The transition between the pass-band and stop-band of a first order filter with cut-off frequency is characterized by the the slope of 20 dB per decade of frequency change. To achieve better selectivity, we can cascade a set of such first order filters to form an nth order filter with a slope of 20n dB per decade.

The FRF of a first-order low-pass filter of unit gain is:

The FRF of such filters in series is (assuming they are well buffered with no loading effect):

The cut-off frequency of this nth order filter can be found by solving the following equation

to get

**Example:** Design an 4th order LP filter with
. The cut-off frequency of the
first-order LP filter can be found to be

The time constant of the first-order filter is . If , then

The *Butterworth filters* have the property that the passing
band is flat. The magnitude of the FRF of an nth order low-pass
Butterworth filter with cut-off frequency is

where is the cut-off frequency at which . The transition between the pass-band and stop-band is controlled by the order . In general, higher order corresponds to more rapid transition. Specially, when , , and , we have

- , is an all-pass filter.
- , the Butterworth filter is the regular first-order filter:

- , the Butterworth filter becomes an ideal low-pass
filter:

Given the nth-order low-pass Butterworth filter defined above, we
can find the corresponding high-pass Butterworth filter by considering

the magnitude of the FRF of an nth order high-pass Butterworth filter with cut-off frequency is

Now we consider the implementation of a Butterworth filter.
For simplicity, in the following we assume the frequency is
normalized by the cut-off frequency, i.e., so that
. Consider the low-pass case:

We need to get its FRF from its magnitude . To do so, we first consider the transfer function (TF) in the s-domain corresponding to the FRF, which is the same as when , i.e., . Now the equation above can be written as

We further find the roots of the denominator, the poles of both and , and separate them so that those on the left s-plane are the poles of (stable and causal), while those on the right s-plane belong to (stable and anti-causal).

The roots of the denominator can be found by solving the equation

Solving these we get the solutions on the unit circle in either of the two different forms depending on whether is even or odd:

- If is even,

These roots form complex conjugate pairs around the unit circle of the s-plane. Corresponding to each of the roots ( ), there is another root that is its complex conjugate:

Also, for to be a pole of , it needs to be on the left s-plane, i.e.,

Now can be found in terms of its poles on the left s-plane:

where is the ceiling of , and we have used the fact that

- If is odd,

These roots contain and , as well as complex conjugate pairs. Corresponding to each root ( ), there is another root that is its complex conjugate:

For to be a pole of , it needs to be on the left s-plane, i.e.,

Now can be found in terms of its poles on the left s-plane:

Specifically, here we find the transfer function of the nth order Butterworth filter for :

- , ,
, the four roots are
():

out of which and on the left s-plane are the roots of :

Note that the coefficient of the first order term is . - , ,
, the six roots are
()

out of which , , and on the left s-plane are the roots of :

Note that the coefficient of the first order term is . - , ,
, the eight roots are
().
Evaluating
for and , we get the
coefficients of the two first order terms
and
.

- , ,
, the 10 roots are

Evaluating for and , we get the coefficients of the two first order terms and , and we get

- , ,
, the 12 roots are

Evaluating for , , and , we get the coefficients of the three first order terms , , and .

The first order filter in the cascade of the Butterworth filter
can be realized by the first order op-amp low-pass circuit shown
above with

where . If we let , we get .

The second order systems in the cascade can be implemented as a
Sallen-Key low-pass filter with

where . If we let for simplicity, we get

Solving these we get

A High-pass Butterworth filter can be similarly implemented with the only
difference that all first and second order systems in the cascade are
high-pass filters

so that the transfer function of the cascade is high-pass filter:

To convert the results obtained above for normalized cut-off frequency to unnormalized cut-off frequency , all we need to do is to scale all capacitances to . The capacitor in the first order filter becomes so that ; while the two capacitors in the second order filter become and so that .