Transfer function block in Simulink - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a transfer function block with numerator 1 and denominator [1 1].

Simulink

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

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

A1

B[1 0]

C0

D[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using a scalar 1 instead of a vector [1] for numerator.

Using an empty vector or zero as numerator.

The numerator of a transfer function must be given as a vector. Here, [1] represents the numerator polynomial.

2fill in blank

medium

Complete the code to create a transfer function with numerator [2] and denominator [1 3 2].

Simulink

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

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

A2

B[2]

C[1 2]

D[3 2]

Attempts:

3 left

💡 Hint

Common Mistakes

Using scalar 2 instead of vector [2].

Using wrong vector length for numerator.

The numerator must be a vector. [2] is the correct vector for numerator 2.

3fill in blank

hard

Fix the error in the code to create a transfer function with numerator [1 0] and denominator [1 2].

Simulink

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

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

A[1 0]

B1 0

C1, 0

D1; 0

Attempts:

3 left

💡 Hint

Common Mistakes

Writing numerator without brackets.

Using commas or semicolons inside numerator.

The numerator must be a vector enclosed in square brackets. [1 0] is correct.

4fill in blank

hard

Fill both blanks to create a transfer function with numerator [1 4] and denominator [1 5 6].

Simulink

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

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

A[1 4]

B[1 5 6]

C[1 4 5]

D[5 6]

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing numerator and denominator vectors.

Using wrong vector lengths.

The numerator is [1 4] and denominator is [1 5 6]. Both must be vectors.

5fill in blank

hard

Fill all three blanks to create a transfer function and assign it to variable sys.

Simulink

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

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

A[3 2]

B[1 4 5]

Csys

D[1 0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable name in disp.

Not using vectors for numerator or denominator.

The numerator is [1 0], denominator is [3 2], and we display the variable sys.

Practice

1. What does the Transfer Function block in Simulink represent?

easy

A. A mathematical model using numerator and denominator polynomials

B. A block that generates random signals

C. A tool for drawing system diagrams

D. A block that stores data values

Solution

1. Step 1: Understand the Transfer Function block purpose and compare options

   The Transfer Function block models system behavior mathematically using polynomials. Only a mathematical model using numerator and denominator polynomials correctly describes the block.

2. Final Answer:

   A mathematical model using numerator and denominator polynomials -> Option A

3. Quick Check:

   Transfer Function = numerator/denominator polynomials [OK]

Hint: Remember: Transfer function = output/input ratio [OK]

Common Mistakes:

• Confusing with signal generator blocks
• Thinking it stores data instead of modeling
• Assuming it draws diagrams

2. Which of the following is the correct way to enter a transfer function with numerator [1 3] and denominator [1 5 6] in the Simulink Transfer Function block?

easy

A. Numerator: [1 3], Denominator: [1 5 6]

B. Numerator: 1/3, Denominator: 1/5/6

C. Numerator: (1,3), Denominator: (1,5,6)

D. Numerator: {1 3}, Denominator: {1 5 6}

Solution

1. Step 1: Recall Simulink syntax for polynomials and match options

   Simulink requires vectors in square brackets for numerator and denominator coefficients. Only Numerator: [1 3], Denominator: [1 5 6] uses square brackets and space-separated coefficients correctly.

2. Final Answer:

   Numerator: [1 3], Denominator: [1 5 6] -> Option A

3. Quick Check:

   Use square brackets for polynomials [OK]

Hint: Use square brackets for coefficient vectors [OK]

Common Mistakes:

• Using slashes or commas instead of spaces
• Using parentheses or braces instead of brackets
• Entering fractions instead of vectors

3. Given a Transfer Function block with numerator [2] and denominator [1 4 4], what is the expected system type?

medium

A. First-order system

B. Third-order system

C. Second-order system

D. Zero-order system

Solution

1. Step 1: Determine system order from denominator degree and match to options

   The denominator polynomial [1 4 4] has 3 coefficients, so degree is 2 (order = degree of denominator). A second-order system corresponds to denominator degree 2.

2. Final Answer:

   Second-order system -> Option C

3. Quick Check:

   Order = degree of denominator polynomial = 2 [OK]

Hint: System order = degree of denominator polynomial [OK]

Common Mistakes:

• Counting number of coefficients instead of degree
• Confusing numerator degree with system order
• Assuming order from numerator length

4. You entered numerator [1 0] and denominator [1 -2 1] in a Transfer Function block, but the simulation shows unexpected oscillations. What is the likely cause?

medium

A. Numerator should have more coefficients

B. Simulation time is too short

C. Transfer Function block does not support negative coefficients

D. Incorrect sign in denominator coefficients causing instability

Solution

1. Step 1: Analyze denominator polynomial for stability and identify cause

   The denominator [1 -2 1] corresponds to (s-1)^2, which has poles at s=1 (right half-plane), causing instability. Incorrect sign in denominator coefficients leads to this.

2. Final Answer:

   Incorrect sign in denominator coefficients causing instability -> Option D

3. Quick Check:

   Poles in right half-plane cause oscillations [OK]

Hint: Check denominator signs for stability [OK]

Common Mistakes:

• Assuming numerator length causes oscillations
• Believing negative coefficients are unsupported
• Ignoring pole locations

5. You want to model a system with transfer function \( \frac{5s + 10}{s^2 + 3s + 2} \) in Simulink. How should you enter the numerator and denominator coefficients in the Transfer Function block?

hard

A. Numerator: [10 5], Denominator: [1 3 2]

B. Numerator: [5 10], Denominator: [1 3 2]

C. Numerator: [5 10], Denominator: [1 2 3]

D. Numerator: [5 10], Denominator: [3 2 1]

Solution

1. Step 1: Identify numerator and denominator coefficients in descending powers of s

   Numerator is 5s + 10, so [5 10]; denominator s^2 + 3s + 2, so [1 3 2].

2. Final Answer:

   Numerator: [5 10], Denominator: [1 3 2] -> Option B

3. Quick Check:

   Coefficients match polynomial terms in order [OK]

Hint: List coefficients from highest to lowest power [OK]

Common Mistakes:

• Reversing coefficient order
• Mixing up numerator and denominator
• Using incorrect coefficient values