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Learn how to simplify Boolean functions using don't care conditions and minimize functions with examples and practical exercises. Explore Product of Sums simplification and Mixed Logic Combinational Circuits.
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CHAPTER 3 Principles of Combinational Logic (Sections 3.4-3.8)
Simplifying Boolean Functions • Exe: F(x,y,z)=∑(0,2,3,4,5,7) F(a,b,c,d)=∑(0,3,4,5,7,11,13,15) F(w,x,y,z)=∑(0,1,4,5,9,11,13,15) F(a,b,c,d)=∑(0,1,2,4,5,6,8,9,12,13,14) F(a,b,c,d)=∑(1,3,4,5,7,8,9,11,15) F(w,x,y,z)=∑(1,5,7,8,9,10,11,13,15)
Don't Care Conditions • There may be a combination of input values which • will never occur • if they do occur, the output is of no concern. • The function value for such combinations is called a don't care. • They are denoted with x or–. Each x may be arbitrarily assigned the value 0 or 1 in an implementation. • Don’t cares can be used to further simplify a function
Minimization using Don’t Cares • Treat don't cares as if they are 1s to generate PIs. • Delete PI's that cover only don't care minterms. • Treat the covering of remaining don't care minterms as optional in the selection process (i.e. they may be, but need not be, covered).
cd Example ab 01 11 00 10 0 1 0 1 00 1 1 0 1 01 • Simplify the function f(a,b,c,d) whose K-map is shown at the right. • f = a’c’d+ab’+cd’+a’bc’ or • f = a’c’d+ab’+cd’+a’bd’ • The middle two terms are EPIs, while the first and last terms are selected tocover the minterms m1, m4, and m5. 0 0 d d 11 10 1 1 d d 0 1 0 1 1 1 0 1 0 0 d d 1 1 d d 0 1 0 1 1 1 0 1 0 0 d d 1 1 d d
Another Example cd ab d 1 0 0 1 d 0 d • Simplify the function g(a,b,c,d) whose K-map is shown at right. • g = a’c’+ abor • g = a’c’+b’d 1 d d 1 0 d d 0 d 1 0 0 1 d 0 d 1 d d 1 0 d d 0 d 1 0 0 1 d 0 d 1 d d 1 0 d d 0
Don't Care Conditions A=f(w,x,y,z)=∑(5,6,7,8,9)+ ∑d(10,11,12,13,14,15) B=f(w,x,y,z)=∑(1.2.3.4.9)+ ∑d(10,11,12,13,14,15) C=f(w,x,y,z)=∑(0,3,4,7,8)+ ∑d(10,11,12,13,14,15) D=f(w,x,y,z)=∑(0,2,4,6,8)+ ∑d(10,11,12,13,14,15)
Don't Care Conditions A=w+xz+xy B=x’y+x’z+xyz’ C=y’z’+yz D=z’
Product of Sums Simplification • Use sum-of-products simplification on the zeros of the function in the K-map to get F’. • Find the complement of F’, i.e. (F’)’ = F • using DeMorgan’s Theorem.
POS Example cd ab 1 1 1 1 1 1 1 0 0 0 1 1 0 0 0 0 • F’(a,b,c,d) = ab’ + ac’ + a’bcd’ • F = (a’+b)(a’+c)(a+b’+c’+d)
Mixed Logic Combinational Circuits Function lable 1 F=A+B F=AB
Mixed Logic Combinational Circuits • P138: • Ex.3.19 • 3.7.2 Conversion to bubble logic • What is mismatch logic? • Convert to bubble logic. (P140: examples)
Mixed Logic Combinational Circuits • Exe. Write the switching expressions for the following logic circuit. H H=D+C(A’+B’) H
AB AB 00 00 01 01 11 11 10 10 1 1 0 0 1 1 1 1 1 1 Multiple Output Function • E.g. F1=f(a,b,c)=∑(2,6,7) F2=f(a,b,c)=∑(1,3,7) C C F2=a’c+bc F2=a’c+abc F1=bc’+abc F1=bc’+ab
Multiple Output Function F2=a’c+bc F2=a’c+abc F1=bc’+abc F1=bc’+ab
Multiple Output Function Page 146 exp. • F1=f(a,b,c)=∑(2,4,5,6) • F2=f(a,b,c)=∑(2,3,6,7) • F3=f(a,b,c)=∑(2,5,6,7)
Multiple Output Function • E.g. F1(a,b,c,d)=∑m(2,3,5,7,8,9,10,11,13,15) F2(a,b,c,d)=∑m(2,3,5,6,7,10,14,15) F3(a,b,c,d)=∑m(2,3,5,7,8,9,10,11,13,15)
Homework • P152: 10. • P153: 11.a, 11.b • P155: 24.b, 24.d