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state of the unused bits cannot be determined, perhaps they have not<br />
actually been constructed in the input port circuit. You can "elimthese<br />
bits by using appropriate software commands. These<br />
commands "mask" these unused bits, so that they become zeros.<br />
Since a 12-bit value may represent decimal values between 0 and<br />
:4095, some means must be found for converting the individual bytes<br />
;that have been input into a single value. We will assume that the<br />
·· ·ght least-significant bits have been input as a single byte from port<br />
312, and that the four most-significant bits have been input from<br />
put port 49313 at bit positions D3-DO. We will further assume that<br />
e unused bits at input port 49313 have been grounded so that they<br />
e logic zeros.<br />
Now that the configuration of the input ports has been defined,<br />
's see how the information is manipulated so that the original value<br />
reconstructed from the two separate bytes of data from the two inf<br />
ports (Fig. 3-13). Since the least-significant bits can represent<br />
ues between 0 and 255 from the 12-bit interface device, these bits<br />
not require any "conversion," since the Apple will simply input<br />
INPUT DATA<br />
DO ---1<br />
Dl ---1<br />
DATA BUS<br />
RD<br />
ADDR<br />
Fig. 3-13. Two-bit input port.<br />
e eight bits and convert them to a value within the range of 0 to<br />
However, if the four most-significant bits are considered apart<br />
the other bits, converting them to decimal will yield values be-<br />
0 and 15, rather than their original positional values of 0, 256,<br />
and so on. These bits have been "offset" by a factor of 256 due<br />
e. fact that the 12-bit data value had to be "split" into smaller<br />
so that it could be input by the Apple. Remember that any<br />
. value that is input into ·the Apple will be automatically con<br />
.d into a decimal number with values in the range of 0 to 255.<br />
en the two values have been input into the Apple, it is a simple<br />
to "reconstruct" the data. If the information from the four<br />
jgnificant bits is multiplied by 256 and then added to the value<br />
e eight least-significant bits, a resulting value will represent<br />
between 0 and 4095, inclusive, the value that was originally<br />
t as a 12-bit binary value at the interlace .device. The complete<br />
re routine is shown in Example 3-3.<br />
55
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