Properties and Synthesis of Passive Networks

Zeros of transmission of a two-port network are the values of complex frequency $s$ at which the transfer function ($z_{21}$, $y_{21}$, $V_2/V_1$ or $I_2/I_1$) is zero, i.e. a finite input produces no output at that frequency. For an LC ladder they lie on the $j\omega$ axis (including $s=0$ and $s=\infty$).

Realization of transmission zeros

In a ladder, a signal is blocked when a series arm is an open circuit (its impedance has a pole) or a shunt arm is a short circuit (its impedance has a zero, i.e. its admittance has a pole). So the transmission zeros of a ladder are the poles of the series-arm impedances and the poles of the shunt-arm admittances.

 zero at w1=1/sqrt(L1C1)   zero at w2=1/sqrt(L2C2)
o--[L1||C1]--+---------o
             |
            [L2]
            [C2]
             |
o------------+---------o

Zeros at $0$ and $\infty$ are obtained by complete removal of poles of $Z$ or $Y$ at $0$ or $\infty$ (Cauer-type steps). A zero at a finite frequency $\omega_z$ is obtained by zero shifting (partial pole removal) followed by complete removal of the pole at $\pm j\omega_z$.

Example. Realize $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$ as a ladder with a transmission zero at $\omega = 2$ rad/s.

The zeros of $Z$ are at $\omega = 1$ and $1.732$; we need one at $\omega=2$, so remove part of the pole at $\infty$: $Z_1(s) = Z(s) - k s$ with $Z_1(j2) = 0$.

$$\begin{aligned} Z(j2) &= \frac{(-4+1)(-4+3)}{j2(-4+2)} = \frac{3}{-j4} = j0.75 \\ Z_1(j2) &= j0.75 - j2k = 0 \;\Rightarrow\; k = \tfrac{3}{8} \\ Z_1(s) &= Z(s) - \tfrac{3}{8}s = \frac{(s^2+4)(5s^2+6)}{8s(s^2+2)} \end{aligned}$$

So series $L_1 = 3/8$ H, and $Z_1$ now has a zero at $s=\pm j2$ (the other zero moved from 1 to 1.095 rad/s). Its reciprocal has a pole at $\pm j2$, removed completely as a shunt branch:

$$\begin{aligned} Y_1(s) &= \frac{8s(s^2+2)}{(s^2+4)(5s^2+6)} = \frac{\tfrac{8}{7}s}{s^2+4} + \frac{16s}{7(5s^2+6)} \\ \tfrac{8}{7}\tfrac{s}{s^2+4} &\Rightarrow L_2 = \tfrac{7}{8}\ \text{H},\ C_2 = \tfrac{2}{7}\ \text{F}\quad (1/\sqrt{L_2C_2} = 2) \\ Z_2(s) &= \frac{7(5s^2+6)}{16s} = \tfrac{35}{16}s + \frac{21}{8s} \Rightarrow L_3 = \tfrac{35}{16}\ \text{H},\ C_3 = \tfrac{8}{21}\ \text{F} \end{aligned}$$

o--[L1=3/8H]--+--[L3=35/16H]--[C3=8/21F]--+
              |                           |
          [L2=7/8H]                       |
          [C2=2/7F]                       |
              |                           |
o-------------+---------------------------+

The shunt branch $L_2$–$C_2$ shorts the signal at $\omega=2$ rad/s, which is the required transmission zero. (Check: rebuilding the ladder gives back the original $Z(s)$.)

Zeros of transmission of a two-port network are the values of complex frequency $s$ at which the transfer function ($z_{21}$, $y_{21}$, $V_2/V_1$ or $I_2/I_1$) is zero, i.e. a finite input produces no output at that frequency. For an LC ladder they lie on the $j\omega$ axis (including $s=0$ and $s=\infty$).

In a ladder, a zero of transmission appears when a series arm has a pole of impedance (open circuit) or a shunt arm has a zero of impedance (short circuit), e.g. a series $L$ gives a zero at $\infty$, a shunt series-LC branch gives a zero at $1/\sqrt{LC}$.

Zero shifting by partial removal of pole

Zero shifting by partial removal of a pole. When a driving-point function is expanded into a ladder, removing a pole completely (its full residue) produces transmission zeros only at $s=0$ or $s=\infty$. To place a zero at a chosen finite frequency $\omega_z$, we remove only part of a pole (at $\infty$ or at $0$) so that the remainder becomes zero at $s = \pm j\omega_z$. The reciprocal of the remainder then has a pole at $\pm j\omega_z$, which is removed completely as a shunt series-LC branch (or a series tank), giving a transmission zero exactly at $\omega_z$.

Key facts:

• Partial removal of a pole shifts the zeros of the remaining function towards the pole removed (removing part of the pole at $\infty$ moves zeros up; part of the pole at 0 moves zeros down).
• Poles of the remainder do not move; only zeros shift.
• The part removed is still a realizable element (e.g. a smaller series inductor), and the remainder stays LC because $k < k_\infty$.

Example. Realize $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$ as a ladder with a transmission zero at $\omega = 2$ rad/s.

The zeros of $Z$ are at $\omega = 1$ and $1.732$; we need one at $\omega=2$, so remove part of the pole at $\infty$: $Z_1(s) = Z(s) - k s$ with $Z_1(j2) = 0$.

$$\begin{aligned} Z(j2) &= \frac{(-4+1)(-4+3)}{j2(-4+2)} = \frac{3}{-j4} = j0.75 \\ Z_1(j2) &= j0.75 - j2k = 0 \;\Rightarrow\; k = \tfrac{3}{8} \\ Z_1(s) &= Z(s) - \tfrac{3}{8}s = \frac{(s^2+4)(5s^2+6)}{8s(s^2+2)} \end{aligned}$$

So series $L_1 = 3/8$ H, and $Z_1$ now has a zero at $s=\pm j2$ (the other zero moved from 1 to 1.095 rad/s). Its reciprocal has a pole at $\pm j2$, removed completely as a shunt branch:

$$\begin{aligned} Y_1(s) &= \frac{8s(s^2+2)}{(s^2+4)(5s^2+6)} = \frac{\tfrac{8}{7}s}{s^2+4} + \frac{16s}{7(5s^2+6)} \\ \tfrac{8}{7}\tfrac{s}{s^2+4} &\Rightarrow L_2 = \tfrac{7}{8}\ \text{H},\ C_2 = \tfrac{2}{7}\ \text{F}\quad (1/\sqrt{L_2C_2} = 2) \\ Z_2(s) &= \frac{7(5s^2+6)}{16s} = \tfrac{35}{16}s + \frac{21}{8s} \Rightarrow L_3 = \tfrac{35}{16}\ \text{H},\ C_3 = \tfrac{8}{21}\ \text{F} \end{aligned}$$

o--[L1=3/8H]--+--[L3=35/16H]--[C3=8/21F]--+
              |                           |
          [L2=7/8H]                       |
          [C2=2/7F]                       |
              |                           |
o-------------+---------------------------+

The shunt branch $L_2$–$C_2$ shorts the signal at $\omega=2$ rad/s, which is the required transmission zero. (Check: rebuilding the ladder gives back the original $Z(s)$.)

Zero shifting by partial removal of a pole. When a driving-point function is expanded into a ladder, removing a pole completely (its full residue) produces transmission zeros only at $s=0$ or $s=\infty$. To place a zero at a chosen finite frequency $\omega_z$, we remove only part of a pole (at $\infty$ or at $0$) so that the remainder becomes zero at $s = \pm j\omega_z$. The reciprocal of the remainder then has a pole at $\pm j\omega_z$, which is removed completely as a shunt series-LC branch (or a series tank), giving a transmission zero exactly at $\omega_z$.

Key facts:

• Partial removal of a pole shifts the zeros of the remaining function towards the pole removed (removing part of the pole at $\infty$ moves zeros up; part of the pole at 0 moves zeros down).
• Poles of the remainder do not move; only zeros shift.
• The part removed is still a realizable element (e.g. a smaller series inductor), and the remainder stays LC because $k < k_\infty$.

Example. Realize $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$ as a ladder with a transmission zero at $\omega = 2$ rad/s.

The zeros of $Z$ are at $\omega = 1$ and $1.732$; we need one at $\omega=2$, so remove part of the pole at $\infty$: $Z_1(s) = Z(s) - k s$ with $Z_1(j2) = 0$.

$$\begin{aligned} Z(j2) &= \frac{(-4+1)(-4+3)}{j2(-4+2)} = \frac{3}{-j4} = j0.75 \\ Z_1(j2) &= j0.75 - j2k = 0 \;\Rightarrow\; k = \tfrac{3}{8} \\ Z_1(s) &= Z(s) - \tfrac{3}{8}s = \frac{(s^2+4)(5s^2+6)}{8s(s^2+2)} \end{aligned}$$

So series $L_1 = 3/8$ H, and $Z_1$ now has a zero at $s=\pm j2$ (the other zero moved from 1 to 1.095 rad/s). Its reciprocal has a pole at $\pm j2$, removed completely as a shunt branch:

$$\begin{aligned} Y_1(s) &= \frac{8s(s^2+2)}{(s^2+4)(5s^2+6)} = \frac{\tfrac{8}{7}s}{s^2+4} + \frac{16s}{7(5s^2+6)} \\ \tfrac{8}{7}\tfrac{s}{s^2+4} &\Rightarrow L_2 = \tfrac{7}{8}\ \text{H},\ C_2 = \tfrac{2}{7}\ \text{F}\quad (1/\sqrt{L_2C_2} = 2) \\ Z_2(s) &= \frac{7(5s^2+6)}{16s} = \tfrac{35}{16}s + \frac{21}{8s} \Rightarrow L_3 = \tfrac{35}{16}\ \text{H},\ C_3 = \tfrac{8}{21}\ \text{F} \end{aligned}$$

o--[L1=3/8H]--+--[L3=35/16H]--[C3=8/21F]--+
              |                           |
          [L2=7/8H]                       |
          [C2=2/7F]                       |
              |                           |
o-------------+---------------------------+

The shunt branch $L_2$–$C_2$ shorts the signal at $\omega=2$ rad/s, which is the required transmission zero. (Check: rebuilding the ladder gives back the original $Z(s)$.)

Significance in two-port synthesis

• Allows transmission zeros at any finite frequency $\omega_z$, not only at $0$ and $\infty$ (needed for elliptic/Cauer and notch filters).
• Keeps the realization a canonical LC ladder with positive elements, using the driving-point function ($z_{11}$ or $y_{11}$) while meeting the zeros of $z_{21}$.
• Several zeros can be produced one after another by repeated partial removal followed by complete removal.
• It is the basic step in Darlington-type synthesis of doubly terminated lossless filters.

Zero shifting by partial removal of a pole. When a driving-point function is expanded into a ladder, removing a pole completely (its full residue) produces transmission zeros only at $s=0$ or $s=\infty$. To place a zero at a chosen finite frequency $\omega_z$, we remove only part of a pole (at $\infty$ or at $0$) so that the remainder becomes zero at $s = \pm j\omega_z$. The reciprocal of the remainder then has a pole at $\pm j\omega_z$, which is removed completely as a shunt series-LC branch (or a series tank), giving a transmission zero exactly at $\omega_z$.

Key facts:

• Partial removal of a pole shifts the zeros of the remaining function towards the pole removed (removing part of the pole at $\infty$ moves zeros up; part of the pole at 0 moves zeros down).
• Poles of the remainder do not move; only zeros shift.
• The part removed is still a realizable element (e.g. a smaller series inductor), and the remainder stays LC because $k < k_\infty$.

Use in two-port synthesis

To synthesize a lossless two-port from $z_{11}$ (or $y_{11}$) with prescribed zeros of $z_{21}$:

1. List the required transmission zeros (at $0$, $\infty$ and finite $\pm j\omega_z$).
2. Zeros at $\infty$ and $0$: remove poles of $Z$ or $Y$ at $\infty$ or $0$ completely (series $L$/shunt $C$ or series $C$/shunt $L$).
3. For each finite zero $\omega_z$: partially remove a pole at $\infty$ or $0$ so that the remainder becomes zero at $s=\pm j\omega_z$.
4. Invert the remainder and remove the new pole at $\pm j\omega_z$ completely as a shunt series-LC branch (or a series tank). This branch blocks transmission at $\omega_z$.
5. Repeat with the remainder until it is fully realized.

Example. Realize $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$ as a ladder with a transmission zero at $\omega = 2$ rad/s.

The zeros of $Z$ are at $\omega = 1$ and $1.732$; we need one at $\omega=2$, so remove part of the pole at $\infty$: $Z_1(s) = Z(s) - k s$ with $Z_1(j2) = 0$.

$$\begin{aligned} Z(j2) &= \frac{(-4+1)(-4+3)}{j2(-4+2)} = \frac{3}{-j4} = j0.75 \\ Z_1(j2) &= j0.75 - j2k = 0 \;\Rightarrow\; k = \tfrac{3}{8} \\ Z_1(s) &= Z(s) - \tfrac{3}{8}s = \frac{(s^2+4)(5s^2+6)}{8s(s^2+2)} \end{aligned}$$

So series $L_1 = 3/8$ H, and $Z_1$ now has a zero at $s=\pm j2$ (the other zero moved from 1 to 1.095 rad/s). Its reciprocal has a pole at $\pm j2$, removed completely as a shunt branch:

$$\begin{aligned} Y_1(s) &= \frac{8s(s^2+2)}{(s^2+4)(5s^2+6)} = \frac{\tfrac{8}{7}s}{s^2+4} + \frac{16s}{7(5s^2+6)} \\ \tfrac{8}{7}\tfrac{s}{s^2+4} &\Rightarrow L_2 = \tfrac{7}{8}\ \text{H},\ C_2 = \tfrac{2}{7}\ \text{F}\quad (1/\sqrt{L_2C_2} = 2) \\ Z_2(s) &= \frac{7(5s^2+6)}{16s} = \tfrac{35}{16}s + \frac{21}{8s} \Rightarrow L_3 = \tfrac{35}{16}\ \text{H},\ C_3 = \tfrac{8}{21}\ \text{F} \end{aligned}$$

o--[L1=3/8H]--+--[L3=35/16H]--[C3=8/21F]--+
              |                           |
          [L2=7/8H]                       |
          [C2=2/7F]                       |
              |                           |
o-------------+---------------------------+

The shunt branch $L_2$–$C_2$ shorts the signal at $\omega=2$ rad/s, which is the required transmission zero. (Check: rebuilding the ladder gives back the original $Z(s)$.)

Checking the functions

An LC function must be an even/odd (or odd/even) ratio with simple poles and zeros on the $j\omega$ axis that alternate, and it must have a pole or zero at both $s=0$ and $s=\infty$ (degrees differ by exactly 1).

So $Z_2(s) = \dfrac{(s^2+3)(s^2+6)}{s(s^2+4)(s^2+9)} = \dfrac{s^4+9s^2+18}{s^5+13s^3+36s}$ is realized.

Foster-I realization

Expand $Z(s)$ in partial fractions (each term is an impedance; the elements are in series):

$$\begin{aligned} k_0 &= \left[s\,Z(s)\right]_{s=0} = \frac{1}{2} \\ 2k_{2} &= \left[\frac{(s^2+4)\,Z(s)}{s}\right]_{s^2=-4} = \frac{1}{10} \\ 2k_{3} &= \left[\frac{(s^2+9)\,Z(s)}{s}\right]_{s^2=-9} = \frac{2}{5} \end{aligned}$$ $$Z(s) = \frac{1}{2 s} + \frac{s}{10 \left(s^{2} + 4\right)} + \frac{2 s}{5 \left(s^{2} + 9\right)}$$

o--[C1]--[L2||C2]--[L3||C3]--o

Cauer-I realization

$Z(s)$ has a zero at $s=\infty$, so $Y(s)=1/Z(s)$ has a pole at infinity and the first element is a shunt capacitor. Divide repeatedly with polynomials in descending powers (continued fraction about $s=\infty$), inverting the remainder each time:

1. $Y_1(s) = \frac{s^{5} + 13 s^{3} + 36 s}{s^{4} + 9 s^{2} + 18}$ → quotient $s$ gives shunt $C_1$ = 1 F; remainder $\frac{2 s \left(2 s^{2} + 9\right)}{\left(s^{2} + 3\right) \left(s^{2} + 6\right)}$.
2. $Z_2(s) = \frac{s^{4} + 9 s^{2} + 18}{4 s^{3} + 18 s}$ → quotient $\frac{s}{4}$ gives series $L_2$ = 1/4 H (≈ 0.25 H); remainder $\frac{9 \left(s^{2} + 4\right)}{4 s \left(2 s^{2} + 9\right)}$.
3. $Y_3(s) = \frac{8 s^{3} + 36 s}{9 s^{2} + 36}$ → quotient $\frac{8 s}{9}$ gives shunt $C_3$ = 8/9 F (≈ 0.8889 F); remainder $\frac{4 s}{9 \left(s^{2} + 4\right)}$.
4. $Z_4(s) = \frac{9 s^{2} + 36}{4 s}$ → quotient $\frac{9 s}{4}$ gives series $L_4$ = 9/4 H (≈ 2.25 H); remainder $\frac{9}{s}$.
5. $Y_5(s) = \frac{s}{9}$ → quotient $\frac{s}{9}$ gives shunt $C_5$ = 1/9 F (≈ 0.1111 F); remainder $0$.

o----+---[L2]----+---[L4]----+
     |           |           |
   [C1]        [C3]        [C5]
     |           |           |
o----+-----------+-----------+

Both networks have five reactive elements, the minimum (canonical) number for this fifth-order function.

Zeros of transmission of a two-port network are the values of complex frequency $s$ at which the transfer function ($z_{21}$, $y_{21}$, $V_2/V_1$ or $I_2/I_1$) is zero, i.e. a finite input produces no output at that frequency. For an LC ladder they lie on the $j\omega$ axis (including $s=0$ and $s=\infty$).

In a ladder the transmission zeros are the poles of the series-arm impedances and the zeros of the shunt-arm impedances (series $L$ or shunt $C$: zero at $\infty$; series $C$ or shunt $L$: zero at 0; series tank or shunt series-LC branch: zero at $1/\sqrt{LC}$).

Steps to realize transmission zeros in a lossless two-port

1. From the specification, find the driving-point function ($z_{11}$ or $y_{11}$) and the required zeros of $z_{21}$ (or $y_{21}$).
2. Check that the driving-point function is a valid LC function.
3. For zeros at $s=\infty$: remove the pole at $\infty$ completely (series $L$ from $Z$ or shunt $C$ from $Y$).
4. For zeros at $s=0$: remove the pole at the origin completely (series $C$ or shunt $L$).
5. For a zero at a finite $\omega_z$: partial removal — subtract only $ks$ (or $k/s$) with $k$ chosen so that the remainder is zero at $s=\pm j\omega_z$ (zero shifting).
6. Invert the remainder; it now has a pole at $\pm j\omega_z$. Remove it completely as a shunt series-LC branch (or series parallel-LC tank) with $1/\sqrt{LC} = \omega_z$.
7. Continue with the remainder until nothing is left; check that all element values are positive.

Example. Realize $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$ as a ladder with a transmission zero at $\omega = 2$ rad/s.

The zeros of $Z$ are at $\omega = 1$ and $1.732$; we need one at $\omega=2$, so remove part of the pole at $\infty$: $Z_1(s) = Z(s) - k s$ with $Z_1(j2) = 0$.

$$\begin{aligned} Z(j2) &= \frac{(-4+1)(-4+3)}{j2(-4+2)} = \frac{3}{-j4} = j0.75 \\ Z_1(j2) &= j0.75 - j2k = 0 \;\Rightarrow\; k = \tfrac{3}{8} \\ Z_1(s) &= Z(s) - \tfrac{3}{8}s = \frac{(s^2+4)(5s^2+6)}{8s(s^2+2)} \end{aligned}$$

So series $L_1 = 3/8$ H, and $Z_1$ now has a zero at $s=\pm j2$ (the other zero moved from 1 to 1.095 rad/s). Its reciprocal has a pole at $\pm j2$, removed completely as a shunt branch:

$$\begin{aligned} Y_1(s) &= \frac{8s(s^2+2)}{(s^2+4)(5s^2+6)} = \frac{\tfrac{8}{7}s}{s^2+4} + \frac{16s}{7(5s^2+6)} \\ \tfrac{8}{7}\tfrac{s}{s^2+4} &\Rightarrow L_2 = \tfrac{7}{8}\ \text{H},\ C_2 = \tfrac{2}{7}\ \text{F}\quad (1/\sqrt{L_2C_2} = 2) \\ Z_2(s) &= \frac{7(5s^2+6)}{16s} = \tfrac{35}{16}s + \frac{21}{8s} \Rightarrow L_3 = \tfrac{35}{16}\ \text{H},\ C_3 = \tfrac{8}{21}\ \text{F} \end{aligned}$$

o--[L1=3/8H]--+--[L3=35/16H]--[C3=8/21F]--+
              |                           |
          [L2=7/8H]                       |
          [C2=2/7F]                       |
              |                           |
o-------------+---------------------------+

The shunt branch $L_2$–$C_2$ shorts the signal at $\omega=2$ rad/s, which is the required transmission zero. (Check: rebuilding the ladder gives back the original $Z(s)$.)

Checking the functions

An RC driving-point admittance $Y_{RC}(s)$ (same form as $Z_{RL}$) must have:

1. Simple poles and zeros on the negative real axis, alternating.
2. The critical frequency nearest the origin is a zero (or $Y(0)$ is a finite constant); there is no pole at $s=0$.
3. The critical frequency farthest from the origin is a pole (it may be at $s=\infty$).
4. $Y_{RC}(\infty) \ge Y_{RC}(0)$.
5. Residues of $Y_{RC}(s)/s$ are real and positive.

Parallel Foster form of (b)

Expand $Y(s)/s$ in partial fractions and multiply back by $s$:

$$\begin{aligned} k_0 &= Y(0) = \frac{2}{3} \\ k_{2} &= \left[\frac{(s+3)Y(s)}{s}\right]_{s=-3} = \frac{2}{3} \\ k_{3} &= \left[\frac{(s+6)Y(s)}{s}\right]_{s=-6} = \frac{5}{3} \end{aligned}$$ $$Y(s) = \frac{2}{3} + \frac{2 s}{3 \left(s + 3\right)} + \frac{5 s}{3 \left(s + 6\right)}$$

o--+-------+-------+
   |       |       |
  [R1]    [R2]    [R3]
   |      [C2]    [C3]
   |       |       |
o--+-------+-------+

Check: $Y(\infty) = \tfrac{2}{3}+\tfrac{2}{3}+\tfrac{5}{3} = 3$ ✓.

RC ladder (Cauer-I) form of (b)

Use $Z(s) = 1/Y(s) = \dfrac{(s+3)(s+6)}{3(s+1)(s+4)}$, which has $Z(\infty) = 1/3$, so the ladder starts with a series resistor.

Divide with descending powers (continued fraction about $s=\infty$). For an RC impedance the quotients alternate between a constant (series $R$) and $ks$ (shunt $C$):

1. $Z_1(s) = \frac{s^{2} + 9 s + 18}{3 s^{2} + 15 s + 12}$ → remove $\frac{1}{3}$, giving series $R_1$ = 1/3 Ω (≈ 0.3333 Ω); remainder $\frac{2 \left(2 s + 7\right)}{3 \left(s + 1\right) \left(s + 4\right)}$.
2. $Y_2(s) = \frac{3 s^{2} + 15 s + 12}{4 s + 14}$ → remove $\frac{3 s}{4}$, giving shunt $C_2$ = 3/4 F (≈ 0.75 F); remainder $\frac{3 \left(3 s + 8\right)}{4 \left(2 s + 7\right)}$.
3. $Z_3(s) = \frac{8 s + 28}{9 s + 24}$ → remove $\frac{8}{9}$, giving series $R_3$ = 8/9 Ω (≈ 0.8889 Ω); remainder $\frac{20}{9 \left(3 s + 8\right)}$.
4. $Y_4(s) = \frac{27 s + 72}{20}$ → remove $\frac{27 s}{20}$, giving shunt $C_4$ = 27/20 F (≈ 1.35 F); remainder $\frac{18}{5}$.
5. $Z_5(s) = \frac{5}{18}$ → remove $\frac{5}{18}$, giving series $R_5$ = 5/18 Ω (≈ 0.2778 Ω); remainder $0$.

o--[R1]----+---[R3]----+---[R5]-+
           |           |        |
         [C2]        [C4]       |
           |           |        |
o----------+-----------+--------+

The final $R_5$ closes the ladder across $C_4$.

Checking the functions

An RC impedance has simple, alternating poles and zeros on the negative real axis, with a pole nearest the origin, a zero farthest from it, and $Z(0) > Z(\infty)$.

Foster-II (parallel) form of (a)

Take $Y(s)=1/Z(s) = \frac{s^{2} + 6 s + 5}{s^{2} + 9 s + 18}$. Expand $Y(s)/s$ in partial fractions and multiply back by $s$:

$$\begin{aligned} k_0 &= Y(0) = \frac{5}{18} \\ k_{2} &= \left[\frac{(s+3)Y(s)}{s}\right]_{s=-3} = \frac{4}{9} \\ k_{3} &= \left[\frac{(s+6)Y(s)}{s}\right]_{s=-6} = \frac{5}{18} \end{aligned}$$ $$Y(s) = \frac{5}{18} + \frac{4 s}{9 \left(s + 3\right)} + \frac{5 s}{18 \left(s + 6\right)}$$

o--+-------+-------+
   |       |       |
  [R1]    [R2]    [R3]
   |      [C2]    [C3]
   |       |       |
o--+-------+-------+

Cauer-I form of (a)

Divide with descending powers (continued fraction about $s=\infty$). For an RC impedance the quotients alternate between a constant (series $R$) and $ks$ (shunt $C$):

1. $Z_1(s) = \frac{s^{2} + 9 s + 18}{s^{2} + 6 s + 5}$ → remove $1$, giving series $R_1$ = 1 Ω; remainder $\frac{3 s + 13}{\left(s + 1\right) \left(s + 5\right)}$.
2. $Y_2(s) = \frac{s^{2} + 6 s + 5}{3 s + 13}$ → remove $\frac{s}{3}$, giving shunt $C_2$ = 1/3 F (≈ 0.3333 F); remainder $\frac{5 \left(s + 3\right)}{3 \left(3 s + 13\right)}$.
3. $Z_3(s) = \frac{9 s + 39}{5 s + 15}$ → remove $\frac{9}{5}$, giving series $R_3$ = 9/5 Ω (≈ 1.8 Ω); remainder $\frac{12}{5 \left(s + 3\right)}$.
4. $Y_4(s) = \frac{5 s + 15}{12}$ → remove $\frac{5 s}{12}$, giving shunt $C_4$ = 5/12 F (≈ 0.4167 F); remainder $\frac{5}{4}$.
5. $Z_5(s) = \frac{4}{5}$ → remove $\frac{4}{5}$, giving series $R_5$ = 4/5 Ω (≈ 0.8 Ω); remainder $0$.

o--[R1]----+---[R3]----+---[R5]-+
           |           |        |
         [C2]        [C4]       |
           |           |        |
o----------+-----------+--------+

Properties of RC driving-point impedance

An RC driving-point impedance $Z_{RC}(s)$ has these properties:

1. All poles and zeros are simple and lie on the negative real axis of the $s$-plane (including the origin).
2. Poles and zeros alternate along the negative real axis.
3. The critical frequency nearest the origin (lowest) is a pole; it may be at $s=0$.
4. The critical frequency farthest from the origin is a zero; it may be at $s=\infty$.
5. $Z_{RC}(0) > Z_{RC}(\infty)$, and $Z_{RC}(\infty)\ge 0$ is a constant (no pole at infinity).
6. Residues of $Z_{RC}(s)$ at its poles are real and positive; on the real axis $dZ(\sigma)/d\sigma<0$.
7. The same form describes an RL admittance $Y_{RL}(s)$.

Checking the functions

Foster form of (a)

Foster-I (series) form: Expand $Z(s)$ in partial fractions; every residue must be positive:

$$\begin{aligned} k_\infty &= Z(\infty) = 1 \\ k_{2} &= [(s+1)Z(s)]_{s=-1} = \frac{5}{2} \\ k_{3} &= [(s+5)Z(s)]_{s=-5} = \frac{1}{2} \end{aligned}$$ $$Z(s) = 1 + \frac{5}{2 \left(s + 1\right)} + \frac{1}{2 \left(s + 5\right)}$$

o--[R1]--[R2||C2]--[R3||C3]--o

Foster-II (parallel) form: Take $Y(s)=1/Z(s) = \frac{s^{2} + 6 s + 5}{s^{2} + 9 s + 18}$. Expand $Y(s)/s$ in partial fractions and multiply back by $s$:

$$\begin{aligned} k_0 &= Y(0) = \frac{5}{18} \\ k_{2} &= \left[\frac{(s+3)Y(s)}{s}\right]_{s=-3} = \frac{4}{9} \\ k_{3} &= \left[\frac{(s+6)Y(s)}{s}\right]_{s=-6} = \frac{5}{18} \end{aligned}$$ $$Y(s) = \frac{5}{18} + \frac{4 s}{9 \left(s + 3\right)} + \frac{5 s}{18 \left(s + 6\right)}$$

o--+-------+-------+
   |       |       |
  [R1]    [R2]    [R3]
   |      [C2]    [C3]
   |       |       |
o--+-------+-------+