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Transmission coefficient; realize 1/(s²+√2s+1)

Short answer

The transmission coefficient t(s) indicates the fraction of the source’s maximum available power that reaches the load, thus defining the passband/stopband magnitude response and transmission zeros. It relates to the reflection coefficient by |t|²+|ρ|²=1. Realization of T(s)=1/(s²+√2 s+1) as LC ladder (1 Ω terminations): series L₁=1.414 H, shunt C₂=1.414 F, R₁=R₂=1 Ω.

What information does the transmission coefficient indicate? Realize the following transfer function using LC Ladder with equal termination of R1 = R2 = 1Ω. T(s) = 1/(s² + √2 s + 1)

Information indicated by the transmission coefficient

t(s)=2R1/R2 V2/V1t(s) = 2\sqrt{R_1/R_2}\,V_2/V_1, and ∣t(jω)∣2=P2/Pmax⁡|t(j\omega)|^2 = P_2/P_{\max} is the fraction of the source's maximum available power (∣V1∣2/4R1|V_1|^2/4R_1) that reaches the load. Hence:

  • it is the magnitude (power) response of the filter: passband where ∣t∣≈1|t| \approx 1, stopband where ∣t∣≈0|t| \approx 0;
  • ∣t∣=1|t| = 1 means perfect matching and maximum power transfer (no reflection, ρ=0\rho = 0);
  • t=0t = 0 at a frequency means a transmission zero;
  • ∣t∣≤1|t| \le 1 always, which limits the gain of a passive ladder (e.g. V2/V1≤0.5V_2/V_1 \le 0.5 for equal terminations);
  • with ∣ρ∣2=1−∣t∣2|\rho|^2 = 1 - |t|^2 it leads to the reflection coefficient used for synthesis.

Realization of T(s)=1s2+2s+1T(s) = \dfrac{1}{s^2+\sqrt2 s+1} with 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=(1−ρ)/(1+ρ)Z_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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