Electronic – All Pass Filter

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.

The key point is that it is 0 in steady state. That is, as \$t\rightarrow\infty\$, \$V_{out}\rightarrow0\$. When you take the differential equation, you're looking at output voltage with respect to time.

Let's try taking a look at an example of this system. I'm going to simplify \$K=1\$ and \$RC=1\$. Then, we are left with: $$ \frac{V_{out}}{V_{in}}\left(j\omega\right)=T\left(j\omega\right)=\frac{1}{1-j/\omega} $$ To simplify our calculations, let's change the form of it and replace \$j\omega\$ with \$s\$. Then, we get: $$ T\left(s\right)=\frac{s}{s+1} $$ Taking the step response of this (that is, \$V_{out}\$ when the input \$V_{in}=u\left(t\right)\$ where \$u\left(t\right)\$ is the Heaviside step function) we get this:

Note that this is in time domain, and in steady state, our output voltage is zero. The bode plot comes out as you'd expect (first order high-pass filter):