[討論] 懸吊系統MATLAB已回收
x1(0)=0=x2(0)=x'2(0) k=50000 b=1500 m=1200
H(s)=X2(s)/X1(s)=(bs+k)/(ms^2+bs+k)
D=10cm L=2m v=5m/s(此懸吊系統往Y方向走)
g(y)=Dsin(2*pi*y/L) when 0<y<L/2 g(y)=0 when L/2<y<L
g(y+L)=g(y) for all y>0
幫我解一下它問把x'2(t)變成(x2[n+1]-x2[n-1])/2*T
x''2(t) (x2[n+1]-2*x2[n]+x2[n-1])/T^2
(1)
Do the approximation similarty for x1(t), The result is a difference equation
that specifies a discrete-time linear time-invariant system
Do the approximation for x1(t) (這句話我看不太懂)=>是叫我變x'1(t) x''(t) 都跟
x2(t)樣的變法嗎? =if yes=>可以解出一等式 但要怎麼求它要問的阿?
(2)
自己決定T(不過我覺得可能要0.01左右) x1[n]=g(y)其中y=vnT 求出H(z)=X2(z)/X1(z)
然後再計算x2[n] in response to the road condition x1[n] for time 0-5s
用MATlab filter()函數解 然後要看畫出來的response的MAX有沒有>D
(3) 找出 pole zero of H(z) 用roots() zplane()
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