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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 14

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

This preview shows document pages 1 - 3. Sign up to view the full document.

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