# High pass filter doesn't get back original signal?

If you simulate this (from https://www.circuitlab.com/editor/#?id=3p73kzybb5r8):

simulate this circuit – Schematic created using CircuitLab

You see this:

The input is a combination of the original signal (sine wave, 1000Hz) and noise (sine wave, 500Hz). There's a 530hz high pass filter designed in that circuit. NODE1 is the output after the active high pass. As you can see it is not the original signal (sine wave, 1000Hz). Why is that? Logically the noise component should go away right?

• What is the attenuation of your filter at 500Hz? Jul 7 at 4:22

## 2 Answers

If you perform a frequency (AC) analysis on this design, you'll find this plot of gain between IN and OUT:

At 500Hz, gain is higher than at 1kHz. That is, your noise is amplified more than the signal!

At 500Hz gain is $$\19.5dBV=\times 9.4\$$. At 1kHz gain is $$\18.5dBV=\times 8.4\$$. Clearly, what you thought was a high pass filter with a cut-off at 530Hz, isn't that at all.

R2 and C2 form a low pass element, and then the combination of OA, R3 and C3 form another low pass filter. There is nothing high-pass about this design, except for C1 & R1, which are responsible for the attenuation below 100Hz.

I believe what you intended was a Sallen-Key filter. A high pass version, would look like the following. Note the positions of resistances R1, R2 and capacitances C1 and C2, which tells you if this is a high or low pass filter:

simulate this circuit – Schematic created using CircuitLab

The gain frequency response of this section, between IN and OUT is:

This is clearly high-pass, since gain drops off at low frequencies. At 500Hz gain is $$\-5.76dBV=\times 0.515\$$, and at 1kHz it's $$\-1.82dBV=\times 0.81\$$.

As you can see, these filters don't have a sudden "cut-off" frequency, below (or above) which frequency components are snuffed out completely. Rather, the response is a gradual "roll-off" (at a rate of 20dB per factor of ten change in frequency, for this 2nd order Sallen-Key design).

In other words, while you may be able to significantly attenuate unwanted frequency components of some signal, you cannot eliminate them completely using these kinds of filters. There are different designs which could perform better in this respect, but yours and mine here aren't that good.

The two frequencies you are trying to distinguish are different only by a factor of two, which means that roll-off of 20dB per decade just isn't steep enough. For such close frequencies you require something that behaves like the fabled "brick-wall" filter, with a vertical roll-off in between them. Such a creature is mythical, akin to dragons, but you can get close with digital (DSP) techniques, or by cascading many 1st or 2nd order filters one after the other, to obtain a higher order response. But there be dragons, too, doing this always has consequences you may not have anticipated.

A waveshape may be decomposed into the sum of a number of sine waves, Fourier analysis. Since the theoretical high-pass did not block all of the "noise", some of it leaked through and you see the sum of 1,00Hz and it's sub-harmonic, 500 Hz.

Practical filters have a roll-off, or gradual diminution of amplitude outside the design frequency. Fiore states a filter can "approach 6 dB per octave per pole," where "pole" refers to a passive R-C pair. This also may be referred to as a first-order filter.

An active filter can have a sharper cutoff by using feedback.