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Arithmetic Circuits I

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Arithmetic Circuits I

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  1. Arithmetic Circuits I

  2. Iterative Combinational Circuits • Like a hierachy, except functional blocks per bit

  3. Adders • Great example of Iterative design • Design a 1-bit adder circuit, then expand to n-bit adder • Look at • Half adder – which is a 2-bit adder • Inputs are bits to be added • Outputs: result and possible carry • Full adder – includes carry in, really a 3-bit adder

  4. Fig. 4-2 Half Adder • S = X  Y • C = XY

  5. Full Adder • Three inputs. Two are operand bits, third is Cin • Two outputs: sum and carry

  6. K Map for S • What is this? In a half adder the sum bit: S = X  Y

  7. K Map for C In a half adder the carry bit: C = XY

  8. Using two half Adders to Build a Full Adder • Full adder functions • Half adder functions S = X  Y C = XY

  9. Fig. 4-4 Two Half Adders (and an OR)

  10. Ripple-Carry Adder • Straightforward – connect full adders • C4 : Chain carry-out to carry-in of FA of bits A4 & B4 • C0 in case this is part of larger chain • otherwise just set to zero

  11. Hierarchical 4-Bit Adder • We can easily use hierarchy here • Design half adder • Use in full adder • Use full adder in 4-bit adder

  12. Binary Subtraction • Example: • Let M = (19)10 ; N = (30)10 ; How is M-N computed? • In decimal notation: (19)10 – (30)10 = - (11)10 • In binary: (10011)2 - (11110)2 = - (01011)2

  13. Using 2’s & 1’s Complement Representation of Signed Numbers • People use complemented interpretation for signed numbers • 1’s complement • 2’s complement

  14. 1’s Complement • Given: binary number N with n digits 1’s complement is defined as (2n – 1) - N • Note that (2n – 1) is a number with n bits, all of them 1 • For n = 4, (2n – 1) = (15)10 = 1111

  15. Example: Find 1’s Complement of N =101 1001 • 1’s Complement of N is (2n – 1) - N • Notice that 1’s complement is complement of each bit

  16. 2’s Complement • Given: binary number N with n digits 2’s complement defined as 2n – N for N  0 0 for N = 0 • Note that since 1’s complement is (2n – 1) – N • 2’s complement is just a 1 added to 1’s complement

  17. Example: Find 2’s Complement of N =101 1001 • 2’s Complement of N is 2n - N • To find 2’s complement of N • Find is 1’s complement of N then add 1

  18. Important Property • Complement of a complement generates original number

  19. Subtraction Using 2s Complement:Compute M-N • Add 2’s complement of N to M • M + (2n – N) = { M – N + 2n } = F • If M  N, this addition will generate a carry, 2n • Discard carry; the result gives the answer of M-N which is positive • If M < N, no carry; M-N is negative; the answer is –(N-M). To compute the answer: • Take 2’s complement of F • 2n – F = 2n - { M – N + 2n } = N-M • Place minus sign in front { - (N-M) }; This is the answer • Hence to compute (M-N): do 1 & 2 if M  N; OR 1 & 3 If M < N

  20. 1710 Compute M-N;Example 1: M > N • M = (84)10= (101 0100)2 • N = (67)10 = (100 0011)2; • 2’s comp N = ( 011 1100 ) + 1 = 011 1101 M > N; Carry generated; Discard carry; Result is positive M - N

  21. Compute M-N; Example 2: M < N;M =6710 = 100 0011; N = 8410 = 101 0100 • M = 100 0011 minus N = 101 0100 • No end carry • Answer: - (2’s complement of Sum) • Answer: - 0010001 We said numbers are unsigned. What does this mean? How is -1710 represented?

  22. Subtractor • Compute M-N • Add 2’s complement of N to M: M + (2n – N) • If M >= N, need only “one adder” and a “complementer of N” to do the subtraction. • If M < N, we also need to take 2’s complement of adder’s output to produce magnitude of result 2n -{ M + (2n – N)} = N-M

  23. S is low for add, high for subtract Inverts each bit of B if S is 1 Add 1 to make 2’s complement Fig. 4-7 Adder-Subtractor Design: (A+B)OR (A-B) • Output is result if A >= B; Discard carry • Output is 2’s complement of result if B > A

  24. Signed Binary • Signed magnitude • Left bit is sign, 0 positive, 1 negative • Other bits are number • 2’s complement • 1’s complement

  25. Example in 8-bit byte • Represent -910 = -(0000 1001) in different ways “Signed magnitude” of - (0000 1001) is 10001001 “Signed 1’s Complement” of - (0000 1001) is 11110110 “Signed 2’s Complement” of - (0000 1001) is 11110111

  26. Observations (assume 4-bit numbers) • 1’s C and Signed Magnitude have two zeros • 2’s C has more negative numbers than positive • All negative numbers have 1 in highest-order bit

  27. Advantages/Disadvantages • Signed magnitude & One’s complement has a positive and negative zero • Two’s complement is most popular • Arithmetic operations easy

  28. Two’s Complement • Addition easy on any combination of positive and negative numbers • To compute A - B • Take 2’s complement of subtrahend B to produce -B • Add to A • This performs A + ( -B), same as A – B

  29. Examples • Assume we use 2’s complement representation of signed numbers • Store each number in 8 bits 6 = (0000 0110) ; 13 = (0000 1101) -6 = (1111 1010) ; -13 = (1111 0011) • Addition • 6 + 13 • -6 + 13 • 6 + (- 13) • (-6) + (-13) • Subtraction • -6 - (-13) • 6 - (- 13)

  30. Overflow • Overflow means that result can not be represented with the number of bits used • Two cases of overflow for addition of signed numbers • Two large positive numbers overflow into sign bit • Not enough bits for result • Two large negative numbers added • Same – Not enough bits for result

  31. Examples:Work on Board Assume 1- We use “2’s complement representation” of signed numbers 2- Store each number in 4 bits : b3 b2 b1 b0

  32. Example 1 • 7 + 7 • Overflow: can not represent 14 using only 4 bits • Generates NO CARRY, C4 = 0

  33. Example 2 • -7 - 7 • Overflow: can not represent -14 using only 4 bits • Generates CARRY, C4 = 1

  34. Example 3 • 4 + 4 • Overflow: can not represent 8 using only 4 bits • Generates NO CARRY, C4 = 0

  35. Example 4 • 7 – 7 • NO Overflow • Generates CARRY, C4 = 1

  36. Fig. 4-8 --------------------------------------------------------- Fig. 5-9 Overflow Detection for Addition and Subtraction Overflow Detection • Condition for overflow: • either Cn-1 or Cn is high, but not both • 7 + 7; only Cn-1 (C3) is high ; overflow • -7 - 7; only Cn (C4)is high ; overflow • 4 + 4; only Cn-1 (C3) is high ; overflow • 7 – 7; BOTH Cn-1 & Cn (C4 & C3) are high; no overflow