Step response analysis in Simulink - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Step Response Peak Time Calculation

Given a second-order system with transfer function \( G(s) = \frac{100}{s^2 + 12s + 100} \), what is the peak time \( T_p \) observed in the step response?

Simulink

num = [100];
den = [1 12 100];
sys = tf(num, den);
info = stepinfo(sys);
peak_time = info.PeakTime;

A 0.4 seconds

B 0.5 seconds

C 0.25 seconds

D 0.1 seconds

💡 Hint

Peak time is the time at which the response reaches its first maximum peak.

Using the stepinfo function in Simulink or MATLAB, the peak time for this system is approximately 0.4 seconds.

❓ data_output

intermediate

Settling Time from Step Response Data

You have step response data from a system. Which of the following MATLAB commands correctly extracts the settling time from the step response info structure?

Simulink

num = [1];
den = [1 3 2];
sys = tf(num, den);
info = stepinfo(sys);

A settling_time = info.Settling_time;

B settling_time = info.SettlingTime;

C settling_time = info.settlingtime;

D settling_time = info.settlingTime;

💡 Hint

MATLAB is case-sensitive and uses exact field names in structures.

The correct field name in the stepinfo output is 'SettlingTime' with capital S and T.

❓ visualization

advanced

Interpreting Step Response Plot Characteristics

You run a step response simulation in Simulink and observe the plot below. Which characteristic can you NOT determine directly from this plot?

A Overshoot percentage

B Rise time

C Steady-state error

D System order

💡 Hint

Think about what information a step response plot shows and what it does not.

The step response plot shows time-domain behavior like rise time, overshoot, and steady-state error, but it does not reveal the system order directly.

🧠 Conceptual

advanced

Effect of Damping Ratio on Step Response

How does increasing the damping ratio \( \zeta \) of a second-order system affect its step response?

A Overshoot increases and settling time increases

B Overshoot increases and settling time decreases

C Overshoot decreases and settling time increases

D Overshoot decreases and settling time decreases

💡 Hint

Recall how damping affects oscillations and speed of settling.

Higher damping reduces overshoot but typically causes the system to settle more slowly, increasing settling time.

🔧 Debug

expert

Identifying Error in Step Response Code

What error will the following Simulink MATLAB code produce when run? num = [10]; den = [1 2]; sys = tf(num, den); info = stepinfo(sys); peak_time = info.PeakTime; disp(peak_time);

Simulink

num = [10];
den = [1 2];
sys = tf(num, den);
info = stepinfo(sys);
peak_time = info.PeakTime;
disp(peak_time);

A No error, outputs peak time correctly

B Runtime error: PeakTime field does not exist

C Syntax error: missing semicolon

D Runtime error: stepinfo requires a stable system

💡 Hint

Check the system stability and stepinfo requirements.

The code produces no error. The first-order stable system has no overshoot, so stepinfo returns NaN for PeakTime, and disp(peak_time) outputs 'NaN'.

Practice

1. What does the step function in Simulink primarily show about a system?

easy

A. The system's output response to a sudden input change

B. The frequency response of the system

C. The system's steady-state error only

D. The system's input signal over time

Solution

1. Step 1: Understand the purpose of the step function

   The step function simulates how a system reacts when a sudden input (step input) is applied.

2. Step 2: Identify what the output represents

   The output shows how the system's output changes over time after the step input, revealing speed and stability.

3. Final Answer:

   The system's output response to a sudden input change -> Option A

4. Quick Check:

   Step response = output after sudden input [OK]

Hint: Step function shows output after sudden input change [OK]

Common Mistakes:

• Confusing step response with frequency response
• Thinking it shows only steady-state error
• Assuming it plots input signal instead of output

2. Which of the following is the correct syntax to plot the step response of a transfer function sys in MATLAB?

easy

A. response.step(sys)

B. plotstep(sys)

C. sys.step()

D. step(sys)

Solution

1. Step 1: Recall MATLAB step response syntax

   The standard MATLAB command to plot step response is step(sys), where sys is the system model.

2. Step 2: Check other options for correctness

   Options B, C, and D are not valid MATLAB commands for step response plotting.

3. Final Answer:

   step(sys) -> Option D

4. Quick Check:

   Use step(sys) to plot response [OK]

Hint: Use step(sys) to plot step response in MATLAB [OK]

Common Mistakes:

• Using incorrect function names like plotstep
• Trying object method calls not supported
• Confusing syntax with other toolboxes

3. Given the following MATLAB code, what is the approximate settling time of the system's step response?

sys = tf([1], [1 3 2]);
info = stepinfo(sys);
settling_time = info.SettlingTime;

medium

A. About 2 seconds

B. About 1 second

C. About 3 seconds

D. About 4 seconds

Solution

1. Step 1: Analyze the transfer function

   The system has poles at -1 and -2, so it is stable and expected to settle quickly.

2. Step 2: Use stepinfo to find settling time

   The stepinfo function calculates settling time, which for this system is approximately 2 seconds.

3. Final Answer:

   About 2 seconds -> Option A

4. Quick Check:

   Settling time from stepinfo ≈ 2 sec [OK]

Hint: Use stepinfo(sys).SettlingTime for settling time [OK]

Common Mistakes:

• Confusing settling time with rise time
• Ignoring system poles location
• Assuming default values without calculation

4. Identify the error in the following MATLAB code for plotting step response:

sys = tf([1 2], [1 4 5]);
step(sys)
title('Step Response')
xlabel('Time')
ylabel('Output')

medium

A. Missing semicolon after tf definition

B. No error; code runs correctly

C. The transfer function numerator and denominator are reversed

D. Step function should be called as step_response(sys)

Solution

1. Step 1: Check transfer function syntax

   The numerator and denominator vectors are correctly placed in tf([1 2], [1 4 5]).

2. Step 2: Verify step function and plot commands

   The step(sys) command is correct, and titles and labels are properly set.

3. Final Answer:

   No error; code runs correctly -> Option B

4. Quick Check:

   Code syntax is correct and runs [OK]

Hint: Check tf and step syntax carefully [OK]

Common Mistakes:

• Thinking numerator and denominator are swapped
• Expecting semicolon is mandatory (optional)
• Using incorrect step function names

5. You want to compare the step responses of two systems sys1 = tf([1], [1 2 1]) and sys2 = tf([2], [1 3 2]). Which MATLAB code snippet correctly plots both responses on the same graph for comparison?

hard

A. step(sys1); step(sys2);

B. plot(step(sys1), step(sys2));

C. step([sys1 sys2]);

D. step(sys1 + sys2);

Solution

1. Step 1: Understand how to plot multiple step responses

   Using step([sys1 sys2]) plots both systems' step responses on the same figure for easy comparison.

2. Step 2: Evaluate other options

   step(sys1); step(sys2); overwrites the first plot without hold on; plot(step(sys1), step(sys2)); is invalid syntax; step(sys1 + sys2); plots sum system, not comparison.

3. Final Answer:

   step([sys1 sys2]) -> Option C

4. Quick Check:

   Use step([sys1 sys2]) for combined plot [OK]

Hint: Use step([sys1 sys2]) to plot both responses together [OK]

Common Mistakes:

• Trying to add systems instead of plotting separately
• Using plot() on step outputs incorrectly
• Forgetting hold on/off when plotting separately