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Laplace is no more difficult!

- 1. There are techniques for finding the system response of a systemdescribed by a differential equation, based on the replacement offunctions of a real variable (usually time or distance) by certainfrequency-dependent representations, or by functions of a complexvariable dependent upon frequency. The equations are converted fromthe time or space domain to the frequency domain through the use ofmathematical transforms.
- 2. Time Domain Frequency Domain Laplace Transform Differential Algebraic equations equations Input Inputexcitation e(t) excitation E(s) Output Output response r(t) response R(s) Inverse Laplace Transform
- 3. Let f(t) be a real function of a real variable t (time) defined for t>0. Thenis called the Laplace transform of f(t). The Laplace transform is afunction of a complex variable s. Often s is separated into its real andimaginary parts: s= +j , where and are real variables.
- 4. After a solution of the transformed problem has been obtained in termsof s, it is necessary to "invert" this transform to obtain the solution interms of the time variable, t. This transformation from the s-domain intothe t-domain is called the inverse Laplace transform.
- 5. Let F(s) be the Laplace transform of a function f(t), t>0. The contour integralwhere c> 0 ( 0 as above) is called the inverse Laplace transform of F(s).
- 6. It is seldom necessary to perform the integration in the Laplace transformor the contour integration in the inverse Laplace transform. Most often,Laplace transforms and inverse Laplace transforms are found using tablesof Laplace transform pairs.
- 7. Time Domain Frequency Domain f(t), t> 0 F(s)1. 12. K K/s3. Kt K/s24. Ke-at K/(s+ a)5. Kte-at K/(s+ a) 26. Ksint K/(s2+ 2)7. Kcost Ks/(s2+ 2)8. Ke-at sint K/((s+ a) 2+ 2))9. Ke-at cost K(s+ a)/ ((s+ a) 2+ 2))
- 8. Time Domain Frequency Domain f(t), t> 0 F(s)10. t s11. f(t) F(s)12. L-1{ F(s)} = f(t) L{ f(t )} = F(s)13. Af 1(t) + Bf 2(t) AF1(s)+ BF2(s)14.15.
- 9. The inverse Laplace transform is usually moredifficult than a simple table conversion. 8( s + 3)( s + 8) X ( s) = s( s + 2)( s + 4)
- 10. If we can break the right-hand side of theequation into a sum of terms and each term is in atable of Laplace transforms, we can get theinverse transform of the equation (partial fractionexpansion). 8( s + 3)( s + 8) K1 K2 K3X ( s) = = + + s( s + 2)( s + 4) s s+2 s+4
- 11. In general, there will be a term on the right-handside for each root of the polynomial in thedenominator of the left-hand side. Multiple rootsfor factors such as (s+2)n will have a term for eachpower of the factor from 1 to n. 8( s + 1) K1 K2 Y ( s) = = + ( s + 2) 2 s + 2 ( s + 2) 2
- 12. Complex roots are common, and they alwaysoccur in conjugate pairs. The two constants inthe numerator of the complex conjugate termsare also complex conjugates. 5.2 K K* Z ( s) = 2 = + s + 2 s + 5 ( s + 1 j 2) ( s + 1 + j 2)where K* is the complex conjugate of K.
- 13. The solution of each distinct (non-multiple)root, real or complex uses a two step process. The first step in evaluating the constant is to multiply both sides of the equation by the factor in the denominator of the constant you wish to find. The second step is to replace s on both sides of the equation by the root of the factor by which you multiplied in step 1
- 14. 8( s + 3)( s + 8) K1 K2 K3X ( s) = = + + s( s + 2)( s + 4) s s+2 s+4 8( s + 3)( s + 8) 8(0 + 3)(0 + 8)K1 = = = 24 ( s + 2)( s + 4) s=0 (0 + 2)(0 + 4) 8( s + 3)( s + 8) 8( 2 + 3)( 2 + 8)K2 = = = 12 s( s + 4 ) s =2 2( 2 + 4 )
- 15. 8( s + 3)( s + 8) 8( 4 + 3)( 4 + 8)K3 = = = 4 s( s + 2 ) s =4 4( 4 + 4) The partial fraction expansion is: 24 12 4 X ( s) = s s+2 s+4
- 16. The inverse Laplace transform is found fromthe functional table pairs to be: 2 t 4 t x (t ) = 24 12e 4e
- 17. Any unrepeated roots are found as before.The constants of the repeated roots (s-a)m arefound by first breaking the quotient into apartial fraction expansion with descendingpowers from m to 0: Bm B2 B1 ++ + (s a) m (s a) 2 (s a)
- 18. The constants are found using one of thefollowing: 1 d m i P( s) Bi = m (m i )! ds m i Q( s ) /( s a1 ) s = a1 P(a )Bm = Q( s) / ( s a ) m s=a
- 19. 8( s + 1) K1 K2Y ( s) = = + ( s + 2) 2 s + 2 ( s + 2) 2 8( s + 1)( s + 2) 2K2 = = 8( s + 1) s=2 = 8 ( s + 2) 2 s =2
- 20. 1 d 8( s + 1) Bi = ( s + 2) 2 /( s + 2) 2 =8 (2 1)! ds s = 2The partial fraction expansion yields: 8 8 Y ( s) = s + 2 ( s + 2) 2
- 21. The inverse Laplace transform derived from the functionaltable pairs yields: 2 t 2 t y (t ) = 8e 8te
- 22. 8( s + 1) K1 K2Y ( s) = = + ( s + 2) 2 s + 2 ( s + 2) 2 8( s + 1) = K1 ( s + 2) + K 2 8s + 8 = K1s + 2 K1 + K 2Equating like terms: 8 = K1 and 8 = 2 K1 + K 2
- 23. 8 = K1 and 8 = 2 K1 + K 28 = 2 8 + K28 16 = 8 = K 2Thus 8 8 Y (s) = s + 2 ( s + 2) 2