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Control outputi.e. Output = Input, despite disturbancesThis is achieved by feedback.
Open loop systems - i.e. without feedback
Process with transfer function P perturbed by a disturbance D.
Suppose P is 10 and disturbance D is 0.If the output O is to be 1, just make input I = 0.1.
But, if P changes by 10% to 11 then O changes by 10% to 1.1.
If disturbance D is 0.1, then O will also change by 0.1.
Feedback control
I OP
DProcessDisturbance
OutputInput
Closed loop system: feedback added!Now consider the ‘closed-loop’ system below: P represents the device being controlled; C is the controller.
Ignoring disturbances (D = 0), by forward over 1 minus loop rule
Let P = 10, as before, and C = 10;
If I is 1, then O is 0.99 i.e. it is within 1% of being 1
If P changes to 11; O = I * 110/111; if I is 1, O is still about 0.99.
Feedback control
I OP
D
C
Controller ProcessDisturbance
OutputInput
O =C * P
1 + C * PI
I 0.99I011
100=I
01* 10 + 1
10 * 10=O
To see the effect of disturbances, assume I is 0. Then
If C = 10, P = 10 and D = 0.1
Negative Feedback•reduces effects on output of disturbances•reduces effects on output of parameter changes
if |closed loop gain| < |open loop gain|
Disturbances control
I OP
D
C
Controller ProcessDisturbance
OutputInput
P * C + 1
1=
Loop-1
Forward
D
O
0.00099011
0.1=0.1
01* 10 + 1
1=O
If CP large: O ~ I + 0 = I
So Feedback • makes output almost same as input, • minimises effects of disturbances and • reduces effect of change in device.This is true because the ‘loop gain’, C * P, is high.
Note, can’t just keep increasing the ‘gain’ of C.Also, need to consider the dynamics of the blocksAlso, there can be a block in the feedback path which we must consider
Principle of Superposition
I OP
D
C
Controller ProcessDisturbance
OutputInput
DP * C + 1
1I
P * C + 1
P*C=O
Specify a motor and resistor for R2-D2?
Assume we walk at 2 m/s & R2-D2’s wheel diameter is 4cm. Therefore, required angular velocity is: 16 revs/s -> 100 radians/s
Input voltage is 4 D cells giving 6V input.
Ke = 6/100 = 0.06 V/rad/s
Weight of R2-D2 gives an inertia torque (J):
J = 0.05 kgm2
Assume current is 1 amp -> R = 1
Want T = 0.2s
KT = 25 Nm/A
We need to form a relationship between input voltage and output velocity:
R2-D2 Motor System
sT
K
I
O
11
te
a
KK
JRT 1
Has a time response to a unit step input:
Unit Step Response of System
Time
Output Output when K = 1, T= 0.01
Input
Output Output when K = 1.6, T= 0.02
Input
Now include armature inductance:
Armature resistor:
Back emf of motor:
Torque proportional to armature current:
Torque is opposed by the inertia torque:
Hint: apply Kirchhoff’s voltage law to the armature circuit
We need to form a relationship between input voltage and output velocity:
R2-D2 Motor System
Rba vvv
sVa sIa
sVb
T
aTiKT eb Kv aaR Riv
dt
dJT
dt
diLv aaL
Include new component:Components of Motor System
sIRsV aR
sKsV eb
sJssT
bV
RV
eK
RI
Js
T
+_
bV
aV RVRba vvv
aTiKT
eb Kv
aaR Riv
dt
dJT
sIKsT TTI TK
ssILsV aL
LVLasIdt
diLv aaL
Reduce block diagram:
bV
aV+_
1
1
2 sKKJR
sKKJL
K
te
a
te
a
e
1.
2.
3.
eKbV
TK Js1
bV
aV+_
eKbV
2sJLsJR
K
aa
T
aV
aIba VV +_ 1/R
sLaaILV
sLR aa 1
Output linked to input:
Previously (inductor = 0)
Can be expressed much more simply!:
Previously (inductor = 0)
Transfer Function of System
1
1
sKKJRK
V
te
a
e
a
11 21
sTsT
K
Va
1
1
2
sKKJR
sKKJL
K
V
te
a
te
a
e
a
11
sT
K
Va
Any system of the form:
Has a time response (depending on input):
Time Response of System
sTsT
K
I
O
21 11
Time
Output Varies!
Has a time response to a unit step input:
Over Damping
Time
Output Output is over damped
Input
Output
Output changes with k, T1 & T2
Input
Has a time response to a unit step input:
Critical Damping
Time
Output
Output is critically damped
Input
Output
Output from unique T1, T2 & k
Input
Has a time response to a unit step input:
Under Damping
Time
Output
Output is under damped
Input
Output
Output changes with k, T1 & T2
Input
d0 = tf(1.0,[1 0 1]) %undamped
d1 = tf(1.0,[1 2 1]) %critically damped
d2 = tf(4.0,[1 2 4]) %under damped
d3 = tf(0.5,[1 2 0.5]) %over damped
T= [0: 0.01: 20];%set up the time increments
[y0,t]=step(d0,T);%step response over one second
[y1,t]=step(d1,T);%step response over one second
[y2,t]=step(d2,T);%step response over one second
[y3,t]=step(d3,T);%step response over one second
stept = 1 + 0*t; %graph to show step response
clf; %clear all graphs
hold on % put each graph on top of each other
plot(t,y0, 'r');
plot(t,y1,'k');
plot(t,y2,'g');
plot(t,y3,'b');
plot(t,stept,'m');
Matlab code
1) Write down the transfer function for the RC circuit when
R= 2kΩ and C = 5mF.
Sketch the response of the system to a unit step input, marking the time constant's position on the time axis and the final value on the other axis.
2) Find the transfer function of the thermal system for which
R = 4 KW-1 and C = 2 JK-1.
Sketch the response of the output if the input is a step change of 2 W
3) For each of the following work out O/I and sketch response if I is a step.
Exercises
I
5s
2
2
5/s OI
3
2/s OO I