Example: Second-Order Butterworth Lowpass
In the second-order case, we have, for the analog prototype,
![$\displaystyle H_a(s) = \frac{1}{(s + a)(s + \overline{a})}
$](http://www.dsprelated.com/josimages_new/filters/img2352.png)
![$ \,$](http://www.dsprelated.com/josimages_new/filters/img94.png)
![$ a = e^{j\pi/4}$](http://www.dsprelated.com/josimages_new/filters/img2353.png)
To convert this to digital form, we apply the bilinear transform
![$\displaystyle s = c\frac{1-z^{-1}}{1+z^{-1}}
$](http://www.dsprelated.com/josimages_new/filters/img2355.png)
![$ \,$](http://www.dsprelated.com/josimages_new/filters/img94.png)
![$\displaystyle c = \cot(\omega_cT/2) \isdef \frac{\cos(\omega_cT/2)}{\sin(\omega_cT/2)}
$](http://www.dsprelated.com/josimages_new/filters/img2356.png)
![$ \omega_c$](http://www.dsprelated.com/josimages_new/filters/img878.png)
![$ \omega_c T = \pi/2$](http://www.dsprelated.com/josimages_new/filters/img2357.png)
![$\displaystyle c = \frac{\cos(\pi/4)}{\sin(\pi/4)} = 1
$](http://www.dsprelated.com/josimages_new/filters/img2358.png)
![]() |
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![]() |
(I.4) |
![]() |
![]() |
(I.5) | |
![]() |
![]() |
(I.6) | |
![]() |
![]() |
(I.7) |
Note that the numerator is
![$ (1+z^{-1})^2$](http://www.dsprelated.com/josimages_new/filters/img2364.png)
![$\displaystyle H_d(1) = \frac{2^2}{2+\sqrt{2} + 2-\sqrt{2}} = 1
$](http://www.dsprelated.com/josimages_new/filters/img2365.png)
![$ H_d(-1) = 0$](http://www.dsprelated.com/josimages_new/filters/img2366.png)
In the analog prototype,
the cut-off frequency is
rad/sec, where,
from Eq.
(I.1), the amplitude response
is
. Since we mapped the cut-off frequency precisely
under the bilinear transform, we expect the digital filter to have
precisely this gain.
The digital frequency response at one-fourth the sampling rate is
and
![$ 20\log_{10}(\left\vert H_d(j)\right\vert)=-3$](http://www.dsprelated.com/josimages_new/filters/img2370.png)
Note from Eq.(I.8) that the phase at cut-off is exactly -90 degrees
in the digital filter. This can be verified against the pole-zero
diagram in the
plane, which has two zeros at
, each
contributing +45 degrees, and two poles at
, each contributing -90
degrees. Thus, the calculated phase-response at the cut-off frequency
agrees with what we expect from the digital pole-zero diagram.
In the plane, it is not as easy to use the pole-zero diagram
to calculate the phase at
, but using Eq.
(I.3), we
quickly obtain
![$\displaystyle H_a(j\cdot 1) = \frac{1}{j^2 + \sqrt{2}j + 1} = -\frac{j}{\sqrt{2}},
$](http://www.dsprelated.com/josimages_new/filters/img2372.png)
![$ H_d(e^{j\pi/2})$](http://www.dsprelated.com/josimages_new/filters/img2373.png)
![$ \,$](http://www.dsprelated.com/josimages_new/filters/img94.png)
A related example appears in §9.2.4.
Previous Section:
Butterworth Lowpass Poles and Zeros