Lecture 141 EEE 302 Electrical Networks II Dr. Keith E. Holbert Summer 2001

Lecture 142 Solving Differential Equations Laplace transform approach automatically includes initial conditions in the solution Exercise: For zero initial conditions, solve

Lecture 143 Inverse Laplace Transform Consider that F(s) is a ratio of polynomial expressions The roots of the denominator, D(s) are called the poles –Poles really determine the response and stability of the system The roots of the numerator, N(s), are called the zeros

Lecture 144 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

Lecture 145 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

Lecture 146 Class Examples Extension Exercise E13.9 Extension Exercise E13.10

Lecture 147 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

Lecture 148 Class Example Extension Exercise E13.11

Lecture 149 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 get the usual exponential multiplied by t's

Lecture 1410 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

Lecture 1411 Class Examples Extension Exercise E13.12 Extension Exercise E13.13