The phase response is just the argument of the transfer function (just as the magnitude response is the absolute value).

The argument of a quotient is the argument of the numerator minus the argument of the denominator, i.e.,

$$
\phi = \operatorname{arg} \frac{j\omega}{j\omega+\frac{1}{RC}} = \operatorname{arg}(j\omega) - \operatorname{arg}(j\omega+\frac{1}{RC}) = \frac{\pi}{2} - \operatorname{atan2}(\omega, 1/RC )
$$
which is essentially just what you wrote.

Note that I did not write \$\operatorname{tan}^{-1}(y/x)\$ but \$\operatorname{atan2(y,x)}\$ because the former is only correct if \$x\$ is positive. It is incorrect when \$x\$ is zero or negative. In your case here, when \$\omega\$ is always positive that makes no difference, but I think it is good practice to use atan2.

1st stage is a 1.9995 kHz bandpass filter using a standard multiple-feedback topology. It has a Q of about 20 and a gain of about 40 dB.

2nd stage is a 2.009 kHz BP filter with a Q of about 25 and a gain of about 40 dB.

Total gain of first stage is 40 dB which makes a 10nV 2 kHz signal into 1 microvolt.

Bandwidth of 1st stage is \$\dfrac{f_C}{Q}\$ = 100Hz and for the purposes of noise calculation you can assume the bandwidth is 1.6x greater at 160 Hz.

\$E_{NOISE}\$ from ADA4004 is 1.8nV /\$\sqrt{Hz}\$. This means noise in a 160 Hz bandwidth is 22.8 nV

This is bigger than your signal (10nV) therefore this isn't going to be a great design.

Even if you took into account the Q of the 2nd stage, the bandwidth would only be halved i.e. a half power point would become a quarter-power point. Noise voltage would be about 16 nV and still a significantly bigger value than your signal. This is made worse by the 2nd stage being 9.5 Hz different to the 1st stage.

## Best Answer

There are several mehods for calculating the closed-loop gain function of an opamp with feedback. I think, for the shown circuit the most simple way is as follows:

From H. Blacks famous formula - assuming ideal opamps with an open-loop gain of infinity - we know that the closed-loop gain can be written as

Acl=-Hf/Hr.with (applying the superposition rule):

Hf(forward function)=Vn/Viif Vout=0, andHr(return function, feedback factor)=Vn/Voutif Vi=0.(Vn:voltage at the negative input terminal),

Both functions Hf and Hr can be found by applying the well-known voltage divider rule.