Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Apr 22, 2012 #1
    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.
  2. jcsd
  3. Apr 22, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    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).
  4. Apr 22, 2012 #3


    User Avatar

    Staff: Mentor

    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.
  5. Apr 22, 2012 #4
    That was a typo:
    z2 should have been
    R2/(jwcR2 + 1)

    But really, I forgot to take:

    Now I have
    R1/(R1 + R2)/[jwcR1R2(1/[R1+R2]) + 1]

    with cutoff frequency (R1 + R2)/(R1 R2)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook