Design of Resistively-Terminated Lossless Filters

Reflection coefficient

The reflection coefficient $\rho(s)$ of a resistively terminated lossless ladder compares the power reflected back towards the source with the maximum power the source can give. Looking into the ladder (input impedance $Z_1$) from a source of resistance $R_1$:

$$\rho(s) = \frac{R_1 - Z_1(s)}{R_1 + Z_1(s)}, \qquad |\rho(j\omega)|^2 = \frac{P_{\text{reflected}}}{P_{\max}}$$

It tells how well the ladder is matched to the source. $\rho = 0$ means perfect matching (all available power goes to the load); $|\rho| = 1$ means total reflection (no power reaches the load, i.e. stopband). It is related to the transmission coefficient by $|\rho|^2 + |t|^2 = 1$, and it is the key step in synthesis because $Z_1 = R_1\dfrac{1-\rho}{1+\rho}$ (or its inverse form) gives the input impedance to be expanded into a ladder.

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

Step 1: Low-pass prototype. Design the normalized 3rd order Butterworth LPF first. From Table 1, $B(s) = s^3+2s^2+2s+1$, so with $R_1 = R_2 = 1$:

$$t(s) = \frac{1}{B(s)}, \quad |\rho|^2 = 1-|t|^2 = \frac{\omega^6}{1+\omega^6}, \quad \rho(s) = \frac{s^3}{s^3+2s^2+2s+1}$$ $$Z_1 = \frac{1+\rho}{1-\rho} = \frac{2s^3+2s^2+2s+1}{2s^2+2s+1} = s + \cfrac{1}{2s + \cfrac{1}{s+1}}$$

Prototype: series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H, $R_2 = 1\ \Omega$.

Step 2: LP to HP transformation. Replace $s$ by $1/s$ ($\omega_c = 1$ rad/s):

• A series inductor $L$ (impedance $sL$) becomes $L/s$: a series capacitor $C = 1/L$.
• A shunt capacitor $C$ (admittance $sC$) becomes $C/s$: a shunt inductor $L = 1/C$.
• Resistors are unchanged.

Step 3: Circuit.

  R1=1    C1=1 F        C3=1 F
 o-/\/\----||----+----||----+
 |               |          |
(~) V1        L2=0.5 H    R2=1
 |               |          |
 o---------------+----------+

The response is $V_2/V_1 = \dfrac{0.5\,s^3}{s^3+2s^2+2s+1}$: zero gain at dc, $-3$ dB at 1 rad/s, and gain 0.5 (the maximum for equal terminations) at high frequency.

Answer (normalized, $\omega_c = 1$ rad/s): $R_1 = 1\ \Omega$, series $C_1 = 1$ F, shunt $L_2 = 0.5$ H, series $C_3 = 1$ F, $R_2 = 1\ \Omega$. For a cutoff $\omega_c$, divide each $L$ and $C$ by $\omega_c$ (and scale impedance if needed).

Method. For a doubly terminated lossless ladder, find the transmission coefficient $t(s)$, get the reflection coefficient from $|\rho|^2 = 1 - |t|^2$, form the input impedance $Z_1 = R_1\dfrac{1+\rho}{1-\rho}$, and expand it as a continued fraction (Cauer-I) to get the ladder elements. Here $t(s) = 2\sqrt{R_1/R_2}\,V_2/V_1$.

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).

Reflection coefficient

The reflection coefficient $\rho(s)$ of a resistively terminated lossless ladder compares the power reflected back towards the source with the maximum power the source can give. Looking into the ladder (input impedance $Z_1$) from a source of resistance $R_1$:

$$\rho(s) = \frac{R_1 - Z_1(s)}{R_1 + Z_1(s)}, \qquad |\rho(j\omega)|^2 = \frac{P_{\text{reflected}}}{P_{\max}}$$

$|\rho|$ lies between 0 (perfect match, all available power delivered) and 1 (total reflection, no power to the load). For a lossless ladder $|\rho|^2 + |t|^2 = 1$.

Design ($R_1 = 1\ \Omega$, $R_2 = 4\ \Omega$)

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).

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 $L' = LR/\omega_c$, $C' = C/(R\omega_c)$.

Transmission coefficient

The transmission coefficient $t(s)$ is the ratio (in power terms) of the power actually delivered to the load $R_2$ to the maximum available power of the source $P_{\max} = |V_1|^2/4R_1$:

$$|t(j\omega)|^2 = \frac{P_2}{P_{\max}} = \frac{4R_1}{R_2}\left|\frac{V_2(j\omega)}{V_1(j\omega)}\right|^2 , \qquad t(s) = 2\sqrt{\frac{R_1}{R_2}}\,\frac{V_2(s)}{V_1(s)}$$

Information obtained from it

• $|t(j\omega)|^2$ is the fraction of the available source power that reaches the load, so it directly gives the filter's magnitude response (passband, stopband, cutoff).
• $|t| = 1$ means maximum power transfer (perfect match); $|t| = 0$ means a transmission zero (no power to the load).
• Since $|t| \le 1$ for a passive lossless ladder, it fixes the maximum possible gain, e.g. $V_2/V_1 \le 1/2$ for equal terminations.
• Through $|\rho|^2 = 1 - |t|^2$ it gives the reflection coefficient, which is used to find $Z_1$ and synthesize the ladder.

Design of 2nd order Butterworth LPF ($R_1 = R_2 = 1\ \Omega$)

Step 1: Transmission coefficient. For $n = 2$ (Table 1), $B(s) = s^2 + \sqrt2 s + 1$. With equal terminations $|t(0)|$ can be 1:

$$t(s) = \frac{1}{s^2+\sqrt2 s+1}, \qquad |t(j\omega)|^2 = \frac{1}{1+\omega^4}$$

(The actual voltage gain is $V_2/V_1 = \tfrac12 t(s)$, since $t = 2\sqrt{R_1/R_2}\,V_2/V_1$.)

Step 2: Reflection coefficient.

$$|\rho(j\omega)|^2 = 1 - \frac{1}{1+\omega^4} = \frac{\omega^4}{1+\omega^4} \;\Rightarrow\; \rho(s)\rho(-s) = \frac{s^4}{B(s)B(-s)}, \quad \rho(s) = \frac{s^2}{s^2+\sqrt2 s+1}$$

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

$$Z_1(s) = \frac{1+\rho}{1-\rho} = \frac{2s^2+\sqrt2 s+1}{\sqrt2 s+1}$$

Step 4: Continued fraction.

$$\begin{aligned} Z_1 &= \sqrt2 s + \frac{1}{\sqrt2 s + 1} \\ &= 1.414\,s + \cfrac{1}{1.414\,s + 1} \end{aligned}$$

So series $L_1 = 1.414$ H, then a shunt $C_2 = 1.414$ F, and the last "1" is $R_2 = 1\ \Omega$.

Step 5: Circuit.

  R1=1  L1=1.414 H
 o-/\/\---UUU----+------+
 |               |      |
(~) V1     C2=1.414 F  R2=1
 |               |      |
 o---------------+------+

Answer: $R_1 = 1\ \Omega$, $L_1 = 1.414$ H, $C_2 = 1.414$ F, $R_2 = 1\ \Omega$ (normalized, $\omega_c = 1$ rad/s). The dual form $Z_1 = (1-\rho)/(1+\rho)$ gives shunt $C_1 = 1.414$ F and series $L_2 = 1.414$ H. These agree with the standard table values (1.414, 1.414).

Transmission coefficient

The transmission coefficient $t(s)$ is the ratio (in power terms) of the power actually delivered to the load $R_2$ to the maximum available power of the source $P_{\max} = |V_1|^2/4R_1$:

$$|t(j\omega)|^2 = \frac{P_2}{P_{\max}} = \frac{4R_1}{R_2}\left|\frac{V_2(j\omega)}{V_1(j\omega)}\right|^2 , \qquad t(s) = 2\sqrt{\frac{R_1}{R_2}}\,\frac{V_2(s)}{V_1(s)}$$

Information obtained: $|t|^2$ is the fraction of available power reaching the load, so it is the filter's power response. $|t| = 1$ means perfect matching and maximum power transfer; $|t| = 0$ marks a transmission zero; and it bounds the gain of a passive ladder ($|t| \le 1$).

Derivation of reflection coefficient

   R1        I1 -->
 o-/\/\----+-------------+------+
 |         |             |      |
(~) V1   Z1(s) ->  Lossless LC  R2  V2
 |         |      ladder  |      |
 o---------+-------------+------+

1. Maximum available power of the source (delivered to a matched load $R_1$):

$$P_{\max} = \frac{|V_1|^2}{4R_1}$$

2. Input power to the ladder with $Z_1 = R_{in} + jX_{in}$ and $I_1 = \dfrac{V_1}{R_1+Z_1}$:

$$P_1 = |I_1|^2 R_{in} = \frac{|V_1|^2\,\text{Re}(Z_1)}{|R_1+Z_1|^2}$$

3. Power not delivered ("reflected") is $P_{\max} - P_1$:

$$\begin{aligned} P_{\max} - P_1 &= \frac{|V_1|^2}{4R_1}\left[1 - \frac{4R_1\,\text{Re}(Z_1)}{|R_1+Z_1|^2}\right] \\ &= \frac{|V_1|^2}{4R_1}\cdot\frac{|R_1+Z_1|^2 - 4R_1\,\text{Re}(Z_1)}{|R_1+Z_1|^2} \end{aligned}$$

4. Using $|R_1+Z_1|^2 - 4R_1 R_{in} = (R_1+R_{in})^2 + X_{in}^2 - 4R_1R_{in} = (R_1-R_{in})^2 + X_{in}^2 = |R_1-Z_1|^2$:

$$\frac{P_{\max}-P_1}{P_{\max}} = \left|\frac{R_1 - Z_1}{R_1+Z_1}\right|^2$$

5. This ratio is defined as $|\rho(j\omega)|^2$, so the reflection coefficient is

$$\rho(s) = \pm\frac{R_1 - Z_1(s)}{R_1 + Z_1(s)}$$

6. Since the ladder is lossless, $P_1 = P_2$, so $|t|^2 = P_1/P_{\max}$ and

$$|\rho(j\omega)|^2 + |t(j\omega)|^2 = 1 \quad \text{(Feldtkeller equation)}$$

7. Solving for $Z_1$ gives the synthesis formula:

$$Z_1(s) = R_1\frac{1 \mp \rho(s)}{1 \pm \rho(s)}$$

Use: given $t(s)$, find $|\rho|^2 = 1-|t|^2$, factor $\rho(s)\rho(-s)$, pick $\rho(s)$ with a Hurwitz denominator, find $Z_1$ and expand it into an LC ladder.

Definitions

The transmission coefficient $t(s)$ is the ratio (in power terms) of the power actually delivered to the load $R_2$ to the maximum available power of the source $P_{\max} = |V_1|^2/4R_1$:

$$|t(j\omega)|^2 = \frac{P_2}{P_{\max}} = \frac{4R_1}{R_2}\left|\frac{V_2(j\omega)}{V_1(j\omega)}\right|^2 , \qquad t(s) = 2\sqrt{\frac{R_1}{R_2}}\,\frac{V_2(s)}{V_1(s)}$$

The reflection coefficient $\rho(s)$ of a resistively terminated lossless ladder compares the power reflected back towards the source with the maximum power the source can give. Looking into the ladder (input impedance $Z_1$) from a source of resistance $R_1$:

$$\rho(s) = \frac{R_1 - Z_1(s)}{R_1 + Z_1(s)}, \qquad |\rho(j\omega)|^2 = \frac{P_{\text{reflected}}}{P_{\max}}$$

Since the ladder is lossless, all power entering it reaches the load, so the two coefficients are linked by Feldtkeller's equation:

$$|\rho(j\omega)|^2 + |t(j\omega)|^2 = 1$$

Realizing finite transmission zeros in a terminated ladder

A transmission zero is a frequency where no signal reaches the load ($t = 0$, $|\rho| = 1$). All-pole filters (Butterworth, Chebyshev) have all zeros at $s = \infty$, made by series inductors and shunt capacitors. Filters such as inverse Chebyshev and elliptic (Cauer) also need finite zeros on the $j\omega$ axis. In a ladder a transmission zero is produced when either:

• a series arm becomes an open circuit: a parallel LC tank in the series arm, resonant at $\omega_z = 1/\sqrt{LC}$; or
• a shunt arm becomes a short circuit: a series LC branch from line to ground, resonant at $\omega_z$.

 series tank (open at wz)    shunt series-LC (short at wz)
 o--+--UUU--+--o             o-----+-----o
    |       |                      |
    +--||---+                     UUU L
                                   |
                                  === C
                                   |
 o-------------o             o-----+-----o

Procedure (zero shifting by partial removal):

1. From $t(s)$ find $\rho(s)$ and $Z_1(s) = R_1\dfrac{1+\rho}{1-\rho}$ as usual. The finite zeros of $t(s)$ are known, say $\pm j\omega_z$.
2. Expand $Z_1$ (or $Y_1$) step by step. Before each finite zero, partially remove a pole at infinity (a series $L$ or shunt $C$), choosing its value so that the remaining function has a zero exactly at $s = j\omega_z$: e.g. $C = Y(s)/s$ evaluated at $s^2 = -\omega_z^2$.
3. The reciprocal of the remainder now has a pole at $\pm j\omega_z$; remove it fully as a series tank (or shunt series-LC branch). This creates the transmission zero.
4. Continue with the next zero; the final remainder must equal the load $R_2$.

Example. For $y = s(s^2+4)/(s^2+1)$ with a required zero at $\omega = 3$: remove shunt $C = y/s|_{s^2=-9} = 5/8$ F, leaving $y' = 3s(s^2+9)/[8(s^2+1)]$, which is zero at $j3$. Then $1/y'$ has a pole at $j3$, removed as a series tank $C = 27/64$ F, $L = 64/243$ H that blocks the signal at $\omega = 3$ rad/s.

The same idea gives the elliptic (Cauer) ladder, where each series inductor has a capacitor in parallel to produce a finite stopband zero.

Significance of reflection coefficient

The reflection coefficient $\rho(s)$ of a resistively terminated lossless ladder compares the power reflected back towards the source with the maximum power the source can give. Looking into the ladder (input impedance $Z_1$) from a source of resistance $R_1$:

$$\rho(s) = \frac{R_1 - Z_1(s)}{R_1 + Z_1(s)}, \qquad |\rho(j\omega)|^2 = \frac{P_{\text{reflected}}}{P_{\max}}$$

• Matching: $\rho = 0$ means $Z_1 = R_1$ (perfect match, maximum power to the load); $|\rho| = 1$ means total reflection (stopband or transmission zero).
• Response: for a lossless ladder $|\rho|^2 = 1 - |t|^2$, so $\rho$ carries the same information as the magnitude response; small $|\rho|$ in the passband means low loss.
• Synthesis: it is the bridge from the transfer function to the circuit, because $Z_1 = R_1\dfrac{1\pm\rho}{1\mp\rho}$ is the impedance that is expanded into the LC ladder.
• Termination limits: with unequal terminations $|\rho(0)| \neq 0$, which shows that the maximum available power cannot be fully delivered at dc.

Derivation of the 3rd order Butterworth ladder ($R_1 = 1\ \Omega$, $R_2 = 4\ \Omega$)

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).

Information from reflection coefficient

$\rho(s) = \dfrac{R_1 - Z_1}{R_1 + Z_1}$ measures the mismatch between the source resistance and the ladder's input impedance. $|\rho(j\omega)|^2$ is the fraction of available power reflected back: $\rho = 0$ means perfect match and maximum power to the load (passband), $|\rho| = 1$ means all power reflected (stopband / transmission zero). Through $|\rho|^2 = 1 - |t|^2$ it gives the magnitude response, and through $Z_1 = R_1(1+\rho)/(1-\rho)$ it gives the impedance to be synthesized.

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 $L' = LR/\omega_c$, $C' = C/(R\omega_c)$.

Meaning of unity transmission coefficient

$|t(j\omega)| = 1$ means $P_2 = P_{\max}$: the maximum available power of the source is delivered to the load at that frequency. Then $|\rho| = 0$ (from $|\rho|^2 + |t|^2 = 1$), i.e. $Z_1(j\omega) = R_1$ and the ladder is perfectly matched with no reflection and no loss. For equal terminations this corresponds to the largest possible voltage gain $V_2/V_1 = 1/2$. In a Butterworth ladder $|t| = 1$ at $\omega = 0$.

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 $L' = LR/\omega_c$, $C' = C/(R\omega_c)$.