First-order system simulation in Simulink - Time & Space Complexity
We want to understand how the time to run a first-order system simulation changes as we increase the simulation length or resolution.
How does the simulation time grow when we use more steps or finer detail?
Analyze the time complexity of the following Simulink model setup.
% Simulink first-order system simulation setup
sys = tf([1],[1 1]); % First-order transfer function
simTime = 10; % total simulation time
stepSize = 0.01; % simulation step size
simOut = sim('first_order_model', 'StopTime', num2str(simTime));
This code simulates a first-order system over a fixed time with a set step size controlling detail.
In this simulation, the main repeating operation is the solver stepping through time.
- Primary operation: Calculating system state at each time step.
- How many times: Number of steps = total simulation time divided by step size.
As we increase the total simulation time or decrease the step size, the number of steps grows, so the simulation takes longer.
| Input Size (number of steps) | Approx. Operations |
|---|---|
| 1000 (10s / 0.01s) | 1000 calculations |
| 10000 (100s / 0.01s) | 10000 calculations |
| 100000 (100s / 0.001s) | 100000 calculations |
Pattern observation: The operations grow directly with the number of steps; doubling steps doubles work.
Time Complexity: O(n)
This means the simulation time grows linearly with the number of time steps taken.
[X] Wrong: "The simulation time stays the same no matter how small the step size is."
[OK] Correct: Smaller step sizes mean more steps, so the solver does more calculations and takes longer.
Understanding how simulation time grows helps you design efficient models and explain performance in real projects.
"What if we switch from a fixed-step solver to a variable-step solver? How would the time complexity change?"
Practice
Solution
Step 1: Understand the purpose of first-order system simulation
It models how the output of a simple system changes over time when given an input.Step 2: Identify what simulation shows in Simulink
Simulink simulates the system response graphically, showing output behavior over time.Final Answer:
How the system output changes over time after an input -> Option AQuick Check:
Simulation = output change over time [OK]
- Confusing simulation with code details
- Thinking simulation shows hardware specs
- Assuming simulation gives exact analytic solutions
Solution
Step 1: Recall first-order transfer function form
A first-order system has denominator of form s + a, so denominator coefficients are [1 a].Step 2: Match numerator and denominator arrays
Numerator is usually constant [1], denominator [1 5] means s + 5, correct for first-order.Final Answer:
Numerator: [1], Denominator: [1 5] -> Option CQuick Check:
Denominator = [1 a] for first-order [OK]
- Swapping numerator and denominator
- Using wrong order of coefficients
- Using zero in denominator leading coefficient
Solution
Step 1: Calculate steady-state gain from transfer function
Steady-state gain = numerator constant / denominator constant = 2 / 3 ≈ 0.67.Step 2: Apply step input magnitude
Step input is 1, so output steady-state = 0.67 * 1 = 0.67.Final Answer:
0.67 -> Option BQuick Check:
Gain = 2/3 = 0.67 [OK]
- Using numerator over denominator reversed
- Ignoring step input magnitude
- Confusing transient with steady-state
Solution
Step 1: Analyze denominator [1 0]
Denominator [1 0] means s + 0 = s, which is an integrator, not a first-order system with constant term.Step 2: Understand simulation behavior
Integrator causes output to ramp, but if input is zero or setup wrong, no visible response occurs.Final Answer:
Denominator [1 0] represents an integrator, not a first-order system -> Option AQuick Check:
Denominator must have constant term for first-order [OK]
- Assuming zero constant term is valid
- Ignoring block connections
- Forgetting to set step input
Solution
Step 1: Write transfer function formula for first-order system
Transfer function = gain / (time_constant * s + 1) = 3 / (4s + 1).Step 2: Convert to Simulink coefficients
Numerator [3], Denominator [1 4] for s + 4.Final Answer:
Numerator: [3], Denominator: [1 4] -> Option DQuick Check:
Denominator = [1 tau] for first-order [OK]
- Swapping numerator and denominator
- Using time constant as denominator constant
- Not normalizing denominator leading coefficient to 1
Ready to prove it?
Pass the test and this pattern moves unknown
