Ideal vs practical filters and scaling

Ideal vs practical filters

An ideal filter passes all frequencies in the passband with constant gain and zero phase distortion and completely blocks all frequencies in the stopband, with a vertical edge between them. It cannot be built, because its impulse response is non-causal and infinitely long. A practical filter only approximates this.

 |T|   ideal              |T|   practical
 1 +------+               1 +~~~~~.  <- ripple (amax)
   |      |                 |      \
   |      |                 |       \  transition
   |      |                 |        `--..__ (amin)
 0 +------+------> w      0 +------+---+------> w
          wc                       wp  ws

Magnitude (impedance) scaling

Every impedance in the network is multiplied by $k_m$, at the same frequency. A voltage transfer function is a ratio of impedances, so it does not change.

$$\begin{aligned} R' &= k_m R \\ sL' = k_m (sL) &\Rightarrow L' = k_m L \\ \frac{1}{sC'} = k_m \frac{1}{sC} &\Rightarrow C' = \frac{C}{k_m} \end{aligned}$$

Frequency scaling

The response at the old frequency $\omega$ must appear at the new frequency $k_f\omega$, where $k_f = \omega_{new}/\omega_{old}$. So each element must have at $k_f\omega$ the same impedance that the old element had at $\omega$:

$$\begin{aligned} j(k_f\omega)L' = j\omega L &\Rightarrow L' = \frac{L}{k_f} \\ \frac{1}{j(k_f\omega)C'} = \frac{1}{j\omega C} &\Rightarrow C' = \frac{C}{k_f} \\ R' &= R \quad (\text{frequency independent}) \end{aligned}$$

The new transfer function is $T'(s) = T(s/k_f)$: the response keeps its shape but is shifted along the frequency axis.

Combined scaling

Applying both:

$$R' = k_m R, \qquad L' = \frac{k_m}{k_f}L, \qquad C' = \frac{C}{k_m k_f}$$

Example: the normalized third-order Butterworth ladder (1 Ω, 1 H, 2 F, 1 H, 1 Ω, $\omega_o$ = 1 rad/s) scaled to $\omega_o = 10^4$ rad/s ($k_f = 10^4$) and 1 kΩ terminations ($k_m = 1000$) gives $R$ = 1 kΩ, $L = 1\times1000/10^4 = 0.1$ H and $C = 2/(1000\times10^4) = 0.2\ \mu$F.