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Sequential Multiplication - CAD Based Logic Design - Lecture Slides, Slides of Computer Science

University of AllahabadComputer Science

These are the Lecture Slides of CAD Based Logic Design which includes Divide and Conquer, Stitching Up, Computation, Design Problem, Leaf Subproblem, Carry-Lookahead Adder, Intricate Techniques, Fast Tree Designs, Associative Operations etc. Key important points are: Sequential Multiplication, Complement Multiplication Example, Booth Multiplication Example, Intricate Techniques, Fast Tree Designs, Associative Operations, Divide and Conquer, Stitching Up

Typology: Slides

2012/2013

Uploaded on 03/21/2013

dheeraj
dheeraj 🇮🇳

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Sequential Multiplication

Regular A&S 2’s Complement Multiplication Example

Multplicand Y = 11001101 = -51, Multiplier X = 01100111 = 103, -Y = 10110011= - Cout-AC-Q = 0| 00000000 | 01100111 M = 11001101 Iter 1: Q0 = 1, perform add AC+Y: 00000000

  • 11001101 0 11001101 = Cout-AC No ovfl, ASR AC-Q: 11100110 | 10110011 Iter 2: Q0 = 1, perform add AC+Y: + 11001101 1 10110011 = Cout-AC No ovfl, ASR AC-Q: 11011001 | 11011001 Iter 3: Q0 = 1, perform add AC+Y: + 11001101 1 10100110 = Cout-AC No ovfl, ASR AC-Q: 11010011 | 01101100 Iter 4: Q0 = 0, only ASR AC-Q: 11101001 | 10110110 Iter 5: Q0 = 0, only ASR AC-Q: 11110100 | 11011011 Iter 6: Q0 = 1, perform add AC+Y: + 11001101 1 11000001 = Cout-AC No ovfl, ASR AC-Q: 11100000 | 11101101 Iter 7: Q0 = 1, perform add AC+Y: + 11001101 1 10101101 = Cout-AC No ovfl, ASR AC-Q: 11010110 | 11110110 Iter 8: Q0 = 0, only ASR AC-Q : 11101011 | 01111011 P = AC-Q = 1110101101111011 = -5253 = (-51) x 103. Needed to perform 5 +’s and 8 shifts

Booth Multiplication Example

Multplicand Y = 01001101 = 77, Multiplier X = 11100111 = -25, -Y = 10110011= -

Cout-AC-Q-q-^ = 0| 00000000 | 11100111 | 0 M = 01001101 (q-^ is the “prev” bit of Q)

Iter 1: q 0 -q-^ = 10, perform sub AC-Y: 00000000

0 10110011 = Cout-AC

No ovfl, ASR AC-Q-q-^ : 11011001 | 11110011 | 1

Iter 2: q 0 -q-^ = 11, only ASR AC-Q-q-^ : 11101100 | 11111001 | 1

Iter 3: q 0 -q-^ = 11, only ASR AC-Q-q-^ : 11110110 | 01111100 | 1

Iter 4: q 0 -q-^ = 01, perform add AC+Y: +

1 01000011 = Cout-AC

No ovfl, ASR AC-Q-q-^ : 00100001 | 10111110 | 0

Iter 5: q 0 -q-^ = 00, only ASR AC-Q-q-^ : 00010000 | 11011111 | 0

Iter 6: q 0 -q-^ = 10, perform sub AC-Y:+ 10110011

0 11000011 = Cout-AC

No ovfl, ASR AC-Q-q-^ : 11100001 | 11101111 | 1

Iter 7: q 0 -q-^ = 11, only ASR AC-Q-q-^ : 11110000 | 11110111 | 1

Iter 8: q 0 -q-^ = 11, only ASR AC-Q-q-^ : 11111000 | 01111011 | 1

P = AC-Q = 1111100001111011 = -1925 = 77 x (-25)

Needed only 3 +/-’s as opposed to 6 +/-’s needed in regular A&S mult.

# of shifts remain the same (8)