sol2-3

sol2-3 - Chapter 2 2.1. The proof is as follows: (x + y )...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

sol2-3 - Chapter 2 2.1. The proof is as follows: (x + y )...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online