Since you have an example of the signal on your scope, the best thing to do is capture the data and transfer it to a PC. Then use a tool like Matlab or Octave to simulate the effect of different filters.

You are looking for a filter, just defined in terms of poles (and maybe zeros) that minimizes the noise, without disturbing the desired features of the signal.

When you have a filter definition, then worry about how to build it.

If a single-pole filter is adequate, a simple RC circuit solves your problem.

For a two-pole filter, the Sallen-Key op-amp circuit is known for having relatively good tolerance for changes in the component values. An LC combination is also possible.

For higher-order filters (which I doubt you need), a cascade of Sallen-Key filters is preferable to a ladder of LC stages, because the op-amp provides buffering that prevents component value shifts in one stage from affecting the characteristics of other stages.

**Edit** In reply to your comment, I'm not a DSP guy, but here's how I'd work out the equivalent continuous time filter:

Your filter function in discrete time is

\$y_n = a x_n + (1-a) y_{n-1}\$

Given an impulse input, the time constant is the time it takes to decay to \$e^{-1}\$ of the value of \$y_0\$.

This is given by

\$(1-0.025)^n = e^{-1}\$

Solving this, *n* is about 39 samples, or 156 us.

So you want to choose R low enough that the input impedance of the ADC doesn't affect the filter performance much, then choose C to give RC = 156 us.

1) Neither, this C - L -C structure is a low pass filter, it blocks high frequency voltage variations from +15V_ISOL to reach the chip.

2) You could calculate a bandwidth for the filter consisting of 6.8 uH and 11.1 uF (the sum of all capacitors) formula: Fc = 1/(2pi*SQRT(L*C)) = 18.4 kHz
So at 18.4 kHz a signal would roughly be halved in amplitude.
How much the most left 10 uF cap filters depends on the impedance of the +15V line.

3) That depends on the expected frequency components on the +15V line, how sensitive the IC is for certain frequencies (it might supress low frequencies already by itself but might not be able to supress high frequencies). Available board space, cost of components, etc...

4) Several in parallel because a 10 uF cap is usually not so effective at higher frequencies, this is where the 0.1 uF comes in, it is more suited for high frequencies. Look at a datasheet of any capacitor and you will see it will only behave as a capacitor within a certain frequency range. By combining several caps, the effective frequency range is extended (the combination gives a better capacitor). The 10 uF at the left also gives some filtering.

5) for small value caps like 0.1uF a 25 V version might not be available or a 50 V version was already used elsewhere in the design. As long as the voltage rating is high enough any rating can be used. The choice depends on price, availability etc.

## Best Answer

1) the reason there is a 16KHz cut off: While our ears do have a hearing range of 20-22KHz the actual range that we use in our every day life for hearing is about 6-10KHz (at most)

If you are listening to music, one of the things to realize is that pitches, musical notes use a

logarithmicscale. so A4 is at 440Hz while A5 880 Hz A6 is at 1760 Hz what this means that there are only a few notes are left to be able to play above 16kHz. Just for reference, the highest note played on a piano would be 7.9kHz which is a Bhttp://peabody.sapp.org/class/st2/lab/notehz/

2) 100K ohm resistors are cheap and common, same-thing with the capacitor. its the most cost effective choice.