Discrete transfer function in Simulink - Mini Project: Build & Apply

Discrete transfer function

📖 Scenario: You are working on a simple digital control system. You want to model how the system responds to inputs over time using a discrete transfer function. This helps you understand how the system behaves step-by-step.

🎯 Goal: Build a discrete transfer function model in MATLAB that represents a system with given numerator and denominator coefficients. Then simulate the system response to a step input.

📋 What You'll Learn

Create a discrete transfer function with specified numerator and denominator coefficients

Set the sample time for the discrete transfer function

Simulate the system response to a step input

Display the output response

💡 Why This Matters

🌍 Real World

Discrete transfer functions are used in digital control systems such as robotics, automotive control, and signal processing to model system behavior over time.

💼 Career

Understanding discrete transfer functions is important for control engineers and data scientists working with time-series data and digital system simulations.

1

Create numerator and denominator coefficient vectors

Create two variables in the MATLAB workspace: num as the numerator coefficients vector [0.5 0.3] and den as the denominator coefficients vector [1 -0.4 0.12].

Simulink

% Create the numerator and denominator coefficient vectors
% Your code here

Hint

Use square brackets to create vectors. For example, num = [0.5 0.3].

2

Set the sample time for the discrete transfer function

Create a variable called Ts and set it to 0.1 to represent the sample time in seconds.

Simulink

num = [0.5 0.3];
den = [1 -0.4 0.12];
% Set the sample time variable Ts
% Your code here

Hint

Sample time is a single number representing seconds between samples.

3

Create the discrete transfer function system

Use the tf function to create a discrete transfer function called sys using the numerator num, denominator den, and sample time Ts. Use the syntax sys = tf(num, den, Ts).

Simulink

num = [0.5 0.3];
den = [1 -0.4 0.12];
Ts = 0.1;
% Create the discrete transfer function sys
% Your code here

Hint

The tf function creates transfer functions. The third argument is the sample time.

4

Simulate and plot the step response

Use the step function to simulate and plot the step response of the discrete transfer function sys. Write step(sys).

Simulink

num = [0.5 0.3];
den = [1 -0.4 0.12];
Ts = 0.1;
sys = tf(num, den, Ts);
% Plot the step response of sys
% Your code here

Hint

The step function shows how the system output changes over time when the input is a step.

Practice

1. What does a discrete transfer function in Simulink represent?

easy

A. The color scheme of the Simulink blocks

B. The hardware configuration of the computer

C. The relationship between input and output in a step-by-step system

D. The file format used to save Simulink models

Solution

1. Step 1: Understand the purpose of discrete transfer functions

   A discrete transfer function models how the output of a system depends on its input at discrete time steps.

2. Step 2: Eliminate unrelated options

   Options about colors, hardware, or file formats do not describe system behavior.

3. Final Answer:

   The relationship between input and output in a step-by-step system -> Option C

4. Quick Check:

   Discrete transfer function = input-output relation [OK]

Hint: Think input-output step relation for discrete systems [OK]

Common Mistakes:

• Confusing transfer function with hardware settings
• Thinking it describes visual aspects
• Mixing file formats with system models

2. Which of the following is the correct way to define a discrete transfer function in Simulink?

easy

A. Using numerator and denominator coefficient vectors with a sample time

B. Using only a numerator vector without denominator

C. Using continuous time parameters without sample time

D. Using a single scalar value without vectors

Solution

1. Step 1: Recall discrete transfer function definition

   It requires numerator and denominator coefficient vectors plus a sample time to define the discrete system.

2. Step 2: Check options for completeness

   Only Using numerator and denominator coefficient vectors with a sample time includes numerator, denominator, and sample time, which is necessary.

3. Final Answer:

   Using numerator and denominator coefficient vectors with a sample time -> Option A

4. Quick Check:

   Numerator + denominator + sample time = correct syntax [OK]

Hint: Remember: numerator, denominator, and sample time needed [OK]

Common Mistakes:

• Omitting denominator coefficients
• Using continuous parameters for discrete blocks
• Using single scalar instead of vectors

3. Given a discrete transfer function with numerator = [0.5 0.5] and denominator = [1 -0.3], and sample time = 1, what is the output at step 1 if the input is 1 at step 0 and 0 afterwards?

medium

A. 0

B. 0.5

C. 0.65

D. 0.15

Solution

1. Step 1: Calculate output at step 1 using difference equation

   Output y[1] = 0.5*u[1] + 0.5*u[0] - (-0.3)*y[0]. Given u[0]=1, u[1]=0, y[0]=0 (initially), y[1] = 0.5*0 + 0.5*1 + 0.3*0 = 0.5.

2. Step 2: Confirm initial conditions and input values

   Initial output y[0] is zero, input at step 1 is zero, so calculation is correct.

3. Final Answer:

   0.5 -> Option B

4. Quick Check:

   y[1] = 0.5*0 + 0.5*1 + 0.3*0 = 0.5 [OK]

Hint: Use difference equation with initial zero output [OK]

Common Mistakes:

• Ignoring initial output conditions
• Using wrong input values for step 1
• Mixing numerator and denominator signs

4. You defined a discrete transfer function with numerator = [1 0] and denominator = [1 -0.5], sample time = 1. The output does not change when input changes. What is the likely error?

medium

A. Denominator coefficients are incorrect length

B. Sample time is set to zero

C. Numerator coefficients do not represent the input properly

D. Input signal is not connected

Solution

1. Step 1: Validate the coefficients

   Numerator [1 0] and denominator [1 -0.5] yield y[n] = u[n] + 0.5 y[n-1], which responds to input changes.

2. Step 2: Identify the likely cause

   In Simulink, if output ignores input despite valid parameters, the input signal is probably not connected to the block.

3. Final Answer:

   Input signal is not connected -> Option D

4. Quick Check:

   No response? Verify input connection [OK]

Hint: If output unresponsive, check input connection first [OK]

Common Mistakes:

• Ignoring numerator vector meaning
• Assuming sample time zero is valid
• Not verifying input connection

5. You want to design a discrete transfer function in Simulink to model a system with a delay of 2 steps and a gain of 0.8. Which numerator and denominator coefficients correctly represent this?

hard

A. Numerator = [0 0 0.8], Denominator = [1]

B. Numerator = [0.8 0 0], Denominator = [1 -0.2]

C. Numerator = [0 0.8 0], Denominator = [1 0]

D. Numerator = [0 0 1], Denominator = [0.8]

Solution

1. Step 1: Understand delay and gain in numerator

   A delay of 2 steps means the gain appears at the third coefficient (index 2), so numerator should be [0 0 0.8].

2. Step 2: Denominator for no feedback is [1]

   Since no feedback is mentioned, denominator is simply [1]. Other options have incorrect placements or values.

3. Final Answer:

   Numerator = [0 0 0.8], Denominator = [1] -> Option A

4. Quick Check:

   Delay = zeros before gain in numerator [OK]

Hint: Delay = zeros before gain in numerator vector [OK]

Common Mistakes:

• Placing gain at wrong numerator index
• Misunderstanding denominator role
• Confusing gain and delay positions