# Are these equations correct? (transfer function)

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

• Please take a look at the appearance of your question. It is almost unreadable...can you fix it? – Elliot Alderson Aug 4 '18 at 13:53
• I'm sure he'll be back LOL – Andy aka Aug 4 '18 at 14:00
• okay fixed it!! hopefully it's better now – arnoldschwarzenegger Aug 4 '18 at 14:14
• BTW what sort of high-pass filter circuit is the TF for? – Andy aka Aug 4 '18 at 14:58
• No one can tell you if the equations are correct since it is unclear what they apply to. You have to include a schematic. – Bimpelrekkie Aug 4 '18 at 15:03

## 1 Answer

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).

• 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 – arnoldschwarzenegger Aug 4 '18 at 16:35
• 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. – a concerned citizen Aug 5 '18 at 6:58