Zagazig University

Faculty of engineering
Electrical Power and Machine Dept.

Power System Control (2)

Root Locus
Lecture (6)
Dr\ M.Fekry
2025-2026

Redistribution is PROHIBITED
Dr. M.Fekry 1
Outline
• The Root-Locus Analysis
• Root-locus Plots of Positive Feedback
• Effects of the Addition of Poles and Zeros
• Designing a Lead Compensator with Root Locus
• Designing a Lag Compensator with Root Locus
• Designing a Lag-Lead Compensator with Root Locus
• Root Locus Using MATLAB

Dr. M.Fekry 2
Root-locus Plots of Positive Feedback
Systems

+
𝑅 𝑠 𝐾𝐺(𝑠)
=
𝑌 𝑠 1 − 𝐾𝐺 𝑠
• The characteristic equation of the closed loop T.F is:
1 − 𝐾𝐺 𝑠 = 0 → 𝐾𝐺 𝑠 = +1 → 𝐾 = +1/𝐺 𝑠
• The equation can be rewritten in polar form as:
𝐾𝐺 𝑠 = 1 → 𝐾 = 1/ 𝐺 𝑠
∠𝐾𝐺 𝑠 = 0 + 360𝑛
Where 𝑛 = 0, ±1, ±2, ±3, …
Any point located at the root locus of the closed loop transfer
Dr. M.Fekry 3
function must satisfy the magnetite and the angle conditions
Root-locus Plots of Positive Feedback
Systems

𝐾 𝑠 + 𝑧1 𝑠 + 𝑧2 𝑠 + 𝑧3 … (𝑠 + 𝑧𝑚 )
𝐾𝐺 𝑠 =
𝑠 + 𝑝1 𝑠 + 𝑝2 𝑠 + 𝑝3 … (𝑠 + 𝑝𝑛 )

𝐾 𝑠 + 𝑧1 𝑠 + 𝑧2 𝑠 + 𝑧3 … 𝑠 + 𝑧𝑚
𝐾𝐺 𝑠 = =1
𝑠 + 𝑝1 𝑠 + 𝑝2 𝑠 + 𝑝3 … 𝑠 + 𝑝𝑛
∠𝑲𝑮 𝒔 = ∠ 𝑠 + 𝑧1 + ∠ 𝑠 + 𝑧2 + ⋯ ∠ 𝑠 + 𝑧𝑚
−∠ 𝑠 + 𝑝Dr. − ∠ 𝑠 + 𝑝2 − ⋯ − ∠ 𝑠 + 𝑝𝑛 = 0
1 M.Fekry 4
Root-locus Plots of Positive Feedback
Systems: Example (6)
Draw the root-locus of positive unity feedback system the following
system
𝐾(𝑠 + 2)
𝐺 𝑠 =
(𝑠 + 3)(𝑠 2 + 2𝑠 + 2)

Dr. M.Fekry 5
Root-locus Plots of Positive Feedback
Systems: Example (6)
Step 1: Write the characteristic equation
1 − 𝐺𝐻 = 0
𝐾 𝑠+2
1− 2
=0
𝑠 + 3 𝑠 + 2𝑠 + 2

𝑠 + 3 𝑠 2 + 2𝑠 + 2 − 𝐾 𝑠 + 2 = 0

Dr. M.Fekry 6
Root-locus Plots of Positive Feedback
Systems: Example (6)
Step 2: Locate the open loop poles and
zeros on the S-plane.
Z=
-2
P=
-3.0000 + 0.0000i
-1.0000 + 1.0000i
-1.0000 - 1.0000i

Dr. M.Fekry 7
Root-locus Plots of Positive Feedback
Systems: Example (6)
Step 3: Locate the segments +1

• Modified as Follows:
𝜽𝒑𝟏
If the total number of real poles and real
-1
zeros to the right of a test point on the real -3 -2

axis is even, then this test point lies on the

-1
root locus.

Dr. M.Fekry 8
Root-locus Plots of Positive Feedback
Systems: Example (6)
Step 4: Determine the number of separate
+1
loci (Branches)
No. of Branches = No. of poles = 3
𝜽𝒑𝟏
Step 5: Modified as Follows:
-1
𝑛𝑝 𝑛 -3 -2
𝑧
σ𝑗=1 −𝑝𝑗 − σ𝑖=1 −𝑧𝑗
𝜎𝐴 = -1
𝑛𝑝 − 𝑛𝑧
±𝑚360
𝜙𝐴 = 𝑚 = 0,1,2, …
𝑛𝑝 − 𝑛𝑧

Dr. M.Fekry 9
Root-locus Plots of Positive Feedback
Systems: Example (6)
Step 5: Modified as Follows:
+1
𝑛𝑝 𝑛
𝑧
σ𝑗=1 −𝑝𝑗 − σ𝑖=1 −𝑧𝑗
𝜎𝐴 =
𝑛𝑝 − 𝑛𝑧 𝜽𝒑𝟏

−3 − 1 + 𝑗1 − 1 − 𝑗1 + 2 -1
𝜎𝐴 = = −1.5 -3 -2
2
±360𝑚 -1
𝜙𝐴 = 𝑚 = 0,1,2, …
𝑛𝑝 − 𝑛 𝑧
𝜙𝐴 = 0, 𝑚 = 0
𝜙𝐴 = 180, 𝑚 = 1

Dr. M.Fekry 10
Root-locus Plots of Positive Feedback
Systems: Example (6)
Step 6: Determine where the locus +1
crosses the imaginary axis
Modified as Follows: 𝜽𝒑𝟏

There are two cases -3 -2

-1

• If K is +ve : There is no intersection

-1
• If K is -ve : There is intersection, The
auxiliary equation is solved to find the
location of intersection

Dr. M.Fekry 11
 Root-locus Plots of Positive Feedback
Systems: Example (6)
Step 6: Determine where the locus +1
crosses the imaginary axis
𝑠 + 3 𝑠 2 + 2𝑠 + 2 − 𝐾 𝑠 + 2 = 0 𝜽𝒑𝟏

𝑠 3 + 5𝑠 2 + 8 − 𝐾 𝑠 − 2𝐾 + 6 = 0 -1
-3 -2
s3 1 8 −𝐾
s2 5 −2𝐾 + 6 -1
34 3𝐾
s1 5
−
5
s0 −2𝐾 + 6

34 3𝐾
− = 0 → 𝐾 = 11.3333 No intersection
Dr. M.Fekry with imaginary axis 12
5 5
Root-locus Plots of Positive Feedback
Systems: Example (6)
Step 7: Determine the break-in point on the real axis. This can be evaluated by
rearranging of the characteristic equation to isolate the multiplying factor K
𝑠 + 3 𝑠 2 + 2𝑠 + 2 − 𝐾 𝑠 + 2 = 0
𝑠 + 3 𝑠 2 + 2𝑠 + 2
𝐾=
𝑠+2
After that, determining the maximum K
𝑑𝐾 2𝑠 3 + 11𝑠 2 + 20𝑠 + 10
= 2
=0
𝑑𝑠 𝑠 + 4𝑠 + 4
𝑠 = −0.8026
𝑠 = −2.3487 ± 𝑗0.8447

Dr. M.Fekry 13
Root-locus Plots of Positive𝑛 Feedback
𝑛 𝑝 𝑧
Systems: Example (6) ෍ 𝜃𝑝 − ෍ 𝜃𝑧 𝑖 = 0 ± 360𝑚 𝑖
𝑖=1 𝑖=1
Step 8: Determine the angle of
departure of the locus from a pole and 𝜃𝑑𝑒𝑝𝑎𝑡𝑢𝑟𝑒 ቚ
𝑝𝑗
the angle of arrival of the locus at a
𝑛𝑝 𝑛𝑧
zero, using the phase angle criterion.
= 0 − ෍ 𝜃𝑝 𝑖 + ෍ 𝜃𝑧 𝑖
Modified as Follows: 𝑖=1 𝑖=1
𝑖≠𝑗
• The sum of all angle of poles and
𝑛𝑝 𝑛𝑧
zeros is zero
𝜃𝑎𝑟𝑟𝑖𝑣𝑎𝑙 ቚ = 0 − ෍ 𝜃𝑝 𝑖 − ෍ 𝜃𝑧 𝑖
𝑧𝑗
𝑖=1 𝑖=1
Dr. M.Fekry 𝑖≠𝑗
14
Root-locus Plots of Positive Feedback
Systems: Example (6)
𝜽𝒅𝟏
Step 8: Determine the angle of departure of
the locus from a pole and the angle of arrival +1
of the locus at a zero, using the phase angle
criterion. 𝜽𝒑𝟏 𝜽𝒛𝟏
𝜽𝒑𝟏
Modified as Follows
𝜃𝑑1 = 0 − 𝜃𝑝1 − 𝜃𝑑2 + 𝜃𝑧1 -3 -2
-1
1 1
𝜃𝑑1 = 0 − tan−1 − 90 + tan−1
2 1
𝜃𝑑1 = −71.5651° = 288.4349° -1
𝜃𝑑2 = 0 − 𝜃𝑝1 − 𝜃𝑑1 + 𝜃𝑧1 𝜽𝒅𝟐
−1 −1
𝜃𝑑2 = 0 − tan−1 − 270 + tan−1
2 1
𝜃𝑑2 = 71.5651° = −288.4349°
Dr. M.Fekry 15
Root-locus Plots of Positive Feedback
Systems: Example (6)

Dr. M.Fekry 16
Effects of the Addition of Poles

• The addition of a pole to the open-loop transfer

function has the effect of:
1. Pulling the root locus to the right,
2. Tending to lower the system’s relative stability
(making the system less stable)
3. Slowing down the settling of the response.

Dr. M.Fekry 17
Effects of the Addition of Poles

Dr. M.Fekry 18
Effects of the Addition of Zeros
• The addition of a zero to the open-loop transfer function has the effect
of:
1. Pulling the root locus to the left.
2. Tending to make the system more stable and
3. Speeding up the settling of the response.

(Physically, the addition of a zero in the feedforward transfer function means the
addition of derivative control to the system. The effect of such control is to introduce
a degree of anticipation into the system and speed up the transient response.)

Dr. M.Fekry 19
Effects of the Addition of Zeros

Dr. M.Fekry 20
Designing a Lead Compensator with Root
Locus
1. From the performance specifications, (rise time,
settling time , damping ratio, and etc.) determine
the desired location for the dominant closed-loop
poles.

Dr. M.Fekry 21
Designing a Lead Compensator with Root
Locus
2. Drawing the root-locus plot of the uncompensated
system (original system),
3. Locate the desired dominant root locations.
4. Check whether or not the gain adjustment alone can
yield the desired closed loop poles.

Dr. M.Fekry 22
Designing a Lead Compensator with Root
Locus

Dr. M.Fekry 23
 Designing a Lead Compensator with Root
Locus
5. If not, Assume the lead compensator 𝑮𝒄 (𝒔) to be
𝐾𝑐 𝑠 + 𝑧
𝐺𝑐 𝑠 = 1
𝑠+𝑝 𝐾𝑐 𝑠 +
Approach 𝐺𝑐 𝑠 = 𝜏
𝐾𝑐 𝑗𝜔 + 𝑧 1
𝐺𝑐 𝑗𝜔 = 𝑠+
𝑗𝜔 + 𝑝 𝛼𝜏
where 𝛼 and 𝜏 are determined from the angle
Where |𝑧| < |𝑝| for Lead Compensator deficiency.
𝐾𝑐 is determined from the requirement of the
• The attenuation factor 𝛼 open-loop gain.
𝑝
𝛼= →0<𝛼<1
𝑧
1 1
𝑝 = 𝛼𝑧, 𝑧 = 𝑎𝑛𝑑 𝑝 = Dr. M.Fekry 24
𝜏 𝛼𝜏
Designing a Lead Compensator with Root
Locus
6. Place the zero of the phase-lead compensator directly
below the desired root location (or to the left of the first
two real poles).
7. Determine the pole location so that the total angle at
the desired root location is ±180°(2𝑚 + 1) and
therefore is on the compensated root locus.

Dr. M.Fekry 25
Designing a Lead Compensator with Root
Locus

Dr. M.Fekry 26
Designing a Lead Compensator with Root
Locus
8. Evaluate the total system gain at the
desired root location and then calculate the
error constant.
9. Repeat the steps if the error constant is not
satisfactory
𝐾𝑣 = lim 𝑠𝐺𝑐 𝑠 𝐺 𝑠 = 𝐾𝑐 𝛼 lim 𝑠𝐺𝑐 𝑠
𝑠→0 𝑠→0
Dr. M.Fekry 27
Designing a Lead Compensator with Root
Locus: Example (7)
Consider a system of open transfer function as follows
10𝐾
𝐺 𝑠 = 2
𝑠
The settling time (with a 2% criterion), 𝑇𝑠 ≤ 4 𝑠;
Percent overshoot for a step input 𝑃. 𝑂. ≤ 35%.
Design the lead compensator that achieves the desired transient
performance

Dr. M.Fekry 28
Designing a Lead Compensator with Root
Locus: Example (7)
−𝜁𝜋
∵ 𝑃. 𝑂. = 100 × 𝑒 1−𝜁 2
∴ 𝜁 ≥ 0.3169
4
∵ 𝑇𝑠 = = 4𝑠
𝜁𝜔𝑛
∴ 𝜁𝜔𝑛 = 1 → 𝜔𝑛 = 2.22 𝑟𝑎𝑑/𝑠
The desired dominant roots are (𝑠 2 + 2𝜁𝜔𝑛 𝑠 + 𝜔𝑛2 )
1
𝑟1 , 𝑟2 = −𝜁𝜔𝑛 ± 2𝜁𝜔𝑛 2 − 4𝜔𝑛2
2
𝑟1 , 𝑟2 = −1 ± 𝑗2.9925

Dr. M.Fekry 29
 Designing a Lead Compensator with Root
Locus: Example (7)
2. Draw the root locus of
uncompensated system.
3. Locate the desired roots
𝑟1,2 = −1 ± 𝑗2.9925
4. It is necessary to design
lead compensator that meet
the desired roots location.

Dr. M.Fekry 30
 Designing a Lead Compensator with Root
Locus: Example (7)
5. Assume the lead
compensator 𝐺𝑐 (𝑠) to be

1
𝐾𝑐 𝑠 +
𝐺𝑐 𝑠 = 𝜏
1
𝑠+
𝛼𝜏

Dr. M.Fekry 31
 Designing a Lead Compensator with Root
Locus: Example (7)
6. Place the zero of the
phase-lead compensator
directly below the desired
root location
𝑍𝑐 = −1 + 𝑗0

Dr. M.Fekry 32
 Designing a Lead Compensator with Root
Locus: Example (7)
7. Determine the pole location so
that the total angle at the desired
root location is 180°
180 = −2𝜃1 + 𝜃𝑧𝑐 − 𝜃𝑝𝑐 𝜽𝒑𝒄
2.9925 𝜽𝟏
180 = −2 180 − tan−1 + 90 − 𝜃𝑝𝑐
1 𝜽𝒛𝒄
2.9925
∴ 𝜃𝑝𝑐 = −2 180 − tan−1 + 90 − 180
1
∴ 𝜃𝑝𝑐 = 53.044°
2.9925
∴ tan 𝜃𝑝𝑐 =
𝑝
2.9925
∴𝑝=− − 1 = 3.2514
tan 𝜃𝑝𝑐 Dr. M.Fekry 33
Designing a Lead Compensator with Root
Locus: Example (7)
8. Evaluate the total system gain at the desired root location and then
calculate the error constant.
10𝐾(𝑠 + 1)
𝐺𝑐 𝐺 = 2
𝑠 𝑠 + 3.2514
10𝐾 𝑠 + 1
𝐺𝑐 𝐺 𝑟1 = 2 = 0.8027 𝐾
𝑠 𝑠 + 3.2514
1
𝐾= = 1.2458
𝐺𝑐 𝐺 𝑟1
𝑆+1
∴ 𝐺𝑐 = 1.2458
𝑠 + 3.2514
Dr. M.Fekry 34
Designing a Lead Compensator with Root
Locus: Example (7)
• The error constants of this system are evaluated. We find that this
system with two open-loop integrations will result in a zero steady-
state error for a step and ramp input signal. The acceleration constant
is
2
10(1.2458)(𝑠 + 1) 10(1.2458)
𝐾𝑎 = lim 𝑠 2
= = 3.8315
𝑠→0 𝑠 𝑠 + 3.2514 3.2514

Dr. M.Fekry 35
Designing a Lead Compensator with Root
Locus: Example (7)

Dr. M.Fekry 36
Designing a Lead Compensator with Root
Locus: Example (8)
Consider the position control system shown in Figure 6–39(a). The
feedforward transfer function is
10
𝐺 𝑠 =
𝑠(𝑠 + 1)
Design a lead compensator 𝐺𝑐 (𝑠) so that the dominant closed-loop poles
have the damping ratio 𝜁 = 0.5 and the undamped natural frequency 𝜔𝑛
= 3 𝑟𝑎𝑑/𝑠

Dr. M.Fekry 37
Designing a Lead Compensator with Root
Locus: Example (8)
1. The desired pole location
𝑠 2 + 2𝜁𝜔𝑛 𝑠 + 𝜔𝑛2 = 0
2
𝑠 + 3𝑠 + 9 = 0
𝑟1,2 = −1.5000 ± 𝑗 2.5981

Dr. M.Fekry 38
Designing a Lead Compensator with Root
Locus: Example (8)
2. Draw the root locus of
uncompensated system.

4. It is necessary to design
lead compensator that
meet the desired roots
location.

Dr. M.Fekry 39
Designing a Lead Compensator with Root
Locus: Example (8)
5. Assume the lead
compensator 𝐺𝑐 (𝑠) to be

1
𝐾𝑐 𝑠 +
𝐺𝑐 𝑠 = 𝜏
1
𝑠+
𝛼𝜏

Dr. M.Fekry 40
Designing a Lead Compensator with Root
Locus: Example (8)
6. Place the zero of the
phase-lead compensator
directly at
𝑍𝑐 = −1 + 𝑗0
To cancel the pole at 𝑠
= −1

Dr. M.Fekry 41
Designing a Lead Compensator with Root
Locus: Example (8)
7. Determine the pole location so that
the total angle at the desired root
location is 180°
180 = −𝜃1 + −𝜃1 + 𝜃𝑧𝑐 + −𝜃𝑝𝑐
2.5981
180 = − 180 − tan −1
𝜽𝒑𝒄 𝜽𝟐 𝜽𝟏
1.5
2.5981
− 180 − tan−1 𝜽𝒛𝒄
0.5
2.5981
+ 180 − tan−1 − 𝜃𝑝𝑐
0.5
∴ 𝜃𝑝𝑐 = 60°
2.5981
∴𝑝=− − 1.5 = −3
tan 𝜃𝑝𝑐

Dr. M.Fekry 42
Designing a Lead Compensator with Root
Locus: Example (8)
8. Evaluate the total system gain at the desired root location and then
calculate the error constant.
10𝐾(𝑠 + 1)
𝐺𝑐 𝐺 =
𝑠(𝑠 + 1)(𝑠 + 3)
10𝐾 𝑠 + 1
𝐺𝑐 𝐺 𝑟1 = 2 = 1.111 𝐾
𝑠 𝑠 + 3.2514
1
𝐾= = 0.9
𝐺𝑐 𝐺 𝑟1
𝑠+1
∴ 𝐺𝑐 = 0.9
𝑠+3
Dr. M.Fekry 43
Designing a Lead Compensator with Root
Locus: Example (8)
• The error constants of this system are evaluated. We find that this
system with one open-loop integrations will result in a zero steady-
state error for a step. The ramp input signal is
10𝐾(𝑠 + 1) 10(0.9)
𝐾𝑎 = lim 𝑠 = =3
𝑠→0 𝑠(𝑠 + 1)(𝑠 + 3) 3

Dr. M.Fekry 44
Designing a Lead Compensator with Root
Locus: Example (8)

Dr. M.Fekry 45