ImpurityConcentration ImpurityConcentration



( )

Time Impurity Concentration Limit

Time Impurity Concentration Limit

Impurity Concentration Limit

(1)

 :

 ( , Setpoint):

 :

 (CV, PV):

 (MV):

 : /

 :

 (DV):

(2)

 :

 ( ):

 :

 :

 ( ): /

 :

 :

 :

(Feedback Control) ( )

 (

)

(Measuring Instrument, Sensor)



Ex:

(Automatic Controller)



(Final Control Element)

(Control Valve)



(Process)



( )

Time

CV ^SP

Time

CV

SP

-

-

-

-

-

(DO)

- O₂

-

-

-

(Ion-specific electrode )

-

Piping and Instrumentation Diagram (P&ID)

Control diagram

P&ID (Engineering flowsheet)

(Sensor)

P&ID

- (Transfer Function)

(TF)

:

( ) ( ) ( )

Process

( )

u t y t

U s → →Y s

u

input ( )

forcing function( )

“cause” ( )

y

output ( ) response ( )

“effect” ( )

G(s)

( ) ( )

( ) ( )

( )

Y s s

G s ≜ U s = 輸出 s 輸入

( ) ( )

( ) ( )

Y s y t U s u t

 

 

 

 

≜

≜ L

L

Both y(t) and u(t) are deviation variables.

(deviation variable = variable – its steady-state value)

1. Steady-State Gain ( )

The steady-state gain of a TF can be used to calculate the steady-state change in an output due to a steady- state change in the input. For example, suppose we

know two steady states for an input, u, and an output, y.

Then we can calculate the steady-state gain, K, from:

,2 ,1

,2 ,1

s s

s s

y y

K u u

= −

−

For a linear system, K is a constant. But for a nonlinear system, K will depend on the operating condition

(

)

( )  K

2. Order of a TF Model ( )

Consider a general n-th order, linear ODE:

1 1

1 1 0 1 1 0

n n m m

n n n n m m m m

d y dy d u d u

a a a y b b b u

dt dt dt dt

− −

− − − −

+ + +… = + + +…

Take , assuming the initial conditions are all zero.

Rearranging gives the TF:

( ) ( )

( )

0

m i

i i

n i

i i

Y s b s

G s n m

U s a s

=

=

= = ≥

∑

∑

( ) n

Note: The order of the TF is equal to the order of the ODE.

L

(Physical realizability)

3. Additive Property ( )

Suppose that an output is influenced by two inputs and that the transfer functions are known:

( ) ( ) ( ) ( )

( ) ( )

1 2

1 2

Y s and Y s

G s G s

U s = U s =

Then the response to changes in both U₁ and U₂ can be written as:

( )

( ) ( )

( ) ( )

Y s = G s U s +G s U s

U₁(s)

U₂(s)

G₁(s)

G₂(s)

Y(s)

The graphical representation (or block diagram) is:

4. Multiplicative Property ( )

Suppose that,

( ) ( ) ( ) ( )

( ) ( )

1

1 2

1 2

Y s and U s

G s G s

U s = U s =

Then,

( )

( ) ( )

( )

( ) ( )

Y s = G s U s and U s = G s U s Substitute,

( )

( ) ( )

( )

Y s = s U s

Or,

( ) ( )

( ) ( )

( )

( )

( ) ( )

2

Y s G s G s U s G s G s Y s

U s =

( )

U1 s

The standard form for a first-order TF is:

where:

Consider the response of this system to a step of magnitude, M:

Step Response of First-Order Process

( ) ( )

Y s K

U s = s +

steady-state gain τ time constant

K ≜

≜

( )

( )

u t M t U s

= ≥ ⇒ = s

( )

(

)

Y s U s

s s s

= = ⇒

+ +

( )

( )

( ) ( ¹

)

y t = KM − e

^{- 1}

• ,

• ,

• ,

y_∞ = KM t → ∞

t =τ y = 0.632KM 5

t = τ y ≈ y_∞

Note: Large means a slow response.

τ

Consider a step change of magnitude M. Then U(s) = M/s and,

Step Response of Integrating Process

Not all processes have a steady-state gain. For example, an

“integrating process” or “integrator” has the transfer function:

( ) ( ) (

)

Y s K

U s = s K =

( )

( )

Y s y t KMt

= s ⇒ =

Thus, y(t) is unbounded and a new steady-state value does not exist.

L-1

h q_i

q

• Standard form:

Step Response of Second-Order Process

( ) ( )

Y s K

U s = s s

+ +

which has three model parameters:

steady-state gain

τ "time constant" [=] time

ζ damping coefficient (dimensionless) K ≜

≜

≜

( )

( )

( )

• The type of behavior that occurs depends on the value of damping coefficient, ζ

• It is convenient to consider three types of behavior:

Complex conjugates Underdamped

Real and = Critically damped

,

Real and ≠ Overdamped

Roots of Type of Response

Damping Coefficient and Transfer Function

ζ >1

ζ =1

0 ≤ <ζ 1

τ2 2s + 2ζτs + =1 0

<

(

)(

)

K

s s

τ + τ +

(

)

K τs +

τ2 2 2ζτ 1 K

s + s +

( )

( )

( )

( )

( )

( )

C h a p te r 5

C h a p te r 5

General Transfer Function Models

• General representation of transfer function:

There are two equivalent representations:

( )

0

m i

i i

n i

i i

b s G s

a s

=

=

=

∑

∑

( ) ( )( ) ( )

(

)(

) ( )

m m

n n

b s z s z s z G s a s p s p s p

− − −

= − − −

…

…

where {z_i} are the “zeros” and {p_i} are the “poles”.

• The dynamic behavior of a transfer function model can be characterized by the value of its poles and zeros.

( :n ≥ m )

Summary: Effects of Pole and Zero Locations

1. Poles

• Pole in “right half plane (RHP)”: results in unstable system (i.e., unstable step responses)

(

)

x

x x

Real axis Imaginary axis

x = unstable pole

• Complex pole : results in oscillatory responses

Real axis Imaginary axis

x

x x = complex poles

2. Zeros

• Pole at the origin (1/s term in TF model): results in an

“integrating process”

Note: Zeros have no effect on system stability.

• Zero in RHP: results in an inverse response to a step change in the input

• Zero in left half plane: may result in “overshoot” during a step response.

x ⇒ _{y 0}

t

inverse response Real

axis

Imaginary axis

Time Delay (Dead Time)

Time delays occur due to:

1. Fluid flow in a pipe 2. Chemical analysis

- Sampling line delay

- Time required to do the analysis (e.g., on-line gas chromatograph)

Mathematical description:

A time delay, , between an input u and an output y results in the following expression:

θ

( ) ( )

0 for t θ

y t  <

=  ⇒

− ≥

( )

( )

Y s = e⁻

Example: Turbulent flow in a pipe

Let, fluid property (e.g., temperature or composition) at point 1

fluid property at point 2 u ≜

y ≜

Fluid In

Fluid Out

Point 1 Point 2

Time Delay:

• First-Order Plus Time Delay (FOPTD) Model

( ) ( )

( ) τ 1

s p

Y s K

G s e

U s s

= = −

+

• Second-Order Plus Time Delay (SOPTD) Model

( ) ( )

( ) τ 2ζτ 1

s p

Y s K

G s e

U s s s

= = −

+ +

( ) ( )

( ) τ 1

s p

Y s K

G s e

U s s

= = −

+

θ ^t

y(t)

KM

0.632KM

τ

θ

u(t) = M

PID

1930

PID

 K_C : ( )



ττττ



ττττ

0

1 ( )

( ) ( ) ( )

where ( ) ( )

t

s c D

I sp

p t p K e t e t dt de t

dt

e t y y t

τ τ

 

= +  + + 

 

= −

∫

e(t) p(t)

(offset)

:

Kc

PB 100%

=

Time Setpoint

1.0

1 2 3

0

Offset

Small K_C Large K_C

PI

Time y

y_sp

p_prop

Time y

y_sp

p_{i nt}

PI K _C ττττ _I

K_C

τ

Time Time Time

K_C K_C

ττττI ττττ_I

( )

Time PID

PI

PID

(parallel) PID (ideal form)

(series) PID (actual form) ( ) 1 1

c c D

I

G s K s

s τ τ

 

=  + + 

 

( ) P s

( )

( ) 1 1 1

c c D

I

G s K s

s τ τ

 

=  +  +

 

1 1

Is

+ τ P s( )

1 ( ) 1 1

1

D

c c

G s K s

s s

τ

τ ατ

   + 

=  +     + 

(with derivative filter) : 0.05 ~ 0.2

α

Consider response of a controlled system after a sustained disturbance occurs

(e.g., step change in the disturbance variable)

y

PID

90% PI

P:

( )

PI:

( )

PID: PI

PID

G_c(s) G_p(s)

G_m(s) +

-

Y_sp(s) E(s) U(s) Y(s)

Y_m(s)

G_L(s) L(s)

+ +

G_v(s)

P(s)

1

p v c

sp p v c m

G G G Y

Y = G G G G +

1

L

p v c m

G Y

L = G G G G +

:

: (Y_sp L)

1

f

i e

Z Z

= ∏

+ ∏

Z Zi

∏ =f

∏ =e

Zi Z

Example

G_p

(BIBO stability)

0

( ) ( ) ( )

1 1

p v c L

sp

p v c m p v c m

G G G G

Y s Y s L s

G G G G G G G G

= +

+ +

1+G_OL = 0 1+G G G G_{p v c m} = 0

：

： ：

： 1+G

(s)=0

( ((

( ))))

(((

( ))))

Example

Solution:

1 0.2 _c 0

s K

⇒ − + + =

which has the single root, s = 1 + 0.2K_c. Thus, the stability requirement is that K_c < -5. This example illustrates the important fact that feedback control can be used to

stabilize a process that is not stable without control.

Consider a process, G_p = 0.2/(-s + 1), and thus is open-loop unstable. If G_v = G_m = 1, determine whether a proportional controller can stabilize the closed-loop system.

The characteristic equation for this system is

1 1 0.2 0

OL 1 c

G K

+ = + s =

− +