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Importance of frequency transformation; obtaining band pass

Short answer

Frequency transformation allows using a single low-pass prototype table for HP, BP, BS designs, reducing complexity. For band-pass, s→(s²+ω₀²)/(B s); series L becomes series L'=L/B, C'=B/(L ω₀²); shunt C becomes parallel C'=C/B, L'=B/(C ω₀²). Example: first-order LP function 1/(s+1) yields BP B s/(s²+B s+ω₀²).

What is the significance (importance) of frequency transformation in filter design? How can a band pass filter be obtained from a (normalized) prototype low pass filter? Explain with example.

Significance of frequency transformation

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.

  • 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)T(s) for given αmax\alpha_{max}, αmin\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 ss) 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.).

Band-pass filter from the normalized prototype low-pass filter

Transformation: replace ss in the low-pass prototype by

s  →  s2+ω02B s,ω0=ω1ω2,B=ω2−ω1s \;\rightarrow\; \frac{s^2 + \omega_0^2}{B\,s}, \qquad \omega_0 = \sqrt{\omega_1\omega_2},\quad B = \omega_2 - \omega_1

On the jωj\omega axis this gives ΩLP=ω2−ω02Bω\Omega_{LP} = \dfrac{\omega^2 - \omega_0^2}{B\omega}. So:

  • ω=ω0⇒ΩLP=0\omega = \omega_0 \Rightarrow \Omega_{LP} = 0 (centre of passband maps to DC of the LP);
  • ω=ω1,ω2⇒ΩLP=∓1\omega = \omega_1, \omega_2 \Rightarrow \Omega_{LP} = \mp1, since ω1ω2=ω02\omega_1\omega_2 = \omega_0^2 and ω2−ω1=B\omega_2 - \omega_1 = B;
  • ω=0\omega = 0 and ω=∞⇒ΩLP=∓∞\omega = \infty \Rightarrow \Omega_{LP} = \mp\infty (stopbands).

The band-pass response is geometrically symmetric about ω0\omega_0, and its order is twice that of the prototype.

Element transformation:

Series inductor LL:

Z=L s2+ω02Bs=LBs+1(BLω02)sZ = L\,\frac{s^2+\omega_0^2}{Bs} = \frac{L}{B}s + \frac{1}{\left(\dfrac{B}{L\omega_0^2}\right)s}

becomes a series LC: L′=LBL' = \dfrac{L}{B}, C′=BLω02C' = \dfrac{B}{L\omega_0^2}.

Shunt capacitor CC:

Y=C s2+ω02Bs=CBs+1(BCω02)sY = C\,\frac{s^2+\omega_0^2}{Bs} = \frac{C}{B}s + \frac{1}{\left(\dfrac{B}{C\omega_0^2}\right)s}

becomes a parallel LC to ground: C′=CBC' = \dfrac{C}{B}, L′=BCω02L' = \dfrac{B}{C\omega_0^2}.

Resistor RR: unchanged (frequency independent). Every new LC pair resonates at ω0\omega_0 (L′C′=1/ω02L'C' = 1/\omega_0^2).

 LP element       BP element
 series L   -->   series L'--C'
 shunt C    -->   shunt  L' || C'
 R          -->   R

Example (transfer function method): first-order prototype TLP(s)=1s+1T_{LP}(s) = \dfrac{1}{s+1}. Substituting s→s2+ω02Bss \to \dfrac{s^2+\omega_0^2}{Bs}:

TBP(s)=1s2+ω02Bs+1=Bss2+Bs+ω02T_{BP}(s) = \frac{1}{\dfrac{s^2+\omega_0^2}{Bs} + 1} = \frac{Bs}{s^2 + Bs + \omega_0^2}

This is the standard second-order band-pass function with centre frequency ω0\omega_0, bandwidth BB and Q=ω0/BQ = \omega_0/B. For ω0=1000\omega_0 = 1000 rad/s and B=100B = 100 rad/s: TBP(s)=100ss2+100s+106T_{BP}(s) = \dfrac{100s}{s^2 + 100s + 10^6}, Q=10Q = 10.

Example (element method): the third-order Butterworth ladder (1 H, 2 F, 1 H) with ω0=2000\omega_0 = 2000 rad/s, B=400B = 400 rad/s gives series arms of 2.5 mH in series with 100 μF and a shunt arm of 5 mF in parallel with 50 μH.

✓ Answer checked by Claude Sonnet 5.5

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