Section 6. Laplace

Section 6. Laplace Transformations

Process Control: Loop Tuning and Analysis

Slide 1

Laplace Transformations

y So why do we need to know anything about Laplace Transforms? y Well, W ll fi firstly, tl it’ it’s nott essential. ti l y I’d estimate that more than 99.9% of the population have never heard of them – never mind use them. y But what about in control engineering? y Well, that’s a different matter. y Pick up virtually any book on the subject and you’re like to be faced with at least one transfer block making use of Laplace Transforms Transforms. y Why? y Well, the fact is that they make life easy for us.

Slide 2

Transfer Functions y As we’ve seen, we make use of block diagrams to represent the p and interconnection of a system. y composition y When used together with Transfer Functions we can represent the cause-and-effect relationships throughout the system. y A Transfer Function is defined simply p y as the relationship p between the input and output signal of a device. Rf Vi

Ri – Vo

Vo = −

Rf Ri

Vi

+

Slide 3

Transfer Functions

y In block diagram g representation p the g gain of the amplifier p is G and the output Vo (ignoring the inversion) is:

Vi

G

Vo

Vo = G ⋅Vi

Slide 4

Transfer Functions

y Supposing we had two amplifiers in series having gains G1 and G2 respectively. respectively Vi

G1

Vo

G2

Vo

y What would be the total gain of the system?

Vo = G1 ⋅ G2 ⋅Vi y Very simple! y Why? y There is no time dependency Slide 5

Transfer Functions y In the integrator circuit the output is time dependent. Cf Vi

Ri – Vo

Vo = −

1 Vi dt + A ∫ Ri ⋅ C f

+

y We no longer have a straightforward relationship between the input and the output y The output (Vo) is now described in a linear Ordinary Differential Equation (ODE). y The most important point to recognize is that the output is now d dependent d t on ti time and d it iis said id tto b be dynamic d i and d operating ti iin th the time domain. Slide 6

Transfer Functions

y The consequence of this is that we can no longer manipulate the block in the same manner as we had done so previously. Cf Vi

Ri – Vo

1 Vo = − Vi dt + A ∫ Ri ⋅ C f

+

y We cannot, for example, perform the simple multiplication of two integrating blocks as we had done previously.

Slide 7

Laplace Transforms y The Laplace Transform is a powerful tool used to solve a wide variety of problems by transforming the difficult differential equations into simple algebraic problems where solutions can be easily obtained. y Applying Laplace Transforms is analogous to using logarithms to simplify certain types of mathematical operations operations. y By taking logarithms, numbers are transformed into powers of 10 (or e – natural logarithms) to allow multiplication and division to be replaced by addition and subtraction respectively. y Similarly, the application of Laplace Transforms to the analysis of systems, systems which can be described by linear ODEs ODEs, enable us to solve ODEs using algebra instead of calculus.

Slide 8

Laplace Transforms y They also provide a straightforward method for handling the mathematical time shift associated with dead time equations. y Thus, complicated analysis can be performed in a straightforward manner. y Once the result from a transformation has been obtained we can then apply the Inverse Laplace Transform f to retrieve solutions to the original problems.

Slide 9

Laplace Transforms

ODE problem Difficult

ODE solution

Slide 10

Laplace Transforms

ODE problem

Laplace Transform

Algebraic problem Very Easy

ODE solution

Inverse Laplace Transform

Algebraic solution

Slide 11

Laplace Transforms y The definition of a Laplace Transform is: ∞

where:

.

L[ f (t)] ≡ ∫ f (t)e−stdt ≡ F(s) O

L[ f (t)] = the symbol for the Laplace Transformation in the brackets; s = complex l variable i bl ((s = a +jb) jb) iintroduced d db by the h transformation.

y Ho-hum….you Ho hum you thought that this was supposed to make things simple?

Slide 12

Laplace Transforms y

Probably the easiest way to look at this ‘s’ operator is that it represents the derivative relative to time:

d s= dt

Slide 13

Laplace Transforms y So, a derivative transformation block having a derivate time Td : sTd

y Continuing along this route, the reciprocal of derivative, 1/s is the integral. y Thus the integral transformation block is: where Ti is the integral time: ti 1 sT Ti

where Ti is the integral time: Slide 14

Transformation blocks

G =1

1 1 + sT

Gain block

First order lag

sTd

1 (1 + sT1 )⋅ (1 + sT2 )

Derivative block

Second order lag g

1 sT Ti

sT 1 + sT 1+ T

Integral block

Lead block

Slide 15

Laplace Transforms y The table below shows a number of standard Transforms for different types of inputs Block type Transform U it iimpulse Unit l Unit step Unit ramp

e − att sin ωt

1 1 s 1 s2 1 s+a

ω s2 +ω 2

y Let’s look at a unit step Slide 16

Laplace Transforms

y If we applied a unit step to a Gain Block what would we expect out? y So, a unit step in….

1

s Step input

G =1

1

s Step output

y ….results in a unit step out

Slide 17

Laplace Transforms y So let’s look at an integral block pp y a step p input p y And let’s apply y The output is given by:

1 1 Output p = ⋅ s s Ti

1 Output p = 2 s Ti

1

s

1 sTi

1

s 2 Ti

Step input Slide 18

Laplace Transforms y If we look to see what this is: y We see it’s a unit ramp Block type Transform Unit impulse p Unit step Unit ramp

e − at sin ωt

1 1 s 1 s2 1 s+a

ω s2 +ω 2 Slide 19

Laplace Transforms y So, for a step input … y ….we we get a ramp output (with time TI) y Which is what we would expect from an integral block

1

s Step input

1 sTi

1

s 2 Ti Ramp output

Slide 20

Laplace Transforms y We can apply the same rationale to a Derivative block y Let Let’s s apply a step input y The output is given by:

1 Output = ⋅ sTd s

Output = 1 ⋅ Td

1

s

sTd

1.Td

Step input Slide 21

Laplace Transforms y If we look to see what this is: y We see it’s a unit impulse

Block type Transform Unit impulse Unit step Unit ramp p

e − at sin ωt

1 1 s 1 s2 1 s+a

ω s2 +ω 2 Slide 22

Laplace Transforms y So, for a step input y ….we we get a impulse output y Which is what we would expect from an derivative block

sTd Step input

Impulse output

Slide 23

Laplace Transforms y So, for a step input y ….we we get a impulse output y Which is what we would expect from an derivative block

sTd Step input

Impulse output

Slide 24

First order lag y The general form of a First Order Lag: K 1+ sτ

K 1+ sτ

τ Response time

Slide 25

Deadtime y The general form of deadtime:

e

− θs

e

− θs

θ Deadtime

Slide 26

FOLPDT y When combined the form is:

K e - θs 1 + sτ

θ Deadtime

K e - θs 1 + sτ

τ Response time

Slide 27