Ece4510 notes07

ECE4510/5510: Feedback Control Systems. 7–1
ROOT-LOCUS CONTROLLER DESIGN
7.1: Using root-locus ideas to design controller
I We have seen how to draw a root locus for given plant dynamics.
I We include a variable gain K in a unity-feedback configuration—we
know this as proportional control.
I Sometimes, proportional control with a carefully chosen value of K is
sufficient for the closed-loop system to meet specifications.
I But, what if the set of closed-loop pole location does not
simultaneously satisfy the geometry that defines the specifications?
I We need to modify the locus itself by adding extra dynamics—a
compensator or controller D(s):
r(t) y(t)K G(s)D(s)
I We redraw the locus and pick K in order to put the poles where we
want them. HOW?
T (s) =
K D(s)G(s)
1 + K D(s)G(s)
. Now, let G(s) = D(s)G(s)
=
K G(s)
1 + K G(s)
« We know how to draw this locus!
I Adding a compensator effectively adds dynamics to the plant.
Lecture notes prepared by and copyright c 1998–2013, Gregory L. Plett and M. Scott Trimboli

ECE4510/ECE5510, ROOT-LOCUS CONTROLLER DESIGN 7–2
Adding a left-half-plane pole or zero
I What types of (1) compensation should we use, and (2) how do we
figure out where to put the additional dynamics?
I In ECE4510/5510, the methods we discuss are “science-inspired art.”
• We need to get a “feel” for how the root locus changes when poles
and zeros are added, to understand what dynamics to use for D(s).
I In more advanced courses, we learn more powerful methods:
• In ECE5520, we learn how to put all closed-loop poles exactly
where we want them (where do we want them?)
• In ECE5530, we learn how to find the optimal set of pole locations.
I But, for us to get started, speaking in generalities, adding a
left-half-plane pole pulls the root locus to the right.
• This tends to lower the system’s relative stability and slow down
the settling of the response.
• But, providing that the closed-loop system is stable, the pole can
also decrease steady-state errors.
• In first plot: The system is stable for all K, responses are smooth.
• In second plot: System also stable for all K, but when poles
become complex, response shows overshoot and oscillations.

• In third plot: The system is stable only for small K, and oscillations
increase as the poles approach the imaginary axis.
• But, steady-state error improves from left to right (assuming the
closed-loop system is stable).
I Again, generally speaking, adding a left-half-plane zero pulls the root
locus to the left.
• This tends to make the system more stable, and speed up the
settling of the response.
• Physically, a zero adds derivative control to the system, introducing
anticipation into the system, speeding up transient response.
• However, steady-state errors can get worse.
• In ﬁrst plot: System is stable only for small K, and oscillates as
poles approach imaginary axis.
• In second plot: System is stable for all K, but still oscillates.
• In third and fourth plots: More stable, less oscillation.
• But, steady-state error degrades from left to right.
I Can’t physically add a zero without a pole: Must put pole very far left
in s-plane so we don’t deteriorate desired impact of zero.

7.2: Reducing steady-state error
I We have a number of options available to us if we wish to reduce
steady-state error.
1) Proportional feedback
D(s) = 1. u(t) = Ke(t)
T (s) =
K G(s)
1 + K G(s)
.
I Same as what we have already looked at.
I Controller consists of only a “gain knob.”
• Increasing gain K often reduces steady-state error, but can
degrade transient response.
• We have to take the locus “as given” since we have no extra
dynamics to modify it.
• Can’t independently choose steady-state error and transient
response. Can design for one or other, not both.
I Usually a very limited approach, but a good place to start.
2) Integral feedback
D(s) =
1
TI s
u(t) =
K
TI
t
0
e(τ) dτ
T (s) =
K
TI
G(s)
s
1 + K
TI
G(s)
s
.
I Usually used to reduce/eliminate steady-state error. i.e., if e(t)
constant, u(t) will become very large and hopefully correct the error.
I Ideally, we would like no error, ess = 0. (Maybe 1 % to 2 % in reality)

ANALYSIS: For a unity-feedback control system, the steady-state error to
a unit-step input is:
ess =
1
1 + K D(0)G(0)
.
I If we make D(s) =
1
TI s
, then as s → 0, D(s) → ∞
ess →
1
1 + ∞
= 0.
I Adding the integrator into the compensator has reduced error from
1
1 + Kp
to zero for systems that do not have any free integrators.
I Adding the integrator increases the system type, but as steady-state
response improves, transient response often degrades.
EXAMPLE: G(s) =
1
(s + a)(s + b)
, a > b > 0.
I Proportional feedback, D(s) = 1, G(0) =
1
ab
, ess =
1
1 + K
ab
.
−a −b
I(s)
R(s)
I We can make ess small by
making K very large, but this
often leads to poorly-damped
behavior and often requires
excessively large actuators.
I Integral feedback, D(s) =
1
TI s
, ess = 0.

−a −b
I(s)
R(s)
I Increasing K to increase the
speed of response pushes
the pole toward the
imaginary axis « oscillatory.
3) Proportional-integral (PI) control
I Now, D(s) = K 1 +
1
TI s
= K
s + (1/TI )
s
. Both a pole and a zero.
−a −b
I(s)
R(s)
I Combination of proportional
and integral (PI) solves many
of the problems with just (I)
integral.
4) Phase-lag control
I The integrator in PI control can cause some practical problems; e.g.,
“integrator windup” due to actuator saturation.
I PI control is often approximated by “lag control.”
D(s) =
(s − z0)
(s − p0)
, |p0| < |z0|.
That is, the pole is closer to the origin than the zero.
I Because |z0| > |p0|, the phase φ added to the open-loop transfer
function is negative. . . “phase lag”
I Pole often placed very close to zero. e.g., p0 ≈ 0.01.

I Zero is placed near pole. e.g., z0 ≈ 0.1. We want |D(s)| ≈ 1 for all s to
preserve transient response (and hence, have nearly the same root
locus as for a proportional controller).
I Idea is to improve steady-state error but to modify the transient
response as little as possible.
• That is, using proportional control, we have pole locations we like
already, but poor steady-state error.
• So, we add a lag controller to minimally disturb the existing good
pole locations, but improve steady-state error.
−a −b
I(s)
R(s)
I Good steady-state error
without overﬂow problems.
Very similar to proportional
control.
I The uncompensated system had loop gain Kbefore = lim
s→0
G(s).
I The lag-compensated system has loop gain
Kafter = lim
s→0
D(s)G(s) = (z0/p0) lim
s→0
G(s).
I Since |z0| > |p0|, there is an improvement in the position/velocity/etc.
error constant of the system, and a reduction in steady-state error.
I Transient response is mostly unchanged, but slightly slower settling
due to small-magnitude slow “tail” caused by lag compensator.

7.3: Improving transient response
I We have a number of options available to us if we wish to improve
transient response
1) Proportional feedback
I Again, we could use a proportional feedback controller.
I It has the same beneﬁts and limitations that we’ve already seen.
2) Derivative feedback
D(s) = TDs, u(t) = K TD ˙e(t).
I Does nothing to help the steady-state error. In fact, it can make it
worse.
I But, derivative control provides feedback that is proportional to the
rate-of-change of e(t) « control response ANTICIPATES future errors.
I Very beneﬁcial—tends to smooth out response, reduce ringing.
EXAMPLE: G(s) =
1
(s + a)(s + b)
, D(s) = TDs.
−a −b
I(s)
R(s)
I No ringing. “Very” stable.
3) Proportional-derivative (PD) control
I Often, proportional control and derivative control go together.
D(s) = 1 + TDs.

−a −b
I(s)
R(s)
I No more zero at s = 0.
I Therefore better steady-state
response.
4) Phase-lead control
I Derivative magniﬁes sensor noise.
I Instead of D-control or PD-control use “lead control.”
D(s) =
(s − z0)
(s − p0)
, |z0| < |p0|.
That is, the zero is closer to the origin than the pole.
I Same form as lag control, but with different intent:
• Lag control does not change locus much since p0 ≈ z0 ≈ 0.
Instead, lag control improves steady-state error.
• Lead control DOES change locus. Pole and zero locations chosen
so that locus will pass through some desired point s = s1.
DESIGN METHOD I: Sometimes, we can be successful by choosing the
value of z0 to cancel a stable pole in the plant.
I Then, we solve for K and p0 such that
[1 + K D(s)G(s)|s=s1
= 0.
I That is, we force one closed-loop pole to be at s = s1.
I This does not ensure that other poles do anything reasonable, so we
must always test design.

I And, what about pole-zero cancelation? Can it occur?
If our zero is too far left If our zero is too far right
p1
p2z0p0
p1
p2z0p0
I Either way, the locus is still okay. (What if we tried to cancel an
unstable pole?)
DESIGN METHOD II: If there is no stable real pole to cancel, we can still
use similar approach.
I Use somewhat modiﬁed version of lead compensator form
D(s) =
a1s + a0
b1s + 1
.
I Choose a0 to get speciﬁed dc gain (e.g., open-loop gain=Kp, Kv, . . .)
a1s + a0
b1s + 1
G(s)
s=0
= dc gain.
|a0||G(0)| = dc gain.
a0 =
Desired dc gain
|G(0)|
.
I a1 and b1 are chosen to make locus go through s = s1,
a1s1 + a0
b1s1 + 1
G(s1) = −1
for that point to be on the root locus.
« Magnitude
a1s1 + a0
b1s1 + 1
|G(s1)| = 1

« Phase
a1s1 + a0
b1s1 + 1
+ G(s1) = 180◦
.
(math happens)
a1 =
sin(β) + a0|G(s1)| sin(β − ψ)
|s1||G(s1)| sin(ψ)
b1 =
sin(β + ψ) + a0|G(s1)| sin(β)
−|s1| sin(ψ)



s1 = |s1|ejβ
G(s1) = |G(s1)|ejψ
.
5) Proportional-integral-derivative (PID) control
I There is a similar design procedure for PID control:
D(s) = K 1 +
1
TI s
+ TDs = Kp +
KI
s
+ Kds.
I Compute: Kp =
− sin(β + ψ)
|G(s1)| sin(β)
−
2KI cos β
|s1|
I Compute: KD =
sin(ψ)
|s1||G(s1)| sin(β)
+
KI
|s1|2
, where s1 = |s1|ejβ
and
G(s1) = |G(s1)|ejψ
for both cases.
I TI chosen to match some design criteria. e.g., steady-state error.
I Convert to first form via K = Kp; TI = K/KI ; TD = Kd/K.
6) Lead-lag control
I If we must satisfy both a transient and steady-state spec:
1. Design a lead controller to meet transient spec first;
2. Include lead controller with plant after its design is final;
3. Design a lag controller (where “plant” = actual plant and lead
controller combined) to meet steady-state spec.

7.4: Examples (a)
EXAMPLE I: We start with the plant G(s) =
1
(s + 1)(s + 3)
.
I The open-loop step response for G(s) is plotted to the left.
I The root locus (assuming proportional control) is plotted to the right.
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
Time (s)
Amplitude
Step response of open−loop plant
−6 −5 −4 −3 −2 −1 0 1
−4
−3
−2
−1
0
1
2
3
4
Real axis
Imaginaryaxis
Root locus
I We see that the open-loop response is smooth (good), slow (bad),
and has very large steady-state error (bad).
I But, root locus shows that proportional control moves pole locations.
I The plot to the right shows step
responses of closed-loop
systems with proportional
control.
I Changing K “shapes” the
transient response. 0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Time (s)
Amplitude
Step response of closed−loop system
K = 1
K = 4
K = 10
K = 30
I Higher values of K speed up the closed-loop response when
compared to the open-loop response (good), decrease steady-state
error (good), but also add ringing to the transient response (bad).

EXAMPLE II: We start with the plant G(s) =
s + 2
(s + 1)(s + 4)
.
I Using proportional control, we wish to solve for the value of K that
places a closed-loop pole at s = −5.
I First, we draw the locus to
ensure that it does pass through
s = −5.
I It does! Looking good so far.
−6 −5 −4 −3 −2 −1 0 1
−1
−0.5
0
0.5
1
Real axis
Imaginaryaxis
Root locus
I Next, we remember that the root-locus “magnitude condition” gives us
K =
1
|G(s)| s=−5
=
(s + 1) (s + 4)
s + 2 s=−5
=
(−4)(−1)
(−3)
=
4
3
.
I We’re done, but we can further double-check that s = −5 is a point on
the root locus using the “angle condition”
[ G(s)|s=−5 = [ (s + 2) − (s + 1) − (s + 4)|s=−5
= 180◦
− 180◦
− 180◦
= −180◦
.
I So, the angle condition is satisﬁed as well (meaning we didn’t have to
draw the root locus to ensure that s = −5 was a valid locus point).

EXAMPLE III: We start with the plant
G(s) =
1
s(10s + 1)
.
I Our goal is to have closed-loop
1. Mp < 16%. This means that ζ ≥ 0.5.
2. ts < 10 secs to 1%. This means that
σ ≥ 0.46.
3. ess for ramp input< 0.01 when slope
of ramp= 0.01. This means that
Kv = 0.01/0.01 = 1.0.
−2 −1.5 −1 −0.5
1
−1
I Since we need to change transient response, we choose to use a
lead controller.
I Since the plant has a stable real pole, we choose D(s) to
approximately cancel plant pole.
D(s) =
10s + 1
s + p0
.
I Initially, choose s1 = −0.5 + j to be a point on the locus. So, we want
1 + K
10s + 1
s + p0
1
s(10s + 1) s=s1
= 0
and
lim
s→0
s K
10s + 1
s + p0
1
s(10s + 1)
≥ 1.
I The steady-state error spec gives K ≥ p0. For simplicity, choose
K = p0.
I The transient spec gives
1 + p0
1
s(s + p0) s=s1
= 0

s1(s1 + p0) + p0 = 0
s2
1 + s1 p0 + p0 = 0
p0(1 + s1) = −s2
1
p0 = −
s2
1
1 + s1
.
I Solving gives p0 = 1.1 − 0.2 j. This is not a feasible design since p0
must be real.
I Modify p0 to p0 = 1.1. This gives
K = 1.1, Kv = 1, and poles at
−0.55 ± 0.893 j.
I This gives ωn ≈ 1 for pole locations,
so tr ≈ 1.8 s.
I Could choose slightly larger K, still achieve transient-response specs,
but have better steady-state response since K ≥ p0.

7.5: Examples (b)
EXAMPLE IV: Consider the plant G(s) =
1
s2
.
I We want to design a compensator
D(s) =
a1s + a0
b1s + 1
so the closed-loop system has a pole at s1 = 2
√
2ej135◦
= −2 + 2 j.
(The point s1 is chosen to achieve ζ = 0.707 and τ = 0.5 s.)
I Here, there is no stable real pole in G(s), so we use the second
design method for a lead compensator.
I Step 1, compute a0: We cannot compute a0 since
1
s2
s=0
→ ∞. So,
arbitrarily choose a0 = 2.
I Step 2, compute a1: Note, β = 135◦
, ψ = −270◦
because
G(s1) =
1
s2
s=2
√
2ej135◦
=
1
8
e− j270◦
.
a1 =
sin(135◦
) + 2(1/8) sin(45◦
)
(2
√
2)(1/8) sin(−270◦)
=
(1/
√
2)(1 + 1/4)
√
2/4
=
5
2
.
I Step 3, compute b1:
b1 =
sin(−135◦
) + 2(1/8) sin(135◦
)
−(2
√
2) sin(−270◦)
=
−(1/
√
2)(1 − 1/4)
−2
√
2
=
3
16
.
I So, the compensator is:
D(s) =
(5/2)s + 2
(3/16)s + 1
.

−6 −5 −4 −3 −2 −1 0 1
−4
−3
−2
−1
0
1
2
3
4
Real Axis
ImagAxis
Example locus passing through (-2,2)
EXAMPLE v: An alternative way to solve the prior problem uses
coefﬁcient matching.
I We have that G(s) =
1
s2
, and have assumed that D(s) =
a1s + 2
b1s + 1
.
I We want two closed-loop poles at s = −2 ± 2 j, but recognize that
there will be a total of three closed-loop poles (because of the added
compensator pole).
I So, we can specify a desired characteristic equation
χd(s) = (s + α)(s + 2 + 2 j)(s + 2 − 2 j)
= (s + α)(s2
+ 4s + 8)
= s3
+ (4 + α)s2
+ (8 + 4α)s + 8α = 0,
where s = −α is the (unknown a priori) location of the third pole.
I The actual characteristic equation is
χa(s) = 1 + D(s)G(s) = 0
= 1 +
a1s + 2
b1s + 1
1
s2
= b1s3
+ s2
+ a1s + 2 = 0.

I The coefficient-matching method forces the polynomial coefficients of
the desired and actual characteristic equations to be the same.
I Looking at the s3
coefficients, we could set b1 = 1, but then we would
have problems because we cannot simultaneously have
4 + α = 1 and 8α = 2.
I So, we divide χa(s) by b1, without changing its meaning:
χa(s) = s3
+
1
b1
s2
+
a1
b1
s +
2
b1
= 0.
I This has given us another degree of freedom when solving. Now, we
have
4 + α =
1
b1
, 8 + 4α =
a1
b1
and 8α =
2
b1
.
I Combining the first and third equations gives
2(4 + α) = 8α
8 = 6α
α =
4
3
.
I With this value of α, we have b1 = 3/16 and a1 = 5/2, as before.
EXAMPLE VI: Consider the compensated system of Example III.
G(s) =
1.1
s(s + 1.1)
.
I We like the transient response (so want to leave it alone), but wish to
improve the steady-state response by a factor of 10.
I This calls for a lag controller. Recall that
Kafter = (z0/p0) Kbefore,
so, we want z0/p0 ≥ 10.

I Choose p0 = 0.001. Then, z0 = 0.01 and D(z) =
s + 0.01
s + 0.001
.
−1 −0.8 −0.6 −0.4 −0.2 0
−1.5
−1
−0.5
0
0.5
1
1.5
Real Axis
ImagAxis
Lag shifts locus slightly to the right
I Plots of error versus time without and with the new lag compensator
(simulated using Simulink):
0 5 10 15 20 25
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
Error
Time (s)
Uncompensated
0 200 400 600 800 1000
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
Error
Time (s)
With lag compensator
I Notice the different time scales: The lag adds a small-amplitude slow
time constant to the output.

7.6: Compensator implementation
I Analog compensators commonly use op-amp circuits.
I See the following pages. . .
R(s)
I(s)
−
K
s
1
K
1
V1 V2
R(s)
I(s)
−Ks
1
K
1
V1 V2
−z1 R(s)
I(s)
−K(s + z1)
K
1
1/(K z1)
V1 V2
−p1 R(s)
I(s)
−K
s + p1
1/K1
K
p1
V1 V2

−p1 R(s)
I(s)
−Ks
s + p1
K
1K
p1
V1 V2
R(s)
I(s)
or
−K
s + z1
s + p1
any p1 and z1
1/K
1
1
z1
K
p1
V1 V2
R(s)
I(s)
or
−K
s + z1
s + p1
K =
K1
K2
any p1 and z1
K1
K2
V1 V2
1
K1z1
1
K2 p1
R(s)
I(s)
s + z1
s + p1
z1 > p1
1
V1 V2
1
z1 − p1
1
p1 LAG
R(s)
I(s)
s + z1
s + p1
z1 > p1
1
V1 V2
1
p1
1
z1 − p1
LAG

R(s)
I(s)
K
s + z1
s + p1
p1 > z1
K =
p1
z1
1
V1 V2
p1 − z1
z1
1
p1 LEAD
R(s)
I(s)
K
s + z1
s + p1
p1 > z1
K =
p1
z1
1
V1 V2
1
p1
p1 − z1
p1
LEAD
−z1 R(s)
I(s)
K(s + z1)
K =
1
z1
1
V1 V2
1
z1
LEAD

Ece4510 notes07

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