Lect11EEE 2021 Inverse Laplace Transformations Dr. Holbert February 27, 2008

Lect11EEE 2022 Inverse Laplace Transform Consider that F(s) is a ratio of polynomial expressions The n roots of the denominator, D(s) are called the poles –Poles really determine the response and stability of the system The m roots of the numerator, N(s), are called the zeros

Lect11EEE 2023 Inverse Laplace Transform We will use partial fractions expansion with the method of 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) 3.repeated roots of same value

Lect11EEE Simple Poles Simple poles are placed in a partial fractions expansion The constants, K i, can be found from (use method of residues) Finally, tabulated Laplace transform pairs are used to invert expression, but this is a nice form since the solution is

Lect11EEE Complex Conjugate Poles Complex poles result in a Laplace transform of the form The K 1 can be found using the same method as for simple poles WARNING: the "positive" pole of the form –  +j  MUST be the one that is used The corresponding time domain function is

Lect11EEE Repeated Poles When F(s) has a pole of multiplicity r, then F(s) is written as Where the time domain function is then That is, we obtain the usual exponential but multiplied by t's

Lect11EEE Repeated Poles (cont’d.) The K 1j terms are evaluated from This actually simplifies nicely until you reach s³ terms, that is for a double root (s+p 1 )² Thus K 12 is found just like for simple roots Note this reverse order of solving for the K values

Lect11EEE 2028 The “Finger” Method Let’s suppose we want to find the inverse Laplace transform of We’ll use the “finger” method which is an easy way of visualizing the method of residues for the case of simple roots (non-repeated) We note immediately that the poles are s 1 = 0 ; s 2 = –2 ; s 3 = –3

Lect11EEE 2029 The Finger Method (cont’d) 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 s 1 =0 This result gives us the constant coefficient for the inverse transform of that pole; here: e –0·t

Lect11EEE The Finger Method (cont’d) Let’s ‘finger’ the 2 nd and 3 rd poles (s 2 & s 3 ) They have inverses of e –2·t and e –3·t The final answer is then

Lect11EEE Initial Value Theorem The initial value theorem states Oftentimes we must use L'Hopital's Rule: –If g(x)/h(x) has the indeterminate form 0/0 or  /  at x=c, then

Lect11EEE Final Value Theorem The final value theorem states The initial and final value theorems are useful for determining initial and 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 transform

Lect11EEE Initial and Final Value Theorems The initial and final value theorems also provide quick ways to somewhat check our answers Example: the ‘finger’ method solution gave Substituting t=0 and t=∞ yields

Lect11EEE Initial and Final Value Theorems What would initial and final value theorems find? First, try the initial value theorem (L'Hopital's too) Next, employ final value theorem This gives us confidence with our earlier answer

Lect11EEE Solving Differential Equations Laplace transform approach automatically includes initial conditions in the solution Example: For zero initial conditions, solve

Lect11EEE Class Examples Find inverse Laplace transforms of Drill Problems P5-3, P5-5 (if time permits)