<急> 誰會工程數學第二移位定理~~?

[匿名]

2015-12-12 12:14 pm

1. f(t)= t-2, 0<=t<16
-1 , t>=16

課本答案: 1/(s^2)-(2/s)-1/(s^2)-(1/s)

求反拉式變換

2. e^(-21s)/[s(s^2+16)]

課本答案: 1/16(1-cos(4(t-2)))H(t-2)

3. e^(-2s)/(s^2+9)

課本答案: (1/3)sin(t-2)H(t-2)

我算出來都跟解答不一樣，誰能給點正確解答~~ 感謝！

回答 (1)

Lopez

2015-12-12 1:33 pm

✔ 最佳答案

1.
f(t)
= (t-2)*[ H(t) - H(t-16) ] + (-1)*H(t-16)
= (t-2)*H(t) + (-t+1)*H((t-16)
= t*H(t) - 2*H(t) - (t-16)*H(t-16) - 15*H(t-16)

利用第二移位定理得:
L { f(t) }
= e^(-0s)*(1/s^2) - 2*e^(-0s)*(1/s) - e^(-16s)*(1/s^2) - 15*e^(-16s)*(1/s)
= [ 1 - 2s - e^(-16s) - 15s*e^(-16s) ] / s^2 ..... Ans

2.
令 1 / [ s(s^2+16) ] = a/s + (bs+c)/(s^2+16)
解得 a = 1/16 , b = - 1/16 , c = 0

L^(-1) { e^(-21s) / [ s(s^2+16) ] }
= L^(-1) { (1/16)*e^(-21s)*[ 1/s - s/(s^2+16) ] }
= (1/16) * L^(-1) { [ e^(-21s)(1/s) - e^(-21s)*s/(s^2+16) ] }
= (1/16) * [ H(t-21) - H(t-21)*cos 4(t-21) ]
= (1/16) * [ 1 - cos 4(t-21) ] * H(t-21) ..... Ans

3.
L^(-1) { e^(-2s) / (s^2+9) }
= (1/3) * L^(-1) { e^(-2s) * 3/(s^2+9) }
= (1/3) * H(t-2) * sin 3(t-2) ..... Ans

👍 0 👎 0