1. Lecture 4 Sums of Product Circuits Simplification
CSCE 211 Digital Design
Lecture 4 Sums of Product Circuits Simplification
Topics
Circuit Simplification
Sums-of-Products
September 3, 2015

2. Overview Last Time: Readings Chapter 2, New: Homework Boolean algebra
Lec 2 slides: two’s complement overflow
Lec 3 slides 1-23
Boolean algebra
Combinational circuits
Algebraic analysis, Truth tables, Logic Diagrams
New:
NAND and NOR gates
Lec 3 slides: The rest of them; we just covered a few
Karnaugh maps
Sums-of-Products and Products-of-Sums
Homework
Chapter 2: 2f, 5c, 11c, Prove n variable Demorgan’s law via Induction (not the version done in class), 13.a,b,e

3. Pop Quiz
What is the representation in 10 bit of the largest unsigned int?
Value?
What is the representation of the largest two’s complement number in 10 bits?
If an IEEE 754 float has exponent 42 what is the exponent field?

4. Lecture 03 slides 23- Algebraic Simpl. Proof by Induction
Universal Sets
Universal Sets (cont)
Combinational Circuit Analysis
Switching Algebra Terminology
Timing Analysis
Circuit Simplification
Sums-of Products
Circuit Simplification
Karnaugh Map
Karnaugh Map Simplification
Karnaugh Map Term. …
… Prime Implicants
Karnaugh Map Simp. Ex
Example
Four Variable Maps

5. Gates again!
NAND Gate
NOR Gate

6. Circuit Simplification
Why would we want to simplify circuits?
To minimize time delays
To minimize costs
To minimize area

7. Boolean Algebra Reformulated
P1a. a + b = b + a P2a. a+(b+c) = (a+b)+c P3a. a + 0 = a P4a. 1 + a = 1 P5a. a + a’ = 1 P6a. a + a = a P7. (a’)’ = a P8a a(b+c) = ab + ac P9a. ab+ab’ = a P10a. a + a’b = a + b

8. Sums-of-Products
What is the delay of sums-of-products circuit?

9. Continuing Examples (CE)
CE1. A system with four inputs, A, B, C, and D, and one output, Z, such that Z=1 if three of the inputs are 1. CE2. A single light (that can be on or off) that can be controlled by any one of three switches. One switch is the master on/off switch. If it is off, the lights are off. When the master switch is on, a change in the position of one of the other switches (from up to down or from down to up) will cause the light to change state. CE3. A system to do 1 bit of binary addition. It has three inputs (the 2 bits to be added plus the carry from the next lower order bit) and produces two outputs, a sum bit and a carry to the next higher order position. CE4. A system that has as its input the code for a decimal digit, and produces as its output the signals to drive a seven-segment display, such as those on most digital watches and numeric displays. CE5. A system with nine inputs, representing two 4-bit binary numbers and a carry input, and one 5-bit output, representing the sum. (Each input number can range from 0 to 15; the input can range from 0 to 31.)

10. Procedure for implementing Combinational Circuits
Step 1: Represent each of the inputs and output in binary. Step 1.5: If necessary, break the problem into smaller subproblems. Step 2: Formalize the design specification either in the form of a truth table or of an algebraic expression. Step 3: Simplify the description. Step 4: Implement the system with the available components, subject to the design objectives and constraints.

11. CE1. A system with four inputs, A, B, C, and D, and one output, Z, such that Z=1 if three of the inputs are 1.

12. CE2. A single light (that can be on or off) that can be controlled by any one of three switches. One switch is the master on/off switch. If it is off, the lights are off. When the master switch is on, a change in the position of one of the other switches (from up to down or from down to up) will cause the light to change state.

13. CE3. Full Adder
CE3. A system to do 1 bit of binary addition. It has three inputs (the 2 bits to be added plus the carry from the next lower order bit) and produces two outputs, a sum bit and a carry to the next higher order position.

14. CE4. A system that has as its input the code for a decimal digit, and produces as its output the signals to drive a seven-segment display, such as those on most digital watches and numeric displays.

15. CE5. 4-bit binary adder
A system with nine inputs, representing two 4-bit binary numbers and a carry input, and one 5-bit output, representing the sum. (Each input number can range from 0 to 15; the input can range from 0 to 31.)

17. Definitions
A literal is the appearance of a variable or its complement.
A product term is one or more literals connected by AND operators.
A standard product term, also minterm is a product term that includes each variable of the problem, either uncomplemented or complemented.
A sum of products expression (often abbreviated SOP) is one or more product terms connected by OR operators.
A canonical sum or sum of standard product terms is just a sum of products expression where all of the terms are standard product terms.

18. A minimum sum of products expression is one of those SOP expressions for a function that has the fewest number of product terms. If there is more than one expression with the fewest number of terms, then minimum is defined as one or more of those expressions with the fewest number of literals.
(1) x´yz´ + x´yz + xy´z´ + xy´z + xyz terms, 15 literals
(2) x´y + xy´ + xyz 3 terms, 7 literals
(3) x´y + xy´ + xz 3 terms, 6 literals
(4) x´y + xy´ + yz 3 terms, 6 literals

21. A sum term is one or more literals connected by OR operators.
A standard sum term, also called a maxterm, is a sum term that includes each variable of the problem, either uncomplemented or complemented.
A product of sums expression (POS) is one or more sum terms connected by AND operators.
A canonical product or product of standard sum terms is just a product of sums expression where all of the terms are standard sum terms.
SOP: x´y + xy´ + xyz
POS: (x + y´)(x´ + y)(x´ + z´)
Both: x´ + y + z or xyz´
Neither: x(w´ + yz) or z´ + wx´y + v(xz + w´)

29. For any two product terms where exactly one variable appears uncomplemented in one and complemented in the other, the consensus is defined as the product of the remaining literals. If no such variable exists or if more than one such variable exists, then the consensus us undefined. If we write one term as at1 and the second as a´t2 (where t1 and t2 represent product terms), then, if the consensus is defined.
at1 ¢ a´t2 = t1t2
P13a. at1 + a´t2 + t1t2 = at1 + a´t2
P13b. (a + t1)(a + t2)(t1 + t2)= (a + t1)(a + t2)

30. G = DE´ + A´B´C + CD´E + ABC´E
G = C´ (A´B´ + ABE) + DE´ + CD´E
G = C´ (B´ + E)(B + A´) + DE´ + CD´E
G = (C´ + D´ E)[C + (B´ + AE)(B + A´)] + DE´

32. Circuit Simplification
Minterms – a product term in which every variable occurs once either complemented or uncomplemented X Y F
minterm 1
X’ . Y’
X’ . Y
X . Y’
X . Y
Sum of minterms form:
F(X,Y) = X’ . Y’ + X . Y’ + X . Y
F(X,Y) = Σ(0, 2, 3) (sum of minterms m, with F(m)=1

33. Karnaugh Maps Tabular technique for simplifying circuits
two variable maps three variable map
XYZ XY X 1 1 Z Y 00 10 01 11
000
010
110
100
001
011
111
101 1 X XY Y 1 Z 2 1 3 1 2 6 4 1 3 7 5

34. Karnaugh Map Simplification
Simplify F(X,Y,Z) = Σ(0,2,6,4,7,5) XY Z 1 1 Z
F(X,Y,Z)=
Minimize ? – here it will mean “fewer gates, fewer inputs”

35. Karnaugh Map Terminology
F(X,Y,Z) = Σ(1,4,5,6,7) XY Z 1 1 Z
Implicant set - rectangular group of size 2i of adjacent containing ones (with wraparound adjacency)
Each implicant set of size 2i corresponds to a product term in which i variables are true and the rest false
Implicant Sets:

36. Karnaugh Map Terminology
F(X,Y,Z) = XY Z 1 Z
Prime implicant – an implicant set that is as large as possible
Implies – We say P implies F if everytime P(X1, X2, … Xn) is true then F (X1, X2, … Xn) is true also.
If P(X1, X2, … Xn) is a prime implicant then P implies F

37. Karnaugh Map Terminology
F(X,Y,Z) = XY Z 1 1 Z
Prime implicants –
If P(X1, X2, … Xn) is a prime implicant then P implies F and if we delete any variable from P this does not imply F.

38. Karnaugh Map Simplification
F(X,Y,Z) = XY Z 1 Z
F(X,Y,Z) =

39. Karnaugh Map Simplification
F(X,Y,Z) = XY Z 1 Z
F(X,Y,Z) =

40. 4 Variable Map Simplification
F(W,X,Y,Z) = X WX YZ
0000
0100
1100
1000
0001
0101
1101
1001
0011
0111
1111
1011
0010
0110
1110
1010 00 01 11 10 Z Y W

41. 4 Variable Map Simplification
F(W,X,Y,Z) = X WX YZ 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 00 01 11 10 Z Y W

46. Products-of-Sums
What is the delay of products-of-sums circuit?

47. PopQuiz
Convert to Octal
Convert to hex

48. Products-of-Sums
What is the delay of products-of-sums circuit?

49. Pop Quiz
F(X,Y,Z) = XY Z 1 Z
F(X,Y,Z) =

52. Boolean? Well almost in C
Logical operators
&& || !
Bitwise operators ~
& | ^
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