Pneumatic circuits: R, CPressure is potential, pressure difference is voltage dropA flow restrictor is a resistanceAn air chamber is a capacitance

Resistance R P+pc

P+pi q

Q

P

Slope R

C

Cpi

Rpc 1

11 1c i i

sCp p psRCR

sC

Position to Pressure transducer/amplifiers

Ps is pressure of supply air from wall pipe or air cylinderX is weighted differential average of input position and the pivot positionX controls the pressure Pb send to the control valve

bP Kx

Cps

R pb

When C is very small,

xb s a

x x

R Rp p p

R R R R

pa

Rx

Pneumatic proportional controller

Ps is const parameter

Diaphragm is fast, so pc and pb are equal

x is weighted diff sum of e and y

b ax e y

a b a b

Notice the direction of x and e:

c bp p Kx c bellow sp A f k y

Pneumatic proportional controller

Since pc=pb

Drawing for this part can be simplified

Proportional + derivative controller

Recall: bellow pressure is related to input pressure by1/(RCs+1)

Pbellow

( ) ( 1)( ) ( ) ( ) ( 1)

( ) ( 1)s s

cs

K a b k RCs bkbP s E s E s RCs

a b KaA a b k RCs aA

Proportional + Integral controller

11

1

1

RCsRCs

RCs

So, in closed-loop TF, replace (RCs+1) by (RCs+1)/RCs:

1( ) ( )

1( ) 1

sc

s

bk RCsP s E s

aA RCsbk

E saA RCs

PID controller

2

1 1

1 1

1 1

1

d i

i d

d i

i

d i i

R Cs RCs

R R Cs

R Cs RCs

RCs

R CRCs RCs

1( ) ( ) 1s

c di

bkP s E s R Cs

aA RCs

Hydraulic servo system

1

1

When =0, at center (=0),

all gaps balanced.

When 0, gap 1 & 3 ,

3 & 4 , , Vol .

( ) ( )

p

x y

x

q

dyA q K xdt

KY s X s

s

Alternative construction

( ) ( )K

Y s X ss

Proportional controller

Same as before

Same as last page

( )

( ) ( )

At low frequency, ( )

b K a b bKTF

a b aK a b s aK a b s

aK a b s

bK bTF

aK a

Dashpot

1 21 2

2

2

Force balance: ( ) ; Flow in restriction:

Conservation of mass: Vol ( )

( )Take L-transform and solve:

( ) 11

P PA P P kz q

Rd d k

A y z q zdt dt AR

A RsZ s Tsk

A RY s Tssk

Proportional + Integral controller

Lastpage

Rest as before

( )( 1)

( ) ( 1)

At low frequency, ( )( 1)

1 (1 )

b K a b TsTF

a b aKTs a b s Ts

aKT a b Ts

bTF

a Ts

Proportional +derivative control

Spring force: k(y-z) instead of kz; new TF=1-oldTF

( )( 1)

( ) ( 1)

At low frequency, ( ) ( 1)

( 1)

b K a b TsTF

a b aK a b s Ts

aK a b s Ts

bTF Ts

a

PID controller

w

Force Balance: k1z=AP1; k2(y-w)=AP1+AP2

1( )kd

A w z zdt AR

2 2 1( )P k kd

A w y w zdt R AR R

2

12 2 2 2

2

1 2 2 1

( )Solve for:

2( ) ( ) 1

A Rs

kZ sA R A R A R A RY s s sk k k k

21 2 1 2

21 1 2 1 2

21 1 2 1 2

1 22

1 1

( )( ( 2 ) 1)

( ) ( ( 2 ) 1)

At low frequency, ( )( ( 2 ) 1)

2 1 ( )

K a b TT s T T sbTF

a b aKT s a b s TT s T T s

aKT a b TT s T T s

T TbTF T s

a T T s

Block diagram representation

e

zw

3 3

3 3 3 3

4 4

4 2 4 2

1 2 21 2

1 1

1 2

1

2

1

2

2

l lz x y x y

l l l l

l lw z x y

l l l l

l l le w y k x k y

l l

K K Ky e k x k y

s s sKk

y xs Kk

Thermal systems

Conservation of heat energy:

iin out i

dC h h h

dt R R

1( ) ( ) ( )

1 1i i

Rs h s s

RCs RCs

a av k v

a a bv Ri v

b mv K

i vq K

i o

dhC q q

dt 1

oo

q hR