# Analisis de un sistema dinamico

Páginas: 2 (256 palabras) Publicado: 9 de junio de 2011
mx ̈ + bx ̇ + kx = 0

ωn = √(k/m) = √(100/25) = 2 rad/s
ωd = ωn √(1-ζ^2 ) ∴ ζ= b/(2√(k m))= 55/(2√((100)(25)))=0.5
ωd= 2 √(1-(0.5)^2 )=1.732050

mx ̈ + bx ̇ + kx = f(t) función escalonx(0)= 0
x’(0)= 0m[s^2 x(s)- sx(0)- x^' (0) ]+ b[sx(s)- x(0) ]+kx(s)=250/5m[s^2]x(s) ┤+ b├ sx(s) ┤+kx(s)=2501/s
x(s)[ms^2]┤+b+k=2501/s

x(s)=2501/s [1/(ms^2 + b+k)] ec.5
255x(s)+505x(s)+100x(s)=250/s
x(s)[ms^2 + b+k]=250/sx(s)=250/s [1/(ms^2 + b+k)]=250/ms [1/(s^2 +b/m s+k/m)]

x(s)=(250/m)(1/s)[1/(s^2 +2ζω_n s +〖ω_n〗^2 )]

x(t)=(250/m)(1/〖ω_n〗^2 )L^(-1) [〖ω_n〗^2/s(s^2 +2ζω_n s +〖ω_n〗^2 ) ] ϕ=〖cos〗^(-1) ζ=60°

x(t)=(250/m)(1/〖ω_n〗^2 )[1-1/√(1-ζ^2 ) ℮^(-ζω_n t)sen(ω_n √(1-ζ^2 ) t+ϕ) ]

x(t)=(10)(1/H)[1-1/√(3/4) ℮^(-(1.73205)(2)t) sen(2 √3/2 t+60°) ]

x(t)=(1/4)[1-2/√3 ℮^(-3.4641t)sen(√3 t+60°) ]

x(t)=(2.5)[1-(2√3)/3 ℮^(-3.4641t) sen(√3 t+60°) ]

x(t)=(2.5)[1-(2√3)/3 ℮^(-3.4641t) sen(√3 t+1.047) ]

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