Discrete transfer function in Simulink - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a discrete transfer function with numerator coefficients.

Simulink

sys = tf([1], [1 0.5 0.25], 0.1);

Drag options to blanks, or click blank then click option'

A[1 0.5]

B[1 0.2]

C[0.1 0.2]

D[0.5 0.25]

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing numerator and denominator coefficients.

Using scalar instead of vector for numerator.

The numerator coefficients define the zeros of the transfer function. Here, [0.1 0.2] is the correct numerator vector.

2fill in blank

medium

Complete the code to specify the sample time for the discrete transfer function.

Simulink

sys = tf([0.1 0.2], [1 0.5 0.25], [1]);

Drag options to blanks, or click blank then click option'

A1

B0

C-1

D0.1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 which means continuous time system.

Using negative sample time which is invalid.

The sample time for a discrete transfer function must be a positive number. 0.1 seconds is a typical sample time.

3fill in blank

hard

Fix the error in the code to correctly create a discrete transfer function with sample time 0.05.

Simulink

sys = tf([0.3 0.4], [1 [1] 0.2], 0.05);

Drag options to blanks, or click blank then click option'

A0.6

B0.5

C0.4

D0.3

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong coefficient causing unstable system.

Mismatching numerator and denominator lengths.

The denominator coefficients must be correctly specified. 0.6 fits the polynomial for the denominator.

4fill in blank

hard

Fill both blanks to create a discrete transfer function with numerator and denominator coefficients and sample time.

Simulink

sys = tf([1], [2], 0.2);

Drag options to blanks, or click blank then click option'

A[0.2 0.1]

B[1 0.4 0.3]

C[1 0.5 0.25]

D[0.3 0.2]

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping numerator and denominator coefficients.

Using incorrect vector lengths.

The numerator is [0.2 0.1] and the denominator is [1 0.5 0.25] for the discrete transfer function.

5fill in blank

hard

Fill all three blanks to create a discrete transfer function with numerator, denominator, and sample time.

Simulink

sys = tf([1], [2], [3]);

Drag options to blanks, or click blank then click option'

A[0.4 0.3]

B0.15

C[1 0.6 0.35]

D[0.5 0.25]

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up numerator and denominator vectors.

Using zero or negative sample time.

The numerator is [0.5 0.25], denominator is [1 0.6 0.35], and sample time is 0.15 seconds.

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