text: 107 449
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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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explanation: The document contains a detailed answer to a question in Electrical Engineering, specifically about designing a 2nd order crossover network with given specifications. The question and the answer are both present.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
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distilabel_metadata.role: system
distilabel_metadata.content: 107 449
  Assignments Done
  96.1%
  Successfully Done
  In May 2023
  
  # 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"
  
  Need a fast expert's response?
  
  Submit order
  
  and get a quick answer at the best price
  
  for any assignment or question with DETAILED EXPLANATIONS!
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed answer to a question in Electrical Engineering, specifically about designing a 2nd order crossover network with given specifications. The question and the answer are both present.
  
  Label: question-and-answer