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Building a frequency dependent circuit- Low pass filter

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David Keeler

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I have a low pass RC filter that looks like the attached picture https://tinypic.com/view.php?pic=1z6bo8p&s=8#.VEMc58mURyN
I have to derive the DC gain, the zero frequency, the pole frequency. I don't quite understand how the resistor R2 acts in this circuit.

The other requirements are
  • the magnitude is 1 for input frequencies below 100Hz
  • The magnitude starts to fall at 20dB/decade at 100Hz
  • a constant magnitude for frequencies above 10kHz
I think there will be two design equations one for the pole and one for the zero, which will give me two equations with 3 unknowns.

I designed the filter using a 1uF capacitor with resistor values of 1.6k ohms and 100 ohms and then I plotted those and found the break frequency to be 93.95Hz


I would like some clarification on how to incorporate the proper design equations for a low pass filter with 2 resistors and one capacitor.

Thank you
 
I'll let you build the equations. However, they should describe this behavior: (normalized for 1Ohm, 1uF, and the ratio of R2 to R1.) Note the low freq gain. Note the high freq gain. Note that the frequency response follows from the ratio of R2 to R1.

75.jpg
 
Last edited:
Okay so this is my attempt at building the equations. I just want to see if I'm making the correct correlations.

A regular usual low pass filter has a transfer function of Vo/Vin= 1/(jωRC+1)

Now my low pass filter has a transfer function of Vo/Vin= R2/Rs (1/jwRsC+1)
where Rs=R1+R2
which would mean the cutoff frequency would be fc=1/2πRsC

Does this mean that once I pass my cutoff frequency my transfer function becomes R2/(R1+R2)
 
Hi,

The function R2/(R1+R2) is a constant function, so this is a little unusual, but assuming that your transfer function is really:
Vo/Vi=(R2/Rs)*(1/(j*w*Rs*C)+1)

then the magnitude is:
(R2/Rs)*sqrt(1/(Rs^2*C^2*w^2)+1)

and we can see the factor:
sqrt(1/(Rs^2*C^2*w^2)+1)

as w gets higher and higher eventually becomes:
sqrt(1)

which of course equals 1, so we end up with:
|Vo/Vi|=R2/Rs=R2/(R1+R2)

At the cutoff frequency however the magnitude is:
(R2/Rs)*sqrt(2)=sqrt(2)*R2/(R1+R2)

and to reach within 1 percent of the final response we have to go to a frequency:
w=100/(sqrt(201)*Rs*C)

and then the response is:
1.01*(R2/(R1+R2))

but that requires a frequency that is:
w2=100/sqrt(201)

which is about 7 times higher than the cutoff frequency.

So yes it does approach that constant response, but only after a certain frequency has been reached.
 
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