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The Boolean expression as a PoS of F implemented in the following circuit is: Note: Assume D3 has the highest priority in the priority encoder Z W XN 3 A₁ A₂ Do 2x4 Decoder D₁ D₂ D3 D₂ D3 B₁ Do 4x2 D₁ Priority Bo Encoder Select one: O F(x,y,z,w) = (x+y)(y+w)(x+z+w) O F(x,y,z,w) = 1 O None of the mentioned choices is true O F(x,y,z,w) = 0 O F(x,y,z,w) = (x+y)(x+y+w)(x'+z+W) O F(x,y,z,w) = x O F(x,y,z,w) = (x+y)(y+w) O F(x,y,z,w) = (x+y+w')(x'+y+w)(x'+z+W) Io 1₁ 12 23 13 S₁ So 4x1 MUX F

The Boolean expression as a PoS of F implemented in the following circuit is: Note: Assume D3 has the highest priority in the priority encoder Z W XN 3 A₁ A₂ Do 2x4 Decoder D₁ D₂ D3 D₂ D3 B₁ Do 4x2 D₁ Priority Bo Encoder Select one: O F(x,y,z,w) = (x+y)(y+w)(x+z+w) O F(x,y,z,w) = 1 O None of the mentioned choices is true O F(x,y,z,w) = 0 O F(x,y,z,w) = (x+y)(x+y+w)(x'+z+W) O F(x,y,z,w) = x O F(x,y,z,w) = (x+y)(y+w) O F(x,y,z,w) = (x+y+w')(x'+y+w)(x'+z+W) Io 1₁ 12 23 13 S₁ So 4x1 MUX F

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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The Boolean expression as a PoS of F implemented in the following circuit is: Note: Assume D, has the highest priority in the priority encoder: Z W XN W 2x4 Do Decoder D₁ D2 D3 B₁ Do 4x2 D, Priority Bo Encoder D₂ D3 Select one: O F(x,y,z,w) = (x+y)(y+w)(x'+z+w) OF (x,y,z,w) = 1 None of the mentioned choices is true OF (x,y,z,w) = 0 O F(x,y,z,w) = (x+y)(x'+y+w)(x'+z+W) OF (x,y,z,w) = x OF (x,y,z,w) = (x+y)(y+w) O F(x,y,z,w) = (x+y+w')(x'+y+w)(x'+Z+W) Io 1₁ 1₂ 13 S₁ So 4x1 MUX F
Transcribed Image Text:The Boolean expression as a PoS of F implemented in the following circuit is: Note: Assume D, has the highest priority in the priority encoder: Z W XN W 2x4 Do Decoder D₁ D2 D3 B₁ Do 4x2 D, Priority Bo Encoder D₂ D3 Select one: O F(x,y,z,w) = (x+y)(y+w)(x'+z+w) OF (x,y,z,w) = 1 None of the mentioned choices is true OF (x,y,z,w) = 0 O F(x,y,z,w) = (x+y)(x'+y+w)(x'+z+W) OF (x,y,z,w) = x OF (x,y,z,w) = (x+y)(y+w) O F(x,y,z,w) = (x+y+w')(x'+y+w)(x'+Z+W) Io 1₁ 1₂ 13 S₁ So 4x1 MUX F
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