of Microprocessors

DIGITAL FILTERING 523
1.0
0.8
0.6
• /INTERPOLATEO Y
0.4
0.2
X
•
OH----+---1--+----1I-----+---+-+-+--I----+---+---+--+-_
• -5 -4 -3 -2 -I 0 \ l
4
6
-0.2 -6
-0.4
-0.6
UNKNOWN X= 0.375 =12/32
-0.8
-1.0
Known X Known Y Table Address Table Entry Y x Table Entry
~6 -.1 204 .0 .0
-5 .3 172 ~.00358 -.00107
-4 .9 140 .01368 .01231
-3 1.0 108 -.03715 -.03715
~2 .4 76 .08349 .03340
-1 .8 44 -.18840 -.15072
o .6 12 .77698 .46619
1 -.2 20 .45854 -.09171
2 -.4 52 -.15138 .06055
3 -.8 84 .06875 -.05500
4 -.6 116 -.02967 .01780
5 .1 148 .01012 .00101
6 .2 180 -.00257 -.00051
.25510 = Interpolated Y
Fig. 14-32. Interpolation table use example
consuming, particularly on a microcomputer. Applying a tailing function
(instead of truncating) takes even longer. The answer to this dilemma is the
maxim: When in doubt use tables! As it turns out, a very simple table
lookup algorithm can be derived that requires N multiplications, table
lookups, and additions, where N is the number of points considered. These
numbers are the same even if a tailing function is used to reduce interpolation
nOIse.
The basic idea is to store the impulse response in a table. Since the
impulse response is symmetrical, the table needs to be only half the size it
would otherwise be. Fig. 14-31 shows how the data is placed in the table.
First, we will assume that N is odd. The table area is then divided into
(N - 1)/2 sections and, fot convenience on a binary computer, each section
contains a power of two number of points. The time span of a section is the
same as the input stream sample period. In the example, N = 13 and there
are six sections with 32 entries each for a total of 192 entries. It is also
necessary to include or imagine 16 zeroes preceding the first and following
the last section.