Transfer function block in Simulink - Time & Space Complexity

Time Complexity: Transfer function block

O(n)

Understanding Time Complexity

We want to understand how the time to compute outputs from a transfer function block changes as the input size grows.

How does the block's processing time grow when we simulate longer or more detailed signals?

Scenario Under Consideration

Analyze the time complexity of the following transfer function block simulation step.

    % Transfer function: H(s) = (b0*s^m + ... + bm) / (a0*s^n + ... + an)
    % Discretized difference equation update:
    y(k) = (b0*u(k) + b1*u(k-1) + ... + bm*u(k-m))
           - (a1*y(k-1) + a2*y(k-2) + ... + an*y(k-n))
    

This code updates the output y at each time step k using past inputs u and outputs y based on the transfer function coefficients.

Identify Repeating Operations

Look at what repeats during simulation.

• Primary operation: Calculating y(k) using sums of past inputs and outputs.
• How many times: Once per time step for the entire input signal length.

How Execution Grows With Input

Each new input point requires a fixed number of calculations based on the order of the transfer function.

Input Size (n)	Approx. Operations
10	10 x (order of transfer function)
100	100 x (order of transfer function)
1000	1000 x (order of transfer function)

Pattern observation: Operations grow linearly with the number of input points.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute outputs grows in direct proportion to the number of input samples.

Common Mistake

[X] Wrong: "The time to compute output grows faster than input size because of the feedback terms."

[OK] Correct: Each output depends on a fixed number of past values, so the work per step stays constant regardless of input length.

Interview Connect

Understanding how simulation time scales with input size helps you design efficient models and explain performance in real projects.

Self-Check

"What if the transfer function order increased with input size? How would the time complexity change?"

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