Residue Integrals (4C)

Transcription

1 Residue Itegrals (4C) Iverse Laplace Trasform

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3 Laplace Trasform Fourier Trasform Laplace Trasform Iverse Laplace Trasform F (s) = f (t)e st d t f (t ) = σ +i 2 πi σ i F (s)e s t d s = {f (t )e xt } e iy t d t s = x + i y, f (t ) = 2π i σ +i σ i F (x+i y)e xt e i y t d s = F (x, y) f (t ) = t < f (t ) = 2π i σ +i σ i F (x, y)e x t e i yt d s g(t) = {f (t)e xt }, for a give x Fourier Trasform F (x, y) = g(t) e iy t d t Iverse Fourier Trasform g(t ) = + 2 π F (x, y)e i y t d y d s = d x + i d y = i d y, for a give x Residue Itegrals (4C) 3

4 Fourier Trasform Laplace Trasform Fourier Trasform F (x, y) = Iverse Fourier Trasform g(t) e iy t d t g(t ) = + 2 π F (x, y)e i y t d y s = x + i y, f (t )e xt = +i 2 π i F ( x, y)e i y t d y f (t ) = t < f (t ) = +i 2π i F (x, y)e + x t e i yt d y g(t) = {f (t)e xt }, for a give x,( x>α) Laplace Trasform Iverse Laplace Trasform F (s) = f (t)e st d t f (t ) = σ +i 2 πi σ i F (s)e s t d s d s = d x + i d y = i d y, for a give x Residue Itegrals (4C) 4

5 Expoetial Order Laplace Trasform Expoetial Order α F (s) = f (t)e st d t a fuctio f has expoetial order α for s > R(s) > the itegral coverges if f(t) does ot grow too rapidly the growth rate of a fuctio f(t) there exist costats M > ad such that for some t > t f (t) M e α t, t > t α f (t) e σt d t < for some σ f (t)e st d t = f (t)e x t e i y t d t = f (t)e xt d t f (t ) expoetial order σ F (s) = < f (t) e σ t f (t)e s t d t d t < for s > σ absolutely coverges for R(s) > σ s > σ Residue Itegrals (4C) 5

6 Covergece of the Laplace Trasform Laplace Trasform F (s) = = f (t)e st d t {f (t )e xt } e iy t d t f st (t )e d t = f (t) xt e d t < ( e st = e xt e iyt = e xt ) f(t) cotiuous o [, ) f(t) = for t < f(t) has expoetial order α f'(t) piecewise cotiuous o [, ) {f (t) e xt } = g(t) absolutely itegrable for x > α Use Fourier Iversio F(s) coverges absolutely for Re(s) > α f s t (t )e d t < Residue Itegrals (4C) 6

7 Fourier Trasform g (t ) = f (t)e xt absolutely itegrable for x > α F (x, y) = F (x, y) = Fourier Trasform { f (t)e xt } e iyt d t g (t) e iyt d t g (t ) = f (t)e xt + g (t ) = F (x, y)e iyt d y 2 π f (t) = + F ( x, y)e xt e iyt d y 2π Iverse Fourier Trasform y = + α s = x + i fixed x s = x + i y d s = i d y f (t ) = + 2π F (x, y)e ( x+ iy)t d y = x+ i F (s)e s t d s 2 πi x i y = (x > α) s = x i = lim y x+ i y 2 πi x i y F (s)e s t d s x i y Residue Itegrals (4C) 7

8 Fourier-Melli Iversio Formula F (x, y) = F (x, y) = Fourier Trasform { f (t)e xt } e iyt d t g (t) e iyt d t g (t ) = f (t)e xt Vertical lie at x : Bromwich lie y = + s = x + i + g (t ) = F (x, y)e iyt d y 2 π f (t) = + F ( x, y)e xt e iyt d y 2π Iverse Fourier Trasform s = x + i y d s = i d y y = α fixed x (x > α) s = x i f (t) = x+ i 2π i x i = lim y 2 π i x i y F (s)e st d s x+ i y F (s)e st d s Complex Iversio Formula (Fourier-Melli Iversio Formula) Residue Itegrals (4C) 8

9 Bromwich Cotour Itegratio C R R s = x + i y 2 πi C F(s)est d s = 2 πi C R F (s)e st d s + 2 πi C L F( s) e st d s α x C L F(s) is aalytic for Re(s) = x > α F(s)all sigularities must lie to the left of Bromwich lie s = x i y Assume F(s) is aalytic for Re(s) = x < α except for havig fiitely may poles z, z 2,, z cotour itegratio o C R cotour itegratio o C L 2 πi C F(s)est d s = k= Res(z k ) 2πi C R R F ( s)e st d s k = 2πi x+ iy st x iy F ( s)e Res(z k ) d s 2 πi C R F (s)e st d s = x+iy 2 πi x iy F (s) e st d s = k= for a give x Res(z k ) Residue Itegrals (4C) 9

10 Growth Restrictio Coditios o F(s) C R R x + i y For s o C R, some p > all R > R F(s) M s p θ x lim C F(s)e st d s = (t > ) R R θ 2 C L F(s) M s p Growth Restrictio x i y F(s) as s a fuctio f has expoetial order α there exist costats M > such that for some t > t ad α f (t) M e α t, t > t Residue Itegrals (4C)

11 Cotour Itegratio o C R (x > ) C R B R A x + i y C R R 2 M π/2 R p θ e Rt ( si ϕ) d ϕ C R F (s)e st d s AB F (s) e st d s C D θ θ 2 x C L x i y E R looks like x C L r θ = l θ = l r 2 M π/2 R p θ c d ϕ 2 M R p l( ÃB) R ÃB, DE R(s) x ÃB F (s) e st d s R e t s e t x = c for a fixed t > DE F(s)e st d s R lim R C R F (s)e st d s = Residue Itegrals (4C)

12 Cotour Itegratio o C R (x < ) C R R s = R e i θ = R(cos θ + i si θ) e st = e Rt (cos θ + isi θ) = e Rt cos θ i Rt si θ e e st Rt cos θ = e x F (s) M s p Growth Restrictio C L C R F (s)e st d s C R F (s) e st d s s = R e i θ θ θ θ 2 ds = i R e iθ d θ ds = R d θ R π θ 3 π 2 2 π/2 3π/2 M R p ert cos θ R d θ M 3 π/2 R p π/2 e Rt cos θ d θ ϕ = θ π/2 = M π R p e Rt ( si ϕ) d ϕ = 2 M π/2 R p e Rt( si ϕ) d ϕ Residue Itegrals (4C) 2

13 y=si(x) ad y=mx m = 2/π C R F (s)e st d s C R F (s) e st d s (π/2, ) 2 M π/2 R p e Rt( si ϕ) d ϕ si ϕ 2 π ϕ ( ϕ π 2 ) 2 M π/2 2 Rt ϕ R p e π d ϕ 2 M [ R π R t ϕ p 2 R t e 2 π ]π/2 = M π R p t ( e R t ) R Residue Itegrals (4C) 3

14 Covergece of the Laplace Trasform Laplace Trasform f(t) cotiuous o [, ) f(t) = for t < f(t) has expoetial order α f'(t) piecewise cotiuous o [, ) F(s) coverges absolutely for Re(s) > α f s t (t )e d t < Iverse Laplace Trasform f(t) cotiuous o [, ) f(t) has expoetial order α o [, ) f'(t) piecewise cotiuous o [, ) F(s) = L{f(t)} for Re(s) > α F(s) M some p > s p for all s sufficietly large F(s) is aalytic except for fiitely may poles at z, z 2,, z covergig F (s) = f (t) e st d t f (t ) = x+i 2π i x i F( s) e st d s = k= for a give x Res(z k ) = {f (t )e xt } e iy t d t Residue Itegrals (4C) 4

15 L - {/s(s-a)} F (s) = s(s a) f (t ) = x+i 2π i x i s 2 a F( s) e st d s = k= for a give x Res(z k ) F (s) 2 s 2 s 2 a ÃB, DE R(s) x e t s e t x = c for a fixed t > s 2 a s a s a s a a s a s a s a a s 2 a F (s) = s s a = 2 a 2 2 (2 a ) 2 ÃB F (s) e st d s R DE F(s)e st d s R Res() = lim s Res(a) = lim s a f (t ) = a (eat ) s e t s F( s) = lim s (s a)e t s F( s) = lim s e t s (s a) = a e t s s = ea t a Residue Itegrals (4C) 5

16 L - {s/(s 2 a 2 )} F (s) = s s 2 a 2 = L (cosh (at)) f (t ) = x+i 2π i x i s 2 a F( s) e st d s = k= for a give x Res(z k ) F (s) = s 2 s 2 a s s 2 a 2 ÃB, DE R(s) x e t s e t x = c for a fixed t > s 2 a s 2 4 a 2 F (s) a 2 a 2 s 2 a 2 s 2 4 s s 2 s 3/4 s = 4/3 2 s s 2 a ÃB F (s) e st d s R DE F(s)e st d s R Res(a) = lim s (s a)e t s F( s) = lim s Res( a) = lim ( s+a)e t s F (s) = lim s a s f (t ) = e+at e at 2 = cosh(at) se t s (s+a) = eat 2 s e t s (s a) = e a t 2 Residue Itegrals (4C) 6

17 L - {/s(s 2 + a 2 ) 2 } F (s) = F (s) F (s) = s(s 2 +a 2 ) 2 (s 2 +a 2 ) s 2 (s 2 +a 2 ) s 2 F (s) = s(s 2 +a 2 ) 2 s 5 M s 5 f (t ) = x+i 2π i x i for a suitably large s (s 2 +a 2 ) = 2 s(s i a) 2 ( s+i a) 2 F( s) e st d s = k= s for a suitably large x Res(z k ) Res() = lim s Res(±i a) = lim s ±i a {s e t s F (s)} = lim s ±ia = d d s {(s i a)2 e t s F (s)} d { e t s } d s s(s±ia) = lim 2 s ±ia e t ( i a) 2 (+i a) 2 = a 4 d { e t s } d s s 3 a 2 s±2 ia s 2 { = lim et s [t (s 3 a 2 s±2 i a s 2 ) (3s 2 a 2 ±4 i a s)] } s ±ia (s 3 a 2 s±2i a s 2 ) 2 { = lim e±i at [t ( ia 3 ia 3 2i a 3 ) ( 3a 2 a 2 4 a 2 )] } s ±ia ( ia 3 i a 3 2 ia 3 ) 2 { = e±i a t [i t( 4 a 3 )+(8 a 2 )] } = { ± it ( 4 a 3 ) 2 4 a 3 e±i at t} 2 a 4 e±ia { f (t ) = + i t 4 a 3 e+ia t } 2 a 4 e+ iat + { it 4 a 3 e i a t t} 2 a 4 e ia + a 4 = a { a 3 2 } t si at cos at Residue Itegrals (4C) 7

18 Iverse Laplace Trasform F (s) = P( s) Q(s) = a s + a s + + a b m s m + b m s m + + b = a + a s + + a s s m (b m + b m s + + b s m ) a a + s b b m m + s s b m b m s m 2 s m a s a + a + + a = c + + b b s m m b m b s s m b m b m b m b m F (s) c /c 2 s m s b m s, b m 2 m b m 2 2 b m = c 2 b m b m 2 b m m, m b 2 b m for a suitably large s P( z) Res( z ) = lim (z z ) z z Q(z) = lim z z { Res(z k ) = e z kt P( z k ) Q ' (z k ) f (t ) = x+i 2π i x i f (t ) = z k P(z) Q (z) Q( z ) z z Res( z k ) = z k } = z e t k P( z k ) Q ' (z k ) P( z ) Q ' (z ) for a suitably large x F( s) e st d s = k= Res(z k ) Residue Itegrals (4C) 8

19 Ifiitely May Poles Assume F(s) has ifiitely may poles to the left of the lie {z k } k= each pole has the coditio R{s}=x > z z 2 z k z k as k Cosider the cotour which ecloses the first poles Γ = C [x iy, x +iy ] x +i y Cauchy Residue Theorem f (t ) = 2π i Γ F (s)e st d s = = x +i y F (s)e st d s + 2 πi x i y If we ca show lim 2 πi C F( s) e st d s = k= 2π i C Res( z k ) F(s)e st d s C C z 3 x +i y y z z z C z 2 C 2 C 3 x i y the we have f (t ) = x +i 2π i x i F (s)e st d s = k= Res( z k ) x i y Residue Itegrals (4C) 9

20 L - {/s( + e as )} (a>) F (s) = s ( + e as ) = e t s G(s )+sg' (s ) = e t s ( + e a s )+a s e a s s = e a s = = e (2 ) π i s = ( 2 a ) π i =, ±, ±2, G(s) = + e a s G(s ) = All are simple pole Res() = lim s Res(s ) = lim s = lim s s e t s F (s) = lim s (s s ) e t s F (s) (s s ) P(s ) Q' (s ) d s G (s) = a ea d s G'( s ) = a < (+e a s ) = 2 P(s) = e a s Q(s) = s G(s) = et s a s e a s ( 2 Res(s ) = et a ) πi (2 )π i Res = 2 = = 2 2 π = f (t ) = 2 π 2 = = et s a s e t ( 2 a ) πi (2 ) π i Cauchy Residue Theorem a s = (2 )π i t { e ( 2 a ) π i e t ( 2 a ) } πi 2i (2 ) (2 ) si {( 2 a ) πt } Residue Itegrals (4C) 2

21 L - {/s(s 2 + a 2 ) 2 } F (s) = F (s) F (s) = s(s 2 +a 2 ) 2 (s 2 +a 2 ) s 2 (s 2 +a 2 ) s 2 F (s) = s(s 2 +a 2 ) 2 s 5 M s 5 f (t ) = x+i 2π i x i for a suitably large s (s 2 +a 2 ) = 2 s(s i a) 2 ( s+i a) 2 F( s) e st d s = k= s for a suitably large x Res(z k ) Res() = lim s Res(±i a) = lim s ±i a {s e t s F (s)} = lim s ±ia = d d s {(s i a)2 e t s F (s)} d { e t s } d s s(s±ia) = lim 2 s ±ia e t ( i a) 2 (+i a) 2 = a 4 d { e t s } d s s 3 a 2 s±2 ia s 2 { = lim et s [t (s 3 a 2 s±2 i a s 2 ) (3s 2 a 2 ±4 i a s)] } s ±ia (s 3 a 2 s±2i a s 2 ) 2 { = lim e±i at [t ( ia 3 ia 3 2i a 3 ) ( 3a 2 a 2 4 a 2 )] } s ±ia ( ia 3 i a 3 2 ia 3 ) 2 { = e±i a t [i t( 4 a 3 )+(8 a 2 )] } = { ± it ( 4 a 3 ) 2 4 a 3 e±i at t} 2 a 4 e±ia { f (t ) = + i t 4 a 3 e+ia t } 2 a 4 e+ iat + { it 4 a 3 e i a t t} 2 a 4 e ia + a 4 = a { a 3 2 } t si at cos at Residue Itegrals (4C) 2

22 L - {/s( + e as )} (a>) s-plae +2 πi poits o the cotour C C R s = R e i θ = 2 π a eiθ ( R = 2 π a ) x assume a = C L s = R e i θ = 2 πe iθ ( R = 2 π ) 2 πi s = +2 πi s = 2 πi θ = +π/2 θ = π/2 Residue Itegrals (4C) 22

23 L - {/s( + e as )} (a>) s-plae +2 πi ζ -plae < x x R R H (s) = ( + e s ) cos(2 π ϵ) + isi (2 π ϵ) x C C L x + i y + e x e i y 2 cos( 2 π +ϵ) + isi ( 2 π +ϵ) y ±2 π 2 πi s = x + i y ζ = + e x e i y < x x y ±2 π as < x x 2 π y +2 π < e x e x e i y = + e s > 2 cos(2 π) i si(2 π) = cos(2 π) + i si(2 π) = Residue Itegrals (4C) 23

24 L - {/s( + e as )} (a>) s-plae +2 πi ζ -plae +2( )π i R ϕ x H (s) = ( + e s ) cos(2 π ϵ) + isi (2 π ϵ) x < 2( ) πi C L 2 x + i y + e x e i y cos( 2 π +ϵ) + x = 2 π cos ϕ isi ( 2 π +ϵ) y = 2 π si ϕ 2 πi s = x + i y ζ = + e x e i y < x x y ±2 π as x < y = +2 π y = +2( ) π e x < e i y = e 2 πcos ϕ cos(2 πsi ϕ) e i y = + e s > e 2 πcos ϕ cos(2 πsi ϕ) > b > Residue Itegrals (4C) 24

25 L - {/s( + e as )} (a>) x = 2 π cos ϕ y = 2 π si ϕ s = x + i y ζ = + e x e i y si ϕ = x < y = +2 π e x < e i y = + si ϕ = ( /4) y = +2 π 2 π e i y = +i si ϕ = ( /2) si ϕ = ( 3/4) y = +2 π π y = +2 π 3 2 π e i y = e i y = i si ϕ = ( ) y = +2 π 2 π e i y = + Residue Itegrals (4C) 25

26 L - {/s( + e as )} (a>) x = 2 π cos ϕ y = 2 π si ϕ s = x + i y ζ = + e x e i y < x x y ±2 π as x < e x < + e s > ϕ = +π/2 y = +2 π y = +(2 ) π e i y = + e i y = i e 2 πcos ϕ cos(2 πsi ϕ) > b > ϕ = π/2 y = +(2 2) π e i y = + m = 2/π e 2 πcos ϕ cos(2 πsi ϕ) si ϕ = ( ) y = +2( ) π = 2 π si ϕ 2, 2 3, 3 4,, 999, π/2 +π/2 Residue Itegrals (4C) 26

27 L - {/s(s 2 + a 2 ) 2 } {z k } k= R{s}=x > z z 2 Γ = C [x iy, x +iy ] f (t ) = x+i 2π i x i F( s) e st d s = k= Res(z k ) k= Res( z k ) = x+i y F(s)e st d s + 2 πi x i y 2 πi C F (s)e st d s lim 2 πi C F (s)e st d s = k= Res( z k ) = x+i y 2 πi F(s)e st d s = x i y k= Res( z k ) Residue Itegrals (4C) 27

28 Refereces [] [2] J.H. McClella, et al., Sigal Processig First, Pearso Pretice Hall, 23 [3] A graphical iterpretatio of the DFT ad FFT, by Steve Ma

29 Refereces [] [2] [3] M.L. Boas, Mathematical Methods i the Physical Scieces [4] E. Kreyszig, Advaced Egieerig Mathematics [5] D. G. Zill, W. S. Wright, Advaced Egieerig Mathematics [6] A. D. Poularikas ad S. Seely, Sigals ad Systems [7] J. H. Mathews ad R. W. Howell, Complex Aalysis: for Mathematics ad Egieerig