Are these equations correct? (transfer function)

Question

I have a given transfer function \$ |H| = \frac{R_2}{\sqrt{R_1^2+\frac{1}{(\omega c)^2}}} \$ where \$ c = 1\mathrm{\mu F} \$ and \$ R_1,R_2 \$ are unknown. I know that \$ |H(j\omega_c)| = 50.1 \$ where \$ \omega_c = 500\mathrm{ rad/s} \$ and that the gain decreases with 20dB/decade for \$\omega<\omega_c\$. The way I've understood the 20dB/decade decrease is that for \$\omega=50\mathrm{rad/s}\$ we should have \$|H|_{dB}=14 \implies |H|=5.012\$ since \$50.1\approx 34\mathrm{dB}\$. By that reasoning I get the following two equations

$$ \frac{R_2}{\sqrt{R_1^2+\frac{1}{(500c)^2}}}=50.1\ $$ and $$ \frac{R_2}{\sqrt{R_1^2+\frac{1}{(50c)^2}}}=5.012\ $$

But they don't give me the right answers, what is wrong with this reasoning?

Here is the schematic

Answer 1

It's difficult to say, given the way you formulated your question, but since you are given the corner frequency \$\omega_c=500\$, then you know that the transfer function -- which is that of a highpass -- has \$\frac{1}{\sqrt{2}}\$ of the maximum magnitude, so then \$H(j\infty)=50.1\sqrt{2}\approx70.8521\$. This is the ratio of the two resistors. Now you have two equations of two unknowns:

$$\frac{R_2}{R_1}=70.8521$$ $$\frac{R_2}{\sqrt{R_1^2+\frac{1}{(500*1\mu)^2}}}=50.1$$

Which gives \$R_1\approx2\text{k}\Omega\$ and \$R_2\approx141.7042\text{k}\Omega\$. Here's a quick check with LTspice:

The voltage is taken after the second, inverting E source (for positive output).