Out () s = ( )( ) In () s Hp () s

Out () s = ( )( ) In () s Hp () s

A 2   A 1  =  s   τ s + 1 

AA 1 = 2 ss ( τ + 1 )

Again using Table 14-2, the inverse Laplace response, or real-time response of the system (Out (t) ), is:

AA  1 e τ ()  t = 1 2 −

Out

where the value of the output will be 0.63A 1 A 2 (63% of A 1 A 2 ) when t = τ . If the input is a unit step (that is, the amplitude A 2 equals 1) and the gain of

Hp (s) equals 1 (A 1 = 1), then the output will be:

Out () t = AA 1 1 e 2 τ −

= ( )( ) 111  − e 

1 e =− 

Out

Process transfer function [ Hp (t) ]

A –t e τ 0.368 1 τ

t 0 t= τ

Figure 14-42. Process transfer function in the time domain.

Industrial Text & Video Company 1-800-752-8398

www.industrialtext.com

S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14

Figure 14-43 graphs the response of the process variable Out (t) for A 1 =A 2 =

1. This curve, representing a first-order response to a step input plus lag, is a function of the system’s transfer function, which is the step value (1) minus

the system’s curve term ( e τ ).

Out Step Input

First-Order Response

to Step Input

e τ (Process)

1– e

Figure 14-43. Process variable’s lag response.

− ts Adding a simple dead time term ( e d ) to a first-order step response with lag generates the Laplace transfer function:

where t d is the dead time. Figure 14-44 shows the graph of this function in the time domain. The value of the output is:

Note that the first output response equation is valid for t values greater than t d , the dead time. Out (t) will be zero for time values before the dead time t d.

0.632 A 1 A 2

t t d t d +1 τ t d +2 τ t d +3 τ t d +4 τ t d +5 τ

Figure 14-44. First-order step response with dead time and lag.

Industrial Text & Video Company 1-800-752-8398

www.industrialtext.com

S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14

E X AM PLE 1 4 -4

A process system has a first-order response with a time constant of 10.8 minutes. (a) Calculate how long it will take for the value of the output V out to be at 90% of the input V in . (b) Calculate the value of V out

at 90% of V in given a 5 minute dead time.

S OLU T I ON

(a)

A first-order system with lag has a response of:

V out = V in 

The value of τ is 10.8 seconds and the required ratio of output over input is 90%, or 0.90; therefore:

Solving for t by taking the natural logarithm (ln) of both sides of the equation yields:

e . 10 8 = . 0 10 −= t

ln . 0 10 . 10 8

−= t ( . )(ln . 10 8 0 10 ) = ( . )( 10 8 − . 2 303 ) = . 24 87 minutes

So, in 24.87 minutes, the value of the output will be at 90% of the value of the input.

(b) The dead time will simply add to the time required to achieve the 90% value. Therefore, with a lag of 5 minutes, the system will reach a value of 90% final output in 29.87 minutes (24.87 min + 5 min).

Industrial Text & Video Company 1-800-752-8398

www.industrialtext.com

S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14