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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a First-Order System Step Response

Consider a first-order system with transfer function H(s) = 1 / (s + 2). You apply a unit step input at time t=0. What is the output value at t = 1 second?

Simulink

sys = tf([1], [1, 2]);
[y, t] = step(sys, 1);
y_end = y(end);

A 0.865

B 0.500

C 0.632

D 1.000

💡 Hint

Recall the step response of a first-order system is 1 - exp(-a*t), where a is the coefficient in the denominator.

The system has a time constant τ = 1/2 = 0.5 seconds. The step response at t=1 is 1 - exp(-2*1) ≈ 0.865.

❓ data_output

intermediate

Number of Data Points in Simulation Output

You simulate a first-order system with a time vector from 0 to 5 seconds with 0.1 second intervals. How many data points will the output vector contain?

Simulink

t = 0:0.1:5;
num_points = length(t);

A 55

B 51

C 52

D 50

💡 Hint

Count how many points are in the range including both start and end.

The vector from 0 to 5 with step 0.1 includes both ends, so number of points = (5 - 0)/0.1 + 1 = 51.

❓ visualization

advanced

Plotting the Step Response of a First-Order System

Which option produces the correct plot of the step response for the system H(s) = 5 / (s + 5) over 0 to 2 seconds?

Simulink

sys = tf([5], [1, 5]);
step(sys, 0:0.01:2);

A A plot starting at 0 and rising smoothly to 5 over 2 seconds

B A plot starting at 0 and rising smoothly to 0.993 over 2 seconds

C A plot starting at 0 and rising smoothly to 0.632 over 2 seconds

D A plot starting at 0 and rising smoothly to 1 over 2 seconds

💡 Hint

Calculate the steady-state value and the time constant.

The steady-state value is the DC gain = 5/5 = 1. The step response reaches 1 - exp(-5*2) ≈ 1 at t=2.

🔧 Debug

advanced

Identify the Error in Simulation Code

What error does the following Simulink MATLAB code produce?

sys = tf([1], [1 0]);
[y, t] = step(sys, 0:0.1:5);

Simulink

sys = tf([1], [1 0]);
[y, t] = step(sys, 0:0.1:5);

A Pole at zero error (system is not stable)

B ZeroDivisionError

C No error, runs successfully

D SyntaxError

💡 Hint

Check the denominator polynomial and system stability.

The denominator [1 0] means the system has a pole at zero, making it unstable and step response undefined.

🚀 Application

expert

Calculate Time Constant from Simulation Data

You have simulation output data of a first-order system step response. The output reaches 63.2% of its final value at 0.8 seconds. What is the system's time constant?

A 1.0 seconds

B 1.25 seconds

C 0.8 seconds

D 0.632 seconds

💡 Hint

The time constant is the time when output reaches 63.2% of final value.

By definition, the time constant τ is the time at which the step response reaches 63.2% of its final value, so τ = 0.8 seconds.

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