Insertion gain/loss and element scaling equations

Filter

A filter is a frequency-selective two-port network that passes signals in a chosen band of frequencies (passband) with little attenuation and attenuates signals at other frequencies (stopband).

Insertion gain and insertion loss

Insertion gain and insertion loss compare the load voltage with and without the filter between a source $V_1$ (resistance $R_s$) and load $R_L$.

 Without filter:  Rs                 With filter:  Rs   +--------+
  V1 o--/\/\--+-- RL  (V20)          V1 o--/\/\--| filter |-- RL (V2)
                                                 +--------+

$$V_{20} = \frac{R_L}{R_s + R_L}V_1, \qquad \text{Insertion gain} = 20\log_{10}\left|\frac{V_2}{V_{20}}\right|\ \text{dB}$$ $$\text{Insertion loss} = 20\log_{10}\left|\frac{V_{20}}{V_2}\right| = -(\text{insertion gain})\ \text{dB}$$

A passive filter normally has a positive insertion loss; an active filter can have insertion gain.

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:

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