Re: [工數] 工數求解
※ 引述《a2626 (Lucas)》之銘言：
: http://miupix.cc/pm-4ND9PO
: 有沒有神人能幫一下31題阿…
: 怎麼算數字都超醜超難轉回來…
: 答案相當簡單是
: http://miupix.cc/pm-AS1A0V
: 但是就是弄不出來orz
: 感謝了
先將 r(t) 表示出來
r(t) = [u(t) - u(t-2π)](3sint - cost) + u(t-2π)(3sin2t - cos2t)
= u(t)(3sint - cost) - u(t-2π)[3sin(t-2π) - cos(t-2π)]
+ u(t-2π)[3sin2(t-2π) - cos2(t-2π)]
取 Laplace Transform
2
=> s Y - s + sY - 1 - 2Y
-0s 3 - s -2πs 3 - s -2πs 6 - s
= e (---------) - e (---------) + e (---------)
s^2 + 1 s^2 + 1 s^2 + 4
(s+2)(s-1)Y
-0s 3 - s -2πs 3 - s -2πs 6 - s
= s + 1 + e (---------) - e (---------) + e (---------)
s^2 + 1 s^2 + 1 s^2 + 4
s + 1 -0s 3 - s
=> Y = ------------ + e [---------------------]
(s+2)(s-1) (s+2)(s-1)(s^2 + 1)
3 2 3 2
-2πs -s - s + 6s + 6 + s + 4s - 3s - 12
+ e [----------------------------------------]
(s+2)(s-1)(s^2 + 1)(s^2 + 4)
2 1 1 1 -0s 1 1 1 1 1
= --- ----- + --- ----- + e (--- ----- - --- ----- - ---------)
3 s-1 3 s+2 3 s-1 3 s+2 s^2 + 1
-2πs 1 1
+ e (--------- - ---------) (不得不說這數字配得神漂亮......)
s^2 + 1 s^2 + 4
取 Inverse Laplace Transform
2 t 1 -2t 1 t -2t
=> y(t) = ---e + ---e + ---u(t)(e - e - 3sint)
3 3 3
1
+ u(t-2π)[sin(t-2π) - ---sin2(t-2π)]
2
取 t ≧ 0
t 1
=> y(t) = e - sint + u(t-2π)(sint - ---sin2t)
2
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