block diagram manipulations the basic control loop
TRANSCRIPT
Block Diagram Manipulations
The Basic Control Loop In this section we will examine methods for simplifying systems of transfer functions to a single function. As the first example, we will consider a generic control loop. In this system,
CG represents
a control algorithm. A
G is the transfer function for the actuator. The function of an actuator is to take the output of the controller (usually a low power electrical signal) and convert it to something that can alter the process; for example, a heater or an agitator.
PG is the transfer function for the process. H is
the transfer function for the sensor (e.g., a thermometer).
This system has two inputs and one output. R is the setpoint (or desired output) for the system. When the output (via the sensor) is subtracted from the setpoint, the resulting error,! , is used as the input for the controller. The second input, U, is the load or disturbance input. This input is usually not under the engineer’s control. Some examples of disturbances might be feedstock entering a reactor or external temperature changes. The output variable, C, is the variable being controlled.
In general, there are two categories of control problems. In “Servo” problems, there is no change in load, but the setpoint changes. In “Regulator” problems, the setpoint remains constant, but the load changes. In real systems both the load and the setpoint may change simultaneously.
R = Setpoint ε = Error U = Load / Disturbance Variable B = Variable Produced by Measuring Element C = Controlled Variable M = Manipulated Variable
Feedback Path
U
R є
—_
- +_ GC GP
H
C
B
+_ +
_ GA M_
Forward Path
Gc = Transfer Function of Controller GA = Transfer Function of Actuator Gp = Transfer Function of Process H = Transfer Function of Measuring Element 1. Determining transfer function for relating R to C, the servo problem.
A. These are “overall” transfer functions (i.e., include entire process), so we are looking for
RC .
B. Remember, these are differences from steady state, not absolute values. If the only
change is in R, U = 0.
Then, by combining
PACGandGG ,, into a single transfer function “
1G ” (
PACGGGG =
1).
R є
—_
- +_ G1
B
C _
M_
H
R є
—_
- +_ GC GP
H
_ _
GA
Went away since U = 0
This loop can be solved using three simultaneous equations:
C = G1 ε I
B = HC II
ε = R - B III III into I yields
C = G1 (R - B) Substitute in II
C = G1 (R - HC)
C = G1R - HG1C
C + HGC = G1R
HG
RGC
1
1
1+=
HG
G
R
C
1
1
1+=
= HGGG
GGG
PAC
PAC
+1
2. Determining the transfer function relating U to C, the regulator problem.
A. As before, we are looking for the overall transfer function relating C to U. So the goal is
UC .
B. In this case, U is changing, not R. Therefore, R is 0 in deviation terms.
G2 = GC GA
( )
PGMUC +=
!2GM = B!="
HCB =
Then
( )PGGUC !
2== ( )( )
PGBGUC !+=
2
( )( )
PGHCGUC !+=
2
U
є
—_
- G2 GP
H
C
B
+_
M_
R = 0 +
U
R є
—_
- +_ GC GP
H
C
B
+_ GA
M_
+
HCGGUGCPP 2
!=
PPUGHCGGC =+
2
( )
PPUGHCGGC =+
21
HGG
GUC
P
P
21+
=
HGGG
GUC
PAC
p
+=
1
HGGG
G
U
C
PAC
P
+=1
If changes occur in both R and U, then we can solve the system by adding the two solutions together. So, if
HGGG
GGG
R
C
PAC
PAC
+=1
And
HGGG
G
U
C
PAC
P
+=1
The response of the system to both changes is
!!"
#$$%
&
++!!"
#$$%
&
+=
HGGG
GU
HGGG
GGGRC
PAC
P
PAC
PAC
11
Or
HGGG
GUGGGRC
PAC
PPAC
+
+=
1
Start by eliminating the innermost loop.
This has the transfer function
22
2
1 HGG
GG
X
Y
AC
AC
+= = Gα
Note: What if this was a positive feedback loop; i.e., Then transfer function
22
2
1 HGG
GG
X
Y
AC
AC
!=
X
—_
+_ GC2
H2
Y
GA
U1
R
—_
+_ GC1
H1
+_ GC2
_
+++ ---_--____-
GA
GP1 GP2
U2
+ H2
C +
+ +
_
+
+
Because we are solving the servo problem, 1
U and 2
U are 0. Now we have: R C This becomes
Where
21 PPABGGGGG !=
11 HG
G
R
C
!
!
+=
So
1211
211
1 HGGGG
GGGG
R
C
PPC
PPC
!
!
+=
R
—_
+_ !G
H1
C
+ GC1 G! GP1
GP2
H1
Some Additional Block Diagram Manipulation Techniques In some cases, simplification of the block diagram will require altering the order of the various elements of a block diagram. Some of the possible operations are described below. 1. Moving a summing junction behind a block X1 X3 ( )
213XXGX ±=
X2
2. Moving a summing junction ahead of a block
X2
G
G X1
X3
X3 = GX1 + GX2
+
+
G
X2
+
X3 = GX1 + X2 +
+
+
G
X1 X3
is equivalent to
3. Altering the order of summing junctions
Is equivalent to
1/G
+
X2
G
X3 = (X1 + (1/G)X2)G = GX1 + 1/G X2 = GX1 + X2
X1 + X3
+ -
X2
+
X3
X4
X4 = (X1 - X2) + X3
= X1 - X2 + X3
+
+
X3
+
X2
-
X4 X1
X4 = (X1 + X3) - X2
= X1 - X2 + X3
+
X1
4. Moving a pick off point ahead of a block
5. Moving a pick off point behind a block
G X1 X2A
X2B
X2A = G X1
X2B = G X1
G X1
G X2B
X2A
X2A = G X1
X2B = G X1
X1A G
X1B
G X2
G1
X1B
X2 = G X1A
X2 = G X1B
X1B = X2
G
X2
6. Parallel forward paths
Becomes
G1 + G2
G1 +
+
G2
A Complex System with Interlaced Control Loops Here is another complex example to practice simplifying block diagrams. The challenging issue is the interlacing of the feedback loops. They need to be converted to nested loops.
In this case, we will manipulate the feedback loop containing
2H to be outside the loop containing
1H .
Then moving the summing junction before 1G , the highlighted area (indicated by dotted line) may be
written:
R +
-
+
+ G1 G2 G3
H2
H1
+
--
C
+
1
2
G
H
G1
-
+
+
+
We can now swap the order of the summing junctions to yield
The transfer function for the boxed area is
121
21
1 HGG
GG
!
Note the sign change (since this is positive feedback loop)
G1 G2
H1
G3 +
-
+ + -
+
121
21
1 HGG
GG
!
G3
1
2
G
H
+
-
+
-
1
2
G
H
Let
121
321
1 HGG
GGG
B
A
!=
Then the transfer function for the highlighted area is
1
21
1
G
H
B
AB
A
+
•
1
21
1
BG
AHBGB
A
+•
21
1
AHBG
BG
B
A
+•
( )23211211
3211
21
1
1 HGGGHGGG
GGGG
AHBG
AG
+!=
+
= E
D
HGGHGG
GGG=
+!232121
321
1
And the diagram becomes
232121
321
1 HGGHGG
GGG
+!
+
-
Note H3 = 1
Then the transfer function becomes
E
D
E
D
+1
So
E
ED
E
D
+
So
ED
E
E
D
+•
And
ED
D
+
Substituting back in for D and E.
321232121
321
1 GGGHGGHGG
GGG
++!
Or
321232121
321
1 GGGHGGHGG
GGG
++!