Goals for today
•
•
•
Block diagrams revisited
– Block diagram components
– Block diagram cascade
– Summing and pick-off junctions
– Feedback topology
– Negative vs positive feedback
Example of a system with feedback
– Derivation of the closed-loop transfer function
– Specification of the transient response by selecting
the feedback gain
The op-amp in feedback configuration
2.004 Fall ’07
Lecture 11 – Monday, Oct. 1
Block diagram components
Transfer Function
C(s)
= G(s).
R(s)
Images removed due to copyright restrictions.
Please see: Fig. 5.2 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 11 – Monday, Oct. 1
Cascading subsystems
Images removed due to copyright restrictions.
Please see: Fig. 5.3 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
Note: to cascade two subsystems, we must ensure that
the second subsystem does not load the first subsystem
2.004 Fall ’07
Lecture 11 – Monday, Oct. 1
Loading and cascade
Images removed due to copyright restrictions.
Please see: Fig. 5.4 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 11 – Monday, Oct. 1
Cascading with an op-amp
R4
I3
R1
+
Vi(s) _
I1
R3
V3
C1
I3
IC
_
V=0
1
R2
V4
+
I2
C2
+
-
V2(s)
Figure by MIT OpenCourseWare.
We can show that
1
1
¶
µ
R1 C 1
R2 C2
V2
R4
×
=
× −
.
Vi
R3
1
1
1
s+
+
s+
R1 C1
R 3 C1
R2 C2
≈
1
R1 C1
s+
2.004 Fall ’07
1
R1 C1
µ
¶
R4
× −
×
R3
1
R2 C 2
s+
1
R2 C 2
Lecture 11 – Monday, Oct. 1
if
R3 À R1 .
Parallel subsystems
Images removed due to copyright restrictions.
Please see: Fig. 5.5 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
2.004 Fall ’07
Lecture 11 – Monday, Oct. 1
Negative feedback
E(s) = R(s) − H(s)C(s)
Plant and
controller
R(s)
Input
E(s)
+
-
Actuating
signal
(error)
C(s)
G(s)
Output
Feedback
Input
⇒
G(s)
1 + G(s)H(s)
C(s)
Output
equivalent
system
Figure by MIT OpenCourseWare.
Figure 5.6
2.004 Fall ’07
C(s)
R(s)
C(s)
R(s)
G(s)H(s)
H(s)
R(s)
C(s) = E(s)G(s)
h
i
⇒ C(s) = R(s) − H(s)C(s) G(s)
Lecture 11 – Monday, Oct. 1
=
G(s)
.
1 + G(s)H(s)
:
Closed—loop TF
:
Open—loop TF.
Positive feedback
E(s) = R(s) + H(s)C(s)
Plant and
controller
R(s)
Input
E(s)
+
+
Actuating
signal
(error)
C(s)
G(s)
Output
Feedback
Input
⇒
C(s)
R(s)
C(s)
R(s)
G(s)H(s)
H(s)
R(s)
C(s) = E(s)G(s)
h
i
⇒ C(s) = R(s) + H(s)C(s) G(s)
G(s)
_
1 G(s)H(s)
=
G(s)
.
1 − G(s)H(s)
:
Closed—loop TF
:
Open—loop TF.
C(s)
Output
equivalent
system
Figure by MIT OpenCourseWare.
Figure 5.6
Generally, positive feedback is dangerous: it may lead
to unstable response (i.e. exponentially increasing) if not used with care
2.004 Fall ’07
Lecture 11 – Monday, Oct. 1
A more general feedback system
Images removed due to copyright restrictions.
Please see: Fig. 5.6 in Nise, Norman S. Control Systems Engineering. 4th ed. Hoboken, NJ: John Wiley, 2004.
Plant: the system we want to control (e.g., elevator plant: input=voltage,
output=elevator position)
Controller: apparatus that produces input to plant (i.e. voltage to elevator’s motor)
Transducers: converting physical quantities so the system can use them
(e.g., input transducer: floor button pushed→voltage;
output transducer: current elevator position →voltage)
Feedback: apparatus that contributes current system state to error signal (e.g.,
in elevator system, error=voltage representing desired position – voltage representing
current position
2.004 Fall ’07
Lecture 11 – Monday, Oct. 1
Transient response of a feedback system
R(s) +
C(s)
25
s(s + 5)
Figure 5.15
Figure by MIT OpenCourseWare.
Plant & Controller
G(s)
Gain
Feedback
K
H(s)
2.004 Fall ’07
Open loop TF
G(s)H(s)
Closed loop TF
G(s)
1 + G(s)H(s)
Lecture 11 – Monday, Oct. 1
25
,
s(s + 5)
= 25,
= 1.
25
=
;
s(s + 5)
25
= 2
.
s + 5s + 25
=
Transient response of a feedback system
R(s) +
C(s)
25
s(s + 5)
Figure 5.15
Figure by MIT OpenCourseWare.
T (s) =
G(s)
1 + G(s)H(s)
Recall
⇒ ωn =
√
25 = 5
=
25
.
s2 + 5s + 25
ωn2
s2 + 2ζωn s + ωn2
rad/sec;
2ζωn = 5 ⇒ ζ = 0.5;
⇒ The feedback system is underdamped.
2.004 Fall ’07
Lecture 11 – Monday, Oct. 1
Transient response of a feedback system
R(s) +
C(s)
25
s(s + 5)
Figure 5.15
Figure by MIT OpenCourseWare.
T (s) =
Peak time
G(s)
1 + G(s)H(s)
Tp
%OS
Settling time
2.004 Fall ’07
Ts
=
25
.
s2 + 5s + 25
π
p
=
= 0.726 sec,
2
ωn 1 − ζ
!
Ã
ζπ
× 100 = 16.3,
= exp − p
2
1−ζ
4
=
= 1.6 sec.
ζωn
Lecture 11 – Monday, Oct. 1
Adjusting the transient by feedback
R(s) +
C(s)
K
s(s + 5)
Figure 5.16
Figure by MIT OpenCourseWare.
T (s) =
G(s)
1 + G(s)H(s)
=
K
.
s2 + 5s + K
We wish to reduce overshoot to %OS=10% or less.
ωn =
√
K;
2ζωn = 5
5
⇒ζ= √ .
2 K
For 10% overshoot or less, we need ζ = 0.591 or more. Therefore,
K = 17.9
2.004 Fall ’07
or less.
Lecture 11 – Monday, Oct. 1
The op-amp as a feedback system
1
Z2 (s)
Z2(s)
Vi(s)
Z1(s)
I1(s)
I2(s)
V1(s)
Ia(s)
-
vo(s)
+
Vi (s)
I1
1
Z1 (s)
+
Figure by MIT OpenCourseWare.
Figure 2.10(c)
I1 =
V i − V1
Z1
≈
Vi
Z1
if
V1 ≈ 0
I2 =
Vo − V1
Z2
≈
Vo
Z2
if
V1 ≈ 0
Vo = −AV1
⇒ Vo
= −AIa Za
µ
AZa
1+
Z2
2.004 Fall ’07
¶
= (I1 + I2 ) Za
¶
µ
Vi
Vo
Za .
≈ −A
+
Z1
Z2
= −AVi
Za
.
Z1
Vo
Vi
I2
+
=
=
≈
Lecture 11 – Monday, Oct. 1
Ia
Za
V1
−A
AZa
−
Z1
− AZa
1−
Z2
1
−
Z1
−
1
1
+
AZa
Z2
Z2
−
Z1
if
A → ∞.
Vo (s)