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# Answer to Question #194292 in Electrical Engineering for John Gly

Question #194292

Electronics engineering: FILTERS

Design a 2nd order crossover network with the following specifications:

• High Pass cutoff frequency = 3100 Hz
• Low Pass cutoff frequency = 310 Hz
• Bandpass frequency = 310Hz to 3100Hz
1
2021-05-31T06:39:54-0400

An Op-Amp based bandpass filter is as shown below

The transfer function "\\frac{V_0(s)}{V_{in}(s)}=-\\frac{Z_0(s)}{Z_{in}(s)}"

"\\frac{V_0(s)}{V_{in}(s)}=-[\\frac{R_2}{(1+sC_2R_2)}]*[\\frac{sc_1}{(1+sC_1R_1)}]=\\frac{sc_1R_2}{(1+s \\tau_1)(1+s \\tau_2)} \\implies \\tau =RC"

The circuit will behave as BPF if "\\tau _1> \\tau_2"

"f_L=\\frac{1}{2 \\pi R_1C_1}"

"f_H=\\frac{1}{2 \\pi R_2C_2}"

"R_1=\\frac{1}{2 \\pi *310*47*10^{-9}}=10.924k \\Omega"

Commercial available resistance nearest to calculated R1 is 11k"\\Omega"

"\\implies f_L=\\frac{1}{2 \\pi R_1C_1} =\\frac{1}{2 \\pi 11*10^3*47*10^{-9}} =307.84 Hz"

Therefore, "R_1=11k \\Omega; C_1=47nF"

"C_2=\\frac{1}{2 \\pi R_2f_H} =\\frac{1}{2 \\pi *3100*20*10^3} = 2.567 nF"

Commercial available capacitor nearest to calculated C2 is 2.7nF

"f_H=\\frac{1}{2 \\pi R_2C_2}=\\frac{1}{2 \\pi*20*10^3*2.7*10^{-9}}=2947.31 Hz"

Therefore, "R_2=20k \\Omega; C_2=2.7nF"

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