Where do RBJ's analog shelving filters come from?
If you remove (for the time being) that leading factor $A$ as a constant gain factor:
$$H(s)=\frac{s^2+\left(\frac{\sqrt{A}}{Q}\right)s + A}{As^2 + \left(\frac{\sqrt{A}}{Q}\right)s + 1}$$
what you get then is a symmetric, but otherwise general shelf that could be equally described as &quot;LowShelf&quot; or &quot;HighShelf&quot;.  In dB, the gain at the low end is the negative of the dB gain at the high end.  And, in log gain (dB) vs. log frequency (octaves or decades), the gain function always has odd symmetry about the &quot;resonant frequency&quot; (more accurately described as the &quot;significant frequency&quot; which is the shelf mid-point frequency, a little different from the biquad resonant frequency), which in the normalized case, is $\Omega_0=1$.  At that shelf mid-point frequency this general low/high shelf has a gain of $1$ (or 0 dB).
$$\begin{align}
\Big|H(s)\Big|^2 &amp;= \left| \frac{s^2+\left(\frac{\sqrt{A}}{Q}\right)s + A}{As^2 + \left(\frac{\sqrt{A}}{Q}\right)s + 1} \right|^2 \\
\\
\Big|H(j\Omega)\Big|^2 &amp;= \left| \frac{(j\Omega)^2+\left(\frac{\sqrt{A}}{Q}\right)(j\Omega) + A}{A(j\Omega)^2 + \left(\frac{\sqrt{A}}{Q}\right)(j\Omega) + 1} \right|^2 \\
\\
 &amp;= \left| \frac{A-\Omega^2+j\frac{\sqrt{A}}{Q}\Omega}{1-A\Omega^2+j\frac{\sqrt{A}}{Q}\Omega} \right|^2 \\
\\
 &amp;= \frac{\Big|A-\Omega^2+j\frac{\sqrt{A}}{Q}\Omega \Big|^2}{\Big|1-A\Omega^2+j\frac{\sqrt{A}}{Q}\Omega\Big|^2}  \\
\\
 &amp;= \frac{(A-\Omega^2)^2+\left(\frac{\sqrt{A}}{Q}\Omega\right)^2}{(1-A\Omega^2)^2+\left(\frac{\sqrt{A}}{Q}\Omega\right)^2}  \\
\\
 &amp;= \frac{ A^2-2A\Omega^2+\Omega^4 + \frac{A}{Q^2} \Omega^2}{1-2A\Omega^2+A^2\Omega^4 + \frac{A}{Q^2} \Omega^2}  \\
\\
 &amp;= \frac{ A+\frac{1}{A}\Omega^4 - (2 - \frac{1}{Q^2}) \Omega^2}{\frac{1}{A}+A\Omega^4 - (2 -\frac{1}{Q^2}) \Omega^2}  \\
\\
 &amp;= \frac{ \frac{A}{\Omega^2}+\frac{\Omega^2}{A} - (2 - \frac{1}{Q^2})}{\frac{1}{A\Omega^2}+A\Omega^2 - (2 -\frac{1}{Q^2})}  \\
\end{align}$$
Remember that the &quot;significant frequency&quot; is set to one.  If you express the normalized frequency in terms of log frequency, then
$$ \Omega \triangleq 2^p $$
where $p$ is the log frequency, in octaves, centered at the &quot;significant frequency&quot;.
$$\begin{align}
 \Big|H(j2^p)\Big|^2 &amp;= \frac{ \frac{A}{2^{2p}}+\frac{2^{2p}}{A} - (2 - \frac{1}{Q^2})}{\frac{1}{A2^{2p}}+A2^{2p} - (2 -\frac{1}{Q^2})}  \\
\\
 &amp;= \frac{2^{2p-\log_2(A)} + 2^{-2p+\log_2(A)} - (2 - \frac{1}{Q^2})}{2^{2p+\log_2(A)} + 2^{-2p-\log_2(A)} - (2 - \frac{1}{Q^2})} \\
\\
 &amp;= \frac{2 \cosh\big(\log(2)(2p-\log_2(A))\big) - (2 - \frac{1}{Q^2})}{2 \cosh\big(\log(2)(2p+\log_2(A))\big) - (2 - \frac{1}{Q^2})} \\
\\
 &amp;= \frac{\cosh\big(2\log(2)p-\log(A)\big) - (1 - \frac{1}{2Q^2})}{\cosh\big(2\log(2)p+\log(A)\big) - (1 - \frac{1}{2Q^2})} \\
\end{align}$$
Now if we express this as log amplitude (let's say in dB) vs. log frequency (in octaves w.r.t. the &quot;significant frequency&quot;), this becomes
$$\begin{align}
\tfrac{10}{\log(10)} \log \left(\Big|H(j2^p)\Big|^2 \right) &amp;= \tfrac{10}{\log(10)} \log \left(\frac{\cosh\big(2\log(2)p-\log(A)\big) - (1-\frac{1}{2Q^2})}{\cosh\big(2\log(2)p+\log(A)\big) - (1-\frac{1}{2Q^2})}\right) \\
\\
&amp;= \tfrac{10}{\log(10)} \log \left(\cosh\big(2\log(2)p-\log(A)\big) - (1-\tfrac{1}{2Q^2})\right) \\
&amp; \qquad - \tfrac{10}{\log(10)} \log \left(\cosh\big(2\log(2)p+\log(A)\big) - (1-\tfrac{1}{2Q^2})\right) \\
\end{align}$$
Since $\cosh(\cdot)$ is an even-symmetry function, I think you can show that the function above is an odd-symmetry function in $p$.  Substitute $-p$ in for $p$ and you will see that the + and - terms swap.
$$\begin{align}
\frac{\mathrm{d}}{\mathrm{d}p} \tfrac{10}{\log(10)} \log \left(\Big|H(j2^p)\Big|^2 \right) &amp;= \tfrac{10}{\log(10)} \frac{\mathrm{d}}{\mathrm{d}p} \log \left(\cosh\big(2\log(2)p-\log(A)\big) - (1-\tfrac{1}{2Q^2})\right) \\
&amp; \qquad - \tfrac{10}{\log(10)} \frac{\mathrm{d}}{\mathrm{d}p} \log \left(\cosh\big(2\log(2)p+\log(A)\big) - (1-\tfrac{1}{2Q^2})\right) \\
\\
&amp;= \tfrac{10}{\log(10)}  \frac{2\log(2)\sinh\big(2\log(2)p-\log(A)\big)}{ \cosh\big(2\log(2)p-\log(A)\big) - (1-\tfrac{1}{2Q^2})} \\
&amp; \qquad - \tfrac{10}{\log(10)} \frac{2\log(2)\sinh\big(2\log(2)p+\log(A)\big)}{ \cosh\big(2\log(2)p+\log(A)\big) - (1-\tfrac{1}{2Q^2})} \\
\end{align}$$
When $p=0$ (at the midshelf) this shelf slope is:
$$\begin{align}
\frac{\mathrm{d}}{\mathrm{d}p} \tfrac{10}{\log(10)} \log \left(\Big|H(j2^p)\Big|^2 \right) \Bigg|_{p=0} &amp;= \tfrac{10}{\log(10)}  \frac{2\log(2)\sinh(-\log(A))}{ \cosh(-\log(A)) - (1-\tfrac{1}{2Q^2})} \\
&amp; \qquad - \tfrac{10}{\log(10)} \frac{2\log(2)\sinh(\log(A))}{\cosh(\log(A)) - (1-\tfrac{1}{2Q^2})} \\
\\
&amp;= - \tfrac{10}{\log(10)} \frac{4\log(2)\sinh(\log(A))}{ \cosh(\log(A)) - (1-\tfrac{1}{2Q^2})} \\
\\
&amp;= - \tfrac{10}{\log(10)} \frac{2\log(2) \left(A - \tfrac{1}{A} \right) }{ \tfrac{1}{2} \left(A + \tfrac{1}{A} \right) - (1-\tfrac{1}{2Q^2})} \\
\\
&amp;= - \tfrac{40\log(2)}{\log(10)} \ \frac{A^2 - 1}{A^2 + 1 - 2A(1-\tfrac{1}{2Q^2})} \\
\\
&amp;= - \tfrac{40\log(2)}{\log(10)} \ \frac{A^2 - 1}{A^2 - 2A + 1 + \tfrac{A}{Q^2}} \\
\\
&amp;= - \tfrac{40\log(2)}{\log(10)} \ \frac{(A-1)(A+1)}{(A-1)^2 + \tfrac{A}{Q^2}} \\
\\
&amp;= - \tfrac{40\log(2)}{\log(10)} \ \frac{(A-1)(A+1)}{(A-1)^2 + \tfrac{A}{Q^2}} \\
\end{align} $$
I'm still not done with this.  I'm doing this from memory.  I can't find my notes from a quarter century ago anywhere.