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 Here is the schematic

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    \$\begingroup\$ Please take a look at the appearance of your question. It is almost unreadable...can you fix it? \$\endgroup\$ Aug 4, 2018 at 13:53
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    \$\begingroup\$ I'm sure he'll be back LOL \$\endgroup\$
    – Andy aka
    Aug 4, 2018 at 14:00
  • \$\begingroup\$ okay fixed it!! hopefully it's better now \$\endgroup\$ Aug 4, 2018 at 14:14
  • \$\begingroup\$ BTW what sort of high-pass filter circuit is the TF for? \$\endgroup\$
    – Andy aka
    Aug 4, 2018 at 14:58
  • \$\begingroup\$ No one can tell you if the equations are correct since it is unclear what they apply to. You have to include a schematic. \$\endgroup\$ Aug 4, 2018 at 15:03

1 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).

  • \$\begingroup\$ Thank you for your answer, my question is why I can't just insert the frequencies and gains in the transfer function and solve \$R_1\$ and \$R_2\$ from there. I mean it's a function with two unknowns and I have two frequencies of which I know the value of the function. Something is obviously wrong, but I don't understand what it is \$\endgroup\$ Aug 4, 2018 at 16:35
  • \$\begingroup\$ You are assuming that the gain is the textbook piece-wise linear across the frequency, but it isn't, as shown in the picture above. There is a soft knee around the corner frequency which means you cannot simply consider a decade higher/lower frequency and deduce the gain. You were given the specific point at fc so that you know that the gain is, specifically, 1/sqrt(2) of the total gain. Because it's a highpass and the transfer function is asymptotic, the gain at infinity is the gain set by R2/R1. Only now you have two, solid equations from which you can deduce the rest. \$\endgroup\$ Aug 5, 2018 at 6:58

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