更新1:
φ= -arctan[w/(1-w^2)]. 定係 φ= -arctan[(1-w^2)/w]. 呀????
求救!!!frequency response 問題
2008-07-21 7:53 am
For the system y" + y' + y = r where r(t) = sinωt is the system input and y(t)the output, the steady-state component of the output is yss(t) = Asin(ωt + φ).Express A and φin terms ω of .
回答 (2)
2008-07-21 4:16 pm
✔ 最佳答案
y=Asin(wt + k)dy/dx =Awcos(wt +k)
d(dy/dx)/dx = -Aw^2sin(wt+k). Therefore,
d(dy/dx)/dx + dy/dx + y = -Aw^2sin(wt+k) + Awcos(wt+k) + Asin(wt+k)
= A(1-w^2)sin(wt+k) + Awcos(wt+k).
Let A(1-w^2) = m , Aw=n and (wt+k)=B. Therefore the expression becomes
msinB + ncosB. From what we learn from a. maths, this can be express as
Rsin(B + h), where
R = sqrt(n^2 + m^2) = sqrt[A^2(1-w^2)^2 + A^2w^2] = Asqrt(1-w^2 + w^4).
h = arctan(n/m) = arctan[Aw/A(1-w^2)]= arctan[w/(1-w^2)].
Therefore,
Asqrt(1-w^2 + w^4)sin{wt+k + arctan[w/(1-w^2)]} = r= sinwt.
That means,
A= 1/sqrt(1-w^2+w^4) and k = -arctan[w/(1-w^2)].
收錄日期: 2021-04-25 22:36:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080720000051KK03103