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Topics for Monday, Jan. 25 - PowerPoint PPT Presentation

Topics for Monday, Jan. 25 Review homework DeMorgan’s Theorem relation to symbols Minterms and Maxterms prove x(x + y) = x 1. (10 points) Hint: Use duality and the proof already shown From Friday, x+xy = x  1 + xy = x(1+y) = x  1 = x Apply duality   , 1  0 x(x+y)

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• Review homework

• DeMorgan’s Theorem

• relation to symbols

• Minterms and Maxterms

prove x(x + y) = x

• 1. (10 points)Hint: Use duality and the proof already shown

• From Friday, x+xy = x 1 + xy = x(1+y) = x 1 = x

• Apply duality   , 1  0

x(x+y)

x(x+y)=(x+0)(x+y)

= x+(0  y)

= x+0 = x

2. (10 points) g = (a + c)(d + ef)

Let x=(a+c), y=d, z=ef and x(y+z) = xy+xz, so

(a+c)(d+ef)=(a+c)d + (a+c)ef

= ad + cd + aef + cef

3. (10 points)h=abc + d + e

Let (d+e)=x, ab = y, c = z and x+yz = (x+y)(x+z), so

abc + d + e = (ab + d + e)(c + d + e)

Let (d+e) = x and ab + d + e looks like x + yz, so

abc + d + e = (a + d + e)(b + d + e)(c + d + e)

• DeMorgan’s Theorem provides the final link in taking any expression, including an expression with NAND or NOR, and converting it to S-o-P or P-o-S form.

• (x+y)’ = x’ • y’(x • y)’ = x’ + y’

(x+y)’ = x’ • y’

f= (x+y)’= x  y

What symbol would express f = x’ • y’?

A logic mismatch between the symbol and the logic convention introduces the NOT operation

(x+y)’ = x’ • y’

(x • y)’ = x’ + y’

• A minterm is composed of every input variable (or its complement) ANDed together

• The truth table for a minterm in an n-variable system has one “1” and 2n - 1 ”0”’s

• A minterm is denoted by a lowercase m and the (equivalent) number of the row on which the “1” occurs

• a’b’c’ has a “1” for inputs 000 and is called minterm 0 and denoted m0

• ab’c has a”1” for inputs 101 and is called minterm 5 and denoted m5

• Minterms provide a way to write the function for any truth table.

• The minterms for every row where a “1” occurs are OR’ed together. ( x + 1 = 1)

1

0

0

0

0

0

0

0

a’bc’

0

0

1

0

0

0

0

0

ab’c

0

0

0

0

0

1

0

0

abc’

0

0

0

0

0

0

1

0

a’b’c’

+ a’bc’

+ ab’c

+ abc’

Write a function

abcf

0001

0010

0101

0110

1000

1011

1101

1110

f =

• A Maxterm is composed of every input variable (or its complement) ORed together

• The truth table for a Maxterm in an n-variable system has one “0” and 2n - 1 ”1”’s

• A Maxterm is denoted by a uppercase M and the (equivalent) number of the row on which the “0” occurs

• a’+b’+c’ has a “0” for inputs 111 and is called Maxterm 7 and denoted M7

• a+b’+c has a”0” for inputs 010 and is called Maxterm 2 and denoted M2

• Maxterms provide a way to write the function for any truth table.

• The Maxterms for every row where a “0” occurs are AND’ed together. ( x • 0 = 0)

1

0

1

1

1

1

1

1

(a+b’+c’)

1

1

1

0

1

1

1

1

(a’+b+c)

1

1

1

1

0

1

1

1

(a’+b’+c’)

1

1

1

1

1

1

1

0

(a+b+c’)

(a+b’+c’)

(a’+b+c)

(a’+b’+c’)

Write a function (Maxterms)

abcf

0001

0010

0101

0110

1000

1011

1101

1110

f =

• The canonical minterm and Maxterm forms list every literal and operator

• f = a’b’c’ + a’bc’ + ab’c + abc’

• f= (a+b+c’)(a+b’+c’)(a’+b+c)(a’+b’+c’)

• A list is a shorthand method of displaying which rows the 0’s and 1’s occupy

• f = m(0,2,5,6) = M(1,3,4,7)

• Problem 1. (10 points)

• a) Write the electrical truth table for the gate symbol shown below

• b) Write the operation for the gate symbol shown below when

• Input=P.A. and Output=N.A

• Input=P.A. and Output=P.A.

• Input=N.A. and Output=N.A.

• Input=N.A. and Output=P.A.

abcdf

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

1101

1110

1111

1

0

1

0

111100000101

Problem 2. (20 pts)

a)Write the minterm expression for the function f(canonical form and list)

b) Write the Maxterm expression for the function f

(canonical form and list)