# UNIT IV: DATA PATH DESIGN

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1

2 UNIT IV: DATA PATH DESIGN

3 Agenda Introduc/on Fixed Point Arithme/c Addi/on Subtrac/on Mul/plica/on & Serial mul/plier Division & Serial Divider Two s Complement (Addi/on, Subtrac/on) Booth s algorithm Design of basic serial adders Serial & parallel(ripple) Subtractors High speed adders Floa/ng Point Arithme/c Addi/on & Subtrac/on Guard, Round & S/cky bits Mul/plica/on & Division Combina/onal & Sequen/al ALU: Assignment 2

4 Data Representa/on

5 Fixed Point Numbers

6 Fixed Point Numbers: Decimal to Binary Conversion

7 Fixed Point Numbers: Unsigned Integers

8 Signed Integers

9 Fixed Point Arithme/c Four basic arithme/c instruc/ons: Addi/on Subtrac/on Mul/plica/on Division

12 Subtrac/on

13 Subtrac/on: Example

14 Mul/plica/on

15 Mul/plica/on Example

16 Mul/plica/on Example Multiplication of two 4-bit unsigned binary integers produces an 8-bit result. Multiplication of two 4-bit signed binary integers produces only a 7-bit result (each operand reduces to a sign bit and a 3-bit magnitude for each operand, producing a sign-bit and a 6-bit result).

17 Mul/plica/on: A serial Mul/plier

18 Mul/plica/on: Example

19 Division: Example

20 Division: Example

21 Division: Example of Base 2 (7 / 3 = 2) 10 with a remainder R of 1. Equivalently, (0111/ 11 = 10) 2 with a remainder R of 1.

22 Sign-Magnitude LeV most bit is sign bit 0 means posi/ve 1 means nega/ve +18 = = Problems Need to consider both sign and magnitude in arithme/c Two representa/ons of zero (+0 and -0)

23 Two s Complement: Signed Integers Numbers are added or subtracted on the number circle by traversing clockwise for addi/on and counterclockwise for subtrac/on. Unlike the number line (a) where overflow never occurs, overflow occurs when a transi/on is made from +3 to -4 while proceeding around the number circle when adding, or from -4 to +3 while subtrac/ng. 2-bits 0,1,2,3 3-bits , (-1)-(-4) 4-bits , (-1)-(-8)

24 Two s Complement: Unsigned Integers

25 Two s Complement

26 Two s Complement

28 Two s Complement: Subtrac/on

29 Two s Complement: Subtrac/on

30 Addi/on and Subtrac/on Normal binary addi/on Monitor sign bit for overflow Take twos compliment of substahend and add to minuend i.e. a - b = a + (-b) So we only need addi/on and complement circuits

31 Hardware for Addi/on and Subtrac/on

32 2's complement of floa/ng point numbers: Example

33 2's complement of floa/ng point numbers To create 2's complement: Take the number given by you Start on the least significant bit and locate the first 1 marked red Then flip every bit aver that first one (1 change to zero and vice verse) >

34 2's complement of floa/ng point numbers: Example Example: = -3.75

35 Conversion of 2 s complement to decimal

36 Booth s Algorithm :Mul/plying Nega/ve Numbers

37 Example of Booth s Algorithm Q 0 Q -1 Opera8on 0 0 ShiVing Multiplicand (M) = 0111 = 7 Multiplier (Q) = 0011 = ShiVing 1 0 A = A M ShiVing 0 1 A = A + M ShiVing No. of cycles = No. of bits

38 Half Adder It adds two 1 bits but has no provision to include the carry output from previous bit posi/on.

39 Full Adder It is a combina/onal circuit that forms the arithme/c sum of three input bits. It consists of three inputs & two outputs. Two input variables as A & B. The third input Cin represents the carry from the previous bit posi/on.

40 Full Subtractor Truth table and schematic symbol for a ripple-borrow subtractor:

41 Serial Adder A serial adder has only a single bit adder. It is used to perform addi/on of two numbers sequen/ally bit by bit. Addi/on of one bit posi/on takes one clock cycle. Thus for n-bit serial adder, n clock cycles are required to complete the addi/on process & get the result. At each cycle, the carry produced by a bit posi/on is stored in a flip-flop & it is given as an input during the next cycle as carry.

44 4-bit Ripple Carry Adder Two binary numbers A and B are added from right to lev, crea/ng a sum and a carry at the outputs of each full adder for each bit posi/on.

46 Ripple-Borrow Subtractor A ripple-borrow subtractor can be composed of a cascade of full subtractors. Two binary numbers A and B are subtracted from right to lev, crea/ng a difference and a borrow at the outputs of each full subtractor for each bit posi/on

47 Ripple Adder/Subtractor (Combined) A single ripple-carry adder can perform both addi/on and subtrac/on. The subtrac/on A-B ca be done by taking two s complement of B & adding it to A. 2 s complement can be obtained by taking 1 s complement & adding one to the least significant pair of bits. 1 s complement can be implemented with inverters & a one can be added to the sum through the input carry to get 2 s complement.

49 Carry Lookahead Adder One method of speeding up this process by elimina/ng inter stage carry delay is called lookahead-carry addi/on. This method look at the lower-order bits of augend & addend to see if a higher order carry is to be generated. It uses two func/ons: carry generate & carry propogate.

51 Carry-Lookahead Addi/on s i = a i b i c i + a i b i c i + a i b i c i + a i b i c i c i+1 = b i c i + a i c i + a i b i c i+1 = a i b i + (a i + b i )c i Carries are represented in terms of G i (generate) and P i (propagate) expressions. c i+1 = G i + P i c i G i = a i b i and P i = a i + b i c 0 = 0 c 1 = G 0 c 2 = G 1 + P 1 G 0 c 3 = G 2 + P 2 G 1 + P 2 P 1 G 0

52 Floa/ng Point Representa/on

53 Floa/ng Point Arithme/c Floating point arithmetic differs from integer arithmetic in that exponents must be handled as well as the magnitudes of the operands. The exponents of the operands must be made equal for addition and subtraction. The fractions are then added or subtracted as appropriate, and the result is normalized. Increase the power of exponent: Right shift Decrease the power of exponent: Left shift

54 Floa/ng Point Arithme/c Ex: Perform the floating point operation: ( ) 2 Start by adjusting the smaller exponent to be equal to the larger exponent, and adjust the fraction accordingly. Thus we have = , rounding off to ( ) 2 4 = = Therefore, losing of precision in the process. The resulting sum is ( ) 2 4 = = , and rounding to three significant digits, , and we have lost another in the rounding process.

55 Floa/ng Point Arithme/c (Cont ) If we simply added the numbers using as much precision as we needed and then applied rounding only in the final normalization step, then the calculation would go like this: Normalizing yields , and rounding to three significant digits using the round to nearest even method yields Which calculation is correct.100 x 2 5 or.101 x 2 5? According to the IEEE 754 standard, the final result should be the same as if the maximum precision needed is used before applying the rounding method, and so the correct result is So what do we do?

56 Rounding using G, R, S 1.BBGRXXX Guard bit: LSB of result Round bit: 1 st bit removed Sticky bit: OR of remaining bits If G=1 & R=1, add 1 to LSB If G=0 & R=0 or 1, no change If G=1 & R=0, look at S If S=1, add 1 to LSB If S=0, round to the nearest even (i.e. if LSB =1, then make it 0 (add 1 to LSB) & if LSB =0 then no change 1 = ODD, 0 = EVEN)

57 Guard, Round, and S/cky Bits This raises the issue of how to compute the intermediate results with sufficient accuracy and without requiring too much hardware, and for this we use guard, round, and sticky bits. For the previous example, applying guard (g) and round (r) bits with the round toward nearest even method, we have:

58 Guard, Round, and S/cky Bits Only one extra bit is needed for this intermediate result, the guard bit (g), but we also show the round bit (r = 0) to locate its position. As we shift the number to normalize, we set a sticky bit (s) if any of the shifted out bits are nonzero. For this case, there are no nonzero bits to the right of the r bit and so s = 0:

59 Guard, Round, and S/cky Bits (Cont ) Now for the rounding step: simply append the sticky bit to the right of the result before rounding. There is no tie as there would be for and so we round up, otherwise we would have rounded down to the closest even number (.100): For this case, the guard, round, and sticky bits changed our previous result. Note that if r is 0 instead of 1, so that the grs combination is 100, we would have rounded down to.100 because.100 is even whereas.101 is not.

60 Examples: Guard, Round, and S/cky Bits Frac/on GRS Incr? Rounded N Y N Y Y Y If G=1 & R=1, add 1 to LSB If G=0 & R=0 or 1, no change If G=1 & R=0, look at S If S=1, add 1 to LSB If S=0, round to the nearest even (i.e. if LSB =1, then make it 0 (add 1 to LSB) & if LSB =0 then no change 1 = ODD, 0 = EVEN)

61 Floa/ng Point Mul/plica/on/Division Floating point multiplication/division are performed in a manner similar to floating point addition/subtraction, except that the sign, exponent, and fraction of the result can be computed separately. Like/unlike signs produce positive/negative results, respectively. Exponent of result is obtained by adding exponents for multiplication, or by subtracting exponents for division. Fractions are multiplied or divided according to the operation, and then normalized. Ex: Perform the floating point operation: ( ) / ( ) 2 The source operand signs are the same, which means that the result will have a positive sign. We subtract exponents for division, and so the exponent of the result is 5 4 = 1. We divide fractions, producing the result: 110/100 = Putting it all together, the result of dividing ( ) by ( ) produces ( ). After normalization, the final result is ( ).

62 FP Arithme/c +/- Check for zeros Align significands (adjus/ng exponents) Add or subtract significands Normalize result

63 FP Addi/on & Subtrac/on Flowchart

64 FP Arithme/c x/ Check for zero Add/subtract exponents Mul/ply/divide significands (watch sign) Normalize Round All intermediate results should be in double length storage

65 Floating Point Multiplication

66 Floating Point Division

67

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