# sol4 - Chapter 4 4.1. SOP form: f = x 1 x 2 + x 2 x 3 POS...

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Unformatted text preview: Chapter 4 4.1. SOP form: f = x 1 x 2 + x 2 x 3 POS form: f = ( x 1 + x 2 )( x 2 + x 3 ) 4.2. SOP form: f = x 1 x 2 + x 1 x 3 + x 2 x 3 POS form: f = ( x 1 + x 3 )( x 1 + x 2 )( x 2 + x 3 ) 4.3. SOP form: f = x 1 x 2 x 3 x 4 + x 1 x 2 x 3 x 4 + x 2 x 3 x 4 POS form: f = ( x 1 + x 4 )( x 2 + x 3 )( x 2 + x 3 + x 4 )( x 2 + x 4 )( x 1 + x 3 ) 4.4. SOP form: f = x 2 x 3 + x 2 x 4 + x 2 x 3 x 4 POS form: f = ( x 2 + x 3 )( x 2 + x 3 + x 4 )( x 2 + x 4 ) 4.5. SOP form: f = x 3 x 5 + x 3 x 4 + x 2 x 4 x 5 + x 1 x 3 x 4 x 5 + x 1 x 2 x 4 x 5 POS form: f = ( x 3 + x 4 + x 5 )( x 3 + x 4 + x 5 )( x 2 + x 3 + x 4 )( x 1 + x 3 + x 4 + x 5 )( x 1 + x 2 + x 4 + x 5 ) 4.6. SOP form: f = x 2 x 3 + x 1 x 5 + x 1 x 3 + x 3 x 4 + x 2 x 5 POS form: f = ( x 1 + x 2 + x 3 )( x 1 + x 2 + x 4 )( x 3 + x 4 + x 5 ) 4.7. SOP form: f = x 3 x 4 x 5 + x 3 x 4 x 5 + x 1 x 4 x 5 + x 1 x 2 x 4 + x 3 x 4 x 5 + x 2 x 3 x 4 + x 2 x 3 x 4 x 5 POS form: f = ( x 3 + x 4 + x 5 )( x 3 + x 4 + x 5 )( x 1 + x 2 + x 3 + x 4 + x 5 ) 4.8. f = ∑ m (0 , 7) f = ∑ m (1 , 6) f = ∑ m (2 , 5) f = ∑ m (0 , 1 , 6) f = ∑ m (0 , 2 , 5) etc. 4.9. f = x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 2 x 3 x 4 4.10. SOP form: f = x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 1 x 2 x 3 + x 1 x 3 x 4 + x 2 x 3 x 4 POS form: f = ( x 1 + x 2 + x 3 )( x 1 + x 2 + x 4 )( x 1 + x 3 + x 4 )( x 2 + x 3 + x 4 )( x 1 + x 2 + x 3 + x 4 ) The POS form has lower cost. 4.11. The statement is false. As a counter example consider f ( x 1 , x 2 , x 3 ) = ∑ m (0 , 5 , 7) . Then, the minimum-cost SOP form f = x 1 x 3 + x 1 x 2 x 3 is unique. But, there are two minimum-cost POS forms: f = ( x 1 + x 3 )( x 1 + x 3 )( x 1 + x 2 ) and f = ( x 1 + x 3 )( x 1 + x 3 )( x 2 + x 3 ) 4-1 4.12. If each circuit is implemented separately: f = x 1 x 4 + x 1 x 2 x 3 + x 1 x 2 x 4 Cost = 15 g = x 1 x 3 x 4 + x 2 x 3 x 4 + x 1 x 3 x 4 + x 1 x 2 x 4 Cost = 21 In a combined circuit: f = x 2 x 3 x 4 + x 1 x 3 x 4 + x 1 x 2 x 3 x 4 + x 1 x 2 x 3 g = x 2 x 3 x 4 + x 1 x 3 x 4 + x 1 x 2 x 3 x 4 + x 1 x 2 x 4 The first 3 product terms are shared, hence the total cost is 31. 4.13. If each circuit is implemented separately: f = x 1 x 2 x 4 + x 2 x 4 x 5 + x 3 x 4 x 5 + x 1 x 2 x 4 x 5 Cost = 22 g = x 3 x 5 + x 4 x 5 + x 1 x 2 x 4 + x 1 x 2 x 4 + x 2 x 4 x 5 Cost = 24 In a combined circuit: f = x 1 x 2 x 4 + x 2 x 4 x 5 + x 3 x 4 x 5 + x 1 x 2 x 4 x 5 g = x 1 x 2 x 4 + x 2 x 4 x 5 + x 3 x 4 x 5 + x 1 x 2 x 4 x 5 + x 3 x 5 The first 4 product terms are shared, hence the total cost is 31. Note that in this implementation f ⊆ g , thus g can be realized as g = f + x 3 x 5 , in which case the total cost is lowered to 28. 4.14. f = ( x 3 ↑ g ) ↑ (( g ↑ g ) ↑ x 4 ) where g = ( x 1 ↑ ( x 2 ↑ x 2 )) ↑ (( x 1 ↑ x 1 ) ↑ x 2 ) 4.15. f = ((( x 3 ↓ x 3 ) ↓ g ) ↓ (( g ↓ g ) ↓ ( x 4 ↓ x 4 )) , where g = (( x 1 ↓ x 1 ) ↓ x 2 ) ↓ ( x 1 ↓ ( x 2 ↓ x 2 )) . Then, f = f ↓ f ....
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## This note was uploaded on 03/26/2008 for the course ECEN 248 taught by Professor Lu during the Spring '08 term at Texas A&M.

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sol4 - Chapter 4 4.1. SOP form: f = x 1 x 2 + x 2 x 3 POS...

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