Zero reflection coefficient; 3rd order LPF equal termination

Meaning of zero reflection coefficient

If $\rho(j\omega) = 0$ at some frequency, then $Z_1(j\omega) = R_1$: the ladder input is perfectly matched to the source. No power is reflected, so the load receives the maximum available power $P_{\max} = |V_1|^2/4R_1$, and $|t(j\omega)| = 1$ (from $|\rho|^2 + |t|^2 = 1$). For a Butterworth ladder with equal terminations this happens at $\omega = 0$, which is why $\rho(s)$ has all its zeros at the origin ($\rho = s^3/B(s)$). In the passband $\rho$ is small; in the stopband $|\rho| \to 1$ (almost all power reflected).

Design of 3rd order Butterworth LPF ($R_1 = R_2 = 1\ \Omega$)

Step 1: Transfer function (Table 1). For $n = 3$, the Butterworth polynomial is $B(s) = (s+1)(s^2+s+1) = s^3+2s^2+2s+1$. With equal terminations the maximum possible $|t(0)| = 1$, so

$$t(s) = \frac{1}{s^3+2s^2+2s+1}, \qquad |t(j\omega)|^2 = \frac{1}{1+\omega^6}$$

Step 2: Reflection coefficient. From $|\rho|^2 = 1 - |t|^2$:

$$|\rho(j\omega)|^2 = \frac{\omega^6}{1+\omega^6} \;\Rightarrow\; \rho(s)\rho(-s) = \frac{-s^6}{1-s^6} = \frac{s^3(-s)^3}{B(s)B(-s)}$$

Keeping the Hurwitz denominator, $\rho(s) = \dfrac{s^3}{s^3+2s^2+2s+1}$.

Step 3: Input impedance. Using $Z_1 = R_1\dfrac{1+\rho}{1-\rho}$ with $R_1 = 1$:

$$Z_1(s) = \frac{2s^3+2s^2+2s+1}{2s^2+2s+1}$$

Step 4: Continued fraction (Cauer-I) expansion.

$$\begin{aligned} Z_1 &= s + \frac{s+1}{2s^2+2s+1} \\ \frac{2s^2+2s+1}{s+1} &= 2s + \frac{1}{s+1} \end{aligned} \;\Rightarrow\; Z_1 = s + \cfrac{1}{2s + \cfrac{1}{s+1}}$$

So: series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H, and the last term "1" is the load $R_2 = 1\ \Omega$ (check: equals the required termination).

Step 5: Circuit.

  R1=1    L1=1 H        L3=1 H
 o-/\/\---UUU----+----UUU---+
 |               |          |
(~) V1        C2=2 F      R2=1
 |               |          |
 o---------------+----------+

Answer: $R_1 = 1\ \Omega$, $L_1 = 1$ H, $C_2 = 2$ F, $L_3 = 1$ H, $R_2 = 1\ \Omega$ (normalized, $\omega_c = 1$ rad/s). Taking $Z_1 = R_1(1-\rho)/(1+\rho)$ instead gives the dual circuit: shunt $C_1 = 1$ F, series $L_2 = 2$ H, shunt $C_3 = 1$ F. For a real cutoff $\omega_c$ and resistance $R$, scale as $L' = LR/\omega_c$, $C' = C/(R\omega_c)$.