Texas Instruments 990 Operation Manual page 157

Prototyping system
Hide thumbs Also See for 990:
Table of Contents

Advertisement

Jd75\ _ _ _ _ _ _ _
~
945255-9701
Assume the beginning memory address is 0000
16 ,
Since the displacement is 0, the first bit string begins
at bit 0 and consists of the first four bits at byte O. The beginning address of the second bit string is
determined by adding the bit increment 10
16
to the previous beginning address of O. The second
four-bit string addressed begins at memory location 0002
16 ,
The bit increment of 10
16
is repeatedly
added to the previous address until 100
16
four-bit strings have been selected from the memory block.
The last bit string will be the first four bits at address 01FE
16 •
In another example, a user may wish to select four bit strings, each four bits wide, from one word of
memory. The parameters needed would be:
Loop level
=
I
Bit increment
(I)
=
4
Number of iterations
(1)
=
4
Initial bit displacement (1)
=
0
Bit string width
=
4
Assume the beginning memory address is 0100
16 ,
Since the displacement is 0, the first bit string will be
the first bits of memory address 0100
16 ,
By adding the bit increment of four, the second bit string is
determined to be the second four bits of memory address 0100
16 ,
The third and fourth bit strings are
determined similarly.
Consider what would happen if the user decided to select similar strings from 32 words of memory
memory beginning with address 0100
16 ,
By changing the number of iterations to 128
(4 X
32) and
beginning with address 0100
16 ,
consecutive four bits of each word could be selected. The parameters
needed would be:
Loop level
Bit increment
Number of iterations
Initial bit displacements
Bit string width
=
1
=4
=
128(80
16 )
=0
=4
By incrementing four bits at a time and selecting four 4-bit strings from each word for 32 words, the
128 four-bit strings can be addressed.
7.4.5.2 Two-Level Looping. The programming sequence in the last example can also be done by
two-level looping, a concept which involves using the parameters ofloop level 2 to determine a number
of beginning addresses. Each address determined by loop level 2 is used as a beginning address for loop
level 1.
In the last exam pIe the parameters would now be:
Loop level
Bit increment
(1)
Number of iterations
(1)
Initial bit displacement
(1)
Bit string width
Loop Level
Bit increment
(2)
Number of iterations (2)
Initial bit displacement (2)
Bit string width
=1
=4
=4
=0
=4
=2
. = 16 (10
16 )
= 32 (20
16 )
=0
=4
7-7
Digital Systems Division .

Advertisement

Table of Contents
loading

Table of Contents