# Magnitude and Frequency Scaling

Magnitude and frequency scaling is very useful when designing filters. Given an existing design (e.g., a low-pass filter with certain low-cutoff frequency, and of course, the values of resistor, capacitor and inductor), one can use scaling and quickly calculate the values of R, C and L for another similar design but with different cut-off frequency.

The scaling relationships are:

$R^{\prime} = s_m R$
$L^{\prime} = \frac{s_m}{s_f} L$
$C^{\prime} = \frac{C}{s_m s_f}$

where R, L, C are the initial values, R’, L’ and C’ are the scaled values, $$s_m$$ is the magnitude scaling factor, and sf is the frequency scaling factor.

How does it work? A general procedure is first set cut-off frequency $$\omega_0$$ to 1 rad/s, and set some component values to easy numbers, such as 1F or 1 Ohm. Calculate the other component values based on that. Finally, scale everything to the ones you want.

There are two examples below:

## Example 1: simple RC filter

Consider a simple RC low-pass filter like this: This is a low-pass filter with 3dB bandwidth of

$\omega_{3dB} = \frac{1}{RC} = 1 \text{rad/s}$

or

$f_{3dB} = \frac{1}{2 \pi RC} = \frac{1}{2 \pi} \text{Hz}$

Now, I need to design a similar filter with 3dB bandwidth of 1kHz, and I want to use a 1nF capacitor instead. The question is: what resistor value should I use? First I calculate the frequency scaling factor:

$s_f = \frac{1000Hz}{f_{3dB}} = 2000\pi$

and then calculate the magnitude scaling factor

$s_m = \frac{C}{s_f C^{\prime}} = \frac{1}{2000\pi 10^{-9}} = 1.59\times 10^5$

At last, the new resistor value should be

$R^{\prime} = s_m R = 1.59 \times 10^5 \Omega$

## Example 2: Sallen-Key active filter

The previous example is just a very simple example with lumped elements. The simple principle can be used in more complex cases, such as Salley-Key active filters. Take the 2nd-order high-pass Butterworth filter for example as below: with transfer function

$H(s) = \frac{H_0 s^2}{s^2 + \frac{\omega_0}{Q} s + \omega_0^2 }$

where

$\omega_0 =\frac{1}{\sqrt{R_1 R2 C_1 C_2}}$

and

$Q = \frac{\sqrt{R_1 R2 C_1 C_2}}{R_2 (C_1 + C_2) + R_1 C_2 (1-H_0)}$

For the convenience, first set the cut-off frequency

$\omega_0 = 1 \text{rad/s}$

and both capacitors C1 and C2 to 1F. Then the Q is reduced to

$Q = \frac{\sqrt{R_1 R_2 }}{(1-H_0) R_1 + 2 R_2 }$

Since Q= $$\frac{\sqrt{2}}{2}$$ for Butterworth filter, we can solve R1 and R2 from the equations:

$\frac{1}{\sqrt{R_1 R2 }} = 1$ $\frac{\sqrt{R_1 R2 }}{(1-H_0) R_1 + 2 R_2 } = \frac{\sqrt{2}}{2}$

The solutions can be easily got like this:

$R_2 = \frac{\sqrt{2} + \sqrt{8 H_0 -6}}{4}$ $R_1 = \frac{4}{\sqrt{2} + \sqrt{8 H_0 -6}}$

Now, if I want to design a Sallen-Key unity-gain (H0=1) Butterworth filter with cut-off frequency of 1kHz, using capacitors 1uF, what are the values of resistors? This can be easily done with scaling techniques as follows.

First find the frequency scaling factor:

$s_f = \frac{1 \text{kHz}}{\frac{1}{2 \pi} \text{Hz}} = 2000 \pi$

Then calculate the magnitude scaling factor:

$s_m = \frac{1 \text{F}}{s_f \times 1 \mu\text{F}} = \frac{500}{\pi}$

Therefore, the two resistor values are:

$R_2 = \frac{500\sqrt{2}}{2 \pi} \approx 112.5 \Omega$ $R_1 = \frac{500\sqrt{2}}{ \pi} \approx 225 \Omega$