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

[Tetraoxygen]

Homework Statement

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.

Homework Equations

Impedance for a capacitor is 1/[jwC], for a resistor is R
Voltage divider for impedances

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.

 

[rude man]

Homework Statement

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.

Homework Equations

Impedance for a capacitor is 1/[jwC], for a resistor is R
Voltage divider for impedances

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]

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]

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!