D E R I V AT I V E L APLACE T RANSFORMS

Laplace transforms replace the derivative terms in both first-order and second-order differential equations with their respective frequency domain s terms. Table 14-3 shows the Laplace transforms for both first- and second-

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LAPLACE S TIME PROCESS E

4 C DESCRIPTION T TRANSFORM FUNCTION RESPONSE

N IO

Any Function f F (s)

Unit Step Input 1 1 u(t ) = 0 t < t 0 s n ss

Step Input

A(t ) At ≥ t 0

lT 0 e

Delay (Dead Time) to

A(t − t d ) = 0 t < t d

Ae t& d Step Input A(t − t )

A(t − t d ) = At ≥ t

w id e First-Order Equations

a m =2 First-Order Response

2 τ =2 m First-Order Response with

t = t d . 1 000 1 000 . A 1 A 2 =1 T First-Order Response plus

3 Dead Time

c tio o n First-Order Response with

A 1 A 2 t = t d . 1 000 0 500 . A 1 A 2 =1 s n se

s Lag plus Dead Time

Table 14-2. Laplace transforms.

DESCRIPTION IO TRANSFORM FUNCTION RESPONSE N

p C First-Order Response to Step

P ro Input ( A 1 s

t = 0 0.000 0.000 A 1 A 2 =1 a lic

/ τ ) with Lag s( τ s + 1) AA

12 ( 1 − e )

= s n e t c 2 0.865 0.632 ss

t = 1 0.632 0.393 tio

A A A τ = 1 τ = 2 For

d First-Order Response to Step Input

t = t d . 1 000 1 000 . A A 1 A 2 A 3

u ( 1 / s s ) with Lag plus Dead Time

Second-Order Equations

A ω 2 A ω n e − ζω n t& t n

Second-Order Transfer Function

w V ( Hp ( s) ) for ζ < 1 (Underdamped)

.in o A − d t C Second-Order Transfer Function At

( Hp ( s) ) for ζ = 1 (Critically Damped)

( τ s + 1)

tri m p

Hp = a Out a Second-Order Transfer Function A A

lte In n y

( e τ 1 − (s) e τ 2 )

( Hp ( s) ) for ζ > 1 (Overdamped) x

Second-Order Step Response

Critically damped ζ =1

0 ( A 1 / s ) for ζ < 1 (Underdamped)

1 ζ 2 Overdamped ζ -7 >1

s(s 2 2 + 2 ζω n s + ω n )

where ψ= tan − 1 −

3 Second-Order Step Response

Underdamped ζ <1 R

sf ss

( / ) for ζ = 1 (Critically Damped)

n s se Second-Order Step Response

1 tio n

/ AA s ) for ζ > 1 (Overdamped)

Table 14-2 continued.

S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14

Time ( t) Domain

Laplace ( s) Domain

dx Second 2

dx ( t = 0 ) 2 () s − s − x ( t dt

sX 2

= 0 Derivative ) dt Integral t Adt

0 s Table 14-3. Derivative Laplace transforms.

order derivative terms. In the frequency domain, a first-order derivative term becomes an s term times the function in the frequency domain minus a constant, which is the value of the function at t = 0 in the time domain. A

second-order derivative becomes an s 2 Laplace term times the Laplace function minus s times the value of the time domain first derivative at t = 0 minus the value of the function at t = 0 in the time domain. Therefore, a simple first-order differential equation of the form:

y () t = + x () t

dt becomes the following equation in Laplace form: Y () s = sX () s − x ( t = 0 ) + X () s

Assuming that the value of the function x (t = 0) is zero, the equation becomes:

Y () s = sX () s + X () s

= X () s ( s + 1 )

If X (s) represents the Laplace output of the process and Y (s) represents the input, as shown in Figure 14-32, the equation for the transfer function of the process (output divided by input) in Laplace form becomes:

Figure 14-32. Process inputs and outputs in Laplace form.

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S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14

This indicates that the Laplace output X (s) is equal to the Laplace input times the transfer function in Laplace form:

() = () s

So, by working in the frequency domain instead of the time domain, the solving of the differential equation is reduced to an algebraic manipulation. In the Laplace domain, the exponent of the s term indicates the order of the transfer function—in this case, a first-order transfer function.

A second-order differential equation of the form: dx 2 dx

becomes the following equation in Laplace form:

 2 dx ( t = 0 Y ) sX

− x ( t = 0 ) + [ sX () s − x ( t = 0 ) ] + X () s

Assuming that the values of the initial parameters are zero, this equation becomes:

() s = sX () s + sX () s + X () s

() s ( s ++ s 1 )

Since Y (s) represents the input and X (s) represents the output, the transfer function for a second-order response in Laplace form is:

X () s

( s ++ s 1 )

() s

Note that the exponent of the s term indicates a second-order transfer function. As a result of the different mathematical formats of first-order and second- order transfer functions, a second-order system will have an oscillating response, while a first-order system will have a smooth response toward its final steady-state value. We will discuss these response curves in more detail later.

The substitutions used in Laplace transforms are the result of computations performed on electrical circuit networks (e.g., resistors, capacitors, and inductors) that create transfer functions. Mechanical and hydraulic systems also use Laplace transforms to represent system transfer functions, primarily because of their similarity in mathematical representation to electrical systems. The following example illustrates the derivation of a transfer function for a resistor/capacitor (R/C) network.

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S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14

E X AM PLE 1 4 -3

Figure 14-33 represents an R/C (resistor/capacitor) electrical network. Find (a) the differential equation that represents this network, (b) the network transfer function in Laplace, and (c) the equivalent closed- loop block diagram. Refer to Appendix G for information about the characteristics of electrical circuit elements.

i = current

current i

V in

C V out

Figure 14-33. R/C electrical network diagram.

S OLU T I ON

The term V C represents the voltage across the capacitor ( C ). This term

is equivalent to V out and is represented by:

V out = V C =

idt

Kirchhoff’s voltage law states that the voltage at the output of an electrical circuit is equal to the input voltage minus the voltage across

the resistor ( V R ), which is equal to the current ( i ) times the resistance

( R ). Therefore: V out = V in − V R

= V in − iR Solving for the current yields:

V out = V in − iR iR V = in − V out

i = ( V in − V out )

 R 

Replacing the current term in the capacitor’s output voltage equation with this value of i produces:

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S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14

V out =

C ∫ 0 i dt

out )

RC in 0 V dt ∫ −

RC ∫ 0

out V dt

∫ 0 in V dt − V dt RC  ∫ 0 out 

(a) To obtain the differential equation of this network, we must take the derivative of both sides of the output voltage equation to eliminate the integral terms:

(b) The Laplace form of this first-order differential equation is: V in( ) s = RCsV out( ) s − V out( t = 0 ) + V out( ) s

Since there is no initial value at t = 0, this equation becomes:

V in( ) s = RCsV out( ) s + V out( ) s

= V out( ) s ( RCs + 1 )

The transfer function is:

V out( ) s

V in( ) s sRC + 1

The equation for a first-order system with lag is represented by:

Hp () s =

Therefore, this electrical network is a first-order lag system where the time constant τ is equal to RC (the resistance times the capacitance).

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S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14

V in( s) +

V in( s) – V out( s)

1 I ( s)

1 V out( s)

– V out( s)

Figure 14-34. Block diagram of the R/C network.

Referring to the hot-water heater example used in Section 14-3, let’s assume that the differential equation that describes the process, otherwise known as the heating system’s enthalpy balance equation, is given by:

Steam flow = A dT + BT () t

where A and B are constants and T is the temperature. Since the steam flow is directly related to the control variable and the temperature is the process variable, we can rewrite this equation as:

dPV

CV () t = A + BPV () t

dt Taking the Laplace transform of this equation, we get: CV () s = AsPV () s − APV ( t = 0 ) + BPV () s

Assuming that the value of PV at t = 0 is zero, this equation becomes:

CV () s = AsPV () s + BPV () s

= PV () s ( As + B )

The process’s transfer function, which is equal to the process’s output divided by its input, is:

PV () s

CV () s As + B

To obtain a standard first-order lag equation (i.e., a fraction whose denominator is τ s + 1), we can divide both the numerator and the denominator by B:

() B

() s

CV As B () + s () B

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S ECTION PLC Process Process Responses C HAPTER 4 Applications

and Transfer Functions 14

Previously, we calculated experimentally that the gain for this process is

0.8 ° C/% and that it took 15 minutes to achieve the final steady-state value. Furthermore, as we will discuss in Section 14-6, it is a given that in a first- order system with lag the output variable will be at 99.33% of the input

variable when t = 5 τ . The observed value of the response at 15 minutes is 100%, which is close to the 99.33% at 5 τ . Therefore, we can approximate the

value of τ as:

Thus, we can represent the process’s transfer function as:

D E R I V AT I V E L APLACE T RANSFORMS