1 CSE20 Lecture 16 Boolean Algebra: Applications Professor CK Cheng CSE Dept. UC San Diego 1.
-
date post
21-Dec-2015 -
Category
Documents
-
view
213 -
download
0
Transcript of 1 CSE20 Lecture 16 Boolean Algebra: Applications Professor CK Cheng CSE Dept. UC San Diego 1.
1
CSE20 Lecture 16Boolean Algebra: Applications
Professor CK Cheng
CSE Dept.
UC San Diego
1
2
Outlines
• Introduction
• Light Switches
• Bit Counter
• Multiplication
• Multiplexer
• Priority Encoder
• Summary
3
Introduction
• Language Description
• Truth Table
• Karnaugh Map
• Minimal Expression
• Digital Logic Networks
4
Light Switches
Given two switches A, B with two states (Up, Down), the switches control one light emitter Y. Initially A=B=Down, and Y=Off. The light Y is turned between Off and On when one of the switch changes its state.
Describe Y as a function of A and B.
4
Assignment: A, B are in {0, 1} (Down, Up)Y is in {0,1} (Off, On)
5
Light SwitchesTruth Table Karnaugh Map
5
Id A B Y
0 0 0 0
1 0 1 1
2 1 0 1
3 1 1 0
Y A=0 A=1
B=0 0 1
B=1 1 0
6
Light Switches
Minimal Expression in sum of products format.
Y(A,B)=A’B+AB’
6
7
Bit CounterInput: Three binary bits (A, B, C)
Output: (S1,S0) that counts the number of 1’s in (A, B, C).
Derive minimal expressions of (S1,S0).
For example:
(A,B,C)=(0,0,0) => (S1,S0)=(0,0) 0: Binary code
(A,B,C)=(1,1,0) => (S1,S0)=(1,0) 2: Binary code
(A,B,C)=(1,1,1) => (S1,S0)=(1,1) 3: Binary code
8
Bit Counter
Id A B C S1 S0
0 0 0 0 0 0
1 0 0 1 0 1
2 0 1 0 0 1
3 0 1 1 1 0
4 1 0 0 0 1
5 1 0 1 1 0
6 1 1 0 1 0
7 1 1 1 1 1
Truth Table Karnaugh Map
8
S1 B,C
0,0
B,C
0,1
B,C
1,1
B,C
1,0
A=0 0 0 1 0
A=1 0 1 1 1
S0 B,C
0,0
B,C
0,1
B,C
1,1
B,C
1,0
A=0 0 1 0 1
A=1 1 0 1 0
9
Bit CounterKarnaugh Map
9
S1 B,C
0,0
B,C
0,1
B,C
1,1
B,C
1,0
A=0 0 0 1 0
A=1 0 1 1 1
S0 B,C
0,0
B,C
0,1
B,C
1,1
B,C
1,0
A=0 0 1 0 1
A=1 1 0 1 0
S1=AB+BC+AC
S0=AB’C’+ABC+A’B’C+A’BC’
10
MultiplicationInput: Two binary numbers (a1,a0), (b1,b0)
Output: (s3,s2,s1,s0) product of the two numbers.
Derive minimal expressions of (s3,s2,s1,s0).
For example: (a1,a0)×(b1,b0) = (s3,s2,s1,s0) (0,0)×(0,0) = (0,0,0,0)
(1,0)×(0,1) = (0,0,1,0)
(1,1)×(1,0) = (0,1,1,0)
(1,1)×(1,1) = (1,0,0,1)
11
Multiplication: Truth Table
Karnaugh Maps are left as exercises.
Id a1 a0 b1 b0 s3 s2 s1 s0
0 0 0 0 0 0 0 0 0
1 0 0 0 1 0 0 0 0
2 0 0 1 0 0 0 0 0
3 0 0 1 1 0 0 0 0
4 0 1 0 0 0 0 0 0
5 0 1 0 1 0 0 0 1
6 0 1 1 0 0 0 1 0
7 0 1 1 1 0 0 1 1
8 1 0 0 0 0 0 0 0
9 1 0 0 1 0 0 1 0
10 1 0 1 0 0 1 0 0
11 1 0 1 1 0 1 1 0
12 1 1 0 0 0 0 0 0
13 1 1 0 1 0 0 1 1
14 1 1 1 0 0 1 1 0
15 1 1 1 1 1 0 0 1 11
12
MultiplicationInput: Two binary numbers (a1,a0), (b1,b0)
Output: (s3,s2,s1,s0) product of the two numbers. A minimal expression:
s3=a1a0b1b0
s2=a1b1b0’+a1a0’b1
s1=a1’a0b1+a0b1b0’+a1a0’b0+a1b1’b0
s0=a0b0
Verification: (a1,a0), (b1,b0) are symmetric in the expressions.
13
MultiplexerInput: Three binary bits S (select), A, B (data)
Output: Y
If S=0, then Y=B; else Y=A.
For example:
(S,A,B)=(0,1,0) => Y= 0.
(S,A,B)=(1,1,0) => Y= 1.
(S,A,B)=(1,0,1) => Y= 0.
14
MultiplexerTruth Table Karaugh Map
Id S A B Y
0 0 0 0 0
1 0 0 1 1
2 0 1 0 0
3 0 1 1 1
4 1 0 0 0
5 1 0 1 0
6 1 1 0 1
7 1 1 1 1
Y AB
0,0
AB
0,1
AB
1,1
AB
1,0
S=0 0 1 1 0
S=1 0 0 1 1
Y=S’B+SA
15
MultiplexerInput: Three binary bits S (select), A, B (data)
Output: Y
If S=0, then Y=B; else Y=A.
Minimal Expression:
Y=S’B+SA
16
Priority EncoderInput: Three binary bits E (Enable), D1, D0 (Devices IDs)Output: A, Y
If E=0, then A=Y=0;
Else if D0=1, A=1, Y=0; (Let Device 0 take higher priority)
Else if D1=1, A=1, Y=1;Else A=0, Y=0.
For example:
(E,D1,D0)=(0,1,0) => Y= 0, A=0.
(E,D1,D0)=(1,0,1) => Y= 0, A=1.
(E,D1,D0)=(1,1,0) => Y= 1, A=1.
(E,D1,D0)=(1,0,0) => Y= 0, A=0.
17
Priority EncoderTruth Table Karaugh Maps
Id E D1 D0 A Y
0 0 0 0 0 0
1 0 0 1 0 0
2 0 1 0 0 0
3 0 1 1 0 0
4 1 0 0 0 0
5 1 0 1 1 0
6 1 1 0 1 1
7 1 1 1 1 0
Y D1D0
0,0
D1D0 0,1
D1D0 1,1
D1D0 1,0
E=0 0 0 0 0
E=1 0 0 0 1
A D1D0
0,0
D1D0 0,1
D1D0 1,1
D1D0 1,0
E=0 0 0 0 0
E=1 0 1 1 1
18
Priority EncoderInput: Three binary bits E (Enable), D1, D0
Output: A, Y
If E=0, then A=Y=0;
Else if D0=1, A=1, Y=0;
Else if D1=1, A=1, Y=1;
Else A=0, Y=0.
Minimal Expression:
Y=ED1D0’
A=ED1+ED0
Quiz
• For the above Priority Encoder, let D1 take higher priority. Derive the minimal expression of the outputs A and Y.
19
2020
Summary
•Language Description•Combinatorial Systems
•Truth Table•#variables <7
•Karnaugh Maps•Two level optimization
•Minimal Expressions•Logic Networks