Discrete transfer function in Simulink - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a discrete transfer function in Simulink?

A discrete transfer function in Simulink represents a system's input-output relationship using difference equations in discrete time steps, often used to model digital control systems.

beginner

How do you specify the sample time for a discrete transfer function in Simulink?

You specify the sample time by setting the 'Sample time' parameter in the block dialog, which defines the time interval between each discrete step.

intermediate

What is the difference between continuous and discrete transfer functions?

Continuous transfer functions use Laplace transform and model systems in continuous time, while discrete transfer functions use Z-transform and model systems at discrete time intervals.

intermediate

In Simulink, what do the numerator and denominator coefficients represent in a discrete transfer function block?

They represent the coefficients of the polynomials in the Z-domain that define the system's behavior, with numerator for zeros and denominator for poles.

intermediate

Why is it important to choose the correct sample time for a discrete transfer function?

Choosing the correct sample time ensures accurate system simulation and stability; too large can miss dynamics, too small can increase computation time.

What does the sample time parameter in a discrete transfer function block control?

A The time interval between discrete steps

B The amplitude of the output signal

C The initial condition of the system

D The continuous time delay

The sample time sets how often the system updates in discrete time steps.

Which transform is used to analyze discrete transfer functions?

A Laplace transform

B Z-transform

C Fourier transform

D Wavelet transform

Discrete transfer functions use the Z-transform to represent systems in discrete time.

In Simulink, what do the numerator coefficients of a discrete transfer function represent?

A Zeros of the system

B Poles of the system

C Sample time

D Initial conditions

Numerator coefficients correspond to the zeros in the system's transfer function.

What happens if the sample time is set too large in a discrete transfer function?

A The system runs faster

B The output amplitude increases

C The system may miss important dynamics

D The system becomes continuous

A large sample time can cause the simulation to miss fast changes in the system.

Which block in Simulink is used to model discrete transfer functions?

A Scope block

B Integrator block

C Gain block

D Transfer Function block with sample time set

The Transfer Function block with a specified sample time models discrete transfer functions.

Explain what a discrete transfer function is and how it differs from a continuous transfer function.

Think about how time is handled differently in discrete vs continuous systems.

Describe how you would set up a discrete transfer function in Simulink and why choosing the right sample time matters.

Consider the block parameters and simulation effects.

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