1. Part One

2. HOMEWORK 1 PART B 1. Compute the following using pipeline method a. e. i.12/4 b. f. J.49/7 c. g.3*10 d. h.91*14 2. Illustrate the generation of 4 variable Dertouzos Table 3. Realize XOR using threshold gates 4. Realize half subtractor as a cascade of threshold gates. 5. Realize full subtractor as a cascade of threshold gates.

3. 6. Realize the following functions in cascade : f1 = X1’X2’X4 f2 = X2X3’X4’ f3 = X1X2X3 7. realize as a cascade of threshold gates: S = A C0 = A’( 8. realize as a cascade of three threshold gates: f = (0,3,4,7,9,13)

4. ANSWER TO QUESTION ONE Rooting a. 10. 00 00 00 01 ---------------------- 01 00 F1=1 Answer : 01 01 ---------------------- 01 00 00 F2 = 0 00 10 01 ---------------------- 00 01 11 00 F3 = 1 00 01 01 01 ---------------------- 00 00 01 11 F4 = 1

5. b. 10. 10 00 00 01 ---------------------- 01.10 F1=1 01. 01 ---------------------- 00. 01 00 F2 = 1 00. 11 01 ---------------------- 00. 01 00 00 F3 = 0 Answer : 00. 01 10 01 ------------------------ 00. 01 00 00 F4 = 0

6. c. 0101. 00 11 00 01 ---------------------- 00 01 F1=1 Answer : 01 01 ---------------------- 00 01. 00 F2 = 0 00 10. 01 ---------------------- 00 01. 00 11 F3 = 0 00 01.00 01 ---------------------- 00 00. 00 10 00 F4 = 1 00 00.10 01 01 ---------------------- 00 00.00 10 00 F5 = 0

7. d. 11 01. 11 01 ---------------------- 10 01 F1=1 Answer : 01 01 ---------------------- 01 00. 11 F2 = 1 00 11.01 ---------------------- 00 01. 10 F3 = 1

8. Squaring e. 1 F1 F2 Answer : 0 1 (F1 = 1) 0 1 0 1 (F2 = 1) ------------ 1 0 0 1

9. f 1 0 1. 1 F1 F2 F3 F4 Answer : 01 F1=1 00 00 F2=0 00 10 01 F3=1 00 01 01. 01 F4=1 ---------------- 01 11 10. 01

10. Multiplication g. 3*10 Answer : Left shift Right shift 1 1 1 1 1 0 1 0 1 0 1 0 -------------- -------------------- 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 ----------------- ---------------------- 1 1 1 1 0

11. H 91*14 1 0 1 1 0 1 1 1 1 1 0 -------------------- 1 0 1 1 0 1 1 0 0 0 0 0 0 0 ------------------------------- 1 0 0 1 1 1 1 1 0 1 0

12. DIVISION I.12/4 12 10 -> 1100 2 4 10 -> 100 2 1100 100 ---------- 100 F1=1 100 ------------ 000 F2=1 Answer: 1100/100 = 11 11 2 -> 3 10

13. J.49/7 110001 111 ------------ 1100 F1=0 Answer : 110001/111 = 0111 0111 0111 -> 7 10 ------------- 010100 F2=1 000101 -------------- 00111100 F3=1 00001101 ---------------- 00101111 F4=1

14. ANSWER TO QUESTION NUMBER TWO Arbitrary weight for W0 <= W1 <= W2 are chosen W0=40, W1=50, W2=60 Atable of weighted sum is built for the possible minterm in 3 variables. Upper threshold limit is selected 150/2 = 75. A list of test threshold that fall between the weighted sums that are less than 75 is constructed : (-1, 20, 45, 55, 75) For each test threshold, a truth table is created, such that a true value is obtained if that minterms weight sum is less than test threshold.

15. To calculate b’s T=-1 b0=(2x0)-8=-8 b= 8, 0, 0, 0 ---(1) T=20 b0=(2x1)-8=-6 b=6, 2, 2, 2 ---(2) b1,2,3=2(0-1)=-2 T=45 b0=2x2-8=-4 b=4, 4, 4, 0 ---(3) b1=2(0-2)=4 b2=2(0-2)=4 b3=2(1-1)=0 T=55 b0=2x3-8=-2 b=2, 6, 2, 2 ---(4) b1=2(0-3)=-6

16. b2=2(1-2)=-2 b3=2(1-2)=-2 T=75 b0=2x4-8=0 b=0, 4, 4, 4 ---(5) b1=2(1-3)=-4 b2=2(1-3)=-4 b3=2(1-3)=-4

17. Each vector is then sorted in decending order and duplicates are eliminated

18. ANSWER TO QUESTION NUMBER THREE Realization of XOR f=x’y+xy’ Take each minterm and realize it with one threshold gate, and then OR them f1=x’y, f2=xy’ f1: 0 1 x -1 -------- 1/2 N(1) 0 1 N(0) 1 0 y 1 1 1/2 T=0.5 x 1 1 1/2 f2: 1 0 -------- y -1 N(1) 1 0 N(0) 0 1 T=0.5

19. ANSWER TO QUESTION NO 4 Realiazation of half subtractor B=x’y D=x’y-xy’=B+xy’ For B=x’y Determine positive function B=xy Find all Minimum True and Maximum False vertices x y The Inequalities: 0 0 Wx + Wy > Wy => Wx > 0 F 0 1 W x + Wy > Wx => Wy > 0 F 1 0 Choose Wx = Wy = 1 Tmin 1 1

20. So, UL = 1x1 + 1x1 = 2 For every input which is complimented in the original function, its weight LL = 1x1 + 0x1 = 1 must be changed to -W and T to T-W T = 3/2 Wx = -1, Wy = 1, T = 1/2

21. For D = B + xy’ Generate the truth table for the 3-variable B, x, y and find the Minimum True and Maxumum False vertices The positive function: D = B + xy Tmin 0 0 1 1 1 0 Fmax 1 0 0 0 1 0

22. The inequalities : W3 > W1 W3 > W2 W1 + W2 > W1 => W2 > 0 W1 + W2 > W2 => W1 > 0 choose W1 = W2 = 1, W3 = 2 UL = 1, LL = 2, T = 3/2 Architecture of Half-Subtractor :

23. REALIZE OF A FULL ADDER S1=A’B’C’+C0 S1= =(A+B+C) C0’ S= (A+B+C) C0’

24. REALIZATION OF FULL SUBTRACTOR B = x’y + x’z + yz D = x’y’z + x’yz’ + xyz + xy’z’ Let, D1 = D + x’yz = x’y’z + x’yz’ + xyz + xy’z’ + x’yz D1 = B + xy’z’ D = (B + xy’z’) - x’yz D = (B + xy’z’)(x + y’ + z’)

25. 0 0 0 1 0 0 1 1 0 1 1 0 0 0 1 0 0 1 Unate 1 0 1 1 1 1 0 1 1 1 N(1) 5 3 3 7 N(0) 3 5 5 1

26. b0 = 2x8 - 16 = 0 b4 b3 b2 b1 b0 b1 = 2(5-3) = 4 12 -4 -4 4 0 b2 = -4 2 -1 -1 1 0 b3 = -4 a4 a3 a2 a1 a0 b4 = 12

27. REALIZATION OF THE FOLLOWING THREE FUNCTION IN CASCADE f1 = x1’ x2’ x4 f2 = x2 x3’ x4’ f3 = x1 x2 x3 1.f1 = x1’ x2’ x4 Positive function f1 = x1 x2 x4 Tmin 1 1 0 1 Fmax 1 1 1 0 1 0 1 1 0 1 1 1 The inequalities: W1 + W2 + W4 > W1 + W2 + W3 => W4 > W2 W1 + W2 + W4 > W1 + W3 + W4 => W2 > W3 W1 + W2 + W4 > W2 + W3 + W4 => W1 > W3 W3 = 0 W1 = W2 = W4 = 1 UL = 3, LL = 2, T = 5/2

28. f2=x2x1’x4’ f3=x1x2x3

29. Now we can cascade the three threshold gates in any order f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

30. REALIZE AS A CASCADE OF THRESHOLD GATES let

31. REALIZE AS A CASCADE OF THREE THRESHOLD GATES 1. Positive function f1=xwz Tmin = 1 0 1 1 Fmax = 1 1 1 0 1 1 0 1 0 1 1 1

32. The iequalities W1+W3+W4 > W1+W2 +W3 => W4 > W2 W1+W3+W4 > W1+W2 +W4 => W4 > W2 W1+W3+W4 > W1+W2 +W3 => W4 > W2 Choose W2 = 0 W1=W3=W4=1 UL=3, LL=2, T=5/2

33. Now, let the cascade gates be in f1 -> f2 -> f3 ->

34. To determine the weight for f1: (-1) (0) (1) (1) x y z w The minterms 0 0 0 0 f1 + 0 > 1.5 & both f1 & f2 0 0 1 1 f1 + 2 > 1.5 0 1 0 0 f1 + 2 > 1.5 0 1 1 1 f1 + 2 > 1.5 min weight for f1=2 for minimum weight for f2: (1) (0) (-1) (1) x y z w 0 0 1 1 f2 + 0 > 1.5 0 1 1 1 f2 + 0 > 1.5 1 0 0 1 f2 + 2 > 1.5 1 1 0 1 f2 + 2 > 1.5 min weight for f2=2

35. Part Two

36. HALF ADDER Create truth table for half adder Expand truth table to three inputs. Terms not found on first table are assigned as don’t care terms for S. S=XC’ + YC’ S=(X + Y)C’

37. FULL ADDER Create truth table for full adder C=XY + YZ + XZ

38. FULL ADDER Expand truth table to four inputs. Terms not found on first table are assigned as don’t care terms for S. S = XC’ + YC’ + ZC’ + XYZ S = (X + Y + Z)C’ + XYZ

39. HALF SUBTRACTOR Create truth table for half subtractor. Expand truth table to three inputs. Terms not found D = X’Y on first table are assigned as don’t care terms for D. D = XY’ + B

40. FULL SUBTRACTOR Create truth table for full subtractor B = X’Y + X’Z + YZ

41. FULL SUBTRACTOR Expand truth table to four inputs. Terms not found on first table are assigned as don’t care terms for D. D = Z’B + XB + Y’B +XY’Z’ D = (X + Y’ + Z’)B + XY’Z’

42. PART THREE

43. Q. Write down the detailed steps for BDD of the majority function Solution: f= Assume: X1=A’B, X2=C, X3=D’E+DF M(A’B, C, D’E+DF)=X1X2+X1X3+X2X3

45. The total result f=

46. Arithmatic cell: BDD for Fi=CoX+PiX’ D=C(B+Fi) Control cell: Fi=CoX+PiX’ E=B(C+C’)+Cfi=BC’+D

48. Implementation: “One-out-of-two” Selector f=(Vv g)(V’ v h) E : Check: E=(B+D)(B’+(C+1)(C’+)) =(B+D)(B’+C’+D) =BC’+D =BC’+C(B+Fi) =B+Cfi D:

49. Fi: Co:

50. S:

51. Interconnection: