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Root locus analysis in Simulink - Interactive Code Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to create a transfer function.

Simulink
sys = tf([1], [1 3 2]);
Drag options to blanks, or click blank then click option'
A1
B[1 5]
C0
Ds
Attempts:
3 left
💡 Hint
Common Mistakes
Using 's' as numerator causes error.
Using a vector instead of scalar for numerator.
2fill in blank
medium

Complete the code to plot the root locus of the system.

Simulink
rlocus([1]);
Drag options to blanks, or click blank then click option'
Aplot
Brootlocus
Csys
Dtf
Attempts:
3 left
💡 Hint
Common Mistakes
Passing a plotting function name instead of system.
Passing transfer function constructor instead of system variable.
3fill in blank
hard

Fix the error in the code to compute the characteristic equation.

Simulink
char_eq = poly([1]);
Drag options to blanks, or click blank then click option'
As
Brlocus
Ctf
Dsys
Attempts:
3 left
💡 Hint
Common Mistakes
Using rlocus instead of system variable.
Using symbolic variable s incorrectly.
4fill in blank
hard

Fill both blanks to create a root locus plot with grid and title.

Simulink
rlocus([1]);
grid [2];
Drag options to blanks, or click blank then click option'
Asys
Bon
Coff
Dplot
Attempts:
3 left
💡 Hint
Common Mistakes
Using grid off disables the grid.
Passing wrong variable to rlocus.
5fill in blank
hard

Fill all three blanks to compute poles, zeros, and plot root locus with title.

Simulink
poles = pole([1]);
zeros = zero([2]);
rlocus([3]);
title('Root Locus Plot');
Drag options to blanks, or click blank then click option'
Asys
Dplot
Attempts:
3 left
💡 Hint
Common Mistakes
Using different variables for poles and zeros.
Using plotting function name instead of system variable.

Practice

(1/5)
1. What does the root locus plot primarily show in control systems analysis?
easy
A. The noise level in the system output
B. The frequency response of the system
C. How the system poles move as the gain changes
D. The time response of the system to a step input

Solution

  1. Step 1: Understand root locus purpose

    The root locus plot shows how the locations of system poles change as the gain parameter varies.
  2. Step 2: Differentiate from other plots

    Frequency response and time response plots show different system behaviors, not pole movement.
  3. Final Answer:

    How the system poles move as the gain changes -> Option C
  4. Quick Check:

    Root locus = pole movement with gain [OK]
Hint: Root locus tracks poles as gain varies [OK]
Common Mistakes:
  • Confusing root locus with frequency response
  • Thinking it shows output noise
  • Mixing it up with time response plots
2. Which Simulink command is used to plot the root locus of a system with transfer function G = tf([1], [1, 3, 2])?
easy
A. rlocus(G)
B. step(G)
C. bode(G)
D. pzmap(G)

Solution

  1. Step 1: Identify root locus command

    In Simulink and MATLAB, rlocus() is the function to plot root locus.
  2. Step 2: Differentiate from other commands

    step() plots step response, bode() plots frequency response, and pzmap() plots poles and zeros but not root locus.
  3. Final Answer:

    rlocus(G) -> Option A
  4. Quick Check:

    Root locus plot = rlocus() [OK]
Hint: Use rlocus() to plot root locus in Simulink [OK]
Common Mistakes:
  • Using step() instead of rlocus()
  • Confusing bode() with root locus
  • Using pzmap() which shows poles but no gain variation
3. Given the transfer function G = tf([2], [1, 4, 5]), what is the number of branches in the root locus plot?
medium
A. 1
B. 2
C. 3
D. 4

Solution

  1. Step 1: Determine number of poles

    The denominator polynomial [1, 4, 5] is second order, so there are 2 poles.
  2. Step 2: Relate poles to root locus branches

    Root locus branches equal the number of poles, so there are 2 branches.
  3. Correction:

    The denominator polynomial [1, 4, 5] is of order 2, so there are 2 poles, but the number of branches in root locus equals the number of poles, which is 2, not 3.
  4. Final Answer:

    2 -> Option B
  5. Quick Check:

    Branches = number of poles = 2 [OK]
Hint: Count denominator order for branches [OK]
Common Mistakes:
  • Counting zeros instead of poles
  • Confusing numerator order with poles
  • Assuming branches equal zeros
4. Identify the error in this Simulink command for root locus: rlocus(tf([1 2], [1 3]))
medium
A. No error, command is correct
B. The numerator vector is invalid
C. The transfer function is not proper
D. The denominator vector length is less than numerator

Solution

  1. Step 1: Check transfer function order

    Numerator order is 1 (degree 1), denominator order is 1 (degree 1). Transfer function is proper if denominator order >= numerator order.
  2. Step 2: Analyze properness

    Here numerator and denominator orders are equal, so transfer function is proper, not improper.
  3. Step 3: Check command validity

    The command is syntactically correct and will run without error.
  4. Final Answer:

    No error, command is correct -> Option A
  5. Quick Check:

    Proper TF and syntax = no error [OK]
Hint: Check numerator and denominator orders for proper TF [OK]
Common Mistakes:
  • Assuming numerator length must be less than denominator
  • Thinking equal order is improper
  • Confusing vector length with polynomial degree
5. You have a system with transfer function G = tf([1], [1, 2, 2, 1]). You want to design a controller gain K so that the system poles lie on the left half of the complex plane for stability. Using root locus analysis, which gain range should you choose?
hard
A. K = 0 only
B. Any positive K value
C. K < 0 only
D. K > 0 and less than the gain where poles cross imaginary axis

Solution

  1. Step 1: Understand root locus stability

    Root locus shows pole movement as gain K changes. Poles in left half-plane mean stable system.
  2. Step 2: Identify gain limits for stability

    Poles start stable at K=0 and move as K increases. Stability is maintained until poles cross imaginary axis at a critical gain.
  3. Step 3: Choose gain range

    Gain K must be positive and less than the critical gain where poles cross imaginary axis to keep system stable.
  4. Final Answer:

    K > 0 and less than the gain where poles cross imaginary axis -> Option D
  5. Quick Check:

    Stable poles = gain below crossing point [OK]
Hint: Pick gain before poles cross imaginary axis [OK]
Common Mistakes:
  • Assuming any positive gain is stable
  • Choosing negative gain for stability
  • Ignoring pole crossing points
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