Root locus analysis in Simulink - Step-by-Step Execution

Concept Flow - Root locus analysis

Start with Open-Loop Transfer Function

↓

Calculate Poles and Zeros

↓

Vary Gain K from 0 to Infinity

↓

Plot Root Locus: Poles Movement

↓

Analyze Stability and Performance

↓

Choose Suitable Gain K

↓

End

Root locus analysis shows how system poles move as gain changes, helping to choose gain for stability.

Execution Sample

Simulink

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

Plot root locus of a system with transfer function 1/(s^2 + 3s + 2).

Execution Table

Step	Gain K	Poles Location	Action	Plot Status
1	0	Poles at -1 and -2	Start root locus plot	Plot initialized with poles marked
2	0.5	[-1.5 + 0.5i, -1.5 - 0.5i]	Calculate poles for K=0.5	Plot updated with new poles
3	1	[-1.5 + 0.87i, -1.5 - 0.87i]	Calculate poles for K=1	Plot updated with new poles
4	5	[-1.5 + 2.18i, -1.5 - 2.18i]	Calculate poles for K=5	Plot updated with new poles
5	10	[-1.5 + 3.12i, -1.5 - 3.12i]	Calculate poles for K=10	Plot updated with new poles
6	Infinity	Poles to infinity along Re=-1.5	Complete root locus plot	Plot shows full root locus path
7	-	-	End of gain range	Root locus plot complete

💡 Gain K reaches infinity, root locus plot fully traced.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
Gain K	0	0.5	1	5	10	Infinity	Infinity
Poles Location	[-1, -2]	[-1.5 + 0.5i, -1.5 - 0.5i]	[-1.5 + 0.87i, -1.5 - 0.87i]	[-1.5 + 2.18i, -1.5 - 2.18i]	[-1.5 + 3.12i, -1.5 - 3.12i]	[to ±j∞ along Re=-1.5]	[to ±j∞ along Re=-1.5]

Key Moments - 3 Insights

Why do poles move on the plot as gain K changes?

Because the root locus shows how the system poles shift with gain K, as seen in execution_table rows 2 to 6 where poles locations update.

What does it mean when poles cross the imaginary axis?

Crossing the imaginary axis means the system becomes unstable.

Why does the root locus end at zeros or infinity?

Because poles move towards zeros or infinity as gain goes to infinity, shown in execution_table row 6 where poles reach zeros or infinity.

Visual Quiz - 3 Questions

Look at the execution_table, what are the poles locations at gain K=1?

A [-1.5 + 0.5i, -1.5 - 0.5i]

B [-1.5 + 0.87i, -1.5 - 0.87i]

C [-0.5+1.5i, -0.5-1.5i]

D [0+2i, 0-2i]

💡 Hint

Check the 'Poles Location' column at Step 3 in execution_table.

At which gain K do poles have the largest imaginary parts?

A 5

B 1

C 10

D 0.5

💡 Hint

Look at execution_table row where poles have largest imaginary parts (Step 5).

If gain K never increases beyond 1, what will the root locus plot show?

A Poles staying near initial locations

B Poles moving towards zeros or infinity

C Poles crossing imaginary axis

D Plot showing instability

💡 Hint

Refer to variable_tracker for poles location at gains up to 1.

Concept Snapshot

Root locus analysis:
- Plots system poles as gain K varies from 0 to infinity.
- Poles start at open-loop poles and move towards zeros or infinity.
- Crossing imaginary axis means instability.
- Used to select gain for desired stability and performance.

Full Transcript

Root locus analysis starts with the open-loop transfer function. We calculate the poles and zeros, then vary the gain K from zero to infinity. At each gain, we find the new pole locations and plot them. This shows how the system's stability changes. Poles moving left mean stable, crossing the imaginary axis means unstable. The root locus ends when poles reach zeros or infinity. This helps us pick a gain that keeps the system stable and performs well.

Practice

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