Laplace Transforms

Case 1:
Solution using Laplace Transform

Mz'' + Bz' + Kz = KU(t)
ML{z''} + BL{z'} + KL{z} = KL{1}
z'(0) = z(0) = 0

Therefore:

Ms''Z(s) + BsZ(s) + KZ(s) = K/s
Z(s)[ Ms² + Bs + K ] = K/s
Z(s) = K/s/( Ms² + Bs + K)

The roots of (Ms² + Bs + K) are calculated above.

r₁ = - 0.14 + 0.87i
r₂ = - 0.14 - 0.87i

Z(s) = K/s(s-r1)(s-r2)
or
Z(s) = K/s(s + 0.14 - 0.87i )*( s + 0.14 + 0.87i )
(K = 2000)

Using partial fraction expansion:

Z(s) = C₁/s + C₂/(s + 0.14 - 0.87i ) + C₃/(s + 0.14 + 0.87i )

C₁=sZ(s)|_s=0= 0
C₂ =(s + 0.14 + 0.87i )Z(s)| _{s=-0.14-0.87i}
C₃ =(s + 0.14 - 0.87i )Z(s)|_{s=-0.14+0.87i}

C₁ = K/(r₁*r₂) = 1
C₂ = K/s(s-r₂) = k/r₁(r₁-r₂) = -1287.8 + 207.24i
C₃ = K/s(s-r₁)=K/r₂(r₂-r₁) = -1287.8 - 207.24i

Write C₁ and C₂ in exponential form.:
C₁ = 1304exp(0.16i)
C₂ = 1304exp(-0.16i)

Z(s) = 2576/s + [1304exp(-0.16i)] / (s + 0.14 - 0.87i ) + [1304exp(0.16i)] / (s + 0.14 + 0.87i )

z(t) = Inverse Laplace Transform of Z(s)

z(t) = 1+ 0.98exp(-0.14t)* cos(0.87t - 0.16)

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