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

by Tetraoxygen
Tags: capacitor, filter, impedance, resistor
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Tetraoxygen
#1
Apr22-12, 01:28 AM
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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rude man
#2
Apr22-12, 05:19 AM
HW Helper
Thanks
PF Gold
rude man's Avatar
P: 4,759
Quote Quote by Tetraoxygen View Post
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).
NascentOxygen
#3
Apr22-12, 06:49 AM
HW Helper
Thanks
P: 5,159
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.

Tetraoxygen
#4
Apr22-12, 07:25 AM
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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