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Express the Boolean function F = xy + x'z as a product of maxterms. First, convert the function into OR terms by using the distributive law: F = xy + x'z = (xy + x')(xy + z) %3D = (x + x')(y + x')(x + z)(y + z) = (x' + y)(x + z)(y + z) The function has three variables: x, y, and z. Each OR term is missing one variable; therefore,

Express the Boolean function F = xy + x'z as a product of maxterms. First, convert the function into OR terms by using the distributive law: F = xy + x'z = (xy + x')(xy + z) %3D = (x + x')(y + x')(x + z)(y + z) = (x' + y)(x + z)(y + z) The function has three variables: x, y, and z. Each OR term is missing one variable; therefore,

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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EXAMPLE 2.5 Express the Boolean function F = xy + x'z as a product of maxterms. First, convert the function into OR terms by using the distributive law: F = xy + x'z = (xy + x')(xy + z) = (x + x')(y + x')(x + z)(y + z) (x' + y)(x + z)(y + z) The function has three variables: x, y, and z. Each OR term is missing one variable; therefore, x' + y = x' + y + zz' (x' + y + z)(x' + y + z') x + z = x + z + yy' = (x + y + z)(x + y' + z) %3D y + z = y + z + xx' (x + y + z)(x' + y + z) %3D Combining all the terms and removing those which appear more than once, we finally obtain F — (х + у z)(x + y' +z)(x' + у + z)(x' + у + z') = M,M2M4M5 F(x, y, z) = II(0, 2, 4, 5)
Transcribed Image Text:EXAMPLE 2.5 Express the Boolean function F = xy + x'z as a product of maxterms. First, convert the function into OR terms by using the distributive law: F = xy + x'z = (xy + x')(xy + z) = (x + x')(y + x')(x + z)(y + z) (x' + y)(x + z)(y + z) The function has three variables: x, y, and z. Each OR term is missing one variable; therefore, x' + y = x' + y + zz' (x' + y + z)(x' + y + z') x + z = x + z + yy' = (x + y + z)(x + y' + z) %3D y + z = y + z + xx' (x + y + z)(x' + y + z) %3D Combining all the terms and removing those which appear more than once, we finally obtain F — (х + у z)(x + y' +z)(x' + у + z)(x' + у + z') = M,M2M4M5 F(x, y, z) = II(0, 2, 4, 5)
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