U 2 (P) F(P)=TrongđóR=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 VớiK=1tacó:a(ω)=20log1=0 b(ω)=π/2 vớiP 10 =-104chọnω o =ω h =104 =>γ’=logω/ω o =logω–log(104)=γ–4 =>a(ω)=0nếuγ’<0a(ω)=0:γ<4 a(ω)=20γ’nếuγ≥0a(ω)=20(γ–4):γ≥4 b(ω)=0khiγ’≤-1γ≤3 b(ω)=π/4khiγ’=0γ=4 b(ω)=π/2khiγ’≥1γ≥5 vớiP 1∞ =-1,1.104chọnω=ω h =1,1.104 γ”=logω/ω o =γ–log(1,1.104)=γ–4 a(ω)=0khiγ”<0<=>γ<4:a(ω)=0khiγ<4 a(ω)=-20γ”khiγ”≥0=>γ≥4:a(ω)=-20(γ-4)khiγ≥4 b(ω)=0khiγ”≤-1=>γ<3 b(ω)=-π/4khiγ”=0=>γ=4 b(ω)=-π/2khiγ”≥1=>γ≥5 Vẽ: . U 2 (P) F(P)=TrongđóR =10 ,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 VớiK=1tacó:a(ω)=20log1=0 b(ω)=π/2 vớiP 10 = -104 chọnω o =ω h = 104 =>γ’=logω/ω o =logω–log (104 )=γ–4 =>a(ω)=0nếuγ’<0a(ω)=0:γ<4 a(ω)=20γ’nếuγ≥0a(ω)=20(γ–4):γ≥4 b(ω)=0khiγ’≤-1γ≤3 b(ω)=π/4khiγ’=0γ=4 b(ω)=π/2khiγ’≥1γ≥5 vớiP 1∞ =-1,1 .104 chọnω=ω h =1,1 .104 γ”=logω/ω o =γ–log(1,1 .104 )=γ–4 a(ω)=0khiγ”<0<=>γ<4:a(ω)=0khiγ<4 a(ω)=-20γ”khiγ”≥0=>γ≥4:a(ω)=-20(γ-4)khiγ≥4 b(ω)=0khiγ”≤-1=>γ<3 b(ω)=-π/4khiγ”=0=>γ=4 b(ω)=-π/2khiγ”≥1=>γ≥5 Vẽ: