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(Q format addition, subtraction): Perform the additions $01001101+11100100$ , and $01111001+10001011$ when the binary numbers are Q-7 format. Also compute $01001101-11100100$ and $10001011-00110111$ . In which cases, do you have overflow?
Intentionally left blank.
Multiplication of two 2's complement numbers is a bit complicated because of the sign bit. Similar to themultiplication of two decimal fractional numbers, the result of multiplying twoQ- $N$ numbers is Q- $2N$ , meaning that we have $2N$ binary digits following the implied binary digit. However, depending on the numbers multiplied, theresult can have either 1 or 2 binary digits before the binary point. We call the digit right before the binarypoint the sign bit and the one proceeding the sign bit (if any) the extended sign bit .
The following is the two examples of binary fractional multiplications:
0.110 0.75 Q-3
X 1.110 -0.25 Q-3--------------------------
00000110
01101010
-------------------------------1110100 -0.1875 Q-6
Above, all partial products are computed and represented
in Q-6 format for summation. For example,
0.110*0.010 =0.01100
in Q-6 for the
second partial product. For the 4th partial product, caremust be taken because in
0.110*1.000
,
1.000
represents
$-1$ , so the product is
-0.110 = 1.01000
(in Q-6 format) that
is 2's complement of
0.11000
. As
noticed in this example, it is important to represent eachpartial product in Q-6 (or in general Q-
$2N$ ) format before adding them together. Another
example makes this point clearer:
1.110 -0.25 Q-3
X 0.110 0.75 Q-3-------------------------
0000111110
111100000
-----------------------------11110100 -0.1875 Q-6
For the second partial product, we need
1.110*0.010
in Q-6 format. This is
obtained as
1111100
in Q-6 (check!).
A simple way to obtain it is to first perform themultiplication in normal fashion as
1110*0010 =
11100
ignoring the binary points, then perform
sign extension by putting enough 1s
(if the result is negative) or 0s (if the result isnonnegative), then put the binary point to obtain a Q-6
number. Also notice that we need to remove the extra signbit to obtain the final result.
In C62x, if we multiply two Q-15 numbers using one of
multiply instruction (for example
MPY
), we obtain 32 bit result in Q-30
format with 2 sign bits. To obtain the result back inQ-15 format, (i) first we remove 15 trailing bits and (ii)
remove the extended sign bit.
(Q format multiplication): Perform the multiplications
01001101*11100100
, and
01111001*10001011
when the binary
numbers are Q-7 format.
Intentionally left blank.
When
A0
and
A1
contain two 16-bit numbers in the Q-15 format, we can
perform the multiplications using
MPY
followed by a right shift.
1 MPY .M1 A0,A1,A2
2 NOP3 SHR .S1 A2,15,A2 ;lower 16 bit contains result
4 ;in Q-15 format
Rather than throwing away the 15 LSBs of the multiplication
result by shifting, you can round up the result by adding
0x4000
before shifting.
1 MPY .M1 A0,A1,A2
2 NOP3 ADDK .S1 4000h,A6
4 SHR .S1 A2,15,A2 ;lower 16 bit contains result5 ;in Q-15 format
Let's suppose we have two 16-bit numbers in Q-15 format,
stored in variable
x
and
y
as follows:
short x = 0x0011; /* 0.000518799 in decimal */
short y = 0xfe12; /* -0.015075684 in decimal */short z; /* variable to store x*y */
The product of
x
and
y
can be
computed and stored in Q-15 format as follows:
z = (x * y) >> 15;
The result of
x*y
is a 32-bit word with
2 sign bits. Right shifting it by 15 bits ignores the last15 bits, and storing the shifted result in
z
that is a
short
variable (16 bit) removes the extended sign bit by taking
only lower 16 bits.
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