November 5, 2001ece309-10-1 November 5, 2001 12.6 Applying Laplace transform to Circuit Example: i(t)L1H12u(t)R1 (0) 0 0Iatt=≤ u(t) is unit step function, so that 1() {()}Us Luts== From the circuit ()() ()Ldi tRi t u tdt+= () 1() ()di t Rit utdt L L+= 11() ()RsI s I sLLs+= ()1() 1Is ss+= 11 1()1(1)Isss ss==++ I(s) can be expressed 111()(1) 1Isss s s=≡−++ To find i(t), we can get inverse laplace transform of I(s). From the table {}1() () ()tit L Is ut e−−==− As a conlusion, The Laplace transform turns a differential equation into a algabraic equation. Therefore , equation comes mor easy to solve. Example:ece309-10-2 v(t)L-R C+Idc If we write the voltage-node equation in this circuit, we will have 0() 1 ()() ()tdcv t dv tvtdt C I utRL dt++=∫ We transform the equation to the s domain () 1 () 1() (0)dcVs VsCsVs v IRLs s−++ −= Let say initil voltahe on the capacitr is zero, and 11 1()dcVs sC IRsL s++ = /()11dcIsVssCRsL=++ 2/()11dcICVsssRC LC=++ To find v(t) we must inverse-transform the expression {}1() ()vt V sL−= Next step is anayzing the inverse-transform.ece309-10-3 12-7. Inverse Laplace Transform: The expressionfor V(s) is a rational function of s. It can be expressed in the form of a ratio of two polynomials in s. In general, we can a rational function in s in the following form 11101110...()()( ) ...nnnnmmmmas a s as aNsFsDs bs b s bs b−−−−++++==++++ The coeffircients a and b are real constants, and the exponnets m and n positive intergers. The ration ()()NsDsis called proper rational function if n<m The ration ()()NsDsis called improper rational function if n>=m Only proper rational function can be expressed as a sum of partial fraction • The roots of N(s)=0 are called the zeros of F(s) • The roots of D(s)=0 are called the poles of F(S) • Use partial fraction expression to break F(s) down into simple terms • Find the inverse transform from the table for each simple term. Example: 26(3)(1)sss s+++ The denominator has four roots. Thus partial fraction expansion of theis thi function 312 4226( 3)( 1) 3 ( 1) 1KKK Ksss s s s s s+≡+ + +++ + + + (≡an identical equation and both side are subject to the same mathematical operation. The both sides of the equation must be same for allvalues of variables s) The key is that we need to recognize the f(t) function of each laplace transform above 11312 4226()( 3)( 1) 3 ( 1) 1KKK Ksftss s s s s sLL−− +≡+++ ++ + + + = 312 3 4()() () () ()tttftKutKeutKteutKeut−−−=+ + + ()31234() ()tttft K Ke Kte Ke ut−−−=+ + + Determine the coeeffients (K1,K2, K3…) then apply into the equation.ece309-10-4 The Problem is to find the roods of D(s). The roods of D(s) can be . 1. real and distinct (Simple poles) such as ()KsaFs+= ! () ()atft Ke ut−= 2. real and repeated (Repeated poles) such as 2()Ksa+ ! () ()atft Kte ut−= 3. complex and distinct (Complex poles) such as *KKsjsjαβαβ++− ++ !() 2 cos( ) ()tft Ke t utαβθ−=+ 4. complex and repeated (Repeated complex poles) such as ()()*22KKsj sjαβαβ++− ++ !() 2 cos( ) ()tft tKe t