sol4 - Chapter 4 4.1 SOP form f = x1 x2 x2 x3 POS form f =(x1 x2(x2 x3 4.2 SOP form f = x1 x2 x1 x3 x2 x3 POS form f =(x1 x3(x1 x2(x2 x3 4.3 SOP form f

sol4 - Chapter 4 4.1 SOP form f = x1 x2 x2 x3 POS form f...

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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 . 4.16. f = ( g k ) (( g g ) ( k k )) , where g = ( x 1 x 1 ) ( x 2 x 2 ) ( x 5 x 5 ) and k = ( x 3 ( x 4 x 4 )) (( x 3 x 3 ) x 4 ) 4.17. f = ( g k ) (( g g ) ( k k )) , where g = x 1 x 2 x 5 and k = (( x 3 x 3 ) x 4 ) ( x 3 ( x 4 x 4 )) . Then, f = f f . 4.18. f = x 1 ( x 2 + x 3 )( x 4 + x 5 ) + x 1 ( x 2 + x 3 )( x 4 + x 5 ) 4.19. f = x 1 x 3 x 4 + x 2 x 3 x 4 + x 1 x 3 x 4 + x 2 x 3 x 4 = ( x 1 + x 2 ) x 3 x 4 + ( x 1 + x 2 ) x 3 x 4 This requires 2 OR and 2 AND gates.

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