Mallard ECE 290_ Computer Engineering I - Spring 2009 - HWK #4 Solution

Mallard ECE 290_ Computer Engineering I - Spring 2009 - HWK #4 Solution

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Unformatted text preview: 3/29/09 8:31 PM Mallard ECE 290: Computer Engineering I - Spring 2009 - HWK #4 Solution Page 1 of 13 https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=ramaitfault&title=HWK%20%234%20Solution&bodyargs=&id=hwk04_s09_soln&num= HWK #4 Solution Problem 4.1 h(A,B,C,D)= OR(m 1 ,m 4 ,m 5 ,m 6 ,m 10 ,m 12 ,m 13 ,d 2 ,d 3 ,d 11 ,d 15 ) h(A,B,C,D)= AND(M ,M 7 ,M 8 ,M 9 ,M 14 ,d 2 ,d 3 ,d 11 ,d 15 ) (b) There are 4 minimal SOP expressions. BC' is an essential prime implicant because no other PI contains m 12 . B'C is an essential prime implicant because no other PI contains m 10 . Hence, any minimal SOP expression will contain BC' + B'C + ... Minterm m 1 is covered by both A'C'D and A'B'D. These PIs are the same size and each covers a second cell that is already covered. Hence, we can equivalently choose A'C'D or A'B'D. Minterm m 6 is covered by both A'CD' and A'BD'. These PIs are the same size and each covers a second cell that is already covered. Hence, we can equivalently choose A'CD' or A'BD'. The min SOP expressions are: BC' + B'C + { A'C'D or A'B'D } + { A'CD' or A'BD' } There are 3 minimal POS expressions: (i) (C' + D') (A' + B' + C') (A + B + D) (A' + B + C) (ii) (C' + D') (A' + B' + C') (B + C + D) (A' + B + C) (iii) (C' + D') (A' + B' + C') (B + C + D) (A' + B + D') 3/29/09 8:31 PM Mallard ECE 290: Computer Engineering I - Spring 2009 - HWK #4 Solution Page 2 of 13 https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=ramaitfault&title=HWK%20%234%20Solution&bodyargs=&id=hwk04_s09_soln&num= (c) Your two expressions in part (b) are equal iff every square covered in your SOP K-map is uncovered in your POS K-map. Why? In the SOP K-map: for every minterm inside a loop, the function is defined to be 1, and for every minterm not inside any loop the function is defined to be 0. The opposite is true in the POS K-map: for every maxterm inside a loop the function is defined to be 0, and for every maxterm not inside a loop the function is defined to be 1. For this example, our min SOP expression does not equal our min POS expression because they differ for d 3 . (Indeed, they also differ for d 11 and maybe for d 2 .) The min SOP has d 3 = 1 and min POS has d 3 = 0. The min SOP has d 11 = 1 and min POS has d 11 = 0. The min SOP has d 2 = 1 and min POS has d 2 = 0 or 1 (0 for expression (i) , 1 for expression (ii) and (iii)). See parts (d) and (e) for additional explanation....
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Mallard ECE 290_ Computer Engineering I - Spring 2009 - HWK #4 Solution

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