Chapter 6: Solutions to Exercises

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1 1 DIGITAL ARITHMETIC Miloš D. Ercegovac and Tomás Lang Morgan Kaufmann Publishers, an imprint of Elsevier Science, c 00 Updated: September 3, 003 With contributions by Elisardo Antelo and Fabrizio Lamberti Exercise 6.1 a) Radix-, s j { 1, 0, 1}, conventional (nonredundant) residual We have x = 1 8 = and ρ = 1. We choose s 0 = 0. Therefore the initialization is w [0] = x s 0 = We use the result-digit selection function for redundant residual but we consider only integer bits since the range of the residual estimate is smaller than in the redundant case. w [0] = ŷ = 1 s 1 = 1 F 1 [0] = F 1 [0] = w [1] = w [1] = ŷ = 1 s = 1 F 1 [1] = F 1 [1] = w [] = w [] = ŷ = 0 s 3 = 1 F 1 [] = F 1 [] = w [3] = w [3] = ŷ = s = 1 F 1 [3] = F 1 [3] = w [] = w [] = ŷ = s 5 = 1 F 1 [] = F 1 [] = w [5] = w [5] = ŷ = s 6 = 1 F 1 [5] = F 1 [5] = w [6] =

2 (continues on next page) w [6] = ŷ = s 7 = 1 F 1 [6] = F 1 [6] = w [7] = w [7] = ŷ = s 8 = 1 F 1 [7] = F 1 [7] = w [8] = w [8] = ŷ = s 9 = 1 F 1 [8] = F 1 [8] = w [9] = We perform 9 iterations to compute the additional bit required for rounding. Since w [9] < 0 the correction step has to be performed. Thus s 9 =. The result is s = = ( ) b) Radix-, s j { 1, 0, 1}, carry-save residual W S [0] = ŷ = 1 s 1 = 1 W C [0] = F 1 [0] = F 1 [0] = W S [1] = W C [1] = W S [1] = ŷ = 1 s = 1 W C [1] = F 1 [1] = F 1 [1] = W S [] = W C [] = W S [] = ŷ = 0 s 3 = 1 W C [] = F 1 [] = F 1 [] = W S [3] = W C [3] = W S [3] = ŷ = s = 1 W C [3] = F 1 [3] = F 1 [3] = W S [] = W C [] = (continues on next page)

3 3 W S [] = ŷ = s 5 = 1 W C [] = F 1 [] = F 1 [] = W S [5] = W C [5] = W S [5] = ŷ = s 6 = 1 W C [5] = F 1 [5] = F 1 [5] = W S [6] = W C [6] = W S [6] = ŷ = s 7 = 1 W C [6] = F 1 [6] = F 1 [6] = W S [7] = W C [7] = W S [7] = ŷ = s 8 = 1 W C [7] = F 1 [7] = F 1 [7] = W S [8] = W C [8] = W S [8] = ŷ = s 9 = 1 W C [8] = F 1 [8] = F 1 [8] = W S [9] = W C [9] = We perform 9 iterations to compute the additional bit required for rounding. Since w [9] < 0 the correction step has to be performed. Thus s 9 =. The result is s = = ( ) c) Radix-, s j {, 1, 0, 1, }, carry-save residual Since ρ = a r 1 = 3 < 1, s 0 should be 1. Therefore w [0] = 1 s 0 =

4 W S [0] = Ŝ = S [0] = 1 W C [0] = ŷ = s 1 = 1 F 1 [0] = S [1] = 0.11 W S [1] = W C [1] = W S [1] = Ŝ = s = 0 W C [1] = ŷ = S [] = W S [] = W C [] = W S [] = Ŝ = s 3 = 0 W C [] = ŷ = S [3] = Since w = 0, the rest of the digits of S are 0. We perform iterations to take into account the generation of the additional bit required for rounding. The radix- digits of the result are s 0 = 1, s 1 = 1, s = 0, s 3 = 0, s = 0 and s 5 = 0. The result is s = ( )

5 5 Exercise 6.3 a) Use S [j] in its original signed digit form In this case it is not necessary the on-the-fly conversion of S [j] for implementing the recurrence. Neverthless the register K [j] is stil necessary. F [j] is computed as ( S j+1 S [j] + S j+1 r (j+1)) which requires a single concatenation of S j+1, and a digit multiplication by S j+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 two s complement representation The conversion is on-the-fly, 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 specifically the term S j+1 S [j] + Sj+1 r (j+1)) is generated in nonredundant form as follows: S j+1 0 Concatenate S j+1 to S [j] in position j +1. Set the most significant digit to one to have a negative operand (the weight of the most significant digit is negative). Then perform digit multiplication. S j+1 < 0 In this case ( S [j] + S j+1 r (j+1)) = ( S [j] r j) + (r S j+1 ) r (j+1) The term S [j] r j is available from the on-the-fly conversion module. The term r S j+1 is precomputed for every digit and is concatenated to ( S [j] r j) in postion j + 1. Finally, the digit multiplication is performed.

6 6 Exercise 6.5 a) Network for digit selection Figure E6.5a shows the network for the selection of s j+1 and s j+ in a radix- square root implementation using two radix- overlapped stages. w[j] F [j] F [j] 1-1 w[j] CSA (w[j]) CSA w[j] SEL SQR SEL SQR SEL SQR SEL SQR s j MUX s j+ 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 t cycle = t SELSQRT () + t buff (1) + t mux (1) + t HA (1) + t reg () = 9t g The latency of the conventional implementation (8 fractional bits) can be computed as 8 t cycle = 8 9t g = 7t g. 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 t SELSQRT () + t buff (1) + t mux (1) + t HA (1) = 7t g Moreover, the delay to produce s j+ (delay of CSA + delay of selection network + delay of 3-1 multiplexer) is t CSA () + t SELSQRT () + t mux (1) + t buff (1) = 8t g

7 7 initial values w[0] j=0 j=0 F 1, F -1, K REGISTERS K[j] WS,WC REGISTERS F [j] -1 K[j+1] F 1, F -1, K MODULE s j+1 F [j] 1 MUX w[j] 3- CSA F [j+] -1 s j+ F [j+] 1 K[j+] F [j+] 1 F [j+] -1 F 1, F -1, K MODULE MUX 3- CSA w[j+1] w[j+] Figure E6.5b: Network to produce the next residual.

8 8 Finally, the delay to produce W [j + ] (delay to produce s j+ + delay of buffer + delay of mux + delay of HA) can be computed as 8t g + 1t g + 1t g + 1t g = 1t g Adding the register delay we get t cycle = 11t g + t g = 13t g. Computing the latency of the overlapped implementation (8 fractional bits) we get t cycle = 13t g = 5t g Exercise 6.8 We compute the radix- square root of x = (53) 10 = ( ). Since n = 8, we perform a right-shift of m = bits and produce x = The number of bits of the integer result is 8 = 3. Consequently, two radix- iterations are necessary. We have S [0] = 1 and w [0] = x 1 = 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: W S [0] = W C [0] = ŷ = s 1 = 1 S [1] = 0.11 F 1 [0] = W S [1] = W C [1] = W S [1] = W C [1] = ŷ = s = S [] = We do not need to compute w []. Therefore the result is s = 3 (0.111) = 111 = (7) 10

9 9 Exercise 6.13 We develop a radix- selection function for J = 3, t = 3 and δ =. k > 0 k 0 ( ) ( 1 min (U k 1 (I i )) = + i k 1 ) 3 ( ) ( 1 max (L k (I i )) = + (i + 1) k ) 3 ( ) ( 1 min (U k 1 (I i )) = + (i + 1) k 1 ) 3 ( ) ( 1 max (L k (I i )) = + i k ) ( + k ) 3 3 L k = max ( L k (I i ) 3 ) m k (i) min ( U k 1 (I i )) 3 3 = Ûk 1 To improve the presentation of results, we use a bound for max (L k (I i )). More specifically, we want an upper bound of the term ( k 3). For k = 0 we have 9 = < For k = 1 we have ( ) 5 3 = 5 30 < 1 6. The selection constants are presented in Table E6.13. Note that we give only half of the table (for Ŝ[j] = 8, 9, 10, 11) since there is an interval Û L 1 that is negative. Consequently, there is no selection function for t = 3 and δ =.

10 10 Ŝ [j] L, Û1 1, 1 1, 1 15, 15 16, 17 m L 1, Û0 3,, 5, 5, 6 m 1 L 0, Û 1 5, 5, 5 6, 5 7, 5 m L 1, Û 13, 13 1, 15 16, 16 18, 17 m 1 13 X Table E6.13: Selection interval and m k constants. Ŝ[j]: real value= shown value/16. L k, Û k 1 and m k : real value = shown value/8.

Chapter 5: Solutions to Exercises

Chapter 5: Solutions to Exercises 1 DIGITAL ARITHMETIC Miloš D. Ercegovac and Tomás Lang Morgan Kaufmann Publishers, an imprint of Elsevier Science, c 2004 Updated: September 9, 2003 Chapter 5: Solutions to Selected Exercises With contributions