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Low Pass Filter Problem

asciid

Mar 19, 2012
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Mar 19, 2012
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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.

Umk2.png
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
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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Perhaps you can consider R2 to be the impedance of the stage following the filter formed by R1 and C1?
 

john monks

Mar 9, 2012
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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.
 

Laplace

Apr 4, 2010
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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.
 

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