First-order system simulation in Simulink - Step-by-Step Execution

Concept Flow - First-order system simulation

Define system parameters

↓

Set initial conditions

↓

Apply input signal

↓

Compute system response step

↓

Update system state

↓

Check simulation time

↓

End simulation

The simulation starts by setting parameters and initial state, then repeatedly computes the system response over time until the simulation ends.

Execution Sample

Simulink

tau = 5;
K = 2;
u = 1;  % step input
x = 0;  % initial state
for t = 0:1:10
  dx = (-(x) + K * u) / tau;
  x = x + dx;
  disp([t, x])
end

This code simulates a first-order system response to a step input over 10 seconds with 1-second steps.

Execution Table

Time (t)	State (x)	Derivative (dx)	Action	Output
0	0	0.4	Calculate dx = (-(0) + 2*1)/5 = 0.4	x updated to 0 + 0.4 = 0.4
1	0.4	0.32	Calculate dx = (-(0.4) + 2*1)/5 = 0.32	x updated to 0.4 + 0.32 = 0.72
2	0.72	0.256	Calculate dx = (-(0.72) + 2*1)/5 = 0.256	x updated to 0.72 + 0.256 = 0.976
3	0.976	0.205	Calculate dx = (-(0.976) + 2*1)/5 = 0.205	x updated to 0.976 + 0.205 = 1.181
4	1.181	0.164	Calculate dx = (-(1.181) + 2*1)/5 = 0.164	x updated to 1.181 + 0.164 = 1.345
5	1.345	0.131	Calculate dx = (-(1.345) + 2*1)/5 = 0.131	x updated to 1.345 + 0.131 = 1.476
6	1.476	0.105	Calculate dx = (-(1.476) + 2*1)/5 = 0.105	x updated to 1.476 + 0.105 = 1.581
7	1.581	0.084	Calculate dx = (-(1.581) + 2*1)/5 = 0.084	x updated to 1.581 + 0.084 = 1.665
8	1.665	0.067	Calculate dx = (-(1.665) + 2*1)/5 = 0.067	x updated to 1.665 + 0.067 = 1.732
9	1.732	0.054	Calculate dx = (-(1.732) + 2*1)/5 = 0.054	x updated to 1.732 + 0.054 = 1.786
10	1.786	0.043	Calculate dx = (-(1.786) + 2*1)/5 = 0.043	x updated to 1.786 + 0.043 = 1.829

💡 Simulation ends after t=10 seconds as loop completes

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	After 8	After 9	After 10
x	0	0.4	0.72	0.976	1.181	1.345	1.476	1.581	1.665	1.732	1.786
dx	0.4	0.32	0.256	0.205	0.164	0.131	0.105	0.084	0.067	0.054	0.043

Key Moments - 3 Insights

Why does the state x not jump immediately to the steady value 2?

Because the system is first-order with time constant tau=5, it changes gradually. Each step adds a fraction of the difference, as shown in execution_table rows where x increases slowly.

Why is dx smaller at later times?

dx depends on the difference between current x and steady input effect. As x gets closer to steady state, the difference shrinks, so dx gets smaller (see execution_table dx values decreasing).

What happens if we increase the time step from 1 to 2 seconds?

The state updates would be larger jumps, possibly less accurate. The simulation steps would be fewer but each dx calculation would cover more time, changing the values in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at time t=3, what is the state x?

A 1.185

B 0.976

C 0.72

D 1.352

💡 Hint

Check the 'State (x)' column at row where 'Time (t)' is 3 in execution_table

At which time step does the derivative dx first drop below 0.1?

A 7

B 5

C 6

D 8

💡 Hint

Look at the 'Derivative (dx)' column in execution_table and find first value less than 0.1

If the input u changes from 1 to 0 at t=5, what happens to dx at t=6?

A dx becomes zero

B dx stays positive

C dx becomes negative

D dx doubles

💡 Hint

Recall dx formula depends on input u; if u=0, dx = (-x + 0)/tau, which is negative if x>0

Concept Snapshot

First-order system simulation:
- Model: dx/dt = ( -x + K*u ) / tau
- Simulate by discrete steps: x += dx * dt
- dx depends on current state and input
- State approaches steady value gradually
- Smaller time steps give smoother results

Full Transcript

This visual execution traces a first-order system simulation in Simulink style. We start by defining parameters tau=5, gain K=2, and a step input u=1. The initial state x is zero. At each time step from 0 to 10 seconds, we calculate the derivative dx as ( -x + K*u ) / tau, then update x by adding dx. The execution table shows how x and dx change over time, with x gradually increasing towards the steady state value of 2. Key moments clarify why x changes slowly and why dx decreases as x approaches steady state. The visual quiz tests understanding of state values at specific times and effects of input changes. The snapshot summarizes the simulation approach and key rules for first-order systems.

Practice

1. What does a first-order system simulation in Simulink typically show?

easy

A. How the system output changes over time after an input

B. The exact solution of differential equations analytically

C. The internal code of Simulink blocks

D. The hardware specifications of the computer running Simulink

Solution

1. 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.

2. Step 2: Identify what simulation shows in Simulink

   Simulink simulates the system response graphically, showing output behavior over time.

3. Final Answer:

   How the system output changes over time after an input -> Option A

4. Quick Check:

   Simulation = output change over time [OK]

Hint: Simulation shows output change over time [OK]

Common Mistakes:

• Confusing simulation with code details
• Thinking simulation shows hardware specs
• Assuming simulation gives exact analytic solutions

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

easy

A. Numerator: [5], Denominator: [1 0]

B. Numerator: [1 5], Denominator: [1]

C. Numerator: [1], Denominator: [1 5]

D. Numerator: [1], Denominator: [5 1]

Solution

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

   A first-order system has denominator of form s + a, so denominator coefficients are [1 a].

2. Step 2: Match numerator and denominator arrays

   Numerator is usually constant [1], denominator [1 5] means s + 5, correct for first-order.

3. Final Answer:

   Numerator: [1], Denominator: [1 5] -> Option C

4. Quick Check:

   Denominator = [1 a] for first-order [OK]

Hint: Denominator starts with 1 for s term [OK]

Common Mistakes:

• Swapping numerator and denominator
• Using wrong order of coefficients
• Using zero in denominator leading coefficient

3. Given the Simulink Transfer Function block with Numerator = [2] and Denominator = [1 3], and a step input of magnitude 1, what is the steady-state output value?

medium

A. 2

B. 0.67

C. 0.5

D. 3

Solution

1. Step 1: Calculate steady-state gain from transfer function

   Steady-state gain = numerator constant / denominator constant = 2 / 3 ≈ 0.67.

2. Step 2: Apply step input magnitude

   Step input is 1, so output steady-state = 0.67 * 1 = 0.67.

3. Final Answer:

   0.67 -> Option B

4. Quick Check:

   Gain = 2/3 = 0.67 [OK]

Hint: Steady-state = numerator/denominator constant term [OK]

Common Mistakes:

• Using numerator over denominator reversed
• Ignoring step input magnitude
• Confusing transient with steady-state

4. You created a first-order system in Simulink with Numerator = [1] and Denominator = [1 0], but the simulation shows no response. What is the likely error?

medium

A. Denominator [1 0] represents an integrator, not a first-order system

B. Numerator should be zero for first-order systems

C. Step input magnitude is missing

D. Scope block is not connected

Solution

1. 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.

2. Step 2: Understand simulation behavior

   Integrator causes output to ramp, but if input is zero or setup wrong, no visible response occurs.

3. Final Answer:

   Denominator [1 0] represents an integrator, not a first-order system -> Option A

4. Quick Check:

   Denominator must have constant term for first-order [OK]

Hint: Denominator must have constant term, not zero [OK]

Common Mistakes:

• Assuming zero constant term is valid
• Ignoring block connections
• Forgetting to set step input

5. You want to simulate a first-order system with time constant 4 seconds and gain 3 in Simulink. Which numerator and denominator coefficients should you use in the Transfer Function block?

hard

A. Numerator: [1], Denominator: [1 0.25]

B. Numerator: [3], Denominator: [4 1]

C. Numerator: [3], Denominator: [1 0.25]

D. Numerator: [3], Denominator: [1 4]

Solution

1. Step 1: Write transfer function formula for first-order system

   Transfer function = gain / (time_constant * s + 1) = 3 / (4s + 1).

2. Step 2: Convert to Simulink coefficients

   Numerator [3], Denominator [1 4] for s + 4.

3. Final Answer:

   Numerator: [3], Denominator: [1 4] -> Option D

4. Quick Check:

   Denominator = [1 tau] for first-order [OK]

Hint: Denominator coefficients = [1 tau] [OK]

Common Mistakes:

• Swapping numerator and denominator
• Using time constant as denominator constant
• Not normalizing denominator leading coefficient to 1