Second-order system simulation in Simulink - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the damping ratio variable.

Simulink

damping_ratio = [1]

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

A10

B2

C-1

D0.5

Attempts:

3 left

💡 Hint

Common Mistakes

Using a negative damping ratio which is not physically meaningful.

Using a damping ratio greater than 1 which represents overdamping.

The damping ratio for a typical second-order system is often between 0 and 1. Here, 0.5 represents an underdamped system.

2fill in blank

medium

Complete the code to define the natural frequency variable.

Simulink

natural_frequency = [1]

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

A0

B-5

C10

D100

Attempts:

3 left

💡 Hint

Common Mistakes

Using zero or negative values which are invalid for frequency.

The natural frequency is a positive value representing how fast the system oscillates without damping. 10 rad/s is a reasonable choice.

3fill in blank

hard

Fix the error in the transfer function definition by completing the blank.

Simulink

sys = tf([natural_frequency**2], [1, 2*[1]*natural_frequency, natural_frequency**2])

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

Adamping_ratio

Bomega

Cfrequency

Ddamping

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined variable names like 'damping' or 'omega'.

The transfer function denominator uses the damping ratio variable to define the system's damping effect.

4fill in blank

hard

Fill both blanks to simulate the step response and plot it.

Simulink

t, y = step(sys, [1])
plot(t, y)
xlabel('[2]')

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

A0:0.01:5

Btime

CTime (seconds)

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using a single number instead of a time vector for simulation.

Using a vague or missing x-axis label.

The step function uses a time vector from 0 to 5 seconds with 0.01 increments. The x-axis label should be descriptive like 'Time (seconds)'.

5fill in blank

hard

Fill all three blanks to create a dictionary of system parameters with keys 'zeta', 'wn', and 'type'.

Simulink

params = {'[1]': damping_ratio, '[2]': natural_frequency, '[3]': 'second_order'}

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

Azeta

Bwn

Ctype

Dorder

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect or inconsistent key names.

Mixing variable names with descriptive strings.

The standard keys for second-order system parameters are 'zeta' for damping ratio, 'wn' for natural frequency, and 'type' for system type.

Practice

1. What does the damping ratio (ζ) affect in a second-order system simulation in Simulink?

easy

A. The color of the simulation plot

B. The speed and overshoot of the system's response

C. The size of the Simulink blocks

D. The input signal amplitude

Solution

1. Step 1: Understand damping ratio role

   The damping ratio controls how oscillations decay in the system response.

2. Step 2: Connect damping ratio to response behavior

   A higher damping ratio reduces overshoot and speeds settling time, affecting speed and overshoot.

3. Final Answer:

   The speed and overshoot of the system's response -> Option B

4. Quick Check:

   Damping ratio = speed and overshoot control [OK]

Hint: Damping ratio controls oscillation and overshoot [OK]

Common Mistakes:

• Confusing damping ratio with input amplitude
• Thinking damping ratio changes block size
• Assuming damping ratio affects plot color

2. Which of the following is the correct way to represent a second-order transfer function in Simulink's Transfer Function block?

easy

A. Numerator: [1 0 0], Denominator: [1 2*zeta*wn wn]

B. Numerator: [1 2*zeta*wn wn^2], Denominator: [1]

C. Numerator: [1], Denominator: [1 2*zeta*wn wn^2]

D. Numerator: [wn^2], Denominator: [1 2*zeta*wn 1]

Solution

1. Step 1: Recall second-order transfer function form

   The standard form is numerator = [1], denominator = [1 2*zeta*wn wn^2].

2. Step 2: Match options to standard form

   Numerator: [1], Denominator: [1 2*zeta*wn wn^2] matches the correct numerator and denominator coefficients for a second-order system.

3. Final Answer:

   Numerator: [1], Denominator: [1 2*zeta*wn wn^2] -> Option C

4. Quick Check:

   Standard form = Numerator: [1], Denominator: [1 2*zeta*wn wn^2] [OK]

Hint: Denominator has 1, 2*zeta*wn, wn squared [OK]

Common Mistakes:

• Swapping numerator and denominator
• Using wrong powers for wn
• Missing damping term 2*zeta*wn

3. Given the Simulink transfer function block with numerator [1] and denominator [1 4 4], what is the natural frequency (wn) and damping ratio (ζ) of the system?

medium

A. wn=2, ζ=1

B. wn=4, ζ=2

C. wn=1, ζ=4

D. wn=2, ζ=0.5

Solution

1. Step 1: Identify denominator coefficients

   Denominator is [1 4 4], which matches [1 2*zeta*wn wn^2].

2. Step 2: Calculate wn and ζ from coefficients

   wn^2 = 4 -> wn = 2; 2*zeta*wn = 4 -> ζ = 4/(2*2) = 1.

3. Final Answer:

   wn=2, ζ=1 -> Option A

4. Quick Check:

   wn=2, ζ=1 from denominator [OK]

Hint: wn is sqrt of last denominator term [OK]

Common Mistakes:

• Confusing wn^2 with wn
• Dividing damping term incorrectly
• Ignoring coefficient order

4. You created a second-order system in Simulink with numerator [1] and denominator [1 0 4], but the simulation shows no oscillations. What is the likely error?

medium

A. The damping ratio is zero, causing no oscillations

B. The input signal is missing

C. The numerator should be [1 0 4]

D. The denominator coefficients are incorrect for oscillations

Solution

1. Step 1: Analyze denominator coefficients

   Denominator [1 0 4] means 2*zeta*wn = 0, so damping ratio ζ = 0.

2. Step 2: Understand effect on oscillations

   Zero damping ratio means undamped oscillations, but the system denominator form is incorrect for natural frequency 2 (since wn^2=4).

3. Step 3: Identify correct denominator form

   Correct denominator should be [1 0 4] only if system is undamped, but zero middle term can cause simulation issues.

4. Final Answer:

   The denominator coefficients are incorrect for oscillations -> Option D

5. Quick Check:

   Incorrect denominator = no oscillations [OK]

Hint: Check denominator matches 1, 2*zeta*wn, wn^2 [OK]

Common Mistakes:

• Assuming zero damping means no oscillations
• Swapping numerator and denominator
• Ignoring input signal presence

5. You want to simulate a second-order system with a natural frequency of 5 rad/s and a damping ratio of 0.3 in Simulink. What should be the denominator coefficients for the Transfer Function block?

hard

A. [1 3 25]

B. [1 3 5]

C. [1 3 15]

D. [1 3 30]

Solution

1. Step 1: Calculate denominator coefficients

   Given wn=5, ζ=0.3, calculate 2*ζ*wn = 2*0.3*5 = 3 and wn^2 = 25.

2. Step 2: Form denominator array

   Denominator coefficients are [1, 3, 25].

3. Final Answer:

   [1 3 25] -> Option A

4. Quick Check:

   2*zeta*wn=3, wn^2=25 [OK]

Hint: Multiply 2*damping*wn and square wn for denominator [OK]

Common Mistakes:

• Using wn instead of wn squared
• Mixing damping ratio with wn
• Incorrect multiplication order