Why is Z(s) = (s^2+4)(s^2+5)/((s^2+2)(s^2+10)) not a valid LC function?

Its zeros (2 and 2.236) are adjacent, so poles and zeros do not alternate, and it has no pole or zero at s = 0 or ∞

It has complex poles off the jω axis

It is odd/even instead of even/odd

It has a negative residue at s = 0

What makes (iii) Z(s) = (s^6+4s^4+8s^2)/(s^3+3s) fail?

The zeros from s^4+4s^2+8=0 are complex (s^2 = -2 ± j2), not on the jω axis, and the degrees differ by 3

The numerator is odd and the denominator is even

It has a double pole at the origin

Its residues are all real but too large

In the Foster-II form of Z(s), what is expanded in partial fractions and how are the resulting terms connected?

Y(s) = 1/Z(s); terms are admittances of branches in parallel, each a series L-C

Y(s) = 1/Z(s); terms are admittances of branches in series

For the valid function (ii), why does the Cauer-II synthesis start with a series capacitor?

Z(s) has a pole at s = 0, so the continued fraction in ascending powers gives a k/s term first

Cauer II always starts with a shunt inductor

Asked 2/26
2080 Bhadra · 3+3+3 marks

Valid lossless function; Foster II and Cauer II

Short answer

Valid lossless: (ii) Z(s)=(s⁴+4s²+3)/(s(s²+2)); poles/zeros on jω axis alternate, pole at 0, pole at ∞. Foster II: parallel branches: series L=2 H, C=0.5 F; series L=2 H, C=1/6 F. Cauer II: series 2/3 F, shunt 1.25 H, series 0.08 F, shunt 5 H.

Which of the following is valid lossless function? State with reason. Pick one of the valid LC lossless functions and synthesize it using Foster II and Cauer II methods. (i) Z(s) = (s²+4)(s²+5)/((s²+2)(s²+10)) (ii) Z(s) = (s⁴+4s²+3)/(s(s²+2)) (iii) Z(s) = (s⁶+4s⁴+8s²)/(s³+3s)

Checking the functions

Tests for a valid LC impedance: (1) even/odd or odd/even ratio, (2) simple poles and zeros only on the $j\omega$ axis, (3) poles and zeros alternate, (4) a pole or a zero at both $s=0$ and $s=\infty$ (degrees differ by exactly 1), (5) positive residues.

Function	Factored form / critical frequencies	Result
(i) $\dfrac{(s^2+4)(s^2+5)}{(s^2+2)(s^2+10)}$	p 1.414, z 2, z 2.236, p 3.162	Even/even with equal degrees (no pole/zero at $0$ , $\infty$ ); zeros adjacent → invalid
(ii) $\dfrac{s^4+4s^2+3}{s(s^2+2)} = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$	p 0, z 1, p 1.414, z 1.732, p ∞	Even/odd, alternate → valid
(iii) $\dfrac{s^6+4s^4+8s^2}{s^3+3s} = \dfrac{s(s^4+4s^2+8)}{s^2+3}$	$s^4+4s^2+8=0$ gives $s^2=-2\pm j2$	Zeros are complex (not on $j\omega$ axis); degrees differ by 3 → invalid

Function (ii) is synthesized.

Foster-II form

Take $Y(s) = 1/Z(s) = \frac{s^{3} + 2 s}{s^{4} + 4 s^{2} + 3}$ and expand $Y(s)$ in partial fractions (each term is an admittance; the branches are in parallel):

\begin{aligned} 2k_{1} &= \left[\frac{(s^2+1)\,Y(s)}{s}\right]_{s^2=-1} = \frac{1}{2} \\ 2k_{2} &= \left[\frac{(s^2+3)\,Y(s)}{s}\right]_{s^2=-3} = \frac{1}{2} \end{aligned}

Y(s) = \frac{s}{2 \left(s^{2} + 1\right)} + \frac{s}{2 \left(s^{2} + 3\right)}

Term	Rule	Element value
$L_1$ – $C_1$ in series ( $\omega^2=1$ )	$L=1/2k$ , $C=2k/\omega^2$	$L_1$ = 2 H, $C_1$ = 1/2 F (≈ 0.5 F)
$L_2$ – $C_2$ in series ( $\omega^2=3$ )	$L=1/2k$ , $C=2k/\omega^2$	$L_2$ = 2 H, $C_2$ = 1/6 F (≈ 0.1667 F)

o--+-------+
   |       |
  [L1]    [L2]
  [C1]    [C2]
   |       |
o--+-------+

Cauer-II form

$Z(s)$ has a pole at $s=0$ , so the first element is a series capacitor. Arrange numerator and denominator in ascending powers of $s$ and divide repeatedly (continued fraction about $s=0$ ); each quotient is $k/s$ :

$Z_1(s) = \frac{3 + 4 s^{2} + s^{4}}{2 s + s^{3}}$ → quotient $\frac{3}{2 s}$ gives series $C_1$ = 2/3 F (≈ 0.6667 F); remainder $\frac{s \left(2 s^{2} + 5\right)}{2 \left(s^{2} + 2\right)}$ .
$Y_2(s) = \frac{4 + 2 s^{2}}{5 s + 2 s^{3}}$ → quotient $\frac{4}{5 s}$ gives shunt $L_2$ = 5/4 H (≈ 1.25 H); remainder $\frac{2 s}{5 \left(2 s^{2} + 5\right)}$ .
$Z_3(s) = \frac{25 + 10 s^{2}}{2 s}$ → quotient $\frac{25}{2 s}$ gives series $C_3$ = 2/25 F (≈ 0.08 F); remainder $5 s$ .
$Y_4(s) = \frac{1}{5 s}$ → quotient $\frac{1}{5 s}$ gives shunt $L_4$ = 5 H; remainder $0$ .

Element	Position	Value
$C_1$	series	2/3 F (≈ 0.6667 F)
$L_2$	shunt	5/4 H (≈ 1.25 H)
$C_3$	series	2/25 F (≈ 0.08 F)
$L_4$	shunt	5 H

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

✓ Answer checked by Claude Sonnet 5.5

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