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

++!