Discrete transfer function in Simulink - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Discrete Transfer Function Block Simulation

Given a discrete transfer function block in Simulink with numerator coefficients [0.5, -0.3] and denominator coefficients [1, -0.8], and a unit step input, what is the output value at the 3rd sample (index 2)? Assume zero initial conditions and a sample time of 1 second.

Simulink

Numerator = [0.5, -0.3];
Denominator = [1, -0.8];
Input = [1, 1, 1, 1, 1];
Output = filter(Numerator, Denominator, Input);
Output(3)

A 0.68

B 0.25

C 0.42

D 0.58

💡 Hint

Recall that the output is computed by filtering the input through the transfer function coefficients.

The output at the 3rd sample is calculated by applying the difference equation defined by the numerator and denominator coefficients to the input sequence. Using the filter function, the value at index 2 is 0.68.

🧠 Conceptual

intermediate

Understanding Poles of a Discrete Transfer Function

Which of the following statements correctly describes the location of poles for a stable discrete transfer function?

A All poles must lie inside the unit circle in the z-plane.

B All poles must lie outside the unit circle in the z-plane.

C Poles can lie anywhere on the real axis for stability.

D Poles must lie exactly on the unit circle for stability.

💡 Hint

Think about how discrete system stability relates to the z-plane.

For a discrete system to be stable, all poles must be inside the unit circle in the z-plane. Poles outside or on the unit circle indicate instability or marginal stability.

🔧 Debug

advanced

Identify the Error in Discrete Transfer Function Definition

A user defines a discrete transfer function in Simulink with numerator [1, -0.5] and denominator [1, 1.2]. After simulation, the output grows without bound. What is the most likely cause?

A The sample time is set to zero, causing simulation errors.

B The numerator polynomial is incorrect and causes output overflow.

C The denominator polynomial has a pole outside the unit circle causing instability.

D Initial conditions are not set to zero, causing output drift.

💡 Hint

Check the denominator roots and their location in the z-plane.

The denominator polynomial roots determine the poles. A pole with magnitude greater than 1 causes the output to grow unbounded, indicating instability.

❓ data_output

advanced

Calculate Output Sequence from Discrete Transfer Function

Given a discrete transfer function with numerator [0.2, 0.1] and denominator [1, -0.5], and input sequence [1, 2, 3, 4], what is the output sequence?

Simulink

num = [0.2, 0.1];
den = [1, -0.5];
input_seq = [1, 2, 3, 4];
output_seq = filter(num, den, input_seq);

A [0.2, 0.6, 1.2, 2.0]

B [0.2, 0.4, 0.6, 0.8]

C [0.2, 0.3, 0.4, 0.5]

D [0.2, 0.6, 1.1, 1.65]

💡 Hint

Use the filter function to compute the output sequence.

Applying the filter with given numerator and denominator on the input sequence yields the output [0.2, 0.6, 1.1, 1.65].

🚀 Application

expert

Designing a Discrete Transfer Function for a Given Response

You want to design a discrete transfer function in Simulink that has a zero at z=0.5 and poles at z=0.3 and z=0.4. Which numerator and denominator coefficient sets correctly represent this system?

A Numerator: [1, -0.5], Denominator: [1, 0.7, -0.12]

B Numerator: [1, -0.5], Denominator: [1, -0.7, 0.12]

C Numerator: [1, 0.5], Denominator: [1, -0.7, 0.12]

D Numerator: [1, 0.5], Denominator: [1, 0.7, -0.12]

💡 Hint

Recall that the numerator polynomial roots are zeros and denominator polynomial roots are poles.

The numerator polynomial with root at 0.5 is (z - 0.5) => coefficients [1, -0.5]. The denominator polynomial with roots at 0.3 and 0.4 is (z - 0.3)(z - 0.4) = z^2 - 0.7z + 0.12 => coefficients [1, -0.7, 0.12].

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