Skip to main content

RC impedance properties; LC Cauer II realization

Short answer

RC impedance: poles/zeros simple on negative real axis, alternating, lowest critical frequency is a pole, highest a zero, Z(0)>Z(∞). Cauer-II for Z(s)=s(s²+3)/[(s²+1)(s²+4)]: first element shunt L1=0.75 H, series C2≈1.222 F, shunt L3≈0.0496 H, series C4=5.5 F.

What are the properties of RC impedance function? Explain with example. Realize the following LC function using Cauer II method. Z(s) = s(s²+3)/((s²+1)(s²+4))

Properties of RC impedance function

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

  1. All poles and zeros are simple and lie on the negative real axis of the ss-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=0s=0.
  4. The critical frequency farthest from the origin is a zero; it may be at s=∞s=\infty.
  5. ZRC(0)>ZRC(∞)Z_{RC}(0) > Z_{RC}(\infty), and ZRC(∞)≥0Z_{RC}(\infty)\ge 0 is a constant (no pole at infinity).
  6. Residues of ZRC(s)Z_{RC}(s) at its poles are real and positive; on the real axis dZ(σ)/dσ<0dZ(\sigma)/d\sigma<0.
  7. The same form describes an RL admittance YRL(s)Y_{RL}(s).

Example: Z(s)=(s+2)(s+4)(s+1)(s+3)Z(s) = \dfrac{(s+2)(s+4)}{(s+1)(s+3)} has poles at −1,−3-1,-3 and zeros at −2,−4-2,-4 (p 1, z 2, p 3, z 4): the lowest is a pole, the highest a zero, they alternate, and Z(0)=8/3>Z(∞)=1Z(0) = 8/3 > Z(\infty) = 1. Its expansion Z=1+1.5s+1+0.5s+3Z = 1 + \dfrac{1.5}{s+1} + \dfrac{0.5}{s+3} has positive residues, giving 1 Ω1\ \Omega in series with 1.5 Ω∥231.5\ \Omega \parallel \tfrac{2}{3} F and 16 Ω∥2\tfrac{1}{6}\ \Omega \parallel 2 F.

Cauer-II realization of the LC function

Z(s)=s(s2+3)(s2+1)(s2+4)=s3+3ss4+5s2+4Z(s) = \dfrac{s(s^2+3)}{(s^2+1)(s^2+4)} = \dfrac{s^3+3s}{s^4+5s^2+4} (z 0, p 1, z 1.732, p 2, z ∞ — valid LC).

Z(s)Z(s) has a zero at s=0s=0, so Y(s)=1/Z(s)Y(s)=1/Z(s) has a pole at the origin and the first element is a shunt inductor. Arrange numerator and denominator in ascending powers of ss and divide repeatedly (continued fraction about s=0s=0); each quotient is k/sk/s:

  1. Y1(s)=4+5s2+s43s+s3Y_1(s) = \frac{4 + 5 s^{2} + s^{4}}{3 s + s^{3}} → quotient 43s\frac{4}{3 s} gives shunt L1L_1 = 3/4 H (≈ 0.75 H); remainder s(3s2+11)3(s2+3)\frac{s \left(3 s^{2} + 11\right)}{3 \left(s^{2} + 3\right)}.
  2. Z2(s)=9+3s211s+3s3Z_2(s) = \frac{9 + 3 s^{2}}{11 s + 3 s^{3}} → quotient 911s\frac{9}{11 s} gives series C2C_2 = 11/9 F (≈ 1.222 F); remainder 6s11(3s2+11)\frac{6 s}{11 \left(3 s^{2} + 11\right)}.
  3. Y3(s)=121+33s26sY_3(s) = \frac{121 + 33 s^{2}}{6 s} → quotient 1216s\frac{121}{6 s} gives shunt L3L_3 = 6/121 H (≈ 0.04959 H); remainder 11s2\frac{11 s}{2}.
  4. Z4(s)=211sZ_4(s) = \frac{2}{11 s} → quotient 211s\frac{2}{11 s} gives series C4C_4 = 11/2 F (≈ 5.5 F); remainder 00.
ElementPositionValue
L1L_1shunt3/4 H (≈ 0.75 H)
C2C_2series11/9 F (≈ 1.222 F)
L3L_3shunt6/121 H (≈ 0.04959 H)
C4C_4series11/2 F (≈ 5.5 F)
o----+---[C2]----+---[C4]-+
     |           |        |
   [L1]        [L3]       |
     |           |        |
o----+-----------+--------+

✓ Answer checked by Claude Sonnet 5.5

All Filter Design chapter 4 questions →