


U
2
(P)

F(P)=TrongđóR=10Ω,L
1
=1mH,L
2
=10mH.

U
1
(P)

Giải:
U
2
(p)((PL
2
/((RPL
1
/(R+PL
1
))+L
2
))U
1
(p)

P
10
=-104

P
1∞
=-1,1.104

VớiK=1tacó:a(ω)=20log1=0
b(ω)=π/2

vớiP
10
=-104chọnω
o
=ω
h
=104

=>γ’=logω/ω
o
=logω–log(104)=γ–4

=>a(ω)=0nếuγ’<0a(ω)=0:γ<4

a(ω)=20γ’nếuγ≥0a(ω)=20(γ–4):γ≥4

b(ω)=0khiγ’≤-1γ≤3

b(ω)=π/4khiγ’=0γ=4