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Transmission and reflection coefficients; 2nd order LPF

Short answer

The transmission coefficient t(s)=(2√(R₁/R₂))V₂/V₁ indicates the fraction of maximum available source power delivered to the load, defining the magnitude response. The reflection coefficient ρ(s) gives the reflected power fraction and input impedance Z₁ for synthesis, with |ρ|²+|t|²=1. For the 2nd-order Butterworth LPF (1 Ω terminations, ωc=1): series L₁=1.414 H, shunt C₂=1.414 F.

What information do you get from transmission coefficient and reflection coefficient? Design a second order Butterworth low pass filter using resistively terminated lossless ladder with equal termination of 1Ω. T(s) = 1/(s² + √2 s + 1)

Information from transmission and reflection coefficients

The transmission coefficient t(s)t(s) tells how much of the source's maximum available power Pmax⁡=∣V1∣2/4R1P_{\max} = |V_1|^2/4R_1 is actually delivered to the load R2R_2:

∣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)}

The reflection coefficient ρ(s)\rho(s) of a resistively terminated lossless ladder compares the power reflected back towards the source with the maximum power the source can give. Looking into the ladder (input impedance Z1Z_1) from a source of resistance R1R_1:

ρ(s)=Z1(s)−R1Z1(s)+R1,∣ρ(jω)∣2=PreflectedPmax⁡,Z1(s)=R11+ρ(s)1−ρ(s)\rho(s) = \frac{Z_1(s) - R_1}{Z_1(s) + R_1}, \qquad |\rho(j\omega)|^2 = \frac{P_{\text{reflected}}}{P_{\max}}, \qquad Z_1(s) = R_1\frac{1+\rho(s)}{1-\rho(s)}
Transmission coefficient ttReflection coefficient ρ\rho
Fraction of available power reaching the loadFraction of available power reflected back
Gives the magnitude response directlyGives the input impedance Z1Z_1 to synthesize
∣t∣=1\lvert t\rvert = 1: maximum power transferρ=0\rho = 0: perfect match
t=0t = 0: transmission zero (stopband)∣ρ∣=1\lvert\rho\rvert = 1: total reflection

For a lossless ladder ∣t∣2+∣ρ∣2=1|t|^2 + |\rho|^2 = 1.

Design of 2nd order Butterworth LPF (R1=R2=1 ΩR_1 = R_2 = 1\ \Omega)

Step 1: Transmission coefficient. For n=2n = 2 (Table 1), B(s)=s2+2s+1B(s) = s^2 + \sqrt2 s + 1. With equal terminations ∣t(0)∣|t(0)| can be 1:

t(s)=1s2+2s+1,∣t(jω)∣2=11+ω4t(s) = \frac{1}{s^2+\sqrt2 s+1}, \qquad |t(j\omega)|^2 = \frac{1}{1+\omega^4}

(The actual voltage gain is V2/V1=12t(s)V_2/V_1 = \tfrac12 t(s), since t=2R1/R2 V2/V1t = 2\sqrt{R_1/R_2}\,V_2/V_1.)

Step 2: Reflection coefficient.

∣ρ(jω)∣2=1−11+ω4=ω41+ω4  ⇒  ρ(s)ρ(−s)=s4B(s)B(−s),ρ(s)=s2s2+2s+1|\rho(j\omega)|^2 = 1 - \frac{1}{1+\omega^4} = \frac{\omega^4}{1+\omega^4} \;\Rightarrow\; \rho(s)\rho(-s) = \frac{s^4}{B(s)B(-s)}, \quad \rho(s) = \frac{s^2}{s^2+\sqrt2 s+1}

Step 3: Input impedance (R1=1R_1 = 1):

Z1(s)=1+ρ1−ρ=2s2+2s+12s+1Z_1(s) = \frac{1+\rho}{1-\rho} = \frac{2s^2+\sqrt2 s+1}{\sqrt2 s+1}

Step 4: Continued fraction.

Z1=2s+12s+1=1.414 s+11.414 s+1\begin{aligned} Z_1 &= \sqrt2 s + \frac{1}{\sqrt2 s + 1} \\ &= 1.414\,s + \cfrac{1}{1.414\,s + 1} \end{aligned}

So series L1=1.414L_1 = 1.414 H, then a shunt C2=1.414C_2 = 1.414 F, and the last "1" is R2=1 ΩR_2 = 1\ \Omega.

Step 5: Circuit.

  R1=1  L1=1.414 H
 o-/\/\---UUU----+------+
 |               |      |
(~) V1     C2=1.414 F  R2=1
 |               |      |
 o---------------+------+

Answer: R1=1 ΩR_1 = 1\ \Omega, L1=1.414L_1 = 1.414 H, C2=1.414C_2 = 1.414 F, R2=1 ΩR_2 = 1\ \Omega (normalized, ωc=1\omega_c = 1 rad/s). The dual form Z1=R1(1−ρ)/(1+ρ)Z_1 = R_1(1-\rho)/(1+\rho) gives shunt C1=1.414C_1 = 1.414 F and series L2=1.414L_2 = 1.414 H. These agree with the standard table values (1.414, 1.414).

✓ Answer checked by Claude Sonnet 5.5

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