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Low Pass Filter Problem

Discussion in 'Electronics Homework Help' started by asciid, Mar 19, 2012.

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  1. asciid


    Mar 19, 2012
    Hi guys, I'm at university and havent touched low pass filters since i was 16 or something and am having a bit of trouble solving an RC LPF problem

    I am trying to find the half power point on this circuit.

    [​IMG] R1 = 1k R2= 10k C1 = 100nF

    I am happy solving a simple LPFs (1C+1R) but I haven't done this kinda thing for a while.

    I know the impedance of the capacitor is 1/(2*pi*f*C) but im getting stuck because of R2 in parallel with C1
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Perhaps you can consider R2 to be the impedance of the stage following the filter formed by R1 and C1?
  3. john monks

    john monks

    Mar 9, 2012
    I hope I'm not making a mistake. So, your Xc must equal your parallel resistance R1*R2/(R1+R2). This is 10/11 or about 909ohms. Therefore your Xc must equal 909ohms. Xc=1/(2pi*f*c). Rewriting f=1/(2piXc) = 17.5KHz.
    Another way to look at this the phase angle at half power is 45 degrees. So when you plot your Xc and totol R on a graph the resultant phase angle will be 45 degrees. You can see then that your Xc will equal your total R and you will have to adjust your frequency accordingly.
  4. Laplace


    Apr 4, 2010
    R1 & R2 form a resistor divider so the question is whether the half-power point is measured from the DC baseline output, or from the initial input voltage. See the attached graph - you could find two different 3dB points.

    Attached Files:

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