Electronic – Designing a Band Pass Filter for Project

If it passes DC it's not a high pass filter.

What you probably want is a high pass filter (or a bandpass filter centred on 36MHz) and a separate DC source, which are combined at the ADC input.

Your filter is probably designed to deliver signal into a typical load impedance (50 ohms, 330 ohms) which is usually orders of magnitude lower than the ADC input impedance, and that load is typically modelled as a simple resistor from the output to ground. See the first circuit below (replace the capacitor with your actual HPF)

^{simulate this circuit – Schematic created using CircuitLab}

In the second circuit, you simply return that resistance to your (clean, low noise) DC voltage of choice, that combines your HPF and DC signals correctly. And as your HPF blocks DC, this does not affect the DC voltages in earlier stages of the circuit. (C3 is a decoupling capacitor, to keep the DC supply clean and noise free)

If you use this circuit on a regular basis I guess it would be worth to calculate its transfer function once and for all, so you can just easily evaluate it given the concrete component values. Doing the math you get (assuming an ideal OP)

$$H(\omega)=\frac{j\omega R_1C_1}{1+j\omega R_1C_1}\left(1+\frac{j\omega R_4C_2}{1+j\omega R_2C_2}\right)\tag{1}$$

A plot of (1) in dB with the specified component values looks like this (|H|/db vs f in Hz):

from which you can see that the -3dB point is at about \$250\,\text{Hz}\$.

If you further assume that

$$R_1C_1=R_2C_2=\tau$$

and if you denote the gain at large frequencies by

$$g=1+\frac{R_4}{R_2}$$

then with a bit more math you get this exact expression for the 3dB cut-off frequency in radians

$$\omega_c=\frac{1}{\tau}\sqrt{1-\frac{1}{g^2}+\sqrt{\left(1-\frac{1}{g^2}\right)^2+1}}\tag{2}$$

which with the given component values gives

$$\omega_c=2\pi\cdot 247.28$$

For a large gain \$g\gg 1\$, formula (2) is closely approximated by

$$\omega_c=\frac{1}{\tau}\sqrt{1+\sqrt{2}}\approx\frac{1.55}{\tau}\tag{3}$$