Discrete Maths 2003 Lecture 32 3 Slides Pp

8/13/2019 Discrete Maths 2003 Lecture 32 3 Slides Pp

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Lecture 32, 9-October-200Discrete Mathematics 2003

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Simplifying the Drawing of a Circuit

For more complicated circuits, it is convenient

to allow AND & OR gates to have more than2 inputs

e.g. Here is a 3-input OR gate:

Exercise: Draw the digital circuit corresponding

to the Boolean expressionxy'+x'+yz

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8.4 Disjunctive Normal Form &

Karnaugh Maps

To date, weve simplified expressions by

recognizing which laws can be used to produce a

simpler expression than the one we started with

However, its often unclear which law might be

appropriate to use, or even whether the

expression can be simplified at all

For these reasons, in this section well take a

more systematic approach to dealing with

Boolean expressions

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Converting a Table to a Boolean

Expression

Firstly look at a method to obtain a Boolean

expression for a Boolean function in table form

Example: Consider the following Booleanfunctionf(x,y)

x y f(x,y)

0 0 1

0 1 0

1 0 1

1 1 0


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Solution to the Example

To express f(x,y,z) in dnf, extend the method

of the previous example

For each row wheref(x,y,z) = 1, form theminterm involving the variablesx,y &z use

the variable itself if the column for the variable

contains a 1; use the complement of the variable

if the column contains a 0

f(x,y,z) is the sum of the resulting minterms

Thus, in the example,

f(x,y,z) =x'y'z +x'yz'+xy'z'+xyz

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Converting an Expression to dnf

Suppose we have a Boolean function that isgiven by an expression that is not in dnf

How can the expression be converted to dnf?

The following example illustrates the process

Example: Supposef(x,y,z) = (x +y'z)(xy)'.Writef(x,y,z) in dnf

Solution: By the 2nd de Morgan's law,

(x +y'z)(xy)' = (x +y'z)(x'+y')

Use the 2nd distrib. law to obtain a sum of terms:

(x +y'z)(x'+y') = xx'+xy'+x'y'z +y'y'z

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Example of Converting to dnf (cont)

Next eliminate repeated occurrences of avariable in a term, by using xx'= 0 & xx =x:

xx'+xy'+x'y'z +y'y'z = xy'+x'y'z +y'z

Each term needs to contain all 3 variables souse x 1 =x & x +x'= 1:

xy'+x'y'z +y'z = xy'(z +z') +x'y'z +y'z(x +x')

= xy'z +xy'z'+x'y'z +xy'z +x'y'z

Finally, any repeated minterms are eliminatedby usingx +x =x:

f(x,y,z) = xy'z +xy'z'+x'y'z


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A Comment on dnf

Note that in the example,xy'z +xy'z'+x'y'z isnot the simplest way of writing f(x,y,z)

i.e. Part-way through the working we obtainedthe simpler expressionxy'+x'y'z +y'z (and theremay be even simpler expressions possible)

The point of dnf is that it provides a standardway of writing a Boolean expression, fromwhich a systematic method for simplifying theexpression can be implemented

This systematic method, based on the use ofKarnaugh maps, will be studied next lecture

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Discrete Maths 2003 Lecture 32 3 Slides Pp