Transmission coefficient; 2nd order Butterworth LPF

Transmission coefficient

The transmission coefficient $t(s)$ is the ratio (in power terms) of the power actually delivered to the load $R_2$ to the maximum available power of the source $P_{\max} = |V_1|^2/4R_1$:

$$|t(j\omega)|^2 = \frac{P_2}{P_{\max}} = \frac{4R_1}{R_2}\left|\frac{V_2(j\omega)}{V_1(j\omega)}\right|^2 , \qquad t(s) = 2\sqrt{\frac{R_1}{R_2}}\,\frac{V_2(s)}{V_1(s)}$$

Information obtained from it

• $|t(j\omega)|^2$ is the fraction of the available source power that reaches the load, so it directly gives the filter's magnitude response (passband, stopband, cutoff).
• $|t| = 1$ means maximum power transfer (perfect match); $|t| = 0$ means a transmission zero (no power to the load).
• Since $|t| \le 1$ for a passive lossless ladder, it fixes the maximum possible gain, e.g. $V_2/V_1 \le 1/2$ for equal terminations.
• Through $|\rho|^2 = 1 - |t|^2$ it gives the reflection coefficient, which is used to find $Z_1$ and synthesize the ladder.

Design of 2nd order Butterworth LPF ($R_1 = R_2 = 1\ \Omega$)

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

$$t(s) = \frac{1}{s^2+\sqrt2 s+1}, \qquad |t(j\omega)|^2 = \frac{1}{1+\omega^4}$$

(The actual voltage gain is $V_2/V_1 = \tfrac12 t(s)$, since $t = 2\sqrt{R_1/R_2}\,V_2/V_1$.)

Step 2: Reflection coefficient.

$$|\rho(j\omega)|^2 = 1 - \frac{1}{1+\omega^4} = \frac{\omega^4}{1+\omega^4} \;\Rightarrow\; \rho(s)\rho(-s) = \frac{s^4}{B(s)B(-s)}, \quad \rho(s) = \frac{s^2}{s^2+\sqrt2 s+1}$$

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

$$Z_1(s) = \frac{1+\rho}{1-\rho} = \frac{2s^2+\sqrt2 s+1}{\sqrt2 s+1}$$

Step 4: Continued fraction.

$$\begin{aligned} Z_1 &= \sqrt2 s + \frac{1}{\sqrt2 s + 1} \\ &= 1.414\,s + \cfrac{1}{1.414\,s + 1} \end{aligned}$$

So series $L_1 = 1.414$ H, then a shunt $C_2 = 1.414$ F, and the last "1" is $R_2 = 1\ \Omega$.

Step 5: Circuit.

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

Answer: $R_1 = 1\ \Omega$, $L_1 = 1.414$ H, $C_2 = 1.414$ F, $R_2 = 1\ \Omega$ (normalized, $\omega_c = 1$ rad/s). The dual form $Z_1 = (1-\rho)/(1+\rho)$ gives shunt $C_1 = 1.414$ F and series $L_2 = 1.414$ H. These agree with the standard table values (1.414, 1.414).