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2.004 Dynamics and Control II
Spring 2008
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Massachusetts Institute of Technology
Department of Mechanical Engineering
2.004 Dynamics and Control II
Spring Term 2008
Lecture 21
Reading:
• Nise: Chapter 1
1
Cruise Control Example (continued from Lecture 1)
From Lecture 1 the model for the car is:
v (t)
F d
Fp (t)
F p (t)
v (t)
v e lo c ity
m a s s
m
Fd (t)
B ( v is c o u s fr ic tio n )
where the propulsion force Fp (t) is proportional to the gas pedal depression θ:
Fp (t) = Kθ(t)
so that
m
dv
+ Bv = Ke θ(t) + Fd (t)
dt
F
e n g in e
q
g a s p e d a l
K
e
F
p
+
d
(t)
+
(t)
m
c a r m o d e l
d v
d t
+ B v = F
V
c a r s p e e d
The “Open-Loop”Dynamic Response:
Let’s examine how the car will respond to
commands at the gas-pedal. Assume Fd (t) = 0, so that we can write the differential equation
as
m
Ke
v̇ + v =
θ
B
B
and compare this to the standard form for a first-order ODE:
τ ẏ + y = f (t)
where the time-constant τ = m/B, and the forcing function f (t) =
1
c D.Rowell 2008
copyright �
2–1
Ke
θ(t).
B
Consider
1) The “coast-down” response from an initial speed v(0) = v0 with θ = 0. The homogeneous
response is
t
B
v(t) = v0 e− τ = v0 e− m t
which has the form
v
v e lo c ity v ( t)
o
e
0 .8 v
-2
e
-4
e
o
0 .6 v
-1
e
-3
= 0
= 0
= 0
= 0
.3 6 7
.1 3 5
.0 4 9
.0 1 8
9
3
8
3
o
0 .4 vo
0 .2 v
o
0
0
t
3 t
2 t
4 t
t (s e c s )
and after a period 4τ we have v < 0.02v0 .
2) The response to a “step” in the command θ(t). Assume that the car is at rest, that is
v(0) = 0, and we ”step on the gas” so that θ(t) = θ. The differential equation is then
m
Ke
v̇ + v =
θ
B
B
with v(0) = 0.
(a) The steady-state (final) speed is
vss =
Ke
θ
B
(found by letting all derivatives to be zero), and
(b) the differential equation may be solved to give the dynamic response:
t
v(t) = vss (1 − e− τ ) =
which is shown below:
2–2
B
Ke θ
(1 − e− m t )
B
v (t)
v
s s
A fte r 4 t th e r e s p o n s e is w ith in
2 % o f th e fin a l v a lu e .
s te a d y -s ta te
r e g io n
tr a n s ie n t r e g io n
0
2
4 t
t
Closed-Loop Control
Now design the controller. Assume we will use error-based control. In other words, given
a desired speed for the car vd , and the measured car speed v(t), we define the error
e(t) as the difference
e(t) = vd (t) − v(t),
and choose a control law that is based on e(t)
θ(t) = Kc e(t) = Kc (vd (t) − v(t))
where Kc is the controller gain. Thus the control effort is proportional to the error, and
the block diagram becomes:
V d (t)
+
e rro r
e (t)
-
F
c o n tr o lle r
K c
q
e n g in e
K e
F
p
+
(t)
d
(t)
+
m
c a r m o d e l
d v
d t
fe e d b a c k p a th
Note that the control law states:
• if v < vd then e > 0 - depress gas pedal.
• if v = vd then e = 0 - set θ = 0 (do nothing).
• if v > vd then e < 0 - set θ < 0 (apply brakes).
The new differential equation is
m
dv
+ Bv = Kc Ke (vd (t) − v) + Fd (t)
dt
and rearranging
m
dv
+ (B + Kc Ke )v = Kc Ke vd (t) + Fd (t)
dt
2–3
+ B v = F
c a r s p e e d
V
which is the closed-loop differential equation. We can write this as
m
dv
Kc Ke
1
+v =
vd (t) +
Fd (t)
B + Kc Ke dt
B + Kc Ke
B + Kc Ke
and compare with the standard first-order form
τ ẏ + y = u(t)
where τ = m/(B + Kc Ke ) is the closed-loop time-constant.
The important thing to note is that feedback has modified the ODE.
3 Questions:
Assume there are no external disturbance forces, that is Fd (t) ≡ 0 for now,
(a) If we command the car to travel at a steady speed vd , what speed will it actually reach?
To find the steady-state speed vss , set dv
= 0 and solve for vss , giving
dt
vss =
Kc Ke
vd
B + Kc Ke
or vss < vd , for B > 0.
Note that as we increase the controller gain so that Kc Ke � B then vss → vd .
(b) How has the dynamic response affected by the feedback?
Consider the first-order ODE
dy
τ
+ y = Ku(t).
dt
The response to a steady input u(t) = 1 with initial condition y(0) = 0 is
t
y(t) = K(1 − e− τ )
y
y (t)
s s
= K
a t t = 4 t , th e r e s p o n s e is a p p r o x .
0 . 9 8 y s s.
s te a d y -s ta te
tr a n s ie n t
0
4 t
2–4
t
For the car under feedback control, this translates to a steady-state speed
vss =
Kc Ke
vd
B + Kc Ke
which is a function of the controller gain Kc , and a closed-loop time constant τ that is
m
τ=
B + Kc Ke
which is also a function of Kc , and as Kc increases, τ decreases, meaning that the car
responds more quickly to changes in the gas pedal.
v
vd
t d e c re a s e s
s te a d y -s ta te s p e e d
a p p ro a c h e s v d.
K c in c r e a s in g
0
t
4 The Effect of an External Disturbance Force
Now consider the car on an incline under closed-loop control:
F d
v (t)
v (t)
Fp (t)
)
F p(t
q )
n (
g s i
m
=
q
( b ) o n a n in c lin e
( a ) o n le v e l g r o u n d
(a) On horizontal ground Fd = 0, and we showed:
vss =
Kc Ke
vd
B + Kc Ke
(b) On the incline there is a constant disturbance force Fd = −mg sin φ acting down the
incline, and from the closed-loop differential equation:
m
dv
Kc Ke
1
+v =
vd (t) +
Fd (t)
B + Kc Ke dt
B + Kc Ke
B + Kc Ke
the steady-state speed will be
vss =
Kc Ke
mg sin φ
vd −
,
B + Kp Ke
B + Kc Ke
2–5
but we note that as the controller gain Kp is increased the impact of the disturbance
is reduced.
Conclusion: Feedback control reduces the effect of external disturbances on the
system behavior.
2–6