Low pass to high pass transformation

Frequency transformation and its applications

Frequency transformation is a change of the complex frequency variable that converts a normalized low-pass prototype filter (cutoff 1 rad/s) into a high-pass, band-pass or band-stop filter (or a low-pass with a different cutoff). The magnitude response of the prototype is mapped onto the new frequency axis, so the same attenuation values appear at the corresponding new frequencies.

Applications in filter design:

• Only one set of approximation tables (low-pass poles and ladder element values) is needed; HP, BP and BS filters are derived from it.
• The difficult approximation step (finding $T(s)$ for given $\alpha_{max}$, $\alpha_{min}$) is done once, for the simpler low-pass case; specifications of the other types are converted to an equivalent LP specification.
• It can be applied either to the transfer function (substitute for $s$) or directly to each element of an LC ladder, so the circuit is obtained without re-deriving it.
• It reduces design time and errors, and keeps the properties of the prototype (equiripple, maximally flat, etc.).

High-pass filter from a normalized low-pass filter

Transformation: replace $s$ in the normalized low-pass prototype by

$$s \;\rightarrow\; \frac{\omega_c}{s} \quad (\text{normalized: } s \to 1/s)$$

On the $j\omega$ axis $\Omega_{LP} = -\omega_c/\omega$: low frequencies map to the LP stopband and high frequencies to the LP passband, and $\omega = \omega_c$ maps to the LP band edge $\Omega = 1$. So the attenuation the LP has at $\Omega$ appears in the HP at $\omega_c/\Omega$.

Element transformation (normalized, $\omega_c = 1$):

• inductor $L$: $Z = sL \to L/s = \dfrac{1}{s(1/L)}$, i.e. a capacitor $C' = 1/L$;
• capacitor $C$: $Y = sC \to C/s = \dfrac{1}{s(1/C)}$, i.e. an inductor $L' = 1/C$;
• resistors unchanged.

Then frequency-scale to $\omega_c$ and magnitude-scale to the required impedance.

Example: third-order Butterworth low-pass prototype ($R_s = R_L = 1\ \Omega$; series $L_1 = 1$ H, shunt $C_2 = 2$ F, series $L_3 = 1$ H). Required: high-pass, $\omega_c = 1000$ rad/s, terminations 1 kΩ.

1. LP → HP (normalized): series $C_1 = 1/1 = 1$ F, shunt $L_2 = 1/2 = 0.5$ H, series $C_3 = 1$ F.
2. Scale with $k_f = 1000$, $k_m = 1000$ ($C' = C/(k_mk_f)$, $L' = Lk_m/k_f$):

$$C_1' = C_3' = \frac{1}{10^3\times10^3} = 1\ \mu\text{F}, \qquad L_2' = \frac{0.5\times10^3}{10^3} = 0.5\ \text{H}$$

  Rs=1k   1uF            1uF
 o-/\/\---||-----+--------||-----+-----o
 |               |               |     +
 Vs            0.5 H           RL=1k   Vo
 |               |               |     -
 o---------------+---------------+-----o

Transfer function check: $T_{LP}(s) = \dfrac{1}{s^3+2s^2+2s+1}$ (for equal terminations the passband gain is 1/2, omitted here); $s \to 1000/s$ gives $T_{HP}(s) = \dfrac{s^3}{s^3 + 2000s^2 + 2\times10^6 s + 10^9}$, a third-order Butterworth high-pass with 3 dB frequency 1000 rad/s.