We can rearrange this equation to get C₂ in terms of C₄:

(28)

We can substitute this expression into the equation for w_n and solve for C₄:

(29)

where R₁ = R₃ = R.

Substituting this result into the equation for C₂ (Equation 28), we get:

(30)

These last two equations are useful when designing a filter of this type. One just selects a resistor value and then uses these two equations to calculate the capacitor values.

Practical examples
Let's consider a couple of practical examples. The first is a simple second order, lowpass filter that we will implement using a "cookbook" approach² where the resistors are of equal value, the capacitors are made equal, and Q is set with the gain. For our example the design parameters are: f_n= 4.8kHz and Q=1.

Choosing 1nF (.001uF) for the capacitors, we can calculate that the resistors should be 33.2k using the closest standard one percent value. Now we must adjust the DC gain to set Q=1. This requires a gain of two. The resulting circuit is shown in Figure 2.

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Now let's implement a filter with the same design goals using our simplified circuit starting with the same 33.2k resistor value. The capacitors are calculated to be; C₂ = 2nF and C₄ = 500pF. The schematic for this circuit is shown in Figure 3:

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