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Derive reflection coefficient for terminated LC ladder

Short answer

Transmission coefficient t(s)=2√(R1/R2)·V2/V1, |t|²=P2/Pmax indicates power response. Derive ρ: from Pmax=|V1|²/(4R1), input power P1, reflected power fraction is (Pmax−P1)/Pmax = |(R1−Z1)/(R1+Z1)|², so ρ = ±(R1−Z1))/(R1+Z1) and |ρ|²+|t|²=1.

What is transmission coefficient? What information do we get from it? Derive the expression for reflection coefficient for a resistively terminated LC ladder circuit.

Transmission coefficient

The transmission coefficient t(s)t(s) is the ratio (in power terms) of the power actually delivered to the load R2R_2 to the maximum available power of the source Pmax⁡=∣V1∣2/4R1P_{\max} = |V_1|^2/4R_1:

∣t(jω)∣2=P2Pmax⁡=4R1R2∣V2(jω)V1(jω)∣2,t(s)=2R1R2 V2(s)V1(s)|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|t|^2 is the fraction of available power reaching the load, so it is the filter's power response. ∣t∣=1|t| = 1 means perfect matching and maximum power transfer; ∣t∣=0|t| = 0 marks a transmission zero; and it bounds the gain of a passive ladder (∣t∣≤1|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 R1R_1):
Pmax⁡=∣V1∣24R1P_{\max} = \frac{|V_1|^2}{4R_1}
  1. Input power to the ladder with Z1=Rin+jXinZ_1 = R_{in} + jX_{in} and I1=V1R1+Z1I_1 = \dfrac{V_1}{R_1+Z_1}:
P1=∣I1∣2Rin=∣V1∣2 Re(Z1)∣R1+Z1∣2P_1 = |I_1|^2 R_{in} = \frac{|V_1|^2\,\text{Re}(Z_1)}{|R_1+Z_1|^2}
  1. Power not delivered ("reflected") is Pmax⁡−P1P_{\max} - P_1:
Pmax⁡−P1=∣V1∣24R1[1−4R1 Re(Z1)∣R1+Z1∣2]=∣V1∣24R1⋅∣R1+Z1∣2−4R1 Re(Z1)∣R1+Z1∣2\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}
  1. Using ∣R1+Z1∣2−4R1Rin=(R1+Rin)2+Xin2−4R1Rin=(R1−Rin)2+Xin2=∣R1−Z1∣2|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:
Pmax⁡−P1Pmax⁡=∣R1−Z1R1+Z1∣2\frac{P_{\max}-P_1}{P_{\max}} = \left|\frac{R_1 - Z_1}{R_1+Z_1}\right|^2
  1. This ratio is defined as ∣ρ(jω)∣2|\rho(j\omega)|^2, so the reflection coefficient is
ρ(s)=±R1−Z1(s)R1+Z1(s)\rho(s) = \pm\frac{R_1 - Z_1(s)}{R_1 + Z_1(s)}
  1. Since the ladder is lossless, P1=P2P_1 = P_2, so ∣t∣2=P1/Pmax⁡|t|^2 = P_1/P_{\max} and
∣ρ(jω)∣2+∣t(jω)∣2=1(Feldtkeller equation)|\rho(j\omega)|^2 + |t(j\omega)|^2 = 1 \quad \text{(Feldtkeller equation)}
  1. Solving for Z1Z_1 gives the synthesis formula:
Z1(s)=R11∓ρ(s)1±ρ(s)Z_1(s) = R_1\frac{1 \mp \rho(s)}{1 \pm \rho(s)}

Use: given t(s)t(s), find ∣ρ∣2=1−∣t∣2|\rho|^2 = 1-|t|^2, factor ρ(s)ρ(−s)\rho(s)\rho(-s), pick ρ(s)\rho(s) with a Hurwitz denominator, find Z1Z_1 and expand it into an LC ladder.

✓ Answer checked by Claude Sonnet 5.5

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