+ + LAPLACE TRANSFORM. Differentiation & Integration of Transforms; Convolution; Partial Fraction Formulas; Systems of DEs; Periodic Functions.

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1 COLOR LAYER red

2 LAPLACE TRANSFORM Differentiation & Integration of Transforms; Convolution; Partial Fraction Formulas; Systems of DEs; Periodic Functions. +

3 Differentiation of Transforms. F (s) e st f(t) dt, F (s) e st (tf(t)) dt L(tf(t)) F (s), L (F (s)) tf(t). Example. Find inverse transforms of s s 2 (s 2 + β 2 ) 2, (s 2 + β 2 ) 2, (s 2 + β 2 ) 2. Differentiating L(sin βt) L(t sin βt) d ds Similarly, β s 2 +β 2, β s 2 + β 2 2βs (s 2 + β 2 ) 2. L(t cos βt) (s2 + β 2 ) 2s 2 (s 2 + β 2 ) 2 s2 β 2 (s 2 + β 2 )

4 Hence, L(t cos βt β sin βt) s2 β 2 (s 2 + β 2 ) 2 s 2 + β 2 Similarly, 2β 2 (s 2 + β 2 ) 2. L(t cos βt + β sin βt) 2s 2 (s 2 + β 2 ) 2. Integration of Transforms. If lim t + f(t)/t exists, then ( ) f(t) L F (σ) dσ t s ( ) L F (σ) dσ f(t), s t because ] F (σ) dσ s s e σt f(t) dt dσ s f(t)e st t ] e σt f(t) dσ dt ( f(t) dt L t ). + 3

5 Note: If G(s) s F (σ) dσ, then F (s) G (s). Example. L ( ln( a 2 /s 2 ) ). First, find F (s) with G(s) s F (σ) dσ. Put G(s) ln( a 2 /s 2 ). F (s) G (s) d ( ) ds ln a2 s 2 L (ln ( 2a2 s(s 2 a 2 ) 2 s s a s + a f(t) L (F ) 2 e at e at a2 s 2 2(cosh at ). )) f(t) t 2( cosh at). t + 4

6 CONVOLUTION. L(fg) L(f)L(g), L (F (s)g(s)) f(t)g(t), but there is a relation between them, which is called convolution: if H(s) F (s)g(s), h(t) (f g)(t) Example. H(s) h(t) 2 cos t sin t 2 t t t f(τ)g(t τ) dτ. 2s (s 2 +) 2 2 s s 2 + cos τ sin(t τ) dτ (sin t + sin(2τ t)) dτ s 2 + τ sin t 2 cos(2τ t) ] t t sin t. Example. H(s) L ( s ), L ( s(s 2 +ω 2 ) s 2 + ω 2 ) ω sin ωt. + 5

7 By convolution, h(t) t ω sin ωt sin ωτ dτ ω ] t ω 2 cos ωτ ω 2 cos ωt. ω2 Proof of convolution: By 2nd shift, e st G(s) L(g(t τ)u(t τ)) e st g(t τ)u(t τ) dt τ e st g(t τ) dt F (s)g(s) e sτ f(τ)g(s) dτ f(τ) e st g(t τ) dt dτ. τ Changing the order of integration, F (s)g(s) t e st f(τ)g(t τ) dτ dt e st (f g)(t) dt L(f g). + 6

8 Differential Equation: y + ay + by r(t), y() y (), Y (s) R(s)Q(s), y(t) r(t) q(t), q(t) L (Q), where Q(s) /(s 2 + as + b) is the transfer function. y(t) Example. t q(t τ)r(τ) dτ. y + y sin 3t, y() y (). Now Q(s) /(s 2 + ), q(t) sin t. y(t) r(t) q(t) y(t) t 2 t sin 3τ sin(t τ) dτ (cos(t 4τ) cos(t + 2τ)) dτ 3 8 sin t sin 3t

9 Integral Equations. Integral equations which are in a convolution form can be solved very easily: y(t) f(t) + (y g)(t) Y (s) F (s) + Y (s)g(s) Y (s) F (s) G(s). Example: y(t) e t 2 t y(τ) cos(t τ) dτ. This is in convolution form y e t 2y cos t Y (s) s 2Y (s) s + s 2 + Y (s) s2 + (s + ) 3 s + 2 (s + ) (s + ) 3 y(t) ( t) 2 e t. + 8

10 Revision of Partial Fractions. Solutions of subsidiary equation appear as Y (s) F (s) G(s), deg(f ) < deg(g), where F, G have real coefficients, and it is necessary to expand the rational function as partial fractions. There are four cases commonly occurring, where G(s) has a factor: Simple factor s a; Repeated factor (s a) m ; Irreducible quadratic (s α) 2 + β 2 ; + 9

11 Simple factor: Y (s) F (s) A has partial fraction G(s) y(t) has a term Ae at. s a. Example. Y (s) 7s s 3 7s + 6 A s + A 2 s 2 + A 3 s + 3 7s A (s 2)(+3) + A 2 (s )(s + 3) +A 3 (s )(s 2). s : 4A 8, A 2. s 2 : 5A 2 5, A 2 3. s 3 : 2A 3 2, A 3. y(t) L (Y ) 2e t 3e 2t + e 3t +

12 Example 2. Repeated factor y + 4y + 4y 2e 2t, y() 3, y (). s 2 Y 3s + + 4(sY 3) + 4Y 2 s + 2, (s 2 +4s+4)Y 3s+2+ 2 s + 2 3s2 + 8s + 6, s + 2 Y (s) F (s) G(s) 3s2 + 8s + 6 (s + 2) 3 A 3 (s + 2) 3 + A 2 (s + 2) 2 + A s s 2 + 8s + 6 A 3 + A 2 (s + 2) + A (s + 2) 2. cft of s 2 : A 3 cft of s : A 2 + 4A 8, A 2 4 const : A 3 + 2A 2 + 4A 6, A 3 2 y(t) 3e 2t 4te 2t + t 2 e 2t. +

13 Irreducible quadratic factor (s α) 2 + β 2 Y (s) has a partial fraction A s + A 2 (s α) 2 + β 2 A (s α) + αa + A 2 (s α) 2 + β 2 A s α (s α) 2 + β 2 + αa + A 2 (s α) 2 + β 2 and the inverse transform is y(t) e αt ( A cos βt + αa + A 2 β sin βt ). + 2

14 Example: y + 4y sinh t, y() y (), s 2 Y sy() y () + 4Y s 2, Y (s) (s 2 + 4)(s )(s + ) A s + A 2 s B s + (A s + A 2 )(s )(s + ) +B(s + )(s 2 + 4) +C(s )(s 2 + 4) s : B, B /. s : C, C /. C s + cft s 3 : A + B + C, A. const : A B 4C, A 2 2/. y(t) sin 2t + (et e t ) + 3

15 Systems of DEs. Example. ] y 3 y Taking the Laplace transform, sy y() AY + s + 2 s + 3 s + 3 ] ] Y s + 2 e 2t, y() ] + ] ] ] Y Y (s) Y 2 (s) s + 3 (s + 2) 2 (s + 4) s + 3 s (s + 2)(s + 4) s + 3 s 2 ] s (s + 2) 2 (s + 4) 3s + 2 s 2 s (s + 2) 2 (s + 4) y (t), 3s + 2 (s + 2) 2 (s + 4) y 2(t). ] 6 2 ] ] ] + 4

16 Y (s) A (s + 2) 2 + B s C s + 4 s 2 s A (s + 4) + B (s + 2)(s + 4) +C (s + 2) 2 s 2 : 4 2A, A 2. s 4 : 4C, C 5/2. cft s 2 : B + C, B 3/2. A 2 Y 2 (s) (s + 2) 2 + B 2 s C 2 s + 4 3s + 2 A 2 (s + 4) + B 2 (s + 2)(s + 4) +C 2 (s + 2) 2 s 2 : 4 2A 2, A 2 2. s 4 : 4C 2, c 2 5/2. cft s 2 : B 2 + C 2, B 2 5/2. y (t) 2te 2t 3e 2t /2 + 5e 4t /2 y 2 (t) 2te 2t + 5e 2t /2 5e 4t /2. + 5

17 Example 2. Two masses on springs. y y ky + k(y 2 y ) 2 k(y 2 y ) ky 2, y () y 2 (), y () 3k, y 2 () 3k. Let Y L(y ), Y 2 L(y 2 ). s 2 Y s 3k ky + k(y 2 Y ) s 2 Y 2 s + 3k k(y 2 Y ) ky 2 (s 2 + 2k)Y ky 2 s + 3k ky + (s 2 + 2k)Y 2 s 3k. + 6

18 Solving for Y, Y 2, Y (s + 3k)(s 2 + 2k) + k(s 3k) (s 2 + 2k) 2 k 2 Y 2 (s2 + 2k)(s 3k) + k(s + 3k) (s 2 + 2k) 2 k 2 (s 2 + 2k) 2 k 2 (s 2 + 2k) k](s 2 + 2k) + k] (s 2 + k)(s 2 + 3k) Y Y 2 s s 2 + k + s s 2 + k 3k s 2 + 3k 3k s 2 + 3k y (t) cos kt + sin 3kt y 2 (t) cos kt sin 3kt + 7

19 Periodic Functions. f(t) defined for all t >, and has period p >, L(f) f(t + p) f(t) t >. p e st f dt + p + 3p 2p p e st f dt 2p e st f dt +... e sτ f(τ) dτ + + p p e s(τ+p) f(τ) dτ e s(τ+2p) f(τ) dτ e sp + e 2sp +...] e ps p e sτ f(τ) dτ. p e sτ f(τ) dτ This is because in the integral (k+)p kp e st f dt, we can substitute t τ + kp to obtain p e s(τ+kp) f(τ +kp) dτ e kp because f(τ + kp) f(τ). p e sτ f(τ) dτ, + 8

20 Example. f(t) { if < t < π, if π < t < 2π, f(t + 2π) f(t). L(f) e 2πs e 2πs s ( π ( e πs ) 2 s( e 2πs ) e st dt + e πs s( + e πs ) s coth 2π π ( 2e πs + e 2πs) ( πs 2 )..e st dt ) + 9

21 Example 2. Halfwave Rectifier: f(t) { sin ωt if < t, π/ω, if π/ω < t < 2π/ω, f(t + 2π/ω) f(t). π/ω L(f) e 2πs/ω e st sin ωt dt. The integral is the imaginary part of π/ω and so e ( s+iω)t dt L(f) e ( s+iω)t ] π/ω s + iω s iω s 2 + ω 2( e sπ/ω ), ω( + e sπ/ω ) (s 2 + ω 2 )( e 2sπ/ω ) ω (s 2 + ω 2 )( e sπ/ω ). + 2

22 Example 3. Rectifier of sin ωt: f(t) sin ωt, t π ω, f(t + π ω ) f(t). L(f) e πs/ω π/ω e st sin ωt dt. π/ω e st e iωt dt π/ω e (s+iω)t dt e (s iω)t s iω π/ω e (s iω)π/ω s iω + e πs/ω s iω ( + e πs/ω )(s + iω) s 2 + ω

23 Since e st sin ωt is the imaginary part of e (s iωt), take the imaginary part of the integral and multiply by /( e πs/ω ) to obtain L(f) ω( + e πs/ω ) e πs/ω s 2 + ω 2 ω(e πs/2ω + e πs/2ω ) e πs/2ω e πs/2ω s 2 + ω 2 ω cosh(πs/2ω) s 2 + ω 2 sinh(πs/2ω). + 22