Ecuaciones

EJERCICIO ECUACIONES ; SOLUCIÓN POR EL HAMMNER(J.P.G.)

Supanga que F(t) es la rectificación de media onda de sen(kt) que aparece en la figura.

Demuestre que L{F(t)} = k/((k^2+s^2 )(1-e^(-πs/k)))

F(t)

t

↑π/k ↑2π/k ↑3π/k

SOLUCION:

CON P=2π/k y F(t)= sen(kt) para 0≤t≤π/k mientras que F(t)=0 para π/k≤t≤2π/k

L{F(t)} = 1/(1-e^ST ) ∫_0^T▒e^(-st) F(t)dt

L{F(t)} = 1/(1-e^ST ) ∫_0^T▒e^(-st) sen(kt)dt

L{F(t)} = 1/(1-e^(-2πs/k) ) ∫_0^(π/k)▒e^(-st) sen(kt) dt

U=sen(kt) dv=e^(-st)

du=k cos(kt) v=-1/s e^(-st)

L{F(t)} = 1/(1-e^(-2πs/k) ) [(-e^(-st) sen(kt))/s |π¦0 + k/s ∫_0^(π/k)▒e^(-st) cos⁡(kt) dt ]

U=cos(kt) dv=e^(-st)

du=k sen(kt) v=- 1/s e^(-st)

L{F(t)} =1/(1-e^(-2πs/k) ) [(-e^(-st) sen(kt))/s |π¦0 + k/s [-1/s e^(-st) cos⁡(kt)+ ∫_0^(π/k)▒〖e^(-st) sen(kt) dt〗 ]

------------------------------∫_0^(π/k)▒〖e^(-st) sen(kt) dt〗 - k^2/s^2 ∫_0^(π/k)▒〖e^(-st) sen(kt) dt〗 = (-e^(-st) sen(kt))/s |π¦0 - k/s^(2 ) e^(-st) cos⁡(kt) |π¦0

[ k^2/s^2 +1] ∫_0^π▒〖e^(-st) sen(kt) dt〗 = (-e^(-st) sen(kt))/s |π¦0 - k/s^(2 ) e^(-st) cos⁡(kt) |π¦0

[ (k^2+ s^2)/s^2 ] ∫_0^π▒〖e^(-st) sen(kt) dt〗 = (-e^(-st) sen(kt))/s |π¦0 - k/s^(2 ) e^(-st) cos⁡(kt) |π¦0

∫_0^π▒〖e^(-st) sen(kt) dt〗 = s^2/(k^2+ s^2 ) [ (-e^(-st) sen(kt))/s |π¦0 - k/s^(2 ) e^(-st) cos⁡(kt) |π¦0 ]

↑Al evaluar se hace cero

∫_0^π▒〖e^(-st) sen(kt) dt〗 = - s^2/(k+ s^2 ) [ k/s^(2 ) e^(-st) cos⁡(kt) |π¦0 ]

∫_0^π▒〖e^(-st) sen(kt) dt〗 = - s^2/(k^2+ s^2 ) k/s^2 [ e^(-πs/k) cos⁡(kπ/k) - e^(-0) cos⁡(k (0)) ]

L{F(t)} = 1/(1-e^(-2πs/k) ) [k/(k^2+s^2 ) (e^(-πs/k)+1) ]

L{F(t)} = (K (〖1+e〗^(-πs/k)) )/((1-e^(-πs/k) )(1+e^(-πs/k)

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