# High Pass filter

1. Homework Statement
''A machineoutputs a 4kHz sine wave signal of amplitude 2 volts peak, plus a 100Hz interference component of amplitude 1 V peak''

Given equation: Vout/Vin = 1/((1+(f0/f^2))^0.5 phase lead = inverse tan (f0/f)

f0 = 1/(2*pi*R*C)

I'm given a 4 kilo-ohm resistor and need to choose a capacitor to reduce the interference (100 Hz signal) to 0.05 V peak; I need to design a high pass RC circuit.

2. Homework Equations

given in the Q

3. The Attempt at a Solution I'm not sure what f0 and f actually is. Equations I can find which are similar to this have 'critical frequencies' mentioned.

gneill
Mentor
I'm not sure what f0 and f actually is. Equations I can find which are similar to this have 'critical frequencies' mentioned.
In the equations given, f would be the frequency of interest that you are trying to suppress, while fo would be the "corner frequency" of the filter which is determined by the choice of R and C for the filter.

Hi thanks for the reply. So would f would be 100 Hz? But I'm not sure how to cut the interference to 0.05 V only. Any ideas?

gneill
Mentor
Hi thanks for the reply. So would f would be 100 Hz? But I'm not sure how to cut the interference to 0.05 V only. Any ideas?
Yes, 100 Hz is what you are trying to suppress from 1 V down to 0.05 V. Looks like a Vo/Vi ratio is close at hand ;)

Vout/Vin = 1/[(1+(f0/f)^2)]^0.5

Hello - at first, I have corrected the expression for the magnitude of a firdt-order highpass (see above).
Because fo must be chosen so that at f=100 hz the damping factor is 1/0.05=20 we have the expression

[(1+(f0/f)^2)]^0.5=20;
(fo/f)^0.5=399~400.
(fo/f)=20
;
fo=20*f=2kHz.

Hence, the highpass corner frequency is fo=2kHz.
Therefore: T=RC=1/wo=1/(2Pi*fo)=1.6 ms.

Now you have to find suitable values for R and C for realizing the time constant T=RC.

Remark: Sorry - if I gave too many details in my answer. Only now I have realized that I am in the homework section.

Last edited:
axe34
Thanks!!!!!!!!!!!