active-filterband passhigh pass filterinverting-amplifierlow pass
I need a band pass filter that only accepts frequencies from 1000 to 1400 Hz and then toggles the transistor. It has to work from .1 Vpp to 1 Vpp and rejects anything else. I have been testing this circuit in different ways and I keep getting a similar result.
In the circuit my source is .0005 Vpp I have changed it to .05 Vpp and only the dB value changed.
There are many types of low pass filter.
The one that gives your 'weird waveform' is the simplest 1st order RC filter. This passes the fundamental signal. However, it not only passes all the higher harmonics, but also gives them a phase shift.
The frequency response of the perfect lowpass filter shown in your image is actually not realisable as a physical filter. One problem is that the infinitely steep transition between passband and stopband requires an infinitely long impulse response, which means that you put a signal into the filter, and the delay is such that you never see it come out!
To this end, we approximate the filter with a shorter, usable impulse response. There are physical component filters, which must be causal (no output occurs before input), and signal processing filters, which can be acausal (in theory, output can occur before input, but any real system that uses them makes them causal by applying a delay).
One option for a second order all-pass filter with one OpAmp is the Delyiannis structure, as shown here (p.2):
The transfer function of this filter is given by
$$H(s)=\frac{R_4}{R_3+R_4}\cdot \frac{s^2-s\frac{2}{R_2C}+\frac{1}{R_1R_2C^2}}{s^2+s\frac{2}{R_2C}+\frac{1}{R_1R_2C^2}}\tag{1}$$
where
$$\frac{R_2R_3}{R_1R_4}=4\tag{2}$$
must be satisfied.
For a given center frequency \$\omega_0\$, a given quality factor \$Q\$, and a chosen value of \$C\$, the resistors \$R_1\$ and \$R_2\$ must be chosen as
$$R_1=\frac{1}{2Q\omega_0C},\quad R_2=\frac{2Q}{\omega_0C}\tag{3}$$
and \$R_3\$ and \$R_4\$ must be chosen to satisfy \$(2)\$. One possibility (as suggested in above document) is \$R_1=R_3\$ and \$R_4=R_2/4\$. (Note there is an error in the definition of \$R_4\$ in the document cited above).
From Eqs \$(2)\$ and \$(3)\$ it is straightforward to show that the filter gain is related to the quality factor \$Q\$ by
$$g=\frac{R_4}{R_3+R_4}=\frac{Q^2}{1+Q^2}\tag{4}$$
Best Answer
A single analog filter that accepts 0.1 Vpp at 1000 Hz, but rejects 1.0 V at 999 Hz would be extremely challenging to design. More so if it needed to work with real world components (that are not all exactly at their nominal values, drift over temperature, etc).
Consider breaking up the problem into two parts.
One circuit measures the signal frequency. Another circuit measures the signal amplitude.
Only if both circuits detect a "good" result is the output enabled.
A useful circuit to measure the frequency could be a frequency-to-voltage converter, available as an IC from various vendors, although you could also roll your own using established techniques.
A useful circuit to compare the output of this circuit with established limits is a window comparator.
You can use google to find numerous designs for either of these circuits.
Realistically, you will need to allow some tolerance in your specs. Accepting a 1000.000 Hz signal and rejecting a 999.999 Hz signal is not even theoretically possible if you don't allow your circuit at least on the order of 1000 s to make the frequency measurement, and the techniques that can do it use complex digital circuits, not analog ones.