Root locus techniques
j ω
5
K=50
4
3
2
K=0
−5
R(s)
+
E(s)
−
K
1
K=25
K=0
σ
0
σ
C(s)
s(s+10)
K=50
•The root locus show the changes in the transient response as the gain ,K, is varied.
K < 25 real poles, overdamped
K = 25 multiple poles, critically damped
K > 25 Complex poles, underdamped
1
• Looking at the underdamped poles (K > 25),
the real parts of the complex poles stay the
same. The setting time is inversely proportional to the real parts for this second order
system so we can say that the settling time will
remain constant for underdamped responses
regardless of the value of K.
•Also, as we increase the gain, the damping
ratio diminishes, and the percent overshoot increases. The damped frequency of oscillator,
equal to the imaginary part of the poles, also
increases with the gain, resulting in a reduction
of the peak time.
•Finally, since the root locaus never passes into
the right half plane, the system is always stable, regardless of the value of K, the gain.
2
Properties of the root locus
Consider the system
R(s)
+
E(s)
C(s)
K G(s)
−
H(s)
R(s)
KG(s)
C(s)
1+KG(s)H(s)
KG(s)
T (s) =
1 + KG(s)H(s)
(1)
3
• A pole exists when the characteristic polynomial in the denominator becomes zero, or
KG(s)H(s) = −1 = 16 (2k + 1)180o,
k = 0, ±1, ±2, ±3, ...
so, |KG(s)H(s)| = 1
6 KG(s)H(s) = 6 (2k + 1)180o
(2)
• The first equation implies that substituting a
value of s into the function KG(s)H(s) yields
a complex number, if this complex number has
an angle which is an odd multiples of 180o, the
value, s, is a system pole for a particular value
of K given by
1
K=
|G(s)H(s)|
(3)
Using the previous example, we already know
closed loop poles exist at −9.47 and −0.53
when K =5. For this system,
K
KG(s)H(s) =
s(s + 10)
(4)
4
Substituting s = −9.47, K = 5
KG(s)H(s) =
5
= −1 (5)
−9.47(−9.47 + 10)
• Applying these concepts to a more complex
system.
R(s)
+
E(s)
−
K (s+3)( s+4)
C(s)
(s+1)(s+2)
The open loop transfer function is
K(s + 3)(s + 4)
(s + 1)(s + 2)
The closed loop transfer function is
KG(s)H(s) =
T (s) =
(6)
K(s + 3)(s + 4)
(1 + K)s2 + (3 + 7K)s+(2 + 12K)
(7)
5
• If a point, s, is a closed loop pole of the
system for a value K, then s must satisfy the
two magnitude and angle equations. This can
be checked graphically by looking at the open
loop poles and zeros.
jω
3
2
L1
L2
θ1
−4
−3
L
3
θ2
−2
L
4
θ3
θ4
1
−1
6
• Consider the point −2 + 3j, if this point is
closed loop pole than the angles of the zeros
minus the angles of poles must equal an odd
multiples of 180o.
θ1 + θ2 − θ3 − θ4 =50.31 + 71.57 − 90 − 108.43 =
−70.55o
therefore, −2 + 3j is not a point on the root
locus, i.e. not a closed loop pole for any gain.
•√Repeat these calculations for a point (−2 +
j 22 ). The angles do add up to 180o. So this
is a point on the root locus for a value of K,
given by,
1
pole lengths
K=
= 1/M = Q
(8)
|G(s)H(s)|
zero lengths
√
2/2 × 1.22
L3 × L 4
K=
=
= 0.33
(9)
L1 × L 2
2.12 × 1.22
Q
√
• Thus the point (−2 + j 22 ) is a point at the
root locus for a gain of 0.33.
7
Example: Given unity feedback system that
has a forward transfer function
K(s + 2)
G(s) = 2
s + 4s + 13
a) Calculate the angles of G(s) at the point
(−3 + j0) by finding the sums of angles of vectors drawn from the zeros and poles of G(s) to
the given point.
b) Determine if this point lies on the root locus.
c)If so, find the gain, K, using the lengths of
the vectors.
8
K(s + 2)
G(s) = 2
s + 4s + 13
a) zero at -2, poles at
−b ±
q
b2 − 4ac
, a = 1, b = 4, c = 13
2a
√
−4 ± 16 − 52
= −2 ± 3j
2
θ2
j ω
3
2
L2
1
θ1
L1
−3
−2
−1
L3
θ3
9
θ1 = 180o, L1 = 1.
o, L =
θ2 = 180o + arctan[ 3
]
=
251.565
2
1
o
θ3 = 180o − arctan[ 3
1 ] = 108.435 , L3 =
√
√
10.
10.
θ = θ1 − θ2 − θ3 = 180 − 257.565 − 108.435 =
−180o
b) This point is on the root locus.
Q
√ √
pole
lengths
c) K = 1/M = Q zero lengths = 101 10 = 10.
10