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.
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)
