THE LAPLACE TRANSFORM THE LAPLACE TRANSFORM IN CIRCUIT
ANALYSISIN CIRCUIT ANALYSISA Resistor in the s DomainA Resistor in
the s DomainR+viv=Ri (Ohms Law).V(s)=RI(sR+VIAn Inductor in the s
DomainAn Inductor in the s Domain+vL iInitial current of I0div
LdtV(s)=L[sI(s)-i(0-)]=sLI(s)-LI0+VsL I+LI00( )( )I V sI ssL s=
++VsLII0sA Caacitor in the s DomainA Caacitor in the s Domain+v
CiInitially charged t V0 vlts.dvi Cdt00( ) [ ( ) (0 )] ( )!( ) ( )I
s C sV s v sCV s CVVV s I ssC s= = = + I+VCV01/sC+V0/s1/sC+VIThe
Natura! Resonse o" an RC CircuitThe Natura! Resonse o" an RC
CircuitRt=0i+vC+V0+VRI 1/sC+V0/s000!00!( ) ( )( )!( ) ( )t tRC
RCVRRCVI s RI ss sCCVI sRCs sVi e u t v Ri V e u tR The Ste Resonse
o" a Para!!e! CircuitThe Ste Resonse o" a Para!!e! Circuit"The Ste
Resonse o" a Para!!e! CircuitThe Ste Resonse o" a Para!!e!
Circuit#! !#! ![ ]dcdcdcICRC LCILCLRC LCI V VsCVR sL sVs sIs s s+ +
==+ +=+ +$# %$&! # #$'!%$' 0#'%( !0( )(000 !) !0 )'%( !0( '#000
#(000)( '#000 #(000)'#000 #(000 '#000 #(000'%( !0#( !0!) !0'%( !0#0
!0 !#).%*( '#000 #(000)( (%000)( ) [#( (0LLLLIs s sIs s j s jK K
KIs s j s jKKj ji t e=+ + =+ + += + ++ + += = = = += +'#000
0cs(#(000 !#).%* )] ( )tt u t mA+Transient Resonse o" aTransient
Resonse o" a Para!!e! RLC CircuitPara!!e! RLC CircuitReplacing the
dc current source with a sinusoidal current source# #!#! !## # #!
!# # #! !cs+ ( )( ) ( )( ) ( )( )( )[ ( ) ( )]( )( )( )[ ( ) (
)]mmmg m gCgRC LCICRC LCILCLRC LCsIi I t I sssV s I ss ssV ss s ssV
sI ssLs s s= =+=+ +=+ + += =+ + +$# % # %& &! ! # #$'!#(
,(0000 -'%( !0( )( !) !0 )( )(000 !) !0 )( )(0000 (0000 '#000 #(000
'#000 #(000'%( !0 ( (0000)*.$ !0.0( %0000)('#000 !)000)('#000
)(000)mLLI mA rad ssI ss s sK K K KI ss j s j s j s jjKj j j= ==+ +
+ = + + + + + + += = + +0$' 0#'#000'%( !0 ( '#000 #(000)!#.$ !0 .0(
'#000 !)000)( '#000 )(000)( (%000)( ) (!$sin(0000 #$ sin#(000 ) (
)!$sin(0000 tLLssjKj j ji t t e t u t mAi t mA += = += =Mesh
Ana!#sisMesh Ana!#sis"Mesh Ana!#sis $cont%&Mesh Ana!#sis
$cont%&! #! #!#'')((# %.( ) (#0 (# (.0 !0 )(0( .) !$ !( !( #)(
!#) # !#!)% * %.( !.(( #)( !#) # !#s I IsI s IsIs s s s s sIs s s s
s s= + = + ++= = + + + += = ++ + + +# !#!# !##!#(!$ !( ) ( ) ,(*
%.( !.( ) ( ) .'')(.0)( ) !$,(#((%)!$((#)( ) * .0t tt ti e e u t Ai
e e u t Ai Ai A The'enin(s TheoremThe'enin(s TheoremUse the
Thevenins theorem to fnd vc(t) "The'enin(s Theorem
$cont%&The'enin(s Theorem $cont%&((((%0- )(0.00# ) (%0#0
0.00#!00.00# (#0) %0( *$00))0#0 0.00#!0ThThs sVsss sZss(( $# )
##$000 $000(%0-( !0 )[%0( *$00) -( !0 )] [(# !0 ) - ]) )!0000 #$ !0
( $000)'0000 )$000( $000)( ) ( '0000 ) ) ( )CCCt tcsIs s ss sIs s
sIssi t te e u t A +=+ + + = =+ + += +++= +$ $# #$ $000! # !0 ) !#
!0( $000) ( $000)( ) !# !0 ( )c CtcsV IsC ss sv t te u t V = = =+
+= MUTUAL INDUCTANCE E)AMPLEMUTUAL INDUCTANCE E)AMPLE ! #)0(0 ) $
,(0 ) 0!#i A i i#(t)=" "MUTUAL INDUCTANCE E)AMPLEMUTUAL INDUCTANCE
E)AMPLE Using the T-equivalent of the inductos! and s-do"ain
equivalent gives the follo#ing cicuit#'##.$ !.#$ !.#$( !)( ') ! '(
) !.#$( ) ( )t tIs s s si t e e u t A ! #! #(' # ) # !0# (!# % )
!0s I sIsI s I+ + =+ + =THE TRANSFER FUNCTIONTHE TRANSFER
FUNCTION/he trans0er 01nctin is de0ined as the rati 0 the
La2lacetrans0rm 0 the 1t21t t the La2lace trans0rm 0 the in21twhen
all the initial cnditins are 3er.( )( )( )Y sH sX s4(s) is the
La2lace trans0rm 0 the 1t21t,5(s) is the La2lace trans0rm 0 the
in21t.THE TRANSFER FUNCTION $cont%&THE TRANSFER FUNCTION
$cont%&!###( ) !( )( ) !-!( ) !( )( )!ggI sHsV s R sL sCsCs LC
RCsV sHsV ss LC RCs= =+ +=+ += =+ +E)AMPLEE)AMPLEFind the transfer
function V0Vg and determine the poles and !eros of
"(s)#"E)AMPLEE)AMPLE00 0)0# )0# )!#!0!000 #$0 0.0$!0!000( $000))000
#$ !0!000( $000)( ))000 #$ !0'000 (000, '000 (000$000gggV VV V sssV
Vs sV sH sVs sp jp jz+ + =++=+ + += =+ + = += = $ssume that
vg(t)%&0tu(t)# Find v0(t)# Identif' the transient and
stead'(state components of v0(t)# 0# ) #&' ! ! ##( 0 (! # '(
'000 00(!000( $000) $0( ) ( ) ( )( )000 #$ !0 )'000 (000 '000 (000$
$ !0 *..* ,!0,( !0[!0 $ !0 cs((000 *..* ) !0 ( !0 ] ( )gtsVs H s V
ss s sK K K Ks j s j ssK K Kv e tt u t V += =+ + = + + ++ + += = =
= ++ The transient component is generated )' the poles of the
transfer function and it is*( '000 0!0 $ !0 cs((000 *..* )te t The
stead'(state components are generated )' the polesof the driving
function (input)*((!0 ( !0 ) ( ) t u tTime In'ariant S#stemsTime
In'ariant S#stemsIf the input dela'ed )' a seconds+ then!( ) ( ) (
)( ) ( ) ( )( ) ( ) ( ) ( )asasL x t a u t a e X sY s H s X s ey t
L Y s y t a u t a === = Therefore+ dela'ing the input )' a seconds
simpl' dela's theresponse function )' a seconds# $ circuit that
e,hi)its thischaracteristic is said to )e time invariant#Imu!se
ResonseImu!se ResonseIf a unit impulse source drives the circuit+
the response of the circuit e-uals the inverse transform of the
transfer function#!( ) ( ) ( ) !( ) ( )( ) ( ) ( )x t t X sY s H sy
t L H s h t$ote that this is also the natual es%onse of the
cicuit&ecause the a%%lication of an i"%ulsive souce is
equivalentto instantaneousl' stoing eneg' in the cicuit.CON*OLUTION
INTE+RALCON*OLUTION INTE+RAL( ) ( ) x t t ='(t)a &t*(t)a
&tCicuit $ is linea #ith no initial stoed eneg'. If we .now the
form of ,(t)+ then how is '(t)descri)ed/ To ans#e this question! #e
need to +no# so"ething a&out $. ,u%%ose #e +no# the i"%ulse
es%onse of the s'ste". $*(t) '(t)$*(t) '(t) (1)t( ) ( ) y t h t =t(
) t Instead of a%%l'ing the unit i"%ulse at t=0! let us su%%ose
that it is a%%lied at t=-. The onl' change in the out%ut is a ti"e
dela'.$( ) h t $e*t! su%%ose that the unit i"%ulse has so"e stength
othe than unit'. Let the stength &e equal to the value of *(t)
#hen t= -. ,ince the cicuit is linea! the es%onse should &e
"ulti%lied &' the sa"e constant *(-)( ) ( ) x t $( ) ( ) x h t
$o# let us su" this latest in%ut ove all %ossi&le values of -
and use the esult as a focing function fo $. .o" the lineait'! the
es%onse is the su" of the es%onses esulting fo" the use of all
%ossi&le values of -( ) ( ) x t d + $( ) ( ) x h t d +.o" the
sifting %o%et' of the unit i"%ulse! #e see that the in%ut is si"%l'
*(t)$( ) ( ) x h t d +/(t)( ) ( ) ( ) ( ) ( ) y t x t z h z dz x t
z h z dz + = = ( ) ( ) & ( ) y t x t h t( ) ( ) ( ) y t x h t
d+0u question is no# ans#eed. 1hen *(t) is +no#n! and h(t)! the
unit i"%ulse es%onse of $ is +no#n! the es%onse is e*%essed &'
This i"%otant elation is +no#n as the convolution integral#It is
often a&&eviated &' "eans of1hee the asteis+ is ead
2convolved #ith3. If #e let 4=t--! then d-=-d4! and the e*%ession
fo '(t) &eco"es( ) ( ) & ( ) ( ) ( ) ( ) ( ) y t x t h t x
z h t z dz x t z h z dz = = = ( ) ( ) & ( ) ( ) ( ) ( ) ( ) y t
x t h t x z h t z dz x t z h z dz = = = Con'o!ution and Rea!i,a-!e
S#stemsCon'o!ution and Rea!i,a-!e S#stems.o a %h'sicall'
eali4a&le s'ste"! the response of the s'stem cannot )egin
)efore the forcing function is applied# ,ince h(t) is the es%onse
of the s'ste" #hen the unit i"%ulse is a%%lied at t=0! h(t) cannot
e*ist fo t50. It follo#s that! in the second integal! the integand
is 4eo #hen 4506 in the 7st integal! the integand is 4eo #hen (t-4)
is negative! o #hen 48t. Theefoe! fo reali!a)le s'stems the
convolution integal &eco"es0( ) ( ) & ( ) ( ) ( ) ( ) ( )ty
t x t h t x z h t z dz x t z h z dz= = = E)AMPLEE)AMPLE( ) # ( )th
t e u th(t)*(t)t'(t)00( ) ( ) ( !)( ) ( ) & ( ) ( ) ( )[ ( ) (
!)][# ( )]zx t u t u ty t x t h t x t z h z dzu t z u t z e u z
dz1+rahica! Method o" Con'o!ution+rahica! Method o" Con'o!ution0( )
# #(! ) 0 !tz ty t e dz e t = = ,ince h(4) does not e*ist %io to
t=0 and vi(t-4) does not e*ist fo 48t! %oduct of these functions
has non4eo values onl' in the inteval of 0545t fo the case sho#n
#hee t51. 1hen t81! the non4eo values fo the %oduct ae o&tained
in the inteval (t-1)545t.!( ) # #( !) !tz tty t e dz e e t = =
>E)AMPLEE)AMPLE9%%l' a unit-ste% function! *(t)=u(t)! as the
in%ut to a s'ste" #hose i"%ulse es%onse is h(t) and dete"ine the
coes%onding out%ut '(t)=*(t):h(t).
"h$t&.u$t&/0u$t/1&2u$t/0&30( ) ! 0 !ty t dz t t =
=
