EECS444, Hui
Design vis Root Locus
Outline
 Use the root locus to design cascade compensators to improve the
steady-state error
 Use the root locus to design cascade compensators to improve the
transient response
 Use the root locus to design cascade compensators to improve both
the steady-state error and the transient response
 Use the root locus to design feedback compensators to improve the
transient response
 Realize the designed compensators physically
EECS444, Hui
Improve system response
 The root locus typically allows us to choose the proper loop gain along
the root locus to meet a transient response specification. Thus, we are
limited to those responses that exist along the root locus.
 Then what if we need to achieve a response that requires the position
of (closed-loop) poles outside the root locus?
 Options:
(1) Replace the system with a new system with desired response: this
would be expensive
(2) Add a compensator with additional poles and zeros: it is a
modification
EECS444, Hui
Improving Transient Response
 For example, moving the pole from A to B on the same damping ratio
line will speed up the response without affecting the percent overshoot.
Since B is not on the root locus, moving a pole from A to B cannot be
accomplished simply by changing the gain K.
 Add additional poles so that the compensated system has a root locus
that goes through point B at some value of gain.
EECS444, Hui
Improving Steady-State Error
 Steady-state error can be improved by adding an open-loop pole at the
origin in the forward path (requiring an integration), thus increasing
the system type and driving the associated steady-state error to zero.
 Increasing gain helps reducing steady-state error but increasing the
percent overshoot
EECS444, Hui
Configurations of compensation
Both compensators
are used at places
with low-power
signals
Plant: can be considered
as a power amplifier
Cascade compensation
Feedback compensation
EECS444, Hui
Some terminologies
 Proportional control system: Feeds the error to the plant with the size of
the correction proportional to the difference between the desired value
and the measured value
 Proportional-Integral Compensator: use an active integrator
 Lag compensator: Does not employ pure (active circuit) integrator
 Derivative control system: Feeds the derivative of the error to the plant
 Proportional-Integral (PI) controller: Ideal integral compensator
EECS444, Hui
Ideal Proportional Integral (PI) Compensation
To improve steady-state error
−θ1 − θ 2 − θ3=
( 2k + 1)180°
−θ1 − θ 2 − θ3 − θ c ≠ ( 2k + 1)180°
Adding an extra
pole at origin
A system operate with desired
transient response with point A
on the root locus
An extra pole is added at the origin to
reduce steady-state error. Then, point A
will no longer be on the root locus (that
means transient response will change)
EECS444, Hui
Improve steady-state error without affecting the transient
response
 Add a zero (𝑠𝑠 = −𝑎𝑎) near the new pole at
origin (the value of 𝑎𝑎 is small)
 The dominate poles are about the same as
those without compensation
 Now the angular contribution of the
compensator zero and compensator pole
approximately cancel each other
 𝜃𝜃𝑝𝑝𝑝𝑝 > 𝜃𝜃𝑧𝑧𝑐𝑐 so 𝜃𝜃𝑧𝑧𝑐𝑐 − 𝜃𝜃𝑝𝑝𝑐𝑐 is negative: lag
compensation (new zero is negative)
 Point A is now on the original root locus
 the system type has been increased
(increased # of poles at origin), so that the
steady-state error is reduced
 The same transient response
−θ1 − θ 2 − θ3 − θ pc + θ zc
( 2k + 1)180°
 This is an ideal integral compensator
EECS444, Hui
Example
This system operates with a
desired damping ratio of 0.174
The addition of an ideal integral compensator reduces the steady-state error to
zero for a step input without appreciably affecting transient response.
 The compensator adds a pole at the origin to increase the system type (so to
reduce steady-state error)
 The compensator adds a zero at −0.1, very close to the pole, to cancel the
angular contribution of the new pole
EECS444, Hui
Root locus for uncompensated system
Open-loop gain G ( s) =
K
( s + 1)( s + 2 )( s + 10 )
Open-loop poles: 𝑝𝑝1 = −1, 𝑝𝑝2 = −2, 𝑝𝑝3 = −10
K
Closed-loop
T (s) =
K + ( s + 1)( s + 2 )( s + 10 )
transfer function
Positions of the 3 closed-loop poles (root locus)
are dependent on the gain 𝐾𝐾
zeros = [1];
poles = poly([-1 -2 -10]);
GH = tf(zeros, poles)
rlocus(GH)
sgrid(0.174, 0)
Blue lines represent root locus
with varying 𝐾𝐾
EECS444, Hui
Root locus for uncompensated system
(1) Damping ratio of 0.174 is represented by a
radial line drawn on the s-plane at 100.02°.
(2) The dominant poles are 𝑠𝑠 = 0.694 ±
𝑗𝑗𝑗.926 for a gain, 𝐾𝐾 = 164.6
(3) With 𝐾𝐾 = 164.6, we find that the third pole is
approximately at 𝑠𝑠 = −11.61.
For this Type 0 system, the
steady-state error is
=
=
K p lim G
(s)
s →0
e=
(∞)
164.6
= 8.23
20
1
1
=
= 0.108
1 + K p 1 + 8.23
EECS444, Hui
Root locus for compensated system
 Adding an ideal integral compensator with a
pole at the origin and a zero at −0.1
 A new section of the root locus is added
between the origin and −0.1
 Other sections of root locus are only slightly
changed
 Intersection with 𝜁𝜁 = 0.174 is now at 𝑠𝑠 =
0.678 ± 𝑗𝑗𝑗.837 for a gain, 𝐾𝐾 = 158.2 (these
two major poles did not move much compared
to uncompensated system)
zeros = poly([-0.1]);
poles = poly([0 -1 -2 -10]);
GH = tf(zeros, poles)
rlocus(GH)
sgrid(0.174, 0)
EECS444, Hui
Root locus for compensated system
 The third pole is now approximately at 𝑠𝑠 =
− 11.55
 The fourth closed-loop pole is found at −0.0902,
close enough to the zero to cause pole-zero
cancellation
 This is now a Type 1 system (with a pole at
origin), the steady-state error is zero
EECS444, Hui
With and without compensation: A comparison
EECS444, Hui
Time-domain transient response comparison
Similar transient response, but zero
steady-state error after compensation
EECS444, Hui
Implementation

K 
K1  s + 2 
K1 
K
Gc ( s ) =K1 + 2 = 
s
s
In this example, choose
𝐾𝐾2 /𝐾𝐾1 = 0.1, 𝐾𝐾1 = 𝐾𝐾
PI: proportional
+ integrator
EECS444, Hui
Implementation
Implementing an ideal integrator requires
active circuit. For example, a Miller integrator
𝑣𝑣
𝑣𝑣𝑖𝑖
𝑇𝑇 𝜔𝜔 = 0 = −
1
𝑗𝑗𝑗𝑗𝐶𝐶
𝑅𝑅
=
−1
−1
=
𝑗𝑗𝑗𝑗𝑅𝑅𝐶𝐶
𝑠𝑠𝑅𝑅𝑅𝑅
R
+
vi
_
_
+
C
A
+
v0
_
If we use passive networks (for example with an RC, or LC circuit), the pole and
zero are moved to the left, not at the origin but can be very close to the origin:
Lag (negative phase) compensator. Still helps reducing steady-state error.
Gc ( s ) =
( s + zc )
( s + pc )
EECS444, Hui
Passive implementation example:
+
1
𝑅𝑅2 +
𝑠𝑠𝐶𝐶
𝑣𝑣0
𝑠𝑠𝑅𝑅2 𝐶𝐶 + 1
𝑇𝑇 𝜔𝜔 =
=
=
𝑣𝑣𝑖𝑖 𝑅𝑅 + 𝑅𝑅 + 1
𝑅𝑅1 + 𝑅𝑅2 𝑠𝑠𝑠𝑠 + 1
1
2
𝑠𝑠𝑠𝑠
1
𝑠𝑠
+
𝑅𝑅2
𝑠𝑠 + 𝑧𝑧𝑐𝑐
𝑅𝑅2 𝐶𝐶
=
= 𝐾𝐾
1
𝑅𝑅1 + 𝑅𝑅2 𝑠𝑠 +
𝑠𝑠 + 𝑝𝑝𝑐𝑐
𝑅𝑅1 + 𝑅𝑅2 𝐶𝐶
𝑃𝑃
𝑅𝑅2
𝐾𝐾 =
𝑅𝑅1 + 𝑅𝑅2
1
𝑧𝑧𝑐𝑐 =
𝑅𝑅2 𝐶𝐶
𝑝𝑝𝑐𝑐 =
1
< 𝑝𝑝𝑐𝑐
𝑅𝑅1 + 𝑅𝑅2 𝐶𝐶
𝜃𝜃𝑧𝑧𝑧𝑧
vi
_
𝜃𝜃𝑝𝑝𝑝𝑝
−𝑧𝑧𝑐𝑐 −𝑝𝑝𝑐𝑐
New zero is on the
left of new pole
𝑗𝑗𝑗𝑗
𝑅𝑅1
𝑅𝑅2
C
+
v0
_
Since the capacitance
C can never be
infinite, 𝑝𝑝𝑐𝑐 cannot be
at the origin
𝜎𝜎
Lag compensator: The new pole is on the right of the new zero
This is because 𝜃𝜃𝑧𝑧𝑧𝑧 − 𝜃𝜃𝑝𝑝𝑐𝑐 < 0 (negative phase)
EECS444, Hui
How can this (non-idea pole-zero pair) reduce steady-state error?
Assume the uncompensated Type 1 system, its ramp response has a static
error constant at 𝑠𝑠 → 0, which is,
K vO =
K z1 z2 
p1 p2 
K v = lim [ sG ( s ) ]
s →0
After adding a lag compensator, static error constant becomes
K vN =
( K z1 z2 )( zc )
( p1 p2 )( pc )
EECS444, Hui
Root locus of
uncompensated system
K z1 z2 
K vO =
p1 p2 
Root locus of
compensated system
K z1 z2 )( zc )
(
Kv =
( p1 p2 )( pc )
Zero and pole
are very close
N
1) In order not to change the major pole at P, we need 𝑧𝑧𝑐𝑐 ≈ 𝑝𝑝𝑐𝑐
𝑧𝑧
𝑝𝑝𝑐𝑐
2) In order for 𝐾𝐾𝑣𝑣𝑣𝑣 ≫ 𝐾𝐾𝑣𝑣0 , we need 𝑐𝑐 ≫ 1
The values of both 𝑧𝑧𝑐𝑐 and 𝑝𝑝𝑐𝑐 need to be very small but their ratio can be
big. For example, 𝑝𝑝𝑐𝑐 = −0.001, 𝑧𝑧𝑐𝑐 = −0.01 so that 𝑧𝑧𝑐𝑐 − 𝑝𝑝𝑐𝑐 = −0.09
(very small value), but 𝑧𝑧𝑐𝑐 /𝑝𝑝𝑐𝑐 = 10 (a reasonably big value)
EECS444, Hui
Ideal Proportional Derivative Compensation (PD)
(To improve transient response)
Purpose: Design a system that has a desirable percent overshoot and a shorter
settling time than the uncompensated system
Add a single zero to the forward path can speed up the transient response
The function of 𝑮𝑮𝒄𝒄 (𝒔𝒔) = 𝒔𝒔 + 𝒛𝒛𝒄𝒄 is called an ideal derivative, or PD controller
Differentiator Pure gain
Why and how an ideal derivative compensation can speed up the response?
Example:
An uncompensated system
has open-loop gain
G (s) =
R( s) +
K
( s + 1)( s + 2)( s + 5)
−
K
1
( s + 1)( s + 2)( s + 5)
C (s)
EECS444, Hui
Example:
K
G (s) =
( s + 1)( s + 2)( s + 5)
zeros = [1];
poles = poly([-1 -2 -5]);
GH = tf(zeros, poles)
rlocus(GH)
sgrid(z, 0)
[K,p] = rlocfind(GH)
R( s) +
−
K
1
( s + 1)( s + 2)( s + 5)
C (s)
Dominant closedloop poles
The root locus crosses with
damping line 𝜁𝜁 = 0.4 at 𝑠𝑠 =
− 0.94 ± 2.15 with 𝐾𝐾 = 23.7
EECS444, Hui
Adding a zero at 𝑧𝑧1 = −4
K ( s + 4)
G (s) =
( s + 1)( s + 2)( s + 5)
zeros = poly([-4]);
poles = poly([-1 -2 -5]);
GH = tf(zeros,poles)
rlocus(GH)
z=0.4;
sgrid(z, 0)
[K,p] = rlocfind(GH)
The root locus crosses
with damping line 𝜁𝜁 = 0.4
at 𝑠𝑠 = −1.9 ± 4.3 with
𝐾𝐾 = 20.8
R( s) +
−
s+4
K
( s + 1)( s + 2)( s + 5)
C (s)
Dominant closedloop poles
EECS444, Hui
Adding a zero at 𝑧𝑧1 = −3
G (s) =
K ( s + 3)
( s + 1)( s + 2)( s + 5)
zeros = poly([-3]);
poles = poly([-1 -2 -5]);
GH = tf(zeros,poles)
rlocus(GH)
z=0.4;
sgrid(z, 0)
[K,p] = rlocfind(GH)
The root locus crosses
with damping line 𝜁𝜁 =
0.4 at 𝑠𝑠 = −2.44 ± 5.6
with 𝐾𝐾 = 35.3
R( s) +
−
s+3
K
( s + 1)( s + 2)( s + 5)
C (s)
Dominant closedloop poles
EECS444, Hui
Adding a zero at 𝑧𝑧1 = −2
K ( s + 2)
G (s) =
( s + 1)( s + 2)( s + 5)
Equivalently this cancels
the pole at 𝑝𝑝2 = −2
zeros = [1];
poles = poly([-1 -5]);
GH = tf(zeros,poles);
rlocus(GH)
z=0.4;
sgrid(z, 0)
[K,p] = rlocfind(GH)
The root locus crosses with
damping line 𝜁𝜁 = 0.4 at 𝑠𝑠 =
− 3 ± 6.87 with 𝐾𝐾 = 51.3
R( s) +
−
s+2
K
( s + 1)( s + 2)( s + 5)
C (s)
Dominant closedloop poles
EECS444, Hui
Impact of ideal derivative
compensation
Moving the new zero from left
to right increases oscillation
frequency 𝜔𝜔𝑑𝑑 , (the imaginary
part of the closed-loop root)
EECS444, Hui
Decreasing 𝑎𝑎 value increases
both 𝜎𝜎𝑑𝑑 and 𝜔𝜔𝑑𝑑 , and thus
decreases settling time.
R( s) +
−
2
s+a
3 𝑎𝑎 = 4
K
( s + 1)( s + 2)( s + 5)
C (s)
No comp
EECS444, Hui
Impact of ideal derivative compensation
 Compensation (moving the new zero toward right) moves the dominant
closed-loop poles farther out along the 0.4 damping ratio line than the
uncompensated system (same percent overshoot)
 Real part of the compensated dominant poles become more negative
(stronger damping and shorter settling times).
 Imaginary parts of the compensated dominant poles become larger
(higher oscillation frequency, and smallest peak time)
 As the zero is placed farther left from the dominant poles (𝑧𝑧 = −2, 𝑧𝑧 =
− 3, and 𝑧𝑧 = −4) the closed-loop, compensated dominant poles move
closer to the origin and to the uncompensated, dominant closed-loop
poles (the impact of compensation is reduced).
EECS444, Hui
Design example
R( s) +
Gc ( s ) K
G (s) =
s ( s + 4)( s + 6)
−
Gc ( s )
K
s ( s + 4)( s + 6)
C (s)
Design an ideal derivative compensator 𝐺𝐺𝑐𝑐 (𝑠𝑠) to yield a 16% overshoot, with a
threefold reduction in settling time
Solution:
=
ζ
For uncompensated system
with16% overshoot, the damping
line can be found.
For 𝜔𝜔𝑛𝑛 = 1.212 + 2.12 = 2.42, the
settling time is,
=
Ts
4
=
ζωn
4
= 3.3
0.504 × 2.42
− ln ( %OS /100 )
=
2
π + ln 2 ( %OS /100 )
zeros = [1];
poles = poly([0 -4 -6]);
GH = tf(zeros,poles)
rlocus(GH)
z=0.504;
sgrid(z, 0)
[K,p] = rlocfind(GH)
− ln ( 0.16 )
= 0.504
π 2 + ln 2 ( 0.16 )
𝑠𝑠 = −1.21 ± 𝑗𝑗𝑗.1
𝐾𝐾 = 43.4
EECS444, Hui
G ( s) =
R( s) +
K ( s + a)
s ( s + 4)( s + 6)
−
s+a
K
s ( s + 4)( s + 6)
C (s)
With derivative compensation by adding a zero at 𝑠𝑠 = −𝑎𝑎
To reduce the settling time by 3-fold, we need 𝑇𝑇𝑠𝑠 = 1.11, the real part 𝜎𝜎𝑑𝑑 of
the dominant 2nd order poles can be found.
=
Ts
4
=
ζωn
4
σd
σ=
d
4
4
=
= 3.61
Ts 1.11
𝜎𝜎𝑑𝑑
= 𝜁𝜁 = 0.504, so that 𝜃𝜃 = 𝜋𝜋 −
𝜔𝜔𝑛𝑛
−1 0.504 = 120.26° , the oscillation
Since 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 =
𝑐𝑐𝑐𝑐𝑐𝑐
frequency 𝜔𝜔𝑑𝑑 can be found as,
ωd
3.613 tan (180=
° − 120.26° ) 6.2
Now we need to design the system with a closed-loop dominant poles at 𝑠𝑠 =
− 3.61 ± 𝑗𝑗𝑗.2 by adding a zero at 𝑠𝑠 = −𝑎𝑎 (we need to find the value of 𝒂𝒂)
EECS444, Hui
R( s) +
K ( s + a)
G ( s) =
s ( s + 4)( s + 6)
−
s+a
Open-loops poles 𝑝𝑝1 = 0, 𝑝𝑝2 = −4,
𝑝𝑝3 = −6
6.2
4−3.61
6.2
𝜃𝜃𝑝𝑝𝑝 = 𝑡𝑡𝑡𝑡𝑡𝑡−1
6−3.61
𝜃𝜃𝑝𝑝𝑝 = 𝑡𝑡𝑡𝑡𝑡𝑡−1
6.2
3.61
= 120.3°
= 86.4°
= 68.9.4
°
θ p1 + θ p 2 + θ p=
120.3 + 86.4 + 68.9
3
C (s)
−3.61 ± 𝑗𝑗𝑗.2
Corresponding angles are
𝜃𝜃𝑝𝑝𝑝 = 180° − 𝑡𝑡𝑡𝑡𝑡𝑡−1
K
s ( s + 4)( s + 6)
𝜃𝜃𝑝𝑝𝑝
𝜃𝜃𝑝𝑝𝑝
𝜃𝜃𝑧𝑧
−𝑎𝑎
𝜃𝜃𝑝𝑝𝑝
275.6 ≠ 180
Adding a new zero at z = −𝑎𝑎, should make
𝜃𝜃𝑧𝑧 − 𝜃𝜃𝑝𝑝𝑝 − 𝜃𝜃𝑝𝑝2 − 𝜃𝜃𝑝𝑝3 = −180° , that is,
θ z = 275.6 − 180= 95.6


6.2
6.2



180 − tan −1 
95.6
=

 − 3.6 ≈ −3
=
a


tan
84.4
 − a − 3.61 

(
) 
EECS444, Hui
G (s) =
K ( s + 3)
s ( s + 4)( s + 6)
zeros = poly([-3]);
poles = poly([0 -4 -6]);
GH = tf(zeros,poles)
rlocus(GH)
z=0.504;
sgrid(z, 0)
[K,p] = rlocfind(GH)
After adding a new
zero at a = −3, 𝑠𝑠 =
− 3.61 ± 𝑗𝑗𝑗.2 is now
on the root locus, and
the dominant closedloop poles are at 𝑠𝑠 =
− 3.61 ± 𝑗𝑗𝑗.2 with
𝐾𝐾 = 47.4
R( s) +
−
K
s ( s + 4)( s + 6)
s+3
C (s)
−3.61 + 𝑗𝑗𝑗.2
−3
−2.77
−3.61 − 𝑗𝑗𝑗.2
EECS444, Hui
Step response comparison
G ( s) =
47.4( s + 3)
s ( s + 4)( s + 6)
G (s) =
43.4
s ( s + 4)( s + 6)
Note:
 Adding a new zero, the closed-loop pole positions will change significantly
 If you want to maintain the closed-loop pole positions, you can add a new
zero/pole pair
EECS444, Hui
Lag compensation vs Lead compensation
In both cases the difference 𝜃𝜃𝑐𝑐𝑐𝑐 − 𝜃𝜃𝑐𝑐𝑐𝑐 provides the required phase compensation to
satisfy ∑ 𝜃𝜃𝑖𝑖 = 2𝑘𝑘 + 1 180° .
𝜁𝜁
𝑃𝑃
𝜁𝜁
𝑗𝑗𝑗𝑗
𝜃𝜃𝑐𝑐𝑐𝑐
𝜃𝜃𝑐𝑐𝑐𝑐
𝜎𝜎
The desired closed-loop pole P is initially not
on the root locus.
Add a new pole-zero pair (or only a new pole)
New zero is on the left of new pole, so that
𝜃𝜃𝑐𝑐𝑐𝑐 − 𝜃𝜃𝑐𝑐𝑐𝑐 < 0, (negative). This is known as
lag compensation
𝜃𝜃𝑐𝑐𝑐𝑐
𝑃𝑃
𝑗𝑗𝑗𝑗
𝜃𝜃𝑐𝑐𝑐𝑐
𝜎𝜎
The desired closed-loop pole P is initially not
on the root locus.
Add a new pole-zero pair (or only a new zero)
New pole is on the left of new zero, so that
𝜃𝜃𝑐𝑐𝑐𝑐 − 𝜃𝜃𝑐𝑐𝑐𝑐 > 0, (positive). This is known as
lead compensation
EECS444, Hui
Lead compensation example
G (s) =
R( s) +
−
Gc ( s ) K
s ( s + 4)( s + 6)
Gc ( s )
K
s ( s + 4)( s + 6)
Design three lead compensators 𝐺𝐺𝑐𝑐 (𝑠𝑠) that reduces
the settling time by a factor of 2 while maintaining
30% overshoot. Compare the system
characteristics between the three designs
Solution:
For uncompensated system with 30%
overshoot, the damping line can be found.
ζ
− ln ( %OS /100 )
=
2
π + ln 2 ( %OS /100 )
− ln ( 0.3)
= 0.358
π 2 + ln 2 ( 0.3)
Settling time,
=
Ts
4
=
ζωn
4
=
σd
4
= 3.972
1.007
EECS444, Hui
C (s)
Gc ( s ) K
G (s) =
s ( s + 4)( s + 6)
R( s) +
−
Gc ( s )
K
s ( s + 4)( s + 6)
For a 2-fold reduction of settling time, 𝑇𝑇𝑠𝑠 = 3.972/2 = 1.986.
Ts =
4
ζωn
ωn
=
4
4
=
= 5.626
ζ Ts 0.358 ×1.986
𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 1 − 𝜁𝜁 2 = 𝜔𝜔𝑛𝑛 1 − 0.3582 = 5.25
 For an ideal PD compensator, 𝐺𝐺𝑐𝑐 𝑠𝑠 = 𝑠𝑠 + 𝑧𝑧𝑐𝑐
 For a general Lead compensator, 𝐺𝐺𝑐𝑐 𝑠𝑠 =
𝑠𝑠+𝑧𝑧𝑐𝑐
𝑠𝑠+𝑝𝑝𝑐𝑐
Now we need to design the system with a closed-loop dominant poles
at 𝑠𝑠 = −2.014 ± 𝑗𝑗𝑗.25 by adding a zero at 𝑠𝑠 = −𝑧𝑧𝑐𝑐 and a new pole at
𝑠𝑠 = −𝑝𝑝𝑐𝑐 (we need to find the values of 𝒛𝒛𝒄𝒄 and 𝒑𝒑𝒄𝒄 )
EECS444, Hui
C (s)
( s − zc ) K
G ( s) =
s ( s + 4)( s + 6)( s + pc )
R( s) +
−
s + zc
s + pc
K
s ( s + 4)( s + 6)
C (s)
With 𝑧𝑧𝑐𝑐 = −5, 𝑝𝑝𝑐𝑐 = −42.96; 𝑧𝑧𝑐𝑐 = −4, 𝑝𝑝𝑐𝑐 = −20.09; 𝑧𝑧𝑐𝑐 = −2, 𝑝𝑝𝑐𝑐 = −8.971
EECS444, Hui
For 𝑧𝑧𝑐𝑐 = −5
Open-loops poles 𝑝𝑝1 = 0, 𝑝𝑝2 = −4, 𝑝𝑝3 = 6
G (s) =
( s + zc ) K
s ( s + 4)( s + 6)( s + pc )
Corresponding angles are
𝜃𝜃𝑝𝑝𝑝 = 180° − 𝑡𝑡𝑡𝑡𝑡𝑡−1
5.25
4−2.014
6.2
𝜃𝜃𝑝𝑝𝑝 = 𝑡𝑡𝑡𝑡𝑡𝑡−1
6−2.014
6.2
𝜃𝜃𝑧𝑧𝑧𝑧 = 𝑡𝑡𝑡𝑡𝑡𝑡−1
5−2.014
𝜃𝜃𝑝𝑝𝑝 = 𝑡𝑡𝑡𝑡𝑡𝑡−1
5.25
2.014
Without adding
new pole at −𝑝𝑝𝑐𝑐
= 110.99°
= 69.28°
= 52.79°
= 60.37°
𝜃𝜃𝑧𝑧𝑧𝑧 − 𝜃𝜃𝑝𝑝𝑝 − 𝜃𝜃𝑝𝑝2 − 𝜃𝜃𝑝𝑝3 =
60.37° − 110.99° − 69.28° − 52.79° =
− 172.69°
−2.014 ± 𝑗𝑗𝑗.25
𝑝𝑝3
𝑧𝑧𝑐𝑐
𝑝𝑝2
𝑝𝑝1
We need to add another pole at −𝑝𝑝𝑐𝑐 to
provide an angle of 𝜃𝜃𝑝𝑝𝑐𝑐 = 180° −
172.69° = 7.31° , and this requires
5.252
= tan 7.31°
pc − 2.014
pc = 42.96
EECS444, Hui
=
G ( s)
s+5
K
⋅
s + 42.96 s ( s + 4)( s + 6)
R( s) +
−
Gc ( s )
zeros = poly([-5]);
poles = poly([0 -4 -6 -42.96]);
GH = tf(zeros,poles)
rlocus(GH)
z=0.358;
sgrid(z, 0)
[K,p] = rlocfind(GH)
−43.8
With 𝐾𝐾 = 1423, the closed-loop poles are at
− 43.8 (which has minimum impact), −5.134
(which approximately cancels with the zero at
− 5, and −2.014 + 𝑗𝑗𝑗.25 (dominant poles)
K
s ( s + 4)( s + 6)
C (s)
−2.014 + 𝑗𝑗𝑗.25
−5.134
−2.014 − 𝑗𝑗𝑗.25
EECS444, Hui
For 𝑎𝑎 = −4
G ( s)
=
s+4
K
⋅
s + 20.09 s ( s + 4)( s + 6)
R( s) +
zeros = poly([-4]);
poles = poly([0 -4 -6 -20.09]);
GH = tf(zeros,poles)
rlocus(GH)
z=0.358;
sgrid(z, 0)
[K,p] = rlocfind(GH)
−22.06
With 𝐾𝐾 = 698, the closed-loop poles are at
− 22.06 (low impact), and −2.014 + 𝑗𝑗𝑗.25
(dominant poles)
−
Gc ( s )
K
s ( s + 4)( s + 6)
C (s)
−2.014 + 𝑗𝑗𝑗.25
−2.014 − 𝑗𝑗𝑗.25
EECS444, Hui
For 𝑎𝑎 = −2
=
G (s)
s+2
K
⋅
s + 8.971 s ( s + 4)( s + 6)
R( s) +
−
zeros = poly([-2]);
poles = poly([0 -4 -6 -8.971]);
GH = tf(zeros,poles)
rlocus(GH)
z=0.358;
sgrid(z, 0)
[K,p] = rlocfind(GH)
Gc ( s )
K
s ( s + 4)( s + 6)
C (s)
−2.014 + 𝑗𝑗𝑗.25
With 𝐾𝐾 = 346, the closed-loop poles are at
−1.642
− 13.3 (low impact), −1.642 (which does not
cancel with the zero at −2), and −2.014 +
𝑗𝑗𝑗.25 (dominant poles)
−2.014 − 𝑗𝑗𝑗.25
The new closed-loop pole at −1.642 will alter
the transient response
EECS444, Hui
Positions of the 3rd and 4th poles and their relative effect upon the 2nd-order approximation.
This effect is measured by their distance from the dominant poles and/or the degree of
cancellation with the closed-loop zero.
Compensation a: new zero at −5
Compensation b: new zero at −4
Compensation c: new zero at −2
EECS444, Hui
Physical Realization of Compensation with
active circuits
Inverting amplifier
V0 ( s )
Z 2 (s)
= −
Vi ( s )
Z1 ( s )
EECS444, Hui
Inverting amplifier
V0 ( s )
Z (s)
= − 2
Vi ( s )
Z1 ( s )
EECS444, Hui
Physical Realization of Compensation with
passive RC circuits
1
1
>
R2C ( R1 + R2 ) C
1
1
1
<
+
R1C
R1C R2C
EECS444, Hui