Doubly terminated ladder for given T(s), R2 = 4 Ω

Given: $T(s) = \dfrac{1}{(s+1)(s^2+s+1)} = \dfrac{1}{s^3+2s^2+2s+1}$ (3rd order Butterworth shape), $R_1 = 1\ \Omega$, $R_2 = 4\ \Omega$.

Method. $t(s) = 2\sqrt{R_1/R_2}\,V_2/V_1$; find $|\rho|^2 = 1 - |t|^2$, then $Z_1 = R_1\dfrac{1+\rho}{1-\rho}$ and expand it into a ladder. The constant in $t(s)$ is limited by the terminations, as shown below.

Step 1: Transmission coefficient. Third order Butterworth (Table 1): $B(s) = (s+1)(s^2+s+1) = s^3+2s^2+2s+1$. At dc the inductors are shorts and capacitors open, so the ladder reduces to a divider: $V_2/V_1 = \frac{R_2}{R_1+R_2} = \frac{4}{5} = 0.8$. Hence the largest possible dc transmission is

$$|t(0)|^2 = \frac{4R_1R_2}{(R_1+R_2)^2} = \frac{4(1)(4)}{(1+4)^2} = \frac{16}{25} = 0.64$$

so $|t(j\omega)|^2 = \dfrac{0.64}{1+\omega^6}$, i.e. $t(s) = \dfrac{0.8}{s^3+2s^2+2s+1}$.

Step 2: Reflection coefficient.

$$|\rho(j\omega)|^2 = 1 - \frac{0.64}{1+\omega^6} = \frac{0.36 + \omega^6}{1+\omega^6} \;\Rightarrow\; \rho(s)\rho(-s) = \frac{0.36 - s^6}{1 - s^6}$$

The zeros satisfy $s^6 = 0.36$, i.e. they lie on a circle of radius $a = (0.36)^{1/6} = 0.8434$ at the Butterworth angles. Choosing the left-half-plane zeros (like the Butterworth poles, scaled by $a$):

$$\rho(s) = \frac{(s+a)(s^2+as+a^2)}{B(s)} = \frac{s^3 + 1.6869 s^2 + 1.4228 s + 0.6}{s^3+2s^2+2s+1}$$

Step 3: Input impedance ($R_1 = 1$):

$$Z_1(s) = R_1\frac{1+\rho(s)}{1-\rho(s)} = \frac{2s^3+3.6869s^2+3.4228s+1.6}{0.3131s^2+0.5772s+0.4}$$

Step 4: Continued fraction expansion (computed by repeated long division):

$$Z_1(s) = 6.387\,s + \cfrac{1}{0.3608\,s + \cfrac{1}{2.170\,s + 4}}$$

So series $L_1 = 6.387$ H, shunt $C_2 = 0.3608$ F, series $L_3 = 2.170$ H, and the remainder is the load $R_2 = 4\ \Omega$, as required (this choice of $\rho$ sign gives the correct $R_2$; the other sign ends in $1/4\ \Omega$ and is rejected).

Step 5: Circuit.

  R1=1  L1=6.387 H     L3=2.170 H
 o-/\/\---UUU----+----UUU---+
 |               |          |
(~) V1     C2=0.3608 F   R2=4
 |               |          |
 o---------------+----------+

Check: analysing this ladder gives $V_2/V_1 = \dfrac{0.8}{s^3+2s^2+2s+1}$, the required Butterworth response.

Answer: $R_1 = 1\ \Omega$, $L_1 = 6.387$ H, $C_2 = 0.3608$ F, $L_3 = 2.170$ H, $R_2 = 4\ \Omega$ (normalized to $\omega_c = 1$ rad/s).