Removing DC Offset

DefaultName

1. The problem statement, all variables and given/known data

I need to remove a DC offset as apart of my research project I'm doing for my CS professor and his research group.

This signal is a sine wave, 10 Hz, riding on a +1V DC offset.

2. Relevant equations

I'm utilizing a high-pass filter using a series RC.

f = 1 / 2*pi*R*C - this is the cutoff frequency (ideal). I realize that practical filters do not cutoff exactly at this frequency, rather over a range of frequency.

3. The attempt at a solution

Input = 10sin(10t) + 1V = x(t)

So, I know that DC is 0 Hz.

Through my research, it seems that choosing the proper R and C values (close to the standard element values that you can buy from an electronics store), it would make sense that to block a 0 Hz signal, you "ideally" need to design the filter so the cutoff frequency is > 0 Hz. However, for practical and lab purposes, what frequency should I select as the 'cutoff frequency'? Do I select a random frequency above 0 Hz, with some room for leeway? I know that it will block it, but is there some sort of "rule" I must follow as to choosing the proper cutoff frequency?

As of now, I could say f = 1 / 2 * pi * R * C.

f = 5 Hz (Ill choose this)
C = 0.1uF
therefore, R = 318kOhm. The closest standard R value is 31.6kOhm.

Is this a good approach?

---

Also, is there a difference in using a simple series RC circuit or a differentiator high pass circuit? I know the gain is modified (inverted), but if I take care of that, is there any *real* advantage of using one?

My connections would be:

Vin -- C -- R -- Inverting Input
GND -- Noninverting Input
Output feedback --- Resistor --- Inverting Pin
and power supply for the OpAmp.

CEL

Any capacitor will block DC. The requirement for the cutoff frequency of your filter is that it should be much less then the frequency of the signal you want to keep, so you should ideally use 1 Hz for your cutoff frequency.
By the way, you should use rd/s instead of Hz in specifying your signal. It should be:
Input = 10sin(2\pi\times10t) + 1V = x(t)
and not as you wrote.