In Z(s) = 2s + 3s/(s^2+1) + s/(s^2+3), what are the elements for the term 3s/(s^2+1)?

L = 3 H in parallel with C = 1/3 F

L = 1/3 H in parallel with C = 3 F

L = 3 H in series with C = 1/3 F

L = 3 H in parallel with C = 3 F

Why does the Cauer-I realization of (iii) start with a series inductor of 2 H?

Z(s) has a pole at s = ∞ (numerator degree is higher), and the first quotient in descending powers is 2s

Cauer I always starts with a shunt capacitor

What is the structure of the Cauer-I ladder obtained for (iii)?

Series L, shunt C, series L, shunt C, series L

Series C and shunt L alternating

Asked 2/26
2080 Bhadra · 2+3+3 marks

Valid lossless function; Foster series and Cauer I

Short answer

Functions (ii) and (iii) are valid lossless. For (iii), Foster-I gives series L=2H, parallel L=3H||C=0.333F and L=0.333H||C=1F; Cauer-I gives series L=2H, shunt C=0.25F, series L=2.667H, shunt C=0.75F, series L=0.667H.

Which of the following is valid lossless function? State with reason. Pick one of the valid LC lossless functions and synthesize it using Foster series and Cauer I methods. i) Z(s) = (s²+4)(s²+5)/((s²+2)(s²+10)) ii) Z(s) = (s⁴+4s²+3)/(s(s²+2)) iii) Z(s) = (2s⁵+12s³+16s)/(s⁴+4s²+3) iv) 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, equal degrees; zeros adjacent → invalid
(ii) $\dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$	p 0, z 1, p 1.414, z 1.732, p ∞	Even/odd, alternate → valid
(iii) $\dfrac{2s(s^2+2)(s^2+4)}{(s^2+1)(s^2+3)}$	z 0, p 1, z 1.414, p 1.732, z 2, p ∞	Odd/even, alternate → valid
(iv) $\dfrac{s(s^4+4s^2+8)}{s^2+3}$	$s^4+4s^2+8=0$ gives $s^2=-2\pm j2$	Complex zeros; degrees differ by 3 → invalid

Valid functions: (ii) and (iii). We synthesize (iii):

Z(s) = \frac{2s^5+12s^3+16s}{s^4+4s^2+3} = \frac{2s(s^2+2)(s^2+4)}{(s^2+1)(s^2+3)}

Foster series (Foster-I) form

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

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

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

Term	Rule	Element value
$L_1$ (series, pole at $\infty$ )	$k_\infty$	2 H
$L_2 \parallel C_2$ (tank, $\omega^2=1$ )	$C=1/2k$ , $L=2k/\omega^2$	$C_2$ = 1/3 F (≈ 0.3333 F), $L_2$ = 3 H
$L_3 \parallel C_3$ (tank, $\omega^2=3$ )	$C=1/2k$ , $L=2k/\omega^2$	$C_3$ = 1 F, $L_3$ = 1/3 H (≈ 0.3333 H)

o--[L1]--[L2||C2]--[L3||C3]--o

Cauer-I form

$Z(s)$ has a pole at $s=\infty$ (numerator degree is higher), so the first element is a series inductor. Divide repeatedly with polynomials in descending powers (continued fraction about $s=\infty$ ), inverting the remainder each time:

$Z_1(s) = \frac{2 s^{5} + 12 s^{3} + 16 s}{s^{4} + 4 s^{2} + 3}$ → quotient $2 s$ gives series $L_1$ = 2 H; remainder $\frac{2 s \left(2 s^{2} + 5\right)}{\left(s^{2} + 1\right) \left(s^{2} + 3\right)}$ .
$Y_2(s) = \frac{s^{4} + 4 s^{2} + 3}{4 s^{3} + 10 s}$ → quotient $\frac{s}{4}$ gives shunt $C_2$ = 1/4 F (≈ 0.25 F); remainder $\frac{3 \left(s^{2} + 2\right)}{4 s \left(2 s^{2} + 5\right)}$ .
$Z_3(s) = \frac{8 s^{3} + 20 s}{3 s^{2} + 6}$ → quotient $\frac{8 s}{3}$ gives series $L_3$ = 8/3 H (≈ 2.667 H); remainder $\frac{4 s}{3 \left(s^{2} + 2\right)}$ .
$Y_4(s) = \frac{3 s^{2} + 6}{4 s}$ → quotient $\frac{3 s}{4}$ gives shunt $C_4$ = 3/4 F (≈ 0.75 F); remainder $\frac{3}{2 s}$ .
$Z_5(s) = \frac{2 s}{3}$ → quotient $\frac{2 s}{3}$ gives series $L_5$ = 2/3 H (≈ 0.6667 H); remainder $0$ .

Element	Position	Value
$L_1$	series	2 H
$C_2$	shunt	1/4 F (≈ 0.25 F)
$L_3$	series	8/3 H (≈ 2.667 H)
$C_4$	shunt	3/4 F (≈ 0.75 F)
$L_5$	series	2/3 H (≈ 0.6667 H)

o--[L1]----+---[L3]----+---[L5]-+
           |           |        |
         [C2]        [C4]       |
           |           |        |
o----------+-----------+--------+

✓ Answer checked by Claude Sonnet 5.5

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