Đồ thị Bode_Câu 10 ppsx




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)


F(p)=——=———————————————
U
1
(p)U
1
(p)

PL
2
10-1+10-5P

↔F(p)=————————=—————————

(RPL
1
/(R+PL
1
))+PL
2
10-2+10-1+10-5P

(-104)(1+P/104)

↔F(p)=—————————

1,1.104(1+P/1,1.104)

→K=(-104)/(-1,1.104)=0,9≈1


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

b(ω)=π/2khiγ’≥1γ≥5

vớiP
1∞
=-1,1.104chọnω=ω
h
=1,1.104

γ”=logω/ω
o
=γ–log(1,1.104)=γ–4

a(ω)=0khiγ”<0<=>γ<4:a(ω)=0khiγ<4

a(ω)=-20γ”khiγ”≥0=>γ≥4:a(ω)=-20(γ-4)khiγ≥4

b(ω)=0khiγ”≤-1=>γ<3

b(ω)=-π/4khiγ”=0=>γ=4

b(ω)=-π/2khiγ”≥1=>γ≥5
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Vẽ: