Analysis of a closed-loop feedback system

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Analysis of a closed-loop
feedback system
This work is licensed under the Creative Commons Attribution- NonCommercial-ShareAlike 3.0 Unported
License. To view a copy of this license, visit http://creativecommons.org/licenses/by-ncsa/3.0/ or send a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California,
94041, USA. Copyright 2012, Andy Packard.
Complete Analysis of simple feedback system #2
Return to our basic block diagram which motivated feedback control, and make
enough
h simplifying
i lif i assumptions
ti
so th
thatt th
the closed-loop
l
dl
system
t
iis governed
db
by a
single, linear differential equation (which we can completely analyze).
d
y
r
P
u
K
ym
x
u
+
1st
+
order controller
x (t )  ax(t )  b1r (t )  b2 ym (t )
u (t )  cx(t )  d1r (t )
n
y (t )   u (t )   d (t )
Constant-gain,
non-dynamic plant
Eliminate u, yM
x (t )   a  b2 c  x(t )   b1  b2 d1  r (t )  b2  d (t )  b2 n(t )
y (t )   cx(t )   d1r (t )   d (t )  0n(t )
u (t )  cx(t )  d1r (t )  0d (t )  0n(t )
1st order closed-loop system
Complete Analysis of simple feedback system #2
Facts: (we know these)
▫ Closed-loop system is stable if
and only if
a  b2 c  0
▫ (if stable) time-constant is
Some Closed-Loop FRFs
d
r
ym
K
x
y
P
u
u
+
1
  a  b2 c 
+
n
Closed-loop steady-state gains
Obtaining
Obt
i i iintegral
t
l control
t l as a
necessary condition to
specifications
This work is licensed under the Creative Commons Attribution- NonCommercial-ShareAlike 3.0 Unported
License. To view a copy of this license, visit http://creativecommons.org/licenses/by-ncsa/3.0/ or send a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California,
94041, USA. Copyright 2012, Andy Packard.
Closed-loop properties and design objectives
Stable if and only if
Time-constant
r
ym
Steady-state gains
d
K
x
y
P
u
u
+
Design Objective:
▫ SSGd→y=0
 choose a=0, which simplifies SSGr→y to…
▫ SSGr→y=1
 choose -b1=b2
 All FRFs, b1 and b2 just multiply c (nowhere else) so: b1=-b2=1
▫ Stability
y and desired time-constant
 Choose c such that:
+
n
Closed-loop properties and design objectives
Design choices
d
r
ym
Controller form becomes
K
y
P
u
u
x
+
+
n
with d1 still free (specifications have not lead to any restrictions on d1).
)
Block diagram of controller is:
r
ym
r
ym
u
-
u
-
For simplicity, take d1=0, and
notate the gain simply as KI
Closed-loop properties and design objectives
Specifications
▫ SSGd→y=0
▫ SSGr→y=1
leads to the architecture
r
ym
d
r
K
ym
x
y
P
u
u
+
u
+
-
The equation governing u, ie., the strategy, is simply
“change u, with rate based on
the mismatch between r and ym”
Contrast this to the proportional strategy,
“set the value of u based on the
values
l
off r and
d ym”
New strategy is an “integral” strategy, usually called integral control.
n
Closed-loop properties and design objectives
Design choices are
Plant parameters
likely uncertain.
Stability:
y
Time Constant:
are
Governing equations:
Closed-loop FRFs
Probably should plot
these, to get some insight,
for different values of the
4 parameters…
Steady-state gains:
Instantaneous gains:
Closed-loop properties and design objectives
Steady-State Gains:
open loop
improvement
closed loop
closed-loop
Keeping track of the effect of n
shows the tradeoff:
open loop
Role of accurate sensor
closed loop
In the context of a cruise
cruise-control
control
system: Vehicle speed sensitivity
is moved from (d)
 geography/hills
which
hi h are external,
t
l tto ((n))
 speedometer
which is a component in the system
Time-Delay “margin”
What is the smallest T>0 such that the closed-loop system is unstable?
Stability (in linear equations) determined solely by free-response
Check this
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