Step response analysis in Simulink - Time & Space Complexity

Time Complexity: Step response analysis

O(n)

Understanding Time Complexity

We want to understand how the time it takes to analyze a step response grows as the system size or simulation length increases.

How does the number of calculations change when we simulate longer or more complex systems?

Scenario Under Consideration

Analyze the time complexity of the following Simulink model snippet that computes a step response.

Step block --> Transfer Function block --> Scope

% Simulink blocks:
% Step: generates a step input signal
% Transfer Function: models system dynamics
% Scope: records output over simulation time
    

This setup simulates how a system reacts over time to a sudden input change.

Identify Repeating Operations

The simulation runs in small time steps repeatedly to calculate system output.

• Primary operation: Calculating system output at each time step.
• How many times: Once per time step, for all time steps in the simulation.

How Execution Grows With Input

As the simulation time or the number of time steps increases, the total calculations increase proportionally.

Input Size (number of time steps)	Approx. Operations
10	10 calculations
100	100 calculations
1000	1000 calculations

Pattern observation: The number of calculations grows directly with the number of time steps.

Final Time Complexity

Time Complexity: O(n)

This means the time to simulate grows linearly with the number of time steps.

Common Mistake

[X] Wrong: "The simulation time stays the same no matter how long we simulate."

[OK] Correct: Each additional time step requires extra calculations, so longer simulations take more time.

Interview Connect

Understanding how simulation time grows helps you design efficient models and explain performance in real projects.

Self-Check

What if we increased the complexity of the Transfer Function block? How would the time complexity change?

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