Me330 lecture4
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1. The document discusses different cases for taking the inverse Laplace transform of a function F(s), depending on the characteristics of the denominator of F(s). 2. If the denominator has real and distinct roots, one multiplies F(s) by the appropriate term and evaluates the residue at each root to find the inverse Laplace transform. 3. If there are real repeated roots, one multiplies F(s) by the appropriate power and differentiates to find all residues needed for the inverse Laplace transform. 4. If there are complex roots, one solves for the residues by balancing coefficients or factorizing, then combines like terms to find the inverse Laplace transform.
![Matlab Rational function of polynomials in s num = [b 0 b 1 … b n ] den = [1 a 1 … a n ] Matlab representation Partial fraction expansion Matlab command [r,p,k] = residue(num,den) r,p,k are vectors whose coefficients define the partial fraction expansion](https://image.slidesharecdn.com/me330lecture4-110223225052-phpapp01/85/Me330-lecture4-8-320.jpg)
Me330 lecture4
- 1. ME 330 Control Systems FA 2010 Lecture 4
- 2. Recall from last lecture Differential equations and forcing functions combine algebraically in the Laplace domain. Time domain representation requires taking the inverse Laplace transform. Simplify the functions F(s) in order to easily find inverse Laplace transforms from the table. Convert original function F(s) into sum of simpler terms
- 3. Case 1: Real Roots If F(s) denominator has real and distinct roots Multiply F(s) by (s+p m ) to find residue K m Set s = -p m to evaluate residue K m 0 0 0
- 4. Case 2: Real Repeated Roots If F(s) denominator has real and repeated roots Multiply F(s) by (s+p 1 ) r to find residue K 1 Differentiate (s+p 1 ) r F(s) to find residues K 2 through K r Set s=-p 1 to find residues K 1 0 0 0 0
- 5. Case 3: Complex Roots If F(s) denominator has complex roots, can be solved with complex K i ’s then recognizing that Solve for K i ’s by balancing the 2 unknown coefficients Or, F(s) denominator has complex roots and Solve K 1 then multiply lowest common denominator 2 nd order 2 nd order Combine like terms
- 6. Case 3: Complex Roots Once K 2 , K 3 are known, complete the square on the quadratic
- 7. Example Assign m = 1 , c = 2 , k = 5 , and f(t) note: recognize:
- 8. Matlab Rational function of polynomials in s num = [b 0 b 1 … b n ] den = [1 a 1 … a n ] Matlab representation Partial fraction expansion Matlab command [r,p,k] = residue(num,den) r,p,k are vectors whose coefficients define the partial fraction expansion
- 9. Example Case 3 If F(s) denominator has complex roots Multiply both sides by the greatest common denominator
- 10. Example Case 3 After polynomial division and partial fraction expansion Completing the square and factoring the numerator
- 11. Next Lecture Derivation of dynamical system models