Root Locus Techniques
A graphic root-solving technique.
Gives information about the stability and transient response of feedback control systems.
Zeros of T(s) are zeros of G(s)
and poles of H(s).
Poles of T(s) depends on gain K
T(s)
Root Locus graphically shows
poles of T(s) as K varies
Closed-loop system
Equivalent Transfer Function
CEN455: Dr. Ghulam Muhammad
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Evaluation of a Complex Function via Vectors
Problem:
Given
F (s) =
( s + 1)
s ( s + 2)
Find F(s) at the point s = -3 + j4
Solution:
For zero ( point -1):
20∠116.6°
For pole at 0:
5∠126.9°
For pole at -2:
17∠104.0°
Vector magnitude,
Vector angle,
Vector
M=
20
= 0.217
5 17
θ = ∠(116.6° − 126.9° − 104.0°)
M∠θ = 0.217∠ − 114.3°
CEN455: Dr. Ghulam Muhammad
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Defining the Root Locus
CEN455: Dr. Ghulam Muhammad
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Defining the Root Locus (contd.)
Gain less than 25, overdamped.
Gain = 25, critically damped.
Gain over 25, underdamped.
Stable system, as no pole on right-hand plane.
During underdamped, real parts are same;
so settling time (which is related to real
part) remains the same.
Damping frequency (imaginary part) increases
with gain, resulting in reduction of peak time.
CEN455: Dr. Ghulam Muhammad
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Properties of Root Locus
-2+j3
If -2+j(√2/2) is on root locus, then gain K:
Find if the point -2+j3 is on root locus:
θ1 + θ 2 − θ 3 − θ 4 = 56.31° + 71.57° − 90° − 108.43° = −70.55°
∑ zero angle - ∑ pole angle
2
(1.22)
Π pole lengths L3 L4
2
K=
=
=
Π zero lengths L1 L2 (2.12)(1.22)
Not a multiple of 180 °. So, -2+j3 is not in root locus.
CEN455: Dr. Ghulam Muhammad
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Sketching the Root Locus
Number of branches:
Equals the number of closed loop poles.
Symmetry:
Symmetrical about the real axis.
Real axis segments:
For K > 0, root locus exists to the left of an odd number real axis
poles and/or zeros.
Start and end points:
The root locus begins at finite or infinite poles and ends at finite and
infinite zeros..
CEN455: Dr. Ghulam Muhammad
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Sketching the Root Locus: Example
Problem:
Solution:
Sketch the Root Locus for the system shown in the following figure.
Calculate asymptotes to find real axis intercept:
σa = ∑
finite poles - ∑ finite zeros (−1 − 2 − 4) − (−3)
4
=−
=
# finite poles - # finite zeros
4 −1
3
The angles of the lines that intersect at -4/3 is given by:
(2k + 1)π
# finite poles - # finite zeros
= π / 3 for k = 0
θa =
=π
for k = 1
= 5π / 3 for k = 2
For higher values of k, the angles would begin to repeat.
CEN455: Dr. Ghulam Muhammad
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Sketching the Root Locus: Example
There are four poles and one finite zero.
Root locus begins at poles and ends at zeros.
Where are the other zeros? At infinity.
π/3
Skill Assessment Exercise: 8.2, 8.3
CEN455: Dr. Ghulam Muhammad
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Refining the Sketch
Real-Axis Breakaway and Break-in Points:
σ1: Breakaway point; σ2: Break-in point.
Breakaway point: at maximum gain
between -2 and -1.
Break-in point: at minimum gain
between +3 and +5.
CEN455: Dr. Ghulam Muhammad
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Finding Breakaway and Break-in Points
n
1
1
=
∑1 σ + z ∑1 σ + p
i
i
m
zi and pi are negative of zero and pole values, respectively.
Example: From figure, we get
1
1
1
1
+
=
+
σ − 3 σ − 5 σ +1 σ + 2
⇒ 11σ 2 − 26σ − 61 = 0
⇒ σ = −1.45, 3.82
-1.45
CEN455: Dr. Ghulam Muhammad
3.82
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Frequency and Gain at Imaginary-Axis Crossing
s(s+1)(s+2)(s+4)+K(s+3)
Closed-loop transfer function:
T (s) =
K ( s + 3)
s 4 + 7 s 3 + 14 s 2 + (8 + K ) s + 3K
We get Routh table as follows:
− K 2 − 65 K + 720 = 0
⇒ K = 9.65
(90 − K ) s 2 + 21K = 80.35s 2 + 202.7 = 0
Gives
A complete row of zeros yields the possibility
for imaginary-axis roots.
For K > 0, only s1 row can be zero.
s = ± j1.59
Thus, the root locus crosses the
imaginary-axis at ±j1.59 at a gain of 9.65
So, the system is stable for 0 ≤ K < 9.65
CEN455: Dr. Ghulam Muhammad
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Angles of Departure and Arrival
Departure: from complex poles.
Arrival: to complex zeros.
Angle of departure:
Complex pole -1+j1
Sum (zero angles) – Sum (pole angles) = (2k+1)180°
⎛1⎞
⎛1⎞
− θ1 − θ 2 + θ 3 − θ 4 = −θ1 − 90° + tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟ = 180°
⎝2⎠
⎝1⎠
⇒ θ1 = −251.6° = 108.4°
Skill Assessment Exercise: 8.4
CEN455: Dr. Ghulam Muhammad
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