Post History

Here is another take at that problem.

First let's define some quantities that will be useful later:

\[
\begin{align}
    A &= \frac {V_2} {V_1}
    \qquad\qquad
    B = \frac {R_L} {R_S}
    \qquad\qquad
    r_1 = \frac {R_1} {R_L}
    \qquad\qquad
    r_2 = \frac {R_2} {R_L}
    \qquad\qquad
    r_3 = \frac {R_3} {R_L}
\end{align}
\]

Since the attenuator port 1 has an input impedance that matches
the source, then the source voltage is split in two at the input.
Moreover, we also apply the definition of attenuation $A$.

\[
\begin{align}
    V_1 &= \frac{V_S}{2}
    \qquad\qquad
    V_2 = A V_1 = A \frac{V_S}{2}
\end{align}
\]

We can therefore compute the currents into the ports as:

\[
\begin{align}
    I_1 &= \frac{V_S}{2R_S}
    \\[2 em]
    I_2 &= -\frac{V_2}{R_L} = -\frac{A}{R_L} \frac{V_S}{2}
\end{align}
\]

Now let's focus on the internal node, the junction between
$R_1,R_2, R_3$ (let's call it node 3). You can use KCL to compute
the current in $R_3$ and then use Ohm's law to obtain $V_3$:

\[
\begin{align}
    I_3 &= I_1 + I_2
    = \frac{V_S}{2R_S} - \frac{A}{R_L} \frac{V_S}{2}
    = \biggl(\frac{1}{R_S} - \frac{A}{R_L} \biggr)  \frac{V_S}{2}
    \\[2 em]
    V_3 &= R_3 I_3
    = R_3 \biggl(\frac{1}{R_S} - \frac{A}{R_L} \biggr)  \frac{V_S}{2}
    = \frac{R_3}{R_L} \biggl(\frac{R_L}{R_S} - A \biggr)  \frac{V_S}{2}
    = r_3 \biggl(B - A \biggr)  \frac{V_S}{2}
\end{align}
\]

Now we can compute $V_3$ also using KVL applied to the input and
output loops, obtaining two more independent equations for $V_3$:

\[
\begin{align}
    V_3 &= V_1 - R_1 I_1
    = \frac{V_S}{2} - R_1 \frac{V_S}{2 R_S}
    = \biggl(1 - \frac{R_1}{R_S} \biggr) \frac{V_S}{2}
    = \biggl(1 - \frac{R_L}{R_S} \cdot \frac{R_1}{R_L} \biggr) \frac{V_S}{2}
    = \biggl(1 - B \cdot r_1 \biggr) \frac{V_S}{2}
    \\[2 em]
    V_3 &= V_2 - R_2 I_2
    = \frac{A V_S}{2} - R_2 \biggl(-\frac{A}{R_L} \frac{V_S}{2} \biggr)
    = A \biggl(1 + \frac{R_2}{R_L} \biggr) \frac{V_S}{2}
    = A \biggl(1 + r_2 \biggr) \frac{V_S}{2}
\end{align}
\]

Now comparing equation 7 with equations 8 and 9, respectively, we
obtain two independent equations in which the only unknown are
$r_1,r_2,r_3$:

\[
\begin{align}
    \left\{
        \begin{aligned}
             r_3 \biggl(B - A \biggr)  \frac{V_S}{2}
             =
             \biggl(1 - B \cdot r_1 \biggr) \frac{V_S}{2}
             \\[2 em]
             r_3 \biggl(B - A \biggr)  \frac{V_S}{2}
             =
            A \biggl(1 + r_2 \biggr) \frac{V_S}{2}
        \end{aligned}
    \right.
    \qquad\qquad &\Leftrightarrow \qquad\qquad
    \left\{
        \begin{aligned}
             r_3 \biggl(B - A \biggr) = \biggl(1 - B \cdot r_1 \biggr)
             \\[2 em]
             r_3 \biggl(B - A \biggr) = A \biggl(1 + r_2 \biggr)
        \end{aligned}
    \right.
    \qquad\qquad \Leftrightarrow \qquad\qquad
    \nonumber\\[2 em]
    \qquad\qquad &\Leftrightarrow \qquad\qquad
    \left\{
        \begin{aligned}
             r_1 = \frac{1 - r_3 \biggl(B - A \biggr)}{B}
             \\[2 em]
             r_2 = \frac{r_3 \biggl(B - A \biggr) - A}{A}
        \end{aligned}
    \right.
\end{align}
\]

Now let's focus on port 2. Looking into it we see an impedance
$R_L$ which can be calculated using the usual methods of
Thevenin's theorem:


\[
\begin{gather}
    R_L
    = R_2 + R_3 \parallel (R_1 + R_S)
    = R_2 + \frac{1}{ \frac{1}{R_3} + \frac{1}{R_1 + R_S} }
    \qquad \Leftrightarrow
    \\[2 em]
    \quad \Leftrightarrow \quad
    R_L - R_2
    = \frac{1}{ \frac{1}{R_3} + \frac{1}{R_1 + R_S} }
    \quad \Leftrightarrow \quad
    \frac{1}{ R_L - R_2 }
    = \frac{1}{R_3} + \frac{1}{R_1 + R_S}
    \quad \Leftrightarrow \quad
    \frac{1}{ 1 - \frac{R_2}{R_L} }
    = \frac{R_L}{R_3} + \frac{1}{\frac{R_1}{R_L} + \frac{R_S}{R_L}}
    \quad \Leftrightarrow
    \nonumber \\[2 em]
    \qquad \Leftrightarrow \qquad
    \frac{1}{ 1 - r_2 }
    = \frac{1}{r_3} + \frac{1}{r_1 + \frac{1}{B}}
\end{gather}
\]

Now let's rewrite equations (9) like this:

\[
\begin{gather}
    \left\{
        \begin{aligned}
             r_1 = \frac{1 - r_3 \biggl(B - A \biggr)}{B}
             \\[2 em]
             r_2 = \frac{r_3 \biggl(B - A \biggr) - A}{A}
        \end{aligned}
    \right.
    \qquad \Leftrightarrow \qquad
    \left\{
        \begin{aligned}
             r_1 + \frac{1}{B} = \frac{2 + r_3 \biggl(A - B \biggr)}{B}
             \\[2 em]
             1 - r_2 = \frac{2A - r_3 \biggl(B - A \biggr)}{A}
        \end{aligned}
    \right.
    \qquad \Leftrightarrow \qquad
    \left\{
        \begin{aligned}
             \frac{1}{ r_1 + \frac{1}{B} } = \frac{B}{2 + r_3 \biggl(A - B \biggr)}
             \\[2 em]
             \frac{1}{ 1 - r_2 } = \frac{A}{2A + r_3 \biggl(A - B \biggr)}
        \end{aligned}
    \right.
\end{gather}
\]

substituting (12) into (11) we get:

\[
\begin{gather}
    \frac{1}{ 1 - r_2 }
    = \frac{1}{r_3} + \frac{1}{r_1 + \frac{1}{B}}
    \qquad \Leftrightarrow \qquad
    \frac{A}{2A + r_3 \biggl(A - B \biggr)}
    = \frac{1}{r_3} + \frac{B}{2 + r_3 \biggl(A - B \biggr)}
\end{gather}
\]

Now we can solve (13) for $r_3$.


\[
\begin{gather}
     \frac{A}{2A + r_3 \biggl(A - B \biggr)}
    = \frac{1}{r_3} + \frac{B}{2 + r_3 \biggl(A - B \biggr)}
    \qquad
    \Leftrightarrow
    \qquad
     \frac{A}{2A + r_3 \biggl(A - B \biggr)}
    = \frac{2 + r_3 \biggl(A - B \biggr) + B r_3}{r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr]}
    \nonumber \\[2 em]
    \Leftrightarrow
    \qquad
     \frac{A}{2A + r_3 \biggl(A - B \biggr)}
    = \frac{2 + A r_3 }{r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr]}
    \qquad
    \Leftrightarrow
    \nonumber \\[2 em]
    \qquad
    A r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr]
     = (2 + A r_3 ) \biggl[ 2A + r_3 \biggl(A - B \biggr) \biggr]
    \qquad
    \Leftrightarrow
    \nonumber \\[2 em]
    \qquad
    2A r_3 + A r_3^2 \biggl(A - B \biggr)
     = 4A + 2r_3 \biggl(A - B \biggr) + 2A^2 r_3 + A r_3^2 \biggl(A - B \biggr)
    \nonumber \\[2 em]
    \qquad
    0
    = 4A - 2Br_3 + 2A^2 r_3
    = 4A + 2 r_3 (A^2 - B )
    \qquad
    \Leftrightarrow
    \qquad
    r_3 (B - A^2 ) = 2A
    \nonumber \\[2 em]
    r_3 = \frac {2A}{ B - A^2 }
\end{gather}
\]

And then substitute $r_3$ back into equations (12) to get $r_1,r_2$:

\[
\begin{gather}
    \left\{
        \begin{aligned}
             r_1
             &= \frac{1 - r_3 \biggl(B - A \biggr)}{B}
             = \frac{1 - \frac {2A}{ B - A^2 } \biggl(B - A \biggr)}{B}
             = \frac{B - A^2 - 2A \biggl(B - A \biggr)}{B (B - A^2)}
             \\[2 em]
             r_2
             &= \frac{r_3 \biggl(B - A \biggr) - A}{A}
             = \frac{ \frac {2A}{ B - A^2 } \biggl(B - A \biggr) - A}{A}
             = \frac{ 2A \biggl(B - A \biggr) - A (B - A^2)}{A (B - A^2)}
        \end{aligned}
    \right.
    \qquad \Leftrightarrow
    \nonumber \\[2 em]
    \Leftrightarrow \qquad
    \left\{
        \begin{aligned}
             r_1
             &= \frac{B - A^2 - 2A \biggl(B - A \biggr)}{B (B - A^2)}
             = \frac{B + A^2 - 2AB}{B (B - A^2)}
             = \frac{B ( 1 + A^2/B - 2A)}{B (B - A^2)}
             \\[2 em]
             r_2
             &= \frac{ 2A \biggl(B - A \biggr) - A (B - A^2)}{A (B - A^2)}
             = \frac{ AB + A^3 - 2A^2}{A (B - A^2)}
             = \frac{ A (B + A^2 - 2A)}{A (B - A^2)}
        \end{aligned}
    \right.
    \qquad \Leftrightarrow
    \nonumber \\[2 em]
    \Leftrightarrow \qquad
    \left\{
        \begin{aligned}
             r_1
             &= \frac{B ( 1 + A^2/B - 2A)}{B (B - A^2)}
             = \frac{ 1 + A^2/B - 2A}{B - A^2}
             \\[2 em]
             r_2
             &= \frac{ A (B + A^2 - 2A)}{A (B - A^2)}
             = \frac{ B + A^2 - 2A}{B - A^2}
        \end{aligned}
    \right.
\end{gather}
\]

Therefore we end up with our solution for $r_1,r_2,r_3$, from
which you can get $R_1,R_2,R_3$:

\[
\begin{gather}
    \left\{
        \begin{aligned}
            r_1 &=  \frac{ 1 + A^2/B - 2A}{B - A^2}
            \\[2 em]
            r_2 &= \frac{ B + A^2 - 2A}{B - A^2}
            \\[2 em]
            r_3 &= \frac {2A}{ B - A^2 }
        \end{aligned}
    \right.
    \qquad
    \Leftrightarrow
    \qquad
    \left\{
        \begin{aligned}
            R_1 &= R_L \cdot \frac{ 1 + A^2/B - 2A}{B - A^2}
            \\[2 em]
            R_2 &= R_L \cdot \frac{ B + A^2 - 2A}{B - A^2}
            \\[2 em]
            R_3 &= R_L \cdot \frac {2A}{ B - A^2 }
        \end{aligned}
    \right.
\end{gather}
\]

which matches your solutions.