Bandpass Filter Homework Statement

AI Thread Summary
A bandpass filter can be constructed by cascading an RC highpass filter with an RC lowpass filter, with an isolation buffer to maintain circuit integrity. For the given audio filter allowing vocal frequencies between 80 Hz and 1100 Hz, the required capacitor values were calculated as approximately 1.99 x 10^-7 F for the highpass filter and 1.45 x 10^-8 F for the lowpass filter. The gain at 10,000 Hz was initially miscalculated but corrected to 0.109 when using the lowpass filter's capacitor value. The discussion emphasizes the importance of accurate calculations in filter design. Understanding the behavior of each filter component is crucial for achieving the desired frequency response.
Lucille
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Homework Statement


A simple way to build a bandpass filter is to filter the output of an RC highpass filter with an RC lowpass filter, as shown on the left in the diagram below. The isolation buffer is a circuit element that keeps the two circuits isolated so they behave as they would on their own. The (highly idealized) response of the circuit is shown on the right, where fL and fH are the low and high cutoff frequencies.

a) Assume fL is given by the equation for the cutoff frequency of a high pass filter, and fH is given by the equation for the cutoff frequency of a low pass filter. If the value of the resistors is 10 kOhm what should the value of the capacitors be if you were building an audio filter to allow vocal frequencies between fL = 80 Hz and fH = 1100Hz.

b) Assuming the high frequency behavior of the filter is described entirely by the response of the low pass filter, what would the gain, Vout/Vin, be at 10 000 Hz?

Image available at http://www.chegg.com/homework-help/questions-and-answers/simple-way-build-bandpass-filter-filter-output-rc-highpass-filter-rc-lowpass-filter-shown--q3608656

Homework Equations



f_c = 1/(RC*2Pi)

Vout/Vin = 1/ sqrt(1+(RwC)^2)

The Attempt at a Solution



a) C = 1/(2pi*f*R) -- do I calculate two different capacitance values?

so C1 = 1.99*10^-7 and C2 = 1.45*10^-8

b) Using C2 and plugging into the equation gives 0.568
 
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Your capacitor values look fine.

I'm not sure about the formula you've used for the gain. What particular setup is it for? The question specifies that you consider only the low pass filter portion's contribution.
 
Last edited:
It is for a low pass filter -- and so I used the value for C = 1.45 * 10^-8 F and subbed it into the equation for the gain of a low pass filter
 
Lucille said:
It is for a low pass filter -- and so I used the value for C = 1.45 * 10^-8 F and subbed it into the equation for the gain of a low pass filter
Okay, that would be correct, but the result you've obtained looks a bit high to me. Can you check you math?
 
Whoops - I got 0.109
 
Lucille said:
Whoops - I got 0.109
Much better :smile:
 
Thank you so so so much! It makes so much more sense to me now.
 
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