Second-order system simulation in Simulink - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a second-order system step response

Given a second-order system with transfer function H(s) = \(\frac{1}{s^2 + 2\zeta\omega_ns + \omega_n^2}\), where \(\zeta = 0.5\) and \(\omega_n = 4\), what is the peak value of the step response in Simulink?

Simulink

Create a Transfer Function block with numerator [1] and denominator [1 4 16]. Connect a Step block with step time 0 and step size 1. Connect the output to a Scope block. Run the simulation and observe the peak value.

A Approximately 0.5

B Exactly 1.0

C Approximately 2.0

D Approximately 1.16

💡 Hint

Recall that the peak overshoot depends on the damping ratio \(\zeta\).

For a damping ratio of 0.5, the system exhibits an overshoot. The peak value is about 1.16 times the final value.

❓ data_output

intermediate

Settling time from simulation data

You simulate a second-order system with \(\zeta = 0.7\) and \(\omega_n = 5\). The step response data is recorded. What is the approximate settling time (time to stay within 2% of final value)?

Simulink

Use a Step block and Transfer Function block with numerator [1] and denominator [1 7 25]. Run simulation for 2 seconds and log output data. Analyze the time when output remains within 0.98 to 1.02.

A About 1.14 seconds

B About 0.4 seconds

C About 2.5 seconds

D About 0.7 seconds

💡 Hint

Settling time for 2% criterion is approximately \(\frac{4}{\zeta \omega_n}\).

Using formula, settling time = 4 / (0.7 * 5) ≈ 1.14 seconds.

❓ visualization

advanced

Plotting the impulse response of a second-order system

You simulate the impulse response of a second-order system with \(\zeta = 0.3\) and \(\omega_n = 3\). Which plot correctly shows the expected oscillatory decay?

Simulink

Use an Impulse block connected to a Transfer Function block with numerator [1] and denominator [1 1.8 9]. Run simulation and plot output versus time.

A A plot showing oscillations that decrease in amplitude over time

B A plot showing a flat line at zero

C A plot showing a monotonic rise to a steady value without oscillations

D A plot showing increasing oscillations over time

💡 Hint

Low damping ratio causes oscillations that fade out.

With \(\zeta = 0.3\), the system is underdamped, so the impulse response oscillates and decays.

🔧 Debug

advanced

Identify the error in the Simulink model for a second-order system

A user models a second-order system with transfer function \(\frac{1}{s^2 + 2s + 1}\) but the simulation output is always zero. The model uses a Transfer Function block with numerator [1] and denominator [1 2]. What is the error?

A Step block is missing from the input

B Numerator should be [0 1] instead of [1]

C Denominator coefficients are incomplete; should be [1 2 1]

D Simulation time is set to zero

💡 Hint

Check the order of the denominator polynomial coefficients.

The denominator must have three coefficients for a second-order system. Using only two causes the system to be invalid and output zero.

🚀 Application

expert

Design a Simulink model to compare overdamped and underdamped responses

You want to compare step responses of two second-order systems: one overdamped with \(\zeta=1.5\), \(\omega_n=3\), and one underdamped with \(\zeta=0.5\), \(\omega_n=3\). Which Simulink setup correctly allows this comparison?

Simulink

Create two Transfer Function blocks with denominators [1 9 9] and [1 3 9] respectively. Connect both to Step blocks with same parameters. Use two Scope blocks to view outputs.

A Use one Transfer Function block with denominator [1 6 9] and change \(\zeta\) dynamically during simulation

B Use two Transfer Function blocks with denominators [1 9 9] and [1 3 9], connect both to Step blocks, and view outputs on separate Scopes

C Use two Transfer Function blocks with numerators [1 0] and [1], connect to Step blocks, and view outputs on one Scope

D Use one Transfer Function block with denominator [1 4.5 9] and simulate twice with different initial conditions

💡 Hint

Overdamped system has \(\zeta > 1\), underdamped has \(\zeta < 1\). Denominator coefficients depend on \(\zeta\) and \(\omega_n\).

Denominator coefficients are \([1, 2\zeta\omega_n, \omega_n^2]\). For \(\zeta=1.5\), denominator is [1 9 9]. For \(\zeta=0.5\), denominator is [1 3 9]. Separate blocks and scopes allow clear comparison.

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