transient responses (laplace transforms)
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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 1st 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.TRANSCRIPT
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 (Laplace Transforms)
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 .
Transient Responses (Laplace Transforms)
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
Transient Responses (Laplace Transforms)
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.
Transient Responses (Laplace Transforms)
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
Transient Responses (Laplace Transforms)
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
Transient Responses (Laplace Transforms)
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
Transient Responses (Laplace Transforms)
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.
Transient Responses (Laplace Transforms)
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
Transient Responses (Laplace Transforms)
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.
Transient Responses (Laplace Transforms)
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
This resource was created by the University of Wales Newport and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme.
© 2009 University of Wales Newport
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Transient Responses (Laplace Transforms)