Digital Signal Processing Reference
In-Depth Information
FIGURE C.3.
Number wheel for fixed-point representation.
itive, its sign bit, which would show up as a negative number (a power of 2), does
not appear. This is an exceptional case, which is treated as an overflow in fractional
representation. Since the fractional representation requires that both operand and
resultant occupy the same range,
-
1
range
<+
1, the operation (
-
1)
¥
(
-
1) pro-
duces an unrepresentable number,
1.
Consider the next larger combination:
+
(
)
-+
(
)
=
P
=-
2
n
-
1
2
n
-
1
1
2
2
n
-
2
-
2
n
-
1
Since the second number subtracts from the first, the product will occupy up to
the (2
n
-
3) bit position, counting from 0. Thus, it is representable in (2
n
-
2) bits.
With the exceptional case ruled out, this makes the bit position (2
n
-
2) available
for the sign bit of the resultant. Therefore, (2
n
-
1) bits are needed to support an
(
n
n
)-bit signed multiplication.
To clarify the preceding equation, consider the 4-bit case, or
¥
(
)
-+
(
)
=-
P
=-
22122
3
3
6
3
The number 2
6
occupies bit position 6. Since the second number is negative, the
summation of the two is a number that will occupy only bit positions less than bit
position 6, or
2
6
-=-==
2
3
64
8
56
00111000
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