fPL = f0/2[√(3–A0)2 +4 –(3–A0)] fPH= f0/2[√(3–A0)2 +4 +(3–A0)]
设 C1=C2=C, A0= (1+R2/R1) 传递函数(推导从略)
Ra ui
C1
Rb C2
Au(s) = UO(s)/Ui(s)
=A0÷[1 + (3–A0) SRC +(SRC)2 ]
令S=jω, f0=1/(2πRC)
电压放大倍数
20lg Au
Au(f) = A0÷[1 – (f/f0)2 + j(3–A0)f/f0] 令 Q =1/(3–A0), 当 f = f0 时, Au(f0) = A0 /(3–A0)= QA0
–20dB/十倍频
2. 简单二阶电路 令 A0=(1+R2/R1) 传递函数
Au(s) = UO(s)/Ui(s) = A0Ub(s)/Ui(s)
Ra ui
C1
Rb C2
+ A
uO
R2
RL
R1
= A0Ub(s)/Ua(s)*Ua(s)/Ui(s), 当 C1= C2= C 时 Ub(s)/Ua(s) =1/(1+SRC), Ua(s)/Ui(s) =[1/SC//(R+1/SC)]÷[R+1/SC//(R+1/SC)] ∴ Au(s) = A0÷[1+3SRC+(SRC)2] 用jω取代 S,且令 f0=1/(2πRC)有: Au(f) = A0÷[1 – (f/f0)2 + j3f/f0]
2. 简单二阶电路(续) Au(f) = A0 ÷[1 – (f/f0)2 + j3f/f0] 截止频率 fP ≈ 0.37 f0