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EEE 302 Electrical Networks II

Learn how to solve differential equations using the Laplace transform approach, which automatically includes initial conditions in the solution. This lecture covers the inverse Laplace transform, including cases with simple poles, complex conjugate poles, and repeated poles.

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EEE 302 Electrical Networks II

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  1. EEE 302Electrical Networks II Dr. Keith E. Holbert Summer 2001 Lecture 14

  2. Solving Differential Equations • Laplace transform approach automatically includes initial conditions in the solution • Exercise: For zero initial conditions, solve Lecture 14

  3. 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 14

  4. 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 14

  5. Simple Poles • Simple poles are placed in a partial fractions expansion • The constants, Ki, 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 14

  6. Class Examples • Extension Exercise E13.9 • Extension Exercise E13.10 Lecture 14

  7. Complex Conjugate Poles • Complex poles result in a Laplace transform of the form • The K1 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 14

  8. Class Example • Extension Exercise E13.11 Lecture 14

  9. 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 14

  10. Repeated Poles (cont’d.) • The K1j terms are evaluated from • This actually simplifies nicely until you reach s³ terms, that is for a double root (s+p1)² • Thus K12 is found just like for simple roots • Note this reverse order of solving for the K values Lecture 14

  11. Class Examples • Extension Exercise E13.12 • Extension Exercise E13.13 Lecture 14

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