0.3, calculate 2*ζ*wn = 2*0.3*5 = 3 and wn^2 = 25.

2. Step 2: Form denominator array

   Denominator coefficients are [1, 3, 25].

3. Final Answer:

   [1 3 25] -> Option A

4. Quick Check:

   2*zeta*wn=3, wn^2=25 [OK]

Hint: Multiply 2*damping*wn and square wn for denominator [OK]

Common Mistakes:

• Using wn instead of wn squared
• Mixing damping ratio with wn
• Incorrect multiplication order