PE Refresher Course Digital Systems and Computers Joanne Degroat - PowerPoint PPT Presentation

PE Refresher Course Digital Systems and Computers Joanne Degroat degroat.1@osu.edu ece.osu.edu/~degroat 3/1/20007 PE Refresher Computer Area - 1 Joanne DeGroat

The Basics � Basic Switching Algebra • Truth Tables of Basic Functions AND and OR AND OR A B C=A*B A B C=A+B 0 0 0 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 3/1/20007 2 PE Refresher Computer Area - Joanne DeGroat

The Basics (cont) � Inversion – The not or NOT - inversion inverter gate A A 0 1 1 0 � Exclusive OR - XOR XOR - exclusive OR Χ=Α ⊕ Β A B 0 0 0 0 1 1 1 0 1 1 1 0 3/1/20007 3 PE Refresher Computer Area - Joanne DeGroat

And two other common gates � The NAND – NOT AND NAND A B C=A*B 1 0 0 0 1 0 1 0 0 1 1 0 � The NOR – NOT OR 3/1/20007 4 PE Refresher Computer Area - Joanne DeGroat

Some Basic Theorems Equality � A*1 = A � A+0 = A Identity � A*0 = 0 � A+1 = 1 Commutative � A*B = B*A � A+B = B+A Distributive � A(B+C) = AB+AC � A+BC = (A+B)(A+C) Involution � A*A’ = 0 � A+A’ = 1 Absorption � A(A+B) = A � A+AB = A 3/1/20007 5 PE Refresher Computer Area - Joanne DeGroat

More Theorems A( Α + B) = AB � A+AB = A + B � (A + B) = AB (AB) = A + B DeMorgan’s Law � A*A = A A + A = A � (A) = A � AB + AC + BC = AB + AC � Dual: (A+B)(A+C)(B+C) = (A+B)(A+C) 3/1/20007 6 PE Refresher Computer Area - Joanne DeGroat

Truth Tables Truth tables can be used to prove equalities Proof of DeMorgan's Law (A+B) A B A+B A B A*B 0 0 0 1 1 1 1 1 0 1 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 0 0 0 = 3/1/20007 7 PE Refresher Computer Area - Joanne DeGroat

Venn Diagram and Karnaugh Maps B=1 A=1 B 100 010 00 01 11 10 110 0 111 101 011 1 A 000 C 001 C=1 3/1/20007 8 PE Refresher Computer Area - Joanne DeGroat

Function Simplification � f(a,b,c) = ab + bc + abc � Simplify using Karnaugh map Start with each term 3/1/20007 9 PE Refresher Computer Area - Joanne DeGroat

Function Simplification � f(a,b,c) = ab + bc + abc A term with a Single element results In 4 ones A term with 2 elements Gives two A term with 3 elements Gives a single 1 3/1/20007 10 PE Refresher Computer Area - Joanne DeGroat

Simplify K maps � Conbining the previous three maps � Can represent what a K map shows by a sum of products � Take the largest group possible • On one line, a square, two BC • Group needs to be a 00 01 11 10 A • 0 1 1 1 1 power of 2 1 1 f(a,b,c) = a + bc 3/1/20007 11 PE Refresher Computer Area - Joanne DeGroat

4 variable functions � Also expressed in minterm notation � f(a,b,c,d) = Σ m(7,13,14,15) • So function is a 1 when abcd = 0111 (7) CD 1101 (13) 00 01 11 10 1110 (14) AB 1 3 2 0 1111 (15) 00 5 7 6 4 01 1 13 15 14 11 12 1 1 1 9 11 10 8 10 3/1/20007 12 PE Refresher Computer Area - Joanne DeGroat

Form into largest groups � For 4 variables form into groups of • 16 – function would be a 0 or a 1 • 8 – 1 variable • 4 – 2 variables • 2 – 3 variables � Simplifies to • F = ABD + BCD + ABC 3/1/20007 13 PE Refresher Computer Area - Joanne DeGroat

Another Example of function simplification x y z F1 F2 � From a Truth Table 0 0 0 1 0 � Step 1 – map to 0 0 1 0 0 0 1 0 1 0 K- maps 0 1 1 1 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 3/1/20007 14 PE Refresher Computer Area - Joanne DeGroat

Step 2 – Generate simplified equation � For F1 � For f2 3/1/20007 15 PE Refresher Computer Area - Joanne DeGroat

Example Problem � Design a circuit having 1 output Z and 4 inputs A B C D which represents a BCD number, such that Z = 1 if the BCD number is greater than 5. � BCD stands for binary coded decimal • Takes 4 binary digits • Only 0 through 9 are used 3/1/20007 16 PE Refresher Computer Area - Joanne DeGroat

Map onto K map � Fill in a 1 whenever output should be a 1 � Here that would be 6, 7, 8 or 9 � And a 0 for blocks 0 through 5 3/1/20007 17 PE Refresher Computer Area - Joanne DeGroat

The K map cont � Filled in � What should other blocks be? � They would be don’t cares as in BCD notation they will never occur � And final K map for simplification is 3/1/20007 18 PE Refresher Computer Area - Joanne DeGroat

Simplified using don’t cares � Form the largest power of 2 grouping from 1’s and don’t cares 1 3 2 0 � Cover all the 1’s but 5 7 6 4 don’t need to cover all 13 15 14 12 the don’t cares 9 11 10 8 � Get Z = A + BC � Gate implementation 3/1/20007 19 PE Refresher Computer Area - Joanne DeGroat

PLAs � PLA – Programmable Logic Array � PAL – Programmable Array Logic – much like a PLA but restricted connections � Formed of an AND plane and an OR plane 3/1/20007 20 PE Refresher Computer Area - Joanne DeGroat

PLAs � PLA – Programmable Logic Array � PAL – Programmable Array Logic – much like a PLA but restricted connections � Formed of an AND plane and an OR plane 3/1/20007 21 PE Refresher Computer Area - Joanne DeGroat

PLA example � Problem: A sequence box to control automatic starting of a jet engine is to be designed. It has the following signals � Problem stated on next slide. 3/1/20007 22 PE Refresher Computer Area - Joanne DeGroat

23 PE Refresher Computer Area - Joanne DeGroat Problem statement 3/1/20007 � xx

Problem statement translated to a table � The table 3/1/20007 24 PE Refresher Computer Area - Joanne DeGroat

25 PE Refresher Computer Area - Joanne DeGroat K maps for the sequencer 3/1/20007

The logic implementation in gates 3/1/20007 26 PE Refresher Computer Area - Joanne DeGroat

And a PLA implementation � You will most like be asked for this in a table representation form like this 3/1/20007 27 PE Refresher Computer Area - Joanne DeGroat

28 PE Refresher Computer Area - Joanne DeGroat The PLA 3/1/20007

Basic Sequential Logic � Flip Flops � Set Reset FF • Q*=S+R’Q � Toggle FF • Q*=Q’ • Q* is next value • Or next state 3/1/20007 29 PE Refresher Computer Area - Joanne DeGroat

Basic Sequential Logic � D F/F • Q* = D � JK F/F • Q*=JQ’+K’Q 3/1/20007 30 PE Refresher Computer Area - Joanne DeGroat

A Simple up/down counter � Start with a state diagram 3/1/20007 31 PE Refresher Computer Area - Joanne DeGroat

A Simple up/down counter � Start with a state diagram � And a state table for T F/Fs 3/1/20007 32 PE Refresher Computer Area - Joanne DeGroat

K Maps for the toggle F/Fs T2 = 1 T1 = x’y2 + x y2’ = x xor y2 3/1/20007 33 PE Refresher Computer Area - Joanne DeGroat

34 CLR Q PE Refresher Computer Area - Joanne DeGroat T D Q SET CLR Q Implementation T D Q SET 3/1/20007