Lecture 13. Inverse Laplace Transformation• Inverse Laplace TransformPl il•Polynomials• Roots, zeros and poles•Complex numbersp• Step & Delta functions1Solving Differential Equationsgq• Example: For zero initial conditions, solve)(4)(30)(11)(22tutydttyddttyd=++• Laplace transform approach automatically includes initial conditions in the solutiondtdtin the solution)0()()(yssdttyd−=⎥⎦⎤⎢⎣⎡YL)0(')0()()(222yysssdttyd−−=⎥⎦⎤⎢⎣⎡⎦⎣YL2Solving Differential Equations (cont’d)gq()• Laplace transform of the equationssYyssYysysYs4)(30)0(11)(11)0(')0()(2=+−+−−)3011(4)11)0('()0()(22+++++=sssysyssY)3011(++sssEasy to solve the differential equation in Laplace space,but needs to transform the solution to real space!pInverse Laplace Transform3pInverse Laplace Transformp• Consider that F(s) is a ratio of polynomial expressions)()()(sssDNF =• The n roots of the denominator, D(s) are called the polesP l ll d t i th d t bilit f th t)(sD–Poles really determine the response and stability of the system• The m roots of the numerator, N(s), are called the zeros4Inverse Laplace Transformp• We will use partial fractions expansion with the method ppof residues to determine the inverse Laplace transform• Three possible cases (need proper rational, i.e., n>m)1. simple poles (real and unequal)2.simple complex roots (conjugate pair)2.simple complex roots (conjugate pair)3. repeated roots of same value51. Simple Polesp• Simple poles are placed in a partial fractions expansion()()()()()nnnmpsKpsKpsKpspspszszsKs++++++=+++++= KLL22112110)(F• The constants, Ki, can be found from (use method of residues)()()()nnpppppp2121• Finally, tabulated Laplace transform pairs are used to invert ipsiispsK−=+= )()( Fy, p pexpression, but this is a nice form since the solution istptptpneKeKeKtf−−−+++=L2121)(6neKeKeKtf+++21)(2. Complex Conjugate Polespjg• Complex poles result in a Laplace transform of the form+++−∠+−+∠=++++−+=)()()()()(11*11βαθβαθβαβαjsKjsKjsKjsKs LF• The K1can be found using the same method as for simple polesβαsjsK−+=)()(1FWARNING: the "positive" pole of the form –α+jβ MUST be the one that is usedβαβαjssjsK+−=+)()(1F• The corresponding time domain function is()L++=−θβαteKtftcos2)(17()++θβteKtfcos2)(13. Repeated Polesp• When F(s) has a pole of multiplicity r, then F(s) is written as()()()LL +++++++=+=rrrpsKpsKpsKpssss112112111111)()()(QPF• Where the time domain function is then()()()11111Th t i bt i th l ti l b t lti li d bt'()LL +−+++=−−−− tprrtptpertKetKeKtf111!1)(111211•That is, we obtain the usual exponential but multiplied by t's83. Repeated Poles (cont’d.)p()• The K1jterms are evaluated from1j()()[])(!111psrjrjrjspsdsdjrK−=−−+−= F• This actually simplifies nicely until you reach s³ terms, that is for a double root (s+p1)²()1ps−=that is for a double root (s+p1)²() ()[]1)()(21112112psspsdsdKspsK−=+=+= FF• Thus K12is found just like for simple roots11pspds−=912jp• Note this reverse order of solving for the K valuesThe “Finger” Methodg• Let’s suppose we want to find the inverse Laplace pp ptransform of)3)(2()1(5)(+++=sssssF• We’ll use the “finger” method which is an easy way of visualizing the method of residues for the case of simple)3)(2(++sssvisualizing the method of residues for the case of simple roots (non-repeated)• We note immediately that the poles ares1= 0 ; s2= –2 ; s3= –310The Finger Method (cont’d)g()• For each pole (root), we will write down the function F(s) and put our finger over the term that caused that particular root, and then substitute that pole (root) value into every other occurrence of ‘s’ in F(s); let’s start with s1=065)3)(2()1(5)30)(20)(()10(5)3)(2()1(5)(==+=+=ssFThi l i h ffi i f h i f6)3)(2()30)(20)(()3)(2()(++++ssss•This result gives us the constant coefficient for the inverse transform of that pole; here: e–0·t11The Finger Method (cont’d)g()• Let’s ‘finger’ the 2ndand 3rdpoles (s2& s3)25)1)(2()1(5)32)(2)(2()12(5)3)(2()1(5)( =−−=+−+−+−=+++=ssssssF310)1)(3()2(5)3)(23)(3()13(5)3)(2()1(5)())(())()(())((−=−−−=++−−+−=+++=ssssssF• They have inverses of e–2·tand e–3·t• The final answer is thentteetf323102565)(−−−+=12Initial Value Theorem• The initial value theorem states)(lim)(lim0s s tfstF∞→→=• Oftentimes we must use L'Hopital's Rule:– If g(x)/h(x) has the indeterminate form 0/0 or ∞/ ∞ at x=c, then0st∞→→)('lim)(limxgxg=)('lim)(limxh xh cxcx →→13Final Value Theorem• The final value theorem states)(lim)(lims s tf 0stF→∞→=• The initial and final value theorems are useful for determining initial and steadystate conditions respectively for transient circuitand steady-state conditions, respectively, for transient circuit solutions when we don’t need the entire time domain answer and we don’t want to perform the inverse Laplace transform14Initial and Final Value Theorems• The initial and final value theorems also provide quick ways to somewhat check our answers• Example: the ‘finger’ method solution gave1055•Substitutingt=0 andt=∞yieldstteetf323102565)(−−−+=Substituting t0 and t yields06201553102565)0(00=−+=−+== eetf652151065)(6326=−+=∞=∞−∞−eetf15Initial and Final Value Theorems• What would initial and final value theorems find?• First, try the initial value theorem (L'Hopital's too))3)(2()1(5lim)(lim)0(∞=+++==ss s f F05525lim65)1(5lim)0()3)(2(2=∞=+=+++=∞++∞→∞→∞→∞→ssssfsssdsddsdsss• Next, employ final value theorem5)1(5)1(5lim)(lim)(==+==∞sssfF• This gives us confidence with our earlier answer6)3)(2()3)(2(lim)(lim)(00==++==∞→→ssss fssF16Class Examplesp• Find Inverse Laplace Transforms