AC and DC Circuit Analysis Using Laplace Transform With Multisim Simulations

8/16/2019 AC and DC Circuit Analysis Using Laplace Transform With Multisim Simulations

1/21

AC and DC Circuit Analysis Using Laplace Transform With

Multisim Simulations

By

Buquis, Merrelyn

Castillo, !mmanuel D

"lagan, May Ann #ose B

$adua, Samuel C

Santoyo, #en Christian !

Umali, "an #ay #

$ro%lem set num%er & su%mitted to !ngr 'ustiano B Menes 'r of the College of

!ngineering Architecture and (ine Arts, Department of !lectrical !ngineering in partial

fulfillment of the requirements for

Ad)anced Mathematics for !lectrical !ngineering

"n

Bachelor of Science in !lectrical !ngineering

Batangas State Uni)ersity Main Campus ""

Alangilan, Batangas City, Batangas

* + -.

1


2/21

TABLE OF CONTENTS

Title $age/////////////////////////////// i

Ta%le of Contents//////////////////////////// ii

"ntroduction////////////////////////////// -

Mathematical #eferences, 0otations, and Definitions///////////// +

Sample Sol)ed !1ercises///////////////////////// .

#eferences////////////////////////////// -.

2


3/21

INTRODUCTION

An #L Series Circuit consists %asically of an inductor of inductance L connected

in series 2ith a resistor of resistance # An #L series circuit is connected across a

constant )oltage source 3the %attery4 and a s2itch Assume that the s2itch, S is open until

it is closed at a time t 5 , and then remains permanently closed producing a 6step

response7 type )oltage input The current, " %egins to flo2 through the circuit %ut does

not rise rapidly to its ma1imum )alue of " ma1 as determined %y the ration of 8 9 # 3:hm;s

La24 After a time, the )oltage source neutrali8L4, is used to define indi)idual )oltage drops that

e1ist around the circuit and then hopefully use it to gi)e an e1pression for the flo2 of

current "n general, transient phenomena occur 2hene)er a circuit is suddenly connectedor disconnected to9from the supply, there is a sudden change in the applied )oltage from

one finite )alue to another, a circuit is short circuited

"n !lectrical !ngineering, a transient response or a natural response is the

electrical response of a system to a change from equili%rium The condition pre)ailing in

an electric circuit %et2een t2o steady=state conditions is ?no2n as the transient state@ it

lasts for a )ery short time The currents and )oltages during the transient state are called

transients A transient state 2ill e1ist in a circuit containing one or more energy storage

elements 2hene)er the energy conditions in the circuit change, until the ne2 steady=state

condition is reached Transients are caused %y changing the applied )oltage or current, or

%y changing any of the circuit elements@ such changes occur due to opening and closing

s2itches "n this paper, such equations are de)eloped analytically using Laplace

transforms for different 2a)eform supply )oltages "n transient analysis, also called time=

domain transient analysis, Multisim computes the circuit;s response as a function of time

This analysis di)ides the time into segments and calculates the )oltage and current le)els

for each gi)en inter)al (inally, the results, current )ersus time, are presented in the

grapher )ie2 Multisim performs transient analysis using the follo2ing process each

input cycle is di)ided into inter)al, a DC and AC operating point analysis is performed

for each time point in the cycle and the solution for the current 2a)eform at a node is

determined %y the )alue of the current at each time point o)er one complete cycle

1


4/21

MATHEMATICAL REFERENCES, NOTATIONS AND DEFINITIONS

The follo2ing theorems and properties are the theorems used in this paper and

their corresponding proofs

Linearity

The linearity property of the Laplace Transform statesa · f (t )+b· g (t ) L

↔a · F (s )+b· G (s)

This is easily pro)en from the definition of the Laplace Transform

L[a· f (t )+b · g (t ) ]= ∫0

∞

[a · f (t )+b · g (t ) ]e− st dt

¿a∫0∞

f (t )e−st

dt +b∫0∞

g (t )e− st

dt

¿a ·F (s )+b ·G (s)

This property can %e easily e1tended to more than t2o functions as sho2n fromthe a%o)e proof With the linearity property, Laplace transform can also %e calledthe linear operator

First Shifting

"f L[f (t ) ]= F (s) , 2hen s >a then, L[eat f (t ) ]= F (s− a )

"n 2ords, the su%stitution s – a for s in the transform corresponds to themultiplication of the original function %y e at .

Proof of First Shifting Pro erty

F (s)= ∫0

∞

e−st f (t )dt

F (s− a )=∫0

∞

e−(s− a)t f (t )dt

F (s− a )=∫0

∞

e−st +at f (t )dt

2


5/21

F (s− a )=∫0

∞

e−st e at f (t )dt

F (s− a )=∫0

∞

eat f (t )dt

Se!on" Shifting Pro erty

"f L[f (t ) ]= F (s ), and g (t )= [f (t − a ) ] ; t >a then, L[g (t ) ]= e− as F (s )

Proof of Se!on" Shifting Pro erty

L[g (t ) ]=∫0

∞

e− st g (t )dt

L[g (t ) ]= ∫0

a

e− st (0 )dt +∫a

∞

e−st f (t − a )dt

L[g (t ) ]=∫0

∞

e− st f (t − a )dt

Let < 5 t a 2hen t 5 a, < 5

t 5 < a 2hen t 5 , < 5

dt 5 d<

L[g (t ) ]=∫0

∞

e− s ( z+a )f ( z)dz

L[g (t ) ]= ∫0

∞

e− sz− sa f ( z)dz

3


6/21


7/21

As f 3t4 is of e1ponential order > and #e3s4 E >, 2e ha)elim

R→ +∞f ( R)e− sR= 0

Fence the preceding equation %ecomes

L(f '

(t ) )=− f (0 )+s∫0∞

f (t )e−st

dt = sF (s )− f (0 )

$ro)ing the theorem

Initia# &a#'e Theore$

"f f3t4 and (3s4 are Laplace transform pairs i e

f (t )← ( L)→F (s )

Then "nitial )alue theorem is gi)en %y

limt → 0 f (t )= lims → ∞ sF (s )= f (0 )

Proof of La #a!e Initia# &a#'e Theore$

Laplace transform of a function f3t4 is

Lf (t )=∫0

∞

e−st f (t )dt = F (s)

Then Laplace transform of its deri)ati)e f;3t4 is

L f ' (t )=∫0

∞

e−st f ' (t )dt = sF (s )− f (0 )…(1 )[¿time differentiationtheorem ]

Consider the integral part first

0 − ¿

0 +¿

e− st f ' (t )dt = ∫¿

0 +¿ e−st f ' (t )dt +∫¿

∞ e− st f ' (t )dt

∫0∞

¿

G lim

s → ∞e− st

is indeterminate and hence it is split into t2o integralsH

5


8/21

0 +¿

¿[f (t ) ]+∫¿

∞e− st f ' (t )dt

Gfor I t I =, e=st 5 -H

+¿0 ¿¿

− ¿0 ¿

¿0 +¿

¿f ¿

Su%stituting 3+4 in 3-4 2e get

+¿0 ¿

¿−¿0 ¿

¿0 +¿

−¿0 ¿

e− st f ' (t )dt = sF (s)− f ¿f ¿

Upon cancelling f3 =4 on %oth sides 2e get

0 +¿

+¿0 ¿

¿e− st f ' (t )dt +f ¿

∫¿

∞ ¿

Considering 3s4 tends to infinity on %oth sides in 3J4

+¿0 ¿

¿f ¿

6


9/21

+¿0 ¿

¿sF (s )= f ¿f (t )= lim

s → ∞¿

t → 0+¿

¿lim¿

¿

DEFINITION OF TERMS

Transient = "t is a momentary %urst of energy induced upon po2er, data, and

communication lines, they are characteri


10/21

L{∂ i∂ t }= s ( I (s ) )− I (0 ) L{10 i}= 10 ( I (s ))

L{10 e− 10 t }= 10s+10

I (s)+10 I (s)= 10

s+10s ¿

I (s) [s+10 ]= 10s+10

Is= 10(s+10 )2

By $artial (raction Decomposition

10

(s +10 )2=

(s +10 )= !

(s+10 )2

10 = (s+10 )+!

s : 0 =

" : 10 = (s +10 )+!

A 5

B 5 -

$lugging %ac? the computed )alues to the pre)ious equation

L−1{ L[ 10(s+10 )2 ]}= L− 1{ 10(s+10 )2 }

I (s)= 10 te− 10 t

8


11/21


12/21


13/21

i= L− 1{ 3770s 2 +142129 }= 10sin (377 t )$lugging the )alue for t

i= 10 sin [377 (0.01 ) ]i= 0.6575

SIMULATION

J "n an #L circuit, >irchhoff;s La2 gi)es the follo2ing relation !5 L di9dt #i 2here

!5 suppy )oltage + 8

#5 resistance + ohms

L5 inductance - Fenry

t 5 time in secondsi 5 current in amperes

if i 5 2hen t5 , find i,2hen t5 + sec

MANUAL SOLUTION

By >irchoff;s 8oltage La2

11


14/21

E− iR− L ∂ i∂ t

= 0

200 − i (20 )− ∂i∂ t

= 0

L(200 )=200

s

20 L(i)= I (s )

L{∂ i∂ t }= s ( I (s ))− I (0 )200

s − 20 I (s )− s

( I (s)

)= 0

200

s = 20 I (s )+s ( I (s ) )

200

s = I (s ) [20 +s ]

I (s)= 200

s(20 +s)


200

s (20 +s )=

s + !

20 +s

200 = (20 +s )+! (s)

200 = 20 + s+!s

s 5 A B

c + 5 + A

A 5 -

B 5 =-

$lugging the )alues %ac?

12


15/21


16/21

L{144 }= 144s

L{36 i}= 36 ( I (s ) )

L{6 didt }= 6 [s ( I (s) )− I (0 ) ]144

s − 36 ( I (s) )− 6 s ( I (s ) )= 0

144

s = I (s)[6 s+36 ]

I (s)= 144

s(6 s+36 )


144

s (6 s +36 )=

s + !

6 s +36

144 = (6 s+36 )+!s

144 = 6 s+36 +!s

s 5 .A B

c -&& 5 J.A

A 5 &

B 5 =+&


L−1{ L[ 144s (6 s +36 ) ]}= L[4s + −246 s +36 ]

14


17/21

L−1{ L[ 144s (6 s +36 ) ]}= L[4s − 4s+6 ]

I (s)= 4 − 4 e− 6 t

$lugging the )alue for t

I (s)= 4 − 4 e− 6 (0.1 )

I (s)= 1.805

SIMULATION

A certain 2elder has a %asic circuit equi)alent to a series #L 2ith #5 - ohm and L5

-mF "t is connected to an AC source 6e7 through a s2itch 6s7 operated %y an automatic

timer, 2hich closes the circuit at the desired point Calculate the magnitude of the

transient current J secs after the s2itch is closed $assing through its pea? )alue of

- )olts

MANUAL SOLUTION

By >irchoff;s 8oltage La2


= 0

100 − i (0.1 )− (0.001 ) ∂ i∂t

= 0

15


18/21

L{100 }= 100s

L{i}= I (s )

L{∂ i∂ t }= 0.001 (s ( I (s ) )− I (0 ))100

s = 0.1 I (s )+0.001 sI (s )

100

s = [0.1 +0.001 ] I (s)

I (s)= 100s (0.1 +0.001 )


100

s (0.1 +0.001 )=

s + !

0.1 +0.001 s

100 = (0.1 +0.001 )+!s

100 = 0.1 +0.001 +!s

s 5 -A B

c - 5 -A

A 5 -

B 5 =-


L−1{ L[ 100s (0.1 +0.001 ) ]}= L− 1[1000s +1000 ( −1s+100 )]

I (s)= 1000 − 1000 e− 100 t


I (s)= 1000 − 1000 e− 100 (0.03 )

16


19/21

I (s)= 950.2129

SIMULATION

. "n an #L circuit, >irchoff;s La2 gi)es the follo2ing relation !5L di9dt #i 2here

! 5 Supply 8oltage 3+ )olts4

# 5 #esistance 3+ ohms4

L 5 "nductance 3- henry4

t 5 time in seconds

i 5 current in amperes

if i 5 2hen t 5 , find " 2hen t 5 -+ secondsMANUAL SOLUTION


= 0

17


20/21


21/21

I (s)= 10 − 10 e− 20 t


I (s)= 10 − 10 e− 20 (0.12 )

I (s)= 9.0928

SIMULATION

REFERENCESG-H http 99222 electronics=tutorials 2s9inductor9lr=circuits html

G+H http 99222 ni com9tutorial9-+KK&9en9

GJH http 99uqu edu sa9files+9tinyNmce9plugins9filemanager9files9&J- JJJ9!lectricalN

CircuitNTheoryNandNTechnologyN+! pdf

1

Documents

AC and DC Circuit Analysis Using Laplace Transform With Multisim Simulations