Transient Responses (Laplace Transforms)

Transient Responses (Laplace Transforms)

Electrical and Electronic Principles

© University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

The following presentation is a part of the level 4 module -- Electrical and Electronic Principles. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1 st

year undergraduate programme.

The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

Contents Transient Responses V is known as the driving function or forcing function What will happen to the capacitor voltage? Features of an Exponential Charge Up Second order equations. Laplace Transformation. Rules Now consider the first order equation for the RC network. Consider the second order equation for the RLC network. Examples Credits

In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see:Green D C, Higher Electrical Principles, Longman 1998 Hughes E , Electrical & Electronic, Pearson Education 2002Hambly A , Electronics 2nd Edition, Pearson Education 2000Storey N, A Systems Approach, Addison-Wesley, 1998


Transient Responses

So far all the calculations we have performed have led to a Steady State solution to a problem i.e. the final value after everything has settled down.

There are in fact two parts to the total response of a system to an input, these are:

The Steady State which lasts indefinitelyand The Transient Response, which decays to zero, leaving only the steady state.

The steady state values can be determined using circuit laws and complex number theory. The transient is more difficult as it involves differential equations.

Consider the simple charge up circuit below.

Vi

R

CWhen the switch is closed we can say that:

VVcRi

The equation for a capacitor is dt

dVcCi

Giving us: VVcdt

dVcCR

This is called a first order linear equation with constant coefficients.If we want to know Vc we cannot solve it using simple circuit theory, but it can be solved using some knowledge of Calculus.Transient Responses (Laplace

Transforms)

The voltage V is known as the driving function or forcing function and is responsible for the final steady state value.

To solve the function we set the forcing function to zero

0Vcdt

dVcCR

We ask the question what form of Vc will allow us to equate it with its derivative?

The form used is

This gives us

mtAeVc

mtmAedt

dVc

mtmAedt

dVc

putting this in the equation gives us:

this is called the Auxiliary Equation.

From which giving us

this is known as the Complementary Function.

This is part of the total solution but we still need to evaluate the constant A. In order to obtain the complete solution we need to add the Particular Integral. This is the steady state value and for our example this is

0 mtmt AeCRmAe

CRm

1

RCt

AeVc

VVc Transient Responses (Laplace

Transforms)

So the complete solution is:

We still need to determine A.

This can be done by using initial conditions:

when t = 0, Vc = 0

So giving

Our solution is therefore

If we start with Vc charged to V and then allow it to discharge we have the same complementary function but the steady state will be 0 and the initial conditions will be Vc = V at t = 0.

A would equal V and we would have:

RCt

AeVVc

AV 0 VA )1( RC

teVVc

RCt

VeVc

Transient Responses (Laplace

Transforms)

If C = 1F and R = 1 M what will happen to the capacitor voltage if a 10v supply is connected across the two components when t = 0?

0

2

4

6

8

10

12

0 1 2 3 4 5 6

time

vo

lta

ge

Some important features of an exponential charge up:

1. The time taken to reach 63.2% of the final value is (tau) the time constant and this is given by R x C.

2. After 4 the value is within 2% of the final value.3. After 5 the value is within 1% of the final value.4. The area above the curve is given by 10, where 10

is the final value voltage.5. If the plot had continued to rise at it’s initial rate it

would have reached the final value in a time .


0

2

4

6

8

10

12

0 1 2 3 4 5 6

time

vo

lta

ge

6.32v

>9.8v

>9.9vArea = 10

4 5


Second order equations.

Consider the diagram below.

Vi

R

CL

For an inductor:

dt

diLVL

For the circuit:

but

giving us: second order.

VVcRidt

diL

dt

dVcCi

VVcdt

dVcRC

dt

VcdLC

2

2

Once again we can use as a solution:

mtAeVc mtmAedt

dVc mtAem

dt

Vcd 22

2

Substituting in and removing the forcing function we get:

012 RCmLCm

This gives us LC

LCCRRCm

2

422

LCL

R

C

Rm

1

42 2

2

This can be solved but begins to become complicated especially if the value in the square root is negative and we end up with complex values.

Solving for transient conditions is therefore not easy. To make the solving of these problems easier we use Laplace Transforms.

Laplace Transformation.

What we are able to do is to take a problem in the time domain (t) and to convert it into the Laplace domain (s).

The conversion is carried out using a simple set of rules.


Rules1. If a function of time is multiplied by a constant

then the Laplace transform is multiplied by the same constant. e.g. a step of 6v to an electrical system is the same as 6 times a unit step and therefore has the value 6/s.

2. If an equation contains the sum of two separate quantities that are functions of time then the transform is the sum of the individual transforms.

3. The Laplace transform of a first derivative of a function is:Transform of

where is the value of the function at t=0

[initial conditions are normally 0]

)0()()( fssFtfdt

d

)0(f


Rules

4. The Laplace transform of a second derivative of a function is:Transform of

where is the value of the derivative of the function at t=0

5. The Laplace transform of an integral of a function is:

Transform of

)0()0()()( 22

2

fdt

dsfsFstf

dt

d

)0(fdt

d

)(1

)( sFs

tft

o


Consider the first order equation for the RC network.

Translating gives us:

which means that Vc is given by:

VVcdt

dVcCR

s

VVcCRsVc

s

VCRsVc )1(

)1(

CRss

VVc

This is now converted back to the time domain using the reverse Laplace transforms. Is there a transform? If not, we must split the function into simpler parts. Use partial fractions:Transient Responses (Laplace

Transforms)

1)1(

CRs

B

s

A

CRss

Vfrom which

BsCRsAV )1(

s = 0 giving

equating s giving

VA

BACR 0 VCRB

CRs

V

s

V

CRs

VCR

s

VVc

11

transform

)1( RCt

RCt

eVVeVVc

What about a ramp input into the network?

Consider the second order equation for the RLC network.

Translating gives us:

which means that Vc is given by:

s

VVcRCsVcVcLCs 2

s

VRCsLCsVc )( 12

)( 12

RCsLCss

VVc

This is now converted back to the time domain using the reverse Laplace transforms. Is there a transform? If not, we must split the function into simpler parts. Use partial fractions. The method depends upon whether the bracket will factorise.

VVcdt

dVcRC

dt

VcdLC

2

2

Example

10vi

290

10F

0.1H

Vc

)( 12

RCsLCss

VVc

V = 10, R = 290

L = 0.1 C = 10 x 10-6

)().( 62

6

626 102900

1010

11010290101010

10

ssssssVc

100000029002 ss a = 1, b = 2900, c = 1000000

2

400000084100002900

2

42

a

acbbs

2

21002900

2

44100002900

s

25002

5000400

2

800

sors

))(()( 4002500

10000000

102900

101062

6

sssss

Vc

Use Partial Fractions to simplify the denominator.

40025004002500

10000000

s

C

s

B

s

A

sss ))((

)()())(( 2500400400250010000000 sCssBsssA

Make s = 0:10

1000000

100000001000000400250010000000 AAA ))((

Make s = -2500:

90515250000

1000000052500002100250010000000 . BBB

Make s = -400:

90511840000

10000000840000210040010000000 .

CCC

400

90511

2500

905110

sssVc

..transform

tt eeVc 4002500 90511905110 ..

When t = 0

Vc = 10 + 1.905 – 11.905 = 0

Over the page shows the plot:

0

1

2

3

4

5

6

7

8

9

10

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014


Example

R is reduced to 100

10vi

100

10F

0.1H

Vc

)( 12

RCsLCss

VVc

V = 10, R = 100

L = 0.1 C = 10 x 10-6

)().( 62

6

626 101000

1010

11010100101010

10

ssssssVc

100000010002 ss a = 1, b = 1000, c = 1000000

2

400000010000001000

2

42

a

acbbs

2

30000001000 s This will not factorise due

to the negative square root.

We need to take the quadratic equation and complete the square – (this then matches one of the transforms): 7500002500001000101000 262 ssss

22262 03866500750000500101000 .)()( ssss

).)(()( 22

6

62

6

03866500

1010

101000

1010

sssssVc

Use Partial Fractions to simplify the denominator.

2222

6

0386650003866500

1010

.)().)((

s

CBs

s

A

ss

)().)(( CBsssA 22 0386650010000000

Make s = 0:

101000000

1000000010000000386650010000000 22 AAA ).(

Equate s values10000100010000 ACCA

Equate s2 values 100 ABBA

2222

6

03866500

100001010

03866500

1010

.)().)((

s

s

sssVc

We now need to ensure that the fraction is in the correct form for inverse transformation.


2222 03866500

100010

03866500

1000010

.)(.)(

s

s

s

s

222222 03866500

50010

03866500

50010

03866500

100010

.)(.)(.)(

ss

s

s

s

2222 03866500

03866

03866

5000

03866500

50010

.)(.

..)(

ss

2222 03866500

038667745

03866500

50010

10

.)(.

..)(

ss

s

sVc

transform

).(.).( tSinetCoseVc tt 038667745038661010 500500 When t = 0 Vc = 10 - 10– 0 = 0Over the page shows the plot:

0

2

4

6

8

10

12

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014


If R was further reduced e.g. 25 this process could be repeated and the following result would be obtained:

).(.).( tSinetCoseVc tt 1699226041169921010 125125

The plots for the three values of R are shown on the next slide.


0

2

4

6

8

10

12

14

16

18

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014

R=290

R=100

R=25

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© 2009 University of Wales Newport

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Transient Responses (Laplace Transforms)

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