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sol2-3

# sol2-3 - Chapter 2 2.1 The proof is as follows(x y(x z = xx...

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Chapter 2 2.1. The proof is as follows: ( x + y ) · ( x + z ) = xx + xz + xy + yz = x + xz + xy + yz = x (1 + z + y ) + yz = x · 1 + yz = x + yz 2.2. The proof is as follows: ( x + y ) · ( x + y ) = xx + xy + x y + y y = x + xy + x y + 0 = x (1 + y + y ) = x · 1 = x 2.3. Proof using Venn diagrams: x y z y x z y x z x x y + x y + ( ) x z + ( ) x y z + x z + y z x y z y x z y x z 2-1

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2.4. Proof of 15 a using Venn diagrams: y x x y x y y x y x y x y x y x y x A similar proof is constructed for 15 b . 2.5. Proof using Venn diagrams: x 1 x 2 + x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 + + x 1 x 2 x 3 x 1 x 2 x 3 + + ( ) x 1 x 2 x 3 + + ( ) x 1 x 2 x 1 x 2 x 3 + + x 3 2-2
2.6. A possible approach for determining whether or not the expressions are valid is to try to manipulate the left and right sides of an expression into the same form, using the theorems and properties presented in section 2.5. While this may seem simple, it is an awkward approach, because it is not obvious what target form one should try to reach. A much simpler approach is to construct a truth table for each side of an expression. If the truth tables are identical, then the expression is valid. Using this approach, we can show that the answers are: (a) Yes (b) Yes (c) No 2.7. Timing diagram of the waveforms that can be observed on all wires of the circuit: x 1 x 2 x 3 A B C D f x 2 x 1 x 3 f A B C D 2-3

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2.8. Timing diagram of the waveforms that can be observed on all wires of the circuit: x 2 x 1 x 3 A B C D f x 2 x 1 x 3 f A B C D 2.9. Starting with the canonical sum-of-products for f get f = x 1 x 2 x 3 + x 1 x 2 x 3 + x 1 x 2 x 3 + x 1 x 2 x 3 + x 1 x 2 x 3 + x 1 x 2 x 3 + x 1 x 2 x 3 = x 1 ( x 2 x 3 + x 2 x 3 + x 2 x 3 + x 2 x 3 ) + x 2 ( x 1 x 3 + x 1 x 3 + x 1 x 3 + x 1 x 3 ) + x 3 ( x 1 x 2 + x 1 x 2 + x 1 x 2 + x 1 x 2 ) = x 1 ( x 2 ( x 3 + x 3 ) + x 2 ( x 3 + x 3 )) + x 2 ( x 1 ( x 3 + x 3 ) + x 1 ( x 3 + x 3 )) + x 3 ( x 1 ( x 2 + x 2 ) + x 1 ( x 2 + x 2 )) = x 1 ( x 2 · 1 + x 2 · 1) + x 2 ( x 1 · 1 + x 1 · 1) + x 3 ( x 1 · 1 + x 1 · 1) = x 1 ( x 2 + x 2 ) + x 2 ( x 1 + x 1 ) + x 3 ( x 1 + x 1 ) = x 1 · 1 + x 2 · 1 + x 3 · 1 = x 1 + x 2 + x 3 2.10. Starting with the canonical product-of-sums for f can derive: f = ( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 ) · ( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 ) = (( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 ))(( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 )) · (( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 ))(( x 1 + x 2 + x 3 )( x 1 + x 2 + x 3 )) = ( x 1 + x 2 + x 3 x 3 )( x 1 + x 2 + x 3 x 3 ) · ( x 1 + x 2 + x 3
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