複數的應用 (請詳細解答)

2010-06-23 7:40 pm
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(請詳細解答)
更新1:

係唔係應該R1(t)+R2(t)=Acos(wt+z)??

回答 (1)

2010-06-24 6:26 pm
✔ 最佳答案
Denote θ1=x, θ2=y, θ3=z
R1(t)+R2(t)=Re[10exp(iwt+ix)+8exp(iwt+iy)]
10exp(iwt+ix)+8exp(iwt+iy)
=exp(iwt)[10exp(ix)+8exp(iy)]
=exp(iwt)[(10cosx+8cosy)+i(10sinx+8siny)]
=exp(iwt)[P+iQ] where P=10cosx+8cosy, Q=10sinx+8siny
=exp(iwt)*A*[cosz+isinz] where A^2=P^2+Q^2, cosz=P/A, sinz=Q/A, tanz=Q/P
=A exp(iwt)*exp(iz)=Aexp[i(wt+z)]
so, R1(t)+R2(t)=Acos(wt-z)

For y=10cos(t+π/6)+8cos(t+π/3)
P=10cos(π/6)+8cos(π/3)=5√3+4
Q=10sin(π/6)+8sin(π/3)=5+4√3
A^2=100+64+80√3= 164+80√3, A=2√(41+20√3)
tanz=Q/P, then z=arctan[(5+4√3)/(5√3+4)] about 43.3 degree.

y=10cos(t+π/6)+8cos(t+π/3)=Acos(t-z)
its graph is a cosinewave
with amplitude A, period 2π and shifting to the right z units.

2010-06-24 13:30:47 補充:
Thks!
R1(t)+R2(t)=Acos(wt+z)
y=Acos(t+z)
shifting to the left z units.


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