Yeah, that looks like it's going to be messy.

I agree with Olin: the problem starts with three variables and three conditions so, no matter how you look at it, you will end up with a system of equations.

However, you can take certain shortcuts (using your 1st picture):

$$\begin{align} R_{23}&=R_3||(R_2+R_o) \tag{1} \\ R_{13}&=R_3||(R_1+R_i) \tag{2} \\ R_i&=R_1+R_{23} \tag{3} \\ R_o&=R_2+R_{13} \tag{4} \end{align}$$

[edit]
[edit 2]
$R_{23}$ is the resistance at point X with the input disconnected, and $R_{13}$ is the rsistance at point X with the output disconnected.

Then, the input source always sees its $R_i$ in series with the equivalent $R_i$, while the output voltage is always $\frac{A}2$ over $R_o||R_o^{eq}$ ($A$ is the attenuation relative to a unity input). Thus:

$$\begin{align} I_1&=\dfrac{1}{2R_i} \tag{5} \\ I_2&=\dfrac{A}{2R_o} \tag{6} \\ I_3&=I_1-I_2=\dfrac12\left(\dfrac{1}{R_i}-\dfrac{A}{R_o}\right) \tag{7} \end{align}$$

Since the biggest unknown is $V_x$, calculate it with (3) in mind:

$$V_x=\dfrac12\dfrac{R_{23}}{R_{23}+R_1}=\dfrac12\dfrac{R_i-R_1}{R_i-R_1+R_1}=\dfrac12\dfrac{R_i-R_1}{R_i}=\dfrac12\left(1-\dfrac{R_1}{R_i}\right) \tag{8}$$

Now calculate $R_2$ and $R_3$ based on (6), (7), and (8), while considering that $V_1=\frac12$ and $V_2=\frac{A}{2}$:

$$\begin{align} R_2&=\dfrac{V_x-V_2}{I_2}=\left(\dfrac{2V_x}{A}-1\right)R_o=\left(\dfrac{1-\dfrac{R_1}{R_i}}{A}-1\right)R_o=\dfrac{\big((1-A)R_i-R_1\big)R_o}{AR_i} \tag{9} \\ R_3&=\dfrac{V_1-V_x}{I_1}=\dfrac{2V_xR_iR_o}{R_o-AR_i}=\dfrac{\left(1-\dfrac{R_1}{R_i}\right)R_iR_o}{R_o-AR_i}=\dfrac{(R_i-R_1)R_o}{R_o-AR_i} \tag{10} \end{align}$$

Since $R_i$, $A$, and $R_o$ are given, both (9) and (10) are dependent on $R_1$, only. Now, use (2) and (4) to derive the expression for $R_1$:

$$R_o\stackrel{(2,4)}{=}R_2+\dfrac{(R_1+R_i)R_3}{R_1+R_3+R_i} \tag{11}$$

After expanding and collecting the terms:

$$R_1=\dfrac{(R_o-R_2)R_i-(R_2+R_i-R_o)R_3}{R_2+R_3-R_o} \tag{12}$$

Substituting (9) and (10) in (12) takes a few lines of simplifications to give:

$$R_1=\dfrac{R_i^2\big(A^2-(2A-1)R_o\big)}{R_o-A^2R_i} \tag{I}$$

At this point, if you want to keep things simple(-ish) then use $(\text{I})$ for $R_1$ and then, sequentially, calculate (9) and (10) through substitution. Otherwise, you'll need a few more lines of simplifications to give:

$$\begin{align} R_2&=\dfrac{R_o^2+(A^2-2A)R_iR_o}{R_o-A^2R_i} \tag{II} \\ R_3&=\dfrac{2AR_iR_o}{R_o-A^2R_i} \tag{III} \
\end{align}$$

I find it interesting that all of them are divided by $R_o-A^2R_i$. Maybe there is something to it, hopefully a simplification but, now, my eyes are getting crossed by the amount of MathJax so, I'll take a break. I'll double check later on but, for now, it looks like I haven't made any mistakes.

A SPICE test with your values confirms both approaches (blue text is active, upper sources output the resistor values, k means $A$):

[edit 3]

This is mostly cosmetic, it certainly doesn't reduce the derivation (it actually adds to it) but, inspired by the formulas from this site (which do not account for correct attenuation), $R_3$ can be calculated as in ($\text{III}$) and then $R_1$ and $R_2$ calculated from (3) and (4), based on it, resulting in slightly more palatable equations:

$$\begin{align} R_1&=\sqrt{\dfrac{\bbox[5px,border:black solid 1px]{R_i}}{R_o}}\sqrt{R_iR_o+R_3^2}-R_3 \tag{IV} \\ R_2&=\sqrt{\dfrac{R_o}{\bbox[5px,border:black solid 1px]{R_i}}}\sqrt{R_iR_o+R_3^2}-R_3 \tag{V} \end{align}$$

posted about 2 years ago

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2y ago

a concerned citizen‭

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