Why FFT of a filtered step increases the lower values of the transform with respect to the unfiltered signal? I'm a bit new into DSP and when performing the FFT of a step filtered with a Notch filter I realized the frequency components arround the design filter frequency were attenuated (as expected), BUT the minimum frecuency increases arround tha design filter frequency. ¿WHY does this happen?
 These points are marked with an arrow and I believe it's just a matter of not fully understanding FFT. I would very much appreciate if someone could give me a hand with this.


I also include the time domain response of the filter (blue is the input, and red the output), and for more details about the filter it is defined as:  

$$
N(z) = \frac{z^2 - 1.939 z + 0.9469}{z^2 - 1.868 z + 0.8764}
$$

(with a lot more decimals for better accuracy).

With a sampling period of 200 microseconds
So taking your filter and looking at its magnitude response we see that the magnitude close to zero will increase but all other magnitudes (on the other side of the dip) should be approximately unity gain (no magnification / increase in amplitude).

The fact that the DC gain of the filter is 0.94 means that the steady-state output for a step input will be 0.94, not 1.  That may be the effect that you're seeing.

Also, the notch is not very deep. You may want to redesign the filter to make it attenuate more.