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Published byKarina Brumit Modified about 1 year ago

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Part One

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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.

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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)

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ANSWER TO QUESTION ONE Rooting a F1=1 Answer : F2 = F3 = F4 = 1

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b F1= F2 = F3 = 0 Answer : F4 = 0

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c F1=1 Answer : F2 = F3 = F4 = F5 = 0

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d F1=1 Answer : F2 = F3 = 1

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

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f F1 F2 F3 F4 Answer : 01 F1= F2= F3= F4=

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Multiplication g. 3*10 Answer : Left shift Right shift

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H 91*

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DIVISION I.12/ > > F1= F2=1 Answer: 1100/100 = > 3 10

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J.49/ F1=0 Answer : /111 = > F2= F3= F4=1

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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.

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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

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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

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

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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 /2 N(1) 0 1 N(0) 1 0 y 1 1 1/2 T=0.5 x 1 1 1/2 f2: y -1 N(1) 1 0 N(0) 0 1 T=0.5

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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

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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

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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 Fmax

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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 :

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

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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’)

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Unate N(1) N(0)

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b0 = 2x = 0 b4 b3 b2 b1 b0 b1 = 2(5-3) = b2 = b3 = -4 a4 a3 a2 a1 a0 b4 = 12

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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 Fmax 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

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f2=x2x1’x4’ f3=x1x2x3

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

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REALIZE AS A CASCADE OF THRESHOLD GATES let

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

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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

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

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

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Part Two

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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’

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

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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

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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

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

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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’

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PART THREE

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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

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The total result f=

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

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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:

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Fi: Co:

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S:

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Interconnection:

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