# Low-pass filter with no inductance, and voltage output over a resistor

by Tetraoxygen
Tags: capacitor, filter, impedance, resistor
 P: 7 1. The problem statement, all variables and given/known data I am supposed to find the circuit design for a first-order low-pass filter. The filter must be made of resistors and capacitors only (no inductances). Furthermore, the output voltage must be measured over a resistor or something in parallel with a resistor. 2. Relevant equations Impedance for a capacitor is 1/[jwC], for a resistor is R Voltage divider for impedances 3. The attempt at a solution i) Attempt at following those constraints: A resistor R1 in series with the signal source and in series with (a resistor R2 in parallel with a capacitor C). Note: I realize this is not a strictly low-pass filter. Impedance over the series resistor z1 = R1 Impedance over the parallel component z2 = R2/[jwCR1 + 1] Equivalent impedance z = (R1 + R2)[jwCR1R2(1/[R1+R2) + 1]/[jwCR2+1] As far as I can tell, this is actually a bandpass (or reject) filter with cutoff frequencies at (R1 + R2)/(C R1 R2) and 1/(C R2). ii) I realize I can make an RC low-pass filter (but then the output voltage is not measured over a resistor) and an RL low-pass filter (but then there is an inductor). Please help. Or, if you know definitively that this is an impossible task, please tell me so. Thanks.
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Thanks
PF Gold
P: 4,827
 Quote by Tetraoxygen 1. The problem statement, all variables and given/known data I am supposed to find the circuit design for a first-order low-pass filter. The filter must be made of resistors and capacitors only (no inductances). Furthermore, the output voltage must be measured over a resistor or something in parallel with a resistor. 2. Relevant equations Impedance for a capacitor is 1/[jwC], for a resistor is R Voltage divider for impedances 3. The attempt at a solution i) Attempt at following those constraints: A resistor R1 in series with the signal source and in series with (a resistor R2 in parallel with a capacitor C). Note: I realize this is not a strictly low-pass filter. Impedance over the series resistor z1 = R1 Impedance over the parallel component z2 = R2/[jwCR1 + 1] Recheck this! Equivalent impedance z = (R1 + R2)[jwCR1R2(1/[R1+R2) + 1]/[jwCR2+1] As far as I can tell, this is actually a bandpass (or reject) filter with cutoff frequencies at (R1 + R2)/(C R1 R2) and 1/(C R2). ii) I realize I can make an RC low-pass filter (but then the output voltage is not measured over a resistor) and an RL low-pass filter (but then there is an inductor). Please help. Or, if you know definitively that this is an impossible task, please tell me so. Thanks.
Your circuit (which I assume is R1 from input to output and R2||C from output to ground) is a low-pass circuit. You can reduce it to the form a/(a+b+jwC).
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P: 5,247
 A resistor R1 in series with the signal source and in series with (a resistor R2 in parallel with a capacitor C).
Sounds okay as a LPF. It will preferentially pass lower frequencies and down to DC.

You can take the output as the voltage across that R∥C combo, or alternatively you could break that R into 2 and take output from across its lower portion.

 P: 7 Low-pass filter with no inductance, and voltage output over a resistor That was a typo: z2 should have been R2/(jwcR2 + 1) But really, I forgot to take: z2/(z1+z2) Now I have R1/(R1 + R2)/[jwcR1R2(1/[R1+R2]) + 1] with cutoff frequency (R1 + R2)/(R1 R2) Thanks!

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