I'm trying to design an LC low pass filter to attenuate 30-35kHz noise on a 3.3V signal. It's preferable to attenuate higher frequencies as well. So I designed a filter with L=10u, C=1u (which is wrong, but this is just an example) - and I found it boosted frequencies near the cut-off point. Is there any way to avoid this? The boost of 22dB corresponded to an increase in 12.5x the voltage. The filter will be powering a microcontroller, which draws 100mA, so this increase in voltage would likely damage it.
In simple terms, add a resistor in series with the inductor. But then you could rather use a RC filter because you're decreasing the quality of your inductor.
Why do you prefer and LC filter? The filters are harder because at a certain frequency, you will get a resonance point. The voltage you put on the filter will be 'amplified'. The amplitude of how much it will shoot up , is called the Q factor. To dampen the Q , you add a series resistance to the coil to dampen it out.
A lot of math stored on: http://en.wikipedia.org/wiki/RLC_circuit
But because you're only operating at 30kHz, I think a RC filter might do the job better/easier.
Edit: Sorry your post was confusing because first you were talking about a 3.3V >signal< and then about a power line.
The answer still stays similar, add a small resistor, but it will reduce the steepness of the filter. Alternatively you can better add a filter which it's cutoffpoint is way less than your noise. I.e. if you have 30kHz noise, add a filter with a cutoff point at 3kHz or something. You won't get the increase in amplitude at the noisy frequency and a pretty good attenuation.
However, as mentioned, patching the source is always better.