ARITHMETIC
Milo 1
DIGITAL D. Ercegovac and Toms Lang
s
a
Morgan Kaufmann Publishers, an imprint of Elsevier Science, c 2004
Updated: September 23, 2003
Chapter 6: Solutions to Exercises
With contributions by Elisardo Antelo and Fabrizio Lamberti
Exercise 6.1
a) Radix-2, sj {1, 0, 1}, conventional (nonredundant) residual
We have x = 144 28 = 0.10010000 and = 1. We choose s0 = 0.
Therefore the initialization is w [0] = x s0 = 0.10010000.
We use the result-digit selection function for redundant residual but we
consider only 2 integer bits since the range of the residual estimate is
smaller than in the redundant case.
2w [0] =
F1 [0] =
w [1] =
001.00100000
11.10000000
00.10100000
y=1
F1 [0] = 11.10000000
s1 = 1
2w [1] =
F1 [1] =
w [2] =
001.01000000
10.11000000
00.00000000
y=1
F1 [1] = 00.11000000
s2 = 1
2w [2] =
F1 [2] =
w [3] =
000.00000000
10.01100000
10.01100000
y=0
F1 [2] = 01.01100000
s3 = 1
2w [3] =
F1 [3] =
w [4] =
100.11000000
01.10110000
10.01110000
y = 4
F1 [3] = 10.00110000
s 4 = 1
2w [4] =
F1 [4] =
w [5] =
100.11100000
01.10011000
10.01111000
y = 4
F1 [4] = 10.01011000
s 5 = 1
2w [5] =
F1 [5] =
w [6] =
100.11110000
01.10001100
1110.01111100
y = 4
F1 [5] = 01.10001100
s 6 = 1
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
2
(continues on next page)
2w [6] =
F1 [6] =
w [7] =
100.11111000
01.10000110
10.01111110
y = 4
F1 [6] = 10.01110110
s7 = 1
2w [7] =
F1 [7] =
w [8] =
100.11111100
01.10000011
10.01111111
y = 4
F1 [7] = 10.01111011
s8 = 1
2w [8] =
F1 [8] =
w [9] =
100.11111110
01.10000001
10.01111111
y = 4
F1 [8] = 10.01111101
s9 = 1
We perform 9 iterations to compute the additional bit required for rounding. Since w [9] < 0 the correction step has to be performed. Thus
s9 = 2. The result is
s = 0.111 = (0.11000000)2
1 11 11 2
b) Radix-2, sj {1, 0, 1}, carry-save residual
2W S [0] =
2W C [0] =
F1 [0] =
W S [1] =
W C [1] =
0001.00100000
0000.00000000
111.10000000
110.10100000
010.00000000
y=1
2W S [1] =
2W C [1] =
F1 [1] =
W S [2] =
W C [2] =
1101.01000000
0100.00000000
110.11000000
111.10000000
000.10000000
y=1
2W S [2] =
2W C [2] =
F1 [2] =
W S [3] =
W C [3] =
1111.00000000
0001.00000000
110.01100000
000.01100000
110.00000000
y=0
2W S [3] =
2W C [3] =
F1 [3] =
W S [4] =
W C [4] =
0000.11000000
1100.00000000
001.10110000
101.01110000
001.00000000
s1 = 1
y = 4
F1 [0] = 111.10000000
s2 = 1
F1 [1] = 000.11000000
s3 = 1
F1 [2] = 001.01100000
s 4 = 1
F1 [3] = 110.00110000
(continues on next page)
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
3
2W S [4] =
2W C [4] =
F1 [4] =
W S [5] =
W C [5] =
1010.11100000
0010.00000000
001.10011000
001.01111000
101.00000000
y = 4
2W S [5] =
2W C [5] =
F1 [5] =
W S [6] =
W C [6] =
0010.11110000
1010.00000000
001.10001100
001.01111100
101.00000000
y = 4
2W S [6] =
2W C [6] =
F1 [6] =
W S [7] =
W C [7] =
0010.11111000
1010.00000000
001.10000110
001.01111110
101.00000000
y = 4
2W S [7] =
2W C [7] =
F1 [7] =
W S [8] =
W C [8] =
0010.11111100
1010.00000000
001.10000011
001.01111111
101.00000000
y = 4
2W S [8] =
2W C [8] =
F1 [8] =
W S [9] =
W C [9] =
0010.11111110
1010.00000000
001.10000001
001.01111111
101.00000000
y = 4
s5 = 1
F1 [4] = 110.01011000
s6 = 1
F1 [5] = 001.10001100
s7 = 1
F1 [6] = 110.01110110
s8 = 1
F1 [7] = 110.01111011
s9 = 1
F1 [8] = 110.01111101
We perform 9 iterations to compute the additional bit required for rounding. Since w [9] < 0 the correction step has to be performed. Thus
s9 = 2. The result is
s = 0.111 = (0.11000000)2
1 11 11 2
c) Radix-4, sj {2, 1, 0, 1, 2}, carry-save residual
a
Since = r1 =
111.10010000.
2
3
< 1, s0 should be 1. Therefore w [0] = 1 s0 =
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
4
4W S [0] =
4W C [0] =
F1 [0] =
W S [1] =
W C [1] =
1110.01000000
0000.00000000
001.11000000
11.10000000
00.10000000
S = 1.0000
y = 1110.010
S [0] = 1
s1 = 1
S [1] = 0.11
4W S [1] =
4W C [1] =
W S [2] =
W C [2] =
1110.00000000
0010.00000000
00.00000000
00.00000000
S = 0.1100
y = 0000.000
s2 = 0
S [2] = 0.1100
4W S [2] =
4W C [2] =
0000.00000000
0000.00000000
S = 0.1100
y = 0000.000
s3 = 0
S [3] = 0.110000
Since w = 0, the rest of the digits of S are 0. We perform 4 iterations to
take into account the generation of the additional bit required for rounding. The radix-4 digits of the result are s0 = 1, s1 = 1 , s2 = 0, s3 = 0,
s4 = 0 and s5 = 0. The result is
s = (0.11000000)2
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
5
Exercise 6.3
a) Use S [j ] in its original signed digit form
In this case it is not necessary the on-the-y conversion of S [j ] for implementing the recurrence. Neverthless the register K [j ] is stil necessary.
F [j ] is computed as
Sj +1 2S [j ] + Sj +1 r(j +1)
which requires a single concatenation of Sj +1 , and a digit multiplication
by Sj +1 . Since F [j ] is represented in signed-digit form, the adder of the
recurrence is more complex, that is, both operands are redundant.
b) Convert S [j ] to twos complement representation
The is conversion on-the-y, and since this conversion is already necessary,
it does not introduce additional complexity. The adder is simpler that
in a) since one operand is in nonredundant form. More specically the
term Sj +1 2S [j ] + Sj +1 r(j +1) is generated in nonredundant form as
follows:
Sj +1 0
Concatenate Sj +1 to 2S [j ] in position j + 1 . Set the most signicant
digit to one to have a negative operand (the weight of the most
signicant digit is negative). Then perform digit multiplication.
Sj +1 < 0
In this case
2S [j ] + Sj +1 r(j +1) = 2 S [j ] r j + (2r Sj +1 ) r(j +1)
The term S [j ] r j is available from the on-the-y conversion module. The term 2r Sj +1 is precomputed for every digit and is concatenated to 2 S [j ] r j in postion j + 1 . Finally, the digit multiplication is performed.
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
6
Exercise 6.5
a) Network for digit selection
Figure E6.5a shows the network for the selection of sj +1 and sj +2 in a
radix-2 square root implementation using two radix-2 overlapped stages.
2
2
2 w[j]
2 w[j]
2F1 [j]
2F-1 [j]
2(2w[j])
CSA
4
4
4
SELSQR
2w[j]
CSA
4
4
SELSQR
4
SELSQR
4
SELSQR
2
2
4
s j+1
2
2
3-1 MUX
2
s j+2
Figure E6.5a: Network for digit selection.
b) Network to produce the next residual
In Figure E6.5b the network producing the next residual is illustrated.
c) Delay analysis
Conventional implementation
Computing the delay in the critical path we have
tcycle = tSELSQRT (4) + tbuf f (1) + tmux (1) + tHA (1) + treg (2) = 9tg
The latency of the conventional implementation (8 fractional bits)
can be computed as 8 tcycle = 8 9tg = 72tg .
Overlapped implementation
Computing the delay in the critical path we have that the delay to
produce W [j + 1] (that is, the delay from W [j ] to W [j + 1]) is
tSELSQRT (4) + tbuf f (1) + tmux (1) + tHA (1) = 7tg
Moreover, the delay to produce sj +2 (delay of CSA + delay of selection network + delay of 3-1 multiplexer) is
tCSA (2) + tSELSQRT (4) + tmux (1) + tbuf f (1) = 8tg
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
7
w[0]
initial values
j=0
j=0
F1 , F-1 , K REGISTERS
WS,WC REGISTERS
K[j]
F-1 [j]
s j+1
F1 [j]
F1 , F-1 , K MODULE
MUX
2w[j]
K[j+1]
3-2 CSA
F-1 [j+2]
s
F1 [j+2]
j+2
F1 , F-1 , K MODULE
MUX
2w[j+1]
K[j+2]
F1 [j+2]
3-2 CSA
F [j+2]
-1
w[j+2]
Figure E6.5b: Network to produce the next residual.
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
8
Finally, the delay to produce W [j + 2] (delay to produce sj +2 +
delay of buer + delay of mux + delay of HA) can be computed as
8tg + 1tg + 1tg + 1tg = 12tg
Adding the register delay we get tcycle = 11tg + 2tg = 13tg . Computing the latency of the overlapped implementation (8 fractional bits)
we get 4 tcycle = 4 13tg = 52tg
Exercise 6.8
We compute the radix-4 square root of x = (53)10 = (00110101)2 . Since
n = 8, we perform a right-shift of m = 2 bits and produce x = .11010100.
The number of bits of the integer result is 82 = 3. Consequently, two radix2
4 iterations are necessary. We have S [0] = 1 and w [0] = x 1 = 11.11010100.
Note that no alignment to digit boundary is needed, since the square root
algorithm does not require to compute a remainder.
The iterations are as follows:
4W S [0] =
4W C [0] =
F1 [0] =
W S [1] =
W C [1] =
1111.01010000
0000.00000000
001.11000000
10.10010000
10.10000000
4W S [1] =
4W C [1] =
1010.01000000
1010.00000000
y = 1111.0101
s1 = 1
S [1] = 0.11
y = 0100.0100
s2 = 2
S [2] = 0.1110
We do not need to compute w [2]. Therefore the result is
s = 23 (0.111) = 111 = (7)10
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
9
Exercise 6.13
We develop a radix-4 selection function for J = 3, t = 3 and = 4.
k>0
min (Uk1 (Ii )) = 2
max (Lk (Ii )) = 2
1
+ i 24
2
k
1
+ (i + 1) 24
2
1
3
k
2
3
k0
min (Uk1 (Ii )) = 2
max (Lk (Ii )) = 2
1
+ (i + 1) 24
2
1
+ i 24
2
k
2
3
k
+ k
Lk = max ( Lk (Ii ) 3 ) mk (i) min ( Uk1 (Ii )) 23
3
2
3
1
3
2
44
= Uk1
To improve the presentation of results, we use a bound for max (Lk (Ii )) .
2
More specically, we want an upper bound of the term k 2 44 . For
3
2
25
4
1
1
1
k = 0 we have 9 44 = 576 < 512 . For k = 1 we have 5 44 = 2304 < 64 .
3
The selection constants are presented in Table E6.13. Note that we give only
half of the table (for S [j ] = 8, 9, 10, 11) since there is an interval U2 L1 that
is negative. Consequently, there is no selection function for t = 3 and = 4.
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises
10
S [j ]
8
9
10
11
L2 , U 1
12, 12
14, 14
15, 15
16, 17
m2
12
14
15
16
L1 , U 0
3, 4
4, 5
4, 5
4, 6
m1
4
4
4
4
L0 , U1
5, 4
5, 5
6, 5
7, 5
m0
4
5
6
6
L1 , U2
13, 13
14, 15
16, 16
18, 17
m1
13
X
16
18
Table E6.13: Selection interval and mk constants.
S [j ]: real value= shown value/16.
Lk , Uk1 and mk : real value = shown value/8.
Digital Arithmetic - Ercegovac & Lang 2004
Chapter 6: Solutions to Exercises