Circuit Cut off Frequency

[rubynithyanand]

How do you calculate the cut-off frequency of the circuit 1?

Wkt, the cut-off frequency of a HPF filter ciruit (circuit 2) is

f= 1/(2*pi*R1*C)

Added after 31 minutes:

Circuit Cut off Frequency : Attachment File

 

[rubynithyanand]

Circuit Cut off Frequency: Attachment

 

[keith1200rs]

If I understand your question correctly, the cut off of the second circuit will be 1/(2*pi*(R1+R2)*C)

R2 will mean that the signal will have a loss, even at low frequencies, but what matters is the 3dB drop from the flat part of the response which will be lower in frequency due to the higher resistance.

Keith.

 

[biff44]

First, let me say that it is extremely important what the source impedance and load impedance is.

But, assuming a zero ohm source impedance, and an infinite load impedance, then

Circuit 1 is a HIGHpass filter, with a pole at f1=2Π(R1)C

Circuit 1 is a HIGHpass filter, with a pole at f1 above, BUT it turns into a fixed insertion loss flat attenuator after the zero frequency of f2=2Π(R2)C. The voltage attenuation value will be R1/(R1+R2)

 

[keith1200rs]

Or, in pictures:

Keith.