Information from reflection coefficient; 3rd order HPF

Information from reflection coefficient

$\rho(s) = \dfrac{R_1 - Z_1}{R_1 + Z_1}$ tells how well the filter input is matched to the source:

• $|\rho(j\omega)|^2$ = fraction of the available power reflected back; $\rho = 0$ means maximum power delivered (passband), $|\rho| \to 1$ means total reflection (stopband).
• With $|\rho|^2 + |t|^2 = 1$ it gives the passband loss and the magnitude response.
• It gives the input impedance $Z_1 = R_1(1+\rho)/(1-\rho)$ that is expanded into the ladder, so it is the starting point of the synthesis.

Design of 3rd order Butterworth HPF ($R_1 = R_2 = 1\ \Omega$)

Step 1: Low-pass prototype. Design the normalized 3rd order Butterworth LPF first. From Table 1, $B(s) = s^3+2s^2+2s+1$, so with $R_1 = R_2 = 1$:

$$t(s) = \frac{1}{B(s)}, \quad |\rho|^2 = 1-|t|^2 = \frac{\omega^6}{1+\omega^6}, \quad \rho(s) = \frac{s^3}{s^3+2s^2+2s+1}$$ $$Z_1 = \frac{1+\rho}{1-\rho} = \frac{2s^3+2s^2+2s+1}{2s^2+2s+1} = s + \cfrac{1}{2s + \cfrac{1}{s+1}}$$

Prototype: series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H, $R_2 = 1\ \Omega$.

Step 2: LP to HP transformation. Replace $s$ by $1/s$ ($\omega_c = 1$ rad/s):

• A series inductor $L$ (impedance $sL$) becomes $L/s$: a series capacitor $C = 1/L$.
• A shunt capacitor $C$ (admittance $sC$) becomes $C/s$: a shunt inductor $L = 1/C$.
• Resistors are unchanged.

Step 3: Circuit.

  R1=1    C1=1 F        C3=1 F
 o-/\/\----||----+----||----+
 |               |          |
(~) V1        L2=0.5 H    R2=1
 |               |          |
 o---------------+----------+

The response is $V_2/V_1 = \dfrac{0.5\,s^3}{s^3+2s^2+2s+1}$: zero gain at dc, $-3$ dB at 1 rad/s, and gain 0.5 (the maximum for equal terminations) at high frequency.

Answer (normalized, $\omega_c = 1$ rad/s): $R_1 = 1\ \Omega$, series $C_1 = 1$ F, shunt $L_2 = 0.5$ H, series $C_3 = 1$ F, $R_2 = 1\ \Omega$. For a cutoff $\omega_c$, divide each $L$ and $C$ by $\omega_c$ (and scale impedance if needed).