This preview shows pages 1–3. Sign up to view the full content.
1
ECE 15A
Fundamentals of Logic Design
Lecture 6
Malgorzata MarekSadowska
Electrical and Computer Engineering Department
UCSB
2
Last time: Implementing Boolean
functions
±
Representations of Boolean functions are not
unique: the same functions can be expressed
in many different ways
±
Different representations yield different
realizations
±
Optimization in terms of gate count, path
length, wiring, power, etc.
3
Last time: Expressions and circuits
±
Any
Boolean expression can be converted into a
circuit
by combining basic
gates in a relatively straightforward way.
±
The diagram below shows the inputs and outputs of each gate.
±
The precedences are explicit in a circuit. Clearly, we have to make sure that
the hardware does operations in the right order!
(x + y’)z + x’
4
Last time example: half adder
x
y
Sum
Carry
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
HA
x
y
Sum
Carry
Sum= x’y + xy’
Carry = xy
x
y
sum
x
y
carry
(dnf: 22 transistors)
(6 transistors)
Dnf realization: 22+6 transistors
5
Different realizations of a half adder
x
y
sum
(14 transistors)
Sum=x’y+y’x = (x+y)(xy)’
x
y
carry
(6 transistors)
Total_2 = 20 transistors
Total_1= 20 transistors
Sum=x’y+y’x = (x+y)(xy)’=
[(x+y)(xy)’]’’=[(x+y)’+(xy)’’]’
x
y
Sum
Carry
Total_3=14 transistors
6
Example
±
abc
c
ab
c
b
a
bc
a
c
b
a
f
+
+
+
+
=
a
b
c
f
Direct representation.
9 gates, 11 nets, 42 transistors.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
7
Simplification
ac
b
ac
b
a
a
ac
ab
b
a
b
b
ac
c
c
ab
c
c
b
a
c
b
a
abc
abc
c
ab
bc
a
c
b
a
abc
c
ab
c
b
a
bc
a
c
b
a
f
+
=
+
+
=
+
+
=
+
+
+
+
+
=
+
+
+
+
+
=
+
+
+
+
=
)
(
)
(
)
(
)
(
f
b
a
c
2 gates, 4 nets, 12 transistors
8
Equivalence check
abc
c
ab
c
b
a
bc
a
c
b
a
f
+
+
+
+
=
1
ac
b
f
+
=
2
?
2
1
f
f
Is
=
a
b
c
f1
f2
0
0
0
0
0
0
0
1
0
0
0
1
0
1
1
0
1
1
1
1
1
0
0
0
0
1
0
1
1
1
1
1
0
1
1
1
1
1
1
1
Both expressions produce the same
truth tables > they are equivalent.
9
Uniting theorem
±
Powerful tool in Boolean simplification
±
Uniting theorem:
AB + AB’ = A (B+B’) = A
±
Example:
ab’c’+ab’c+abc’+abc = ab’(c’+c) + ab(c + c’)=
ab’+ab = a(b+b’) = a
10
Boolean Cubes
Visual technique for
identifying when the
Uniting Theorem can be applied
1cube
0
1
x
2cube
01
11
00
10
xy
011
111
110
001
000
100
010
101
y
x
z
3cube
±
Subcubes of on nodes can be used for simplification.
²
Onset  filled in nodes, offset  empty nodes
This is the end of the preview. Sign up
to
access the rest of the document.
 Winter '08
 M

Click to edit the document details