Feedback Control System

Lecture 19
Chapter 8: Root Locus Techniques

Prepared By: Muhammad Abdullah

muhammad.abdullah@nu.edu.pk , uetian.09@gmail.com

National University FAST (CFD Campus)

My YouTube Channel:
https://www.youtube.com/@BlueFish506

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Sketching a Root Locus with Asymptotes

The root locus approaches straight lines as asymptotes as the locus approaches
infinity. Further, the equation of the asymptotes is given by the real-axis intercept,
𝜎𝑎 and angle, 𝜃𝑎 as follows:
∑ finite poles − ∑ finite zeros
𝜎𝑎 =
# finite poles − # finite zeros
(2𝑘 + 1)𝜋
𝜃𝑎 =
# finite poles − # finite zeros
where 𝑘 = 0, ±1, ±2, ±3 and the angle is given in radians with respect to the
positive extension of the real axis.

Example 1
Sketch the root locus for the system shown in Figure 1.

Figure 1: Root locus and Asymptotes for the System of Figure 1

Solution:
Real-axis intercept, 𝜎𝑎 is calculated as:
∑ finite poles − ∑ finite zeros
𝜎𝑎 =
# finite poles − # finite zeros
(−1 − 2 − 4) − (−3) 4
𝜎𝑎 = =−
4−1 3
The angles of the lines that intersect at −4/3 are given as;
(2𝑘 + 1)𝜋 (2𝑘 + 1)𝜋
𝜃𝑎 = =
# finite poles − # finite zeros 3

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 𝜋/3 for 𝑘 = 0
𝜃𝑎 = { 𝜋 for 𝑘 = 1
5𝜋/3 for 𝑘 = 2

Figure 2

Refining the Sketch

Purpose of Refinement
• To accurately locate key points on the root locus such as:
o Real-axis breakaway and break-in points
o jω-axis crossings

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 o Angles of departure/arrival from complex poles and zeros
• Helps determine gain values and stability behavior more precisely.

Real-Axis Breakaway and Break-In Points

Concept
• Occur where root loci:
o Leave the real axis (breakaway)
o Return to the real axis (break-in)
• Common when poles/zeros transition between real and complex locations.
Identification
• Breakaway happens between two poles, break-in between two zeros or as
complex poles become real again.

Figure 3: Root locus showing real-axis breakaway (−𝝈𝟏 ) and break-in points (𝝈𝟐 )

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Example:
• Root locus leaves real axis between -1 and -2
• Returns between +3 and +5
Angle at Breakaway/Break-In Points
• At these points, the branches form an angle of.
180∘
𝑛
where n = number of closed-loop poles arriving or departing.
• For 2 poles, the angle is:
180∘
= 90∘
2
• This helps in estimating the direction of the root locus at critical transition
points.
Finding Breakaway and Break-In Points – Gain Method

Key Concept

• Breakaway and break-in points occur at extrema of gain (K) along the real
axis.
• These points correspond to local maximum or minimum values of gain.

Method 1: Using Calculus on the Gain Function

For all points on the root locus: 1 + 𝐾𝐺(𝑠)𝐻(𝑠) = 0
1
𝐾=−
𝐺(𝑠)𝐻(𝑠)
For points specifically along the real-axis segment, where 𝑠 = 𝜎 :
1
𝐾=−
𝐺(𝜎)𝐻(𝜎)
This equation represents a curve of 𝐾 vs. 𝜎, where:
• 𝜎 is a real value along the axis.

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 • Points of maximum/minimum K indicate potential breakaway or break-in.
Procedure

1. Evaluate 𝐾(𝜎) using Equation given above

2. Differentiate 𝐾(𝜎) with respect to 𝜎.
𝑑𝐾
3. Set derivative = 0 to find critical points.
𝑑𝜎
4. Solve for values of 𝜎 where extrema occur.
5. These values correspond to breakaway or break-in points.
Example 2
Find the breakaway and break-in points for the root locus of Figure 3.
The open-loop transfer function is:
𝐾(𝑠 − 3)(𝑠 − 5) 𝐾(𝑠 2 − 8𝑠 + 15)
𝐾𝐺(𝑠)𝐻(𝑠) = =
(𝑠 + 1)(𝑠 + 2) 𝑠 2 + 3𝑠 + 2
On the root locus, we use the condition:
𝐾𝐺(𝑠)𝐻(𝑠) = −1
Restricting to the real-axis segment (𝑠 = 𝜎), we write:
𝐾(𝜎 2 − 8𝜎 + 15)
= −1
𝜎 2 + 3𝜎 + 2
Step 1: Solve for Gain 𝐾
−(𝜎 2 + 3𝜎 + 2)
𝐾= 2
𝜎 − 8𝜎 + 15
Step 2: Differentiate 𝐾 with respect to 𝜎
𝑑𝐾 11𝜎 2 − 26𝜎 − 61
= =0
𝑑𝜎 (𝜎 2 − 8𝜎 + 15)2
Step 3: Solve the Numerator Equation
Solving: 11𝜎 2 − 26𝜎 − 61 = 0
Gives: 𝜎 = −1.45 and 𝜎 = 3.82

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 • Breakaway point: 𝜎 = −1.45
• Break-in point: 𝜎 = 3.82
These points are where the root locus leaves and re-enters the real axis.

Method 2: Transition Method (No Differentiation)

𝑚 𝑛
1 1
∑ =∑
𝜎 + 𝑧𝑖 𝜎 + 𝑝𝑖
𝑖=1 𝑖=1

• Where 𝑧𝑖 and 𝑝𝑖 are negatives of the system zeros and poles respectively.
• Solve for 𝜎 to find real-axis breakaway and break-in points.
Example 3
Repeat Example 2 without differentiating
Solution:
Using Equation for Transition method, gives
1 1 1 1
+ = +
𝜎−3 𝜎−5 𝜎+1 𝜎+2
From LHS:
1 1 (𝜎 − 5) + (𝜎 − 3) 2𝜎 − 8
+ = =
𝜎−3 𝜎−5 (𝜎 − 3)(𝜎 − 5) (𝜎 − 3)(𝜎 − 5)
Similarly, for RHS:
1 1 (𝜎 + 2) + (𝜎 + 1) 2𝜎 + 3
+ = =
𝜎+1 𝜎+2 (𝜎 + 1)(𝜎 + 2) (𝜎 + 1)(𝜎 + 2)
Equating Both Sides
2𝜎 − 8 2𝜎 + 3
=
(𝜎 − 3)(𝜎 − 5) (𝜎 + 1)(𝜎 + 2)
(2𝜎 − 8)(𝜎 + 1)(𝜎 + 2) = (2𝜎 + 3)(𝜎 − 3)(𝜎 − 5)
Simplifying both sides yields:
11𝜎 2 − 26𝜎 − 61 = 0
Solving the quadratic equation:

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 𝜎 = −1.45 and 𝜎 = 3.82
Same breakaway and break-in points as in the previous (calculus-based) example—
method verified!

jω-Axis Crossings in Root Locus Analysis

Definition and Importance

• The jω-axis crossing is where the root locus intersects the imaginary axis.
• It marks the boundary between stable and unstable behavior in a feedback
system.
• Crossing into the RHP signifies instability as poles move from left-half to
right-half.
Behavior Insight from Root Locus
• Below a critical gain value, poles remain in the left-half plane → system is
stable.
• At a specific gain, poles move to the right-half plane → system becomes
unstable.
• The frequency (ω) at the jω-axis crossing indicates oscillation frequency.
• The gain (K) at this crossing is the maximum stable gain for the system.
Finding jω-Axis Crossings
• Use the Routh-Hurwitz criterion:
1. Construct the Routh table from the characteristic polynomial.
2. A row of zeros in the table indicates jω-axis crossing.
3. Go back one row to get the auxiliary polynomial.
4. Solve for roots to find the imaginary-axis frequency (ω).

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Example 4
For the system of Figure 4, find the frequency and gain, 𝐾, for which the root locus
crosses the imaginary axis. For what range of 𝐾 is the system stable?

Figure 4

Solution:
The closed-loop transfer function for the system of Figure 4 is;
𝐾(𝑠 + 3)
𝑇(𝑠) =
𝑠 4 + 7𝑠 3 + 14𝑠 2 + (8 + 𝐾)𝑠 + 3𝐾
Routh’s table of the system is drawn as;

For the system to be stable, there should be no sign change in the first column of
Routh’s table, so;
−𝐾 2 − 65𝐾 + 720
>0
90 − 𝐾 > 0 90 − 𝐾 21𝐾 > 0
𝐾 < 90 −𝐾 2 − 65𝐾 + 720 > 0 𝐾>0
𝐾 2 + 65𝐾 − 720 < 0

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 (𝐾 + 74.65)(𝐾 − 9.65) < 0
Or
𝐾 < −74.65
𝐾 < 9.65
So, stability range for given system is: 0 < 𝐾 < 9.65. So, for gain 𝐾 = 9.65, the
root locus branches cross the 𝑗𝜔-axis.
With 𝐾 = 9.65, the 𝑠 2 row becomes the auxiliary polynomial:
(90 − 𝐾)𝑠 2 + 21𝐾 = 80.35𝑠 2 + 202.7 = 0
Solving:
𝑠 = ±𝑗1.59
• 𝑗𝜔-axis crossing frequency: 𝜔 = 1.59 rad/s
Angles of Departure and Arrival
Purpose: To improve the accuracy of root locus sketches by calculating:
• Angle of departure from complex poles
• Angle of arrival at complex zeros

Angle of Departure (from Complex Poles)

Rule: The sum of angles from all finite poles and zeros to a point ε close to a complex
pole is an odd multiple of 180°.
• Let 𝜃1 be the unknown angle (departure from the complex pole).
• Known angles 𝜃2 , 𝜃3 , … come from other poles and zeros.

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Using the angle criterion:

∑(angles from zeros) − ∑(angles from poles) = (2𝑘 + 1)180∘

𝜃2 + 𝜃3 + 𝜃6 − 𝜃4 − 𝜃5 − 𝜃1 = (2𝑘 + 1)180∘
𝜃1 = 𝜃2 + 𝜃3 − 𝜃4 − 𝜃5 + 𝜃6 − (2𝑘 + 1)180∘

Angle of Arrival (at Complex Zeros)

Rule: The sum of angles from all finite poles and zeros to a point ε close to a complex
zero is an odd multiple of 180°.

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Using the angle criterion:

∑(angles from zeros) − ∑(angles from poles) = (2𝑘 + 1)180∘

𝜃2 + 𝜃3 + 𝜃6 − 𝜃1 − 𝜃4 − 𝜃5 = (2𝑘 + 1)180∘
𝜃2 = 𝜃1 − 𝜃3 + 𝜃4 + 𝜃5 − 𝜃6 + (2𝑘 + 1)180∘

Example 5
Given the unity feedback system of Figure 5, find the angle of departure from the
complex poles and sketch the root locus.

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 Figure 5

Solution:

From above root locus plot, we get (Assume 𝑘 = 0);

∑(angles from zeros) − ∑(angles from poles) = (2𝑘 + 1)180∘ = 180∘

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𝜃3 − 𝜃1 − 𝜃2 − 𝜃4 = 180∘
1 1
tan−1 ( ) − 𝜃1 − 90∘ − tan−1 ( ) = 180∘
1 2
45∘ − 𝜃1 − 90∘ − 26.56∘ = 180∘
−𝜃1 − 71.56∘ = 180∘
𝜃1 = −180∘ − 71.56∘ = −251.56∘ or
𝜃1 = −251.56∘ + 360∘ = 108.44∘

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