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# Part One - PowerPoint PPT Presentation

Part One. 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

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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.

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)

Rooting

a.

10. 00 00 00

01

----------------------

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

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

0101. 00 11 00

01

----------------------

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

11 01 . 11

01

----------------------

01 01

----------------------

01 00 . 11 F2 = 1

00 11 .01

----------------------

00 01 . 10 F3 = 1

e.

1 1

0 1 (F1 = 1)

0 1 0 1 (F2 = 1)

------------

1 0 0 1

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

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 1 1 1 1 0

1 0 1 1 0 1 1

1 1 1 0

--------------------

1 0 1 1 0 1 1

1 0 1 1 0 1 1

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

I.12/4

1210 -> 11002

410 -> 1002

1100

100

----------

100 F1=1

100

------------

000 F2=1

1100/100 = 11

112 -> 310

110001

111

------------

1100 F1=0 Answer : 110001/111 = 0111

0111 0111 -> 710

-------------

010100 F2=1

000101

--------------

00111100 F3=1

00001101

----------------

00101111 F4=1

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.

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

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

ANSWER TO QUESTION NUMBER THREE are eliminated

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

ANSWER TO QUESTION NO 4 are eliminated

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

So, are eliminated

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

For D = B + xy’ are eliminated

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

The inequalities : are eliminated

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 :

REALIZE OF A FULL ADDER are eliminated

S1=A’B’C’+C0

S1= =(A+B+C) C0’

S= (A+B+C) C0’

REALIZATION OF FULL SUBTRACTOR are eliminated

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’)

0 0 0 1 are eliminated

0 0 1 1

0 1 0 1

1 0 0 0

1 0 0 1 Unate

1 0 1 1

1 1 0 1

1 1 1 1

N(1) 5 3 3 7

N(0) 3 5 5 1

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

REALIZATION OF THE FOLLOWING THREE FUNCTION IN CASCADE b4 b3 b2 b1 b0

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

f2=x2x1’x4’ b4 b3 b2 b1 b0

f3=x1x2x3

REALIZE AS A CASCADE OF THRESHOLD GATES f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

let

REALIZE AS A CASCADE OF THREE THRESHOLD GATES f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

1.

Positive function f1=xwz

Tmin = 1 0 1 1

Fmax = 1 1 1 0

1 1 0 1

0 1 1 1

The iequalities f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

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

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

To determine the weight for f1: f1->f2->f3 or f3->f1->f2-> or f2->f3->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

Part Two f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

HALF ADDER f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

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’

FULL ADDER f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

Create truth table for full adder

C=XY + YZ + XZ

FULL ADDER f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

table are assigned as don’t care terms for S.

S = XC’ + YC’ + ZC’ + XYZ

S = (X + Y + Z)C’ + XYZ

HALF SUBTRACTOR f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

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

FULL SUBTRACTOR f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

Create truth table for full subtractor

B = X’Y + X’Z + YZ

FULL SUBTRACTOR f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

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’

PART THREE f1->f2->f3 or f3->f1->f2-> or f2->f3->f1

Solution: f=

Assume: X1=A’B, X2=C, X3=D’E+DF

M(A’B, C, D’E+DF)=X1X2+X1X3+X2X3

The total result function

f=

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

Control cell:

Fi=CoX+PiX’

E=B(C+C’)+Cfi=BC’+D

Implementation: BDD for Fi=CoX+PiX’ D=C(B+Fi)

“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:

Fi: BDD for Fi=CoX+PiX’ D=C(B+Fi)

Co:

S: BDD for Fi=CoX+PiX’ D=C(B+Fi)