Digital Signal Processing Reference
In-Depth Information
In order to use WFilter to determine the normalized transfer functions, we
will assume a passband edge frequency of 1 rad/sec (0.159154943092 Hz) and a
stopband edge frequency of 2 rad/sec (0.318309886184 Hz). (We must enter the
frequencies into WFilter using the hertz values and they must have more
significant digits than we require in our answer.) Twelve significant digits were
used to enter the edge frequencies, but they are displayed on the coefficient screen
with only ten significant digits. Rest assured that they are stored internally with
the higher accuracy, but the display is set for the more typical requirements of
filter frequency. The coefficient values determined are shown in Figure 2.4.
Butterworth 3rd-Order Normalized Lowpass
Selectivity: Lowpass
Approximation: Butterworth
Implementation: Analog
Passband gain (dB): -1.0
Stopband gain (dB): -12.0
Passband freq (Hz): 0.1591549431
Stopband freq (Hz): 0.3183098862
Filter Length/Order: 03
Overall Filter Gain: 1.00000000000E+00
Numerator Coefficients
QD [S^2 + S + 1 ]
== =========================================
01 0.0 0.00000000000E+00 1.25257638818E+00
02 0.0 0.00000000000E+00 1.56894760823E+00
Denominator Coefficients
QD [S^2 + S + 1 ]
== =========================================
01 0.0 1.00000000000E+00 1.25257638818E+00
02 1.0 1.25257638818E+00 1.56894760823E+00
Figure 2.4
Butterworth normalized third-order coefficients from WFilter.
Example 2.3 Butterworth Fourth-Order Normalized Transfer Function
Problem:
Determine the order, pole locations, and transfer function
coefficients for a Butterworth filter to satisfy the following specifications:
a
pass
= −1 dB,
a
stop
= −18 dB, ω
pass
= 1 rad/sec, and ω
stop
= 2 rad/sec
Solution:
First, we determine the constants needed from (2.7)-(2.9):
ε = 0.508847
n
= 3.95 (4th order)
R
= 1.184004
Next, we find the locations of the two complex poles in the second quadrant
from (2.10)-(2.13). A graph of the pole locations is shown in Figure 2.5.
θ
0
= 5π/8
σ
0
= −0.453099
ω
0
= +1.093877
θ
1
= 7π/8
σ
1
= −1.093877
ω
1
= +0.453099
