WI12-pp2-sol

WI12-pp2-sol - UNIVERSITY OF CALIFORNIA-DAVIS DEPARTMENT OF...

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UC Davis 1 Hussain Al-Asaad U NIVERSITY OF C ALIFORNIA —D AVIS D EPARTMENT OF E LECTRICAL OMPUTER E NGINEERING EEC180A—D IGITAL S YSTEMS I Winter 2012 S OLUTIONS OF P RACTICE P ROBLEMS — S ET 2 1. N UMBER R EPRESENTATION Using the Karnaugh maps below, we can get the following equations: X 3 X 2 X 1 X 0 Y 3 Y 2 Y 1 Y 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 0 0 0 0 00 01 11 10 X 3 X 2 X 1 X 0 0 0 1 0 0 0 1 0 1 0 1 1 1 1 1 0 00 01 11 10 00 01 11 10 X 3 X 2 X 1 X 0 0 0 1 0 1 1 1 1 1 0 1 0 0 0 0 0 00 01 11 10 00 01 11 10 X 3 X 2 X 1 X 0 1 0 0 1 1 0 0 1 1 0 1 0 0 1 1 0 00 01 11 10 00 01 11 10 X 3 X 2 X 1 X 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 0 0 00 01 11 10 Y 3 Y 2 Y 1 Y 0 Y 3 X 3 X 2 X 3 X 1 X 3 X 0 + + X 3 X 2 X 1 X 0 = = Y 2 X 2 X 3 X 2 X 1 X 2 X 0 X 3 X 2 X 1 X 0 + + + X 2 X 3 X 1 X 0 X 3 X 2 X 1 X 0 + X 2 X 3 X 1 X 0 = = = Y 1 X 1 X 3 X 1 X 0 X 3 X 1 X 0 + + X 1 X 3 X 0 X 3 X 1 X 0 + X 1 X 3 X 0 = = = Y 0 X 0 X 3 X 3 X 0 + X 3 X 0 = =
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UC Davis 2 Hussain Al-Asaad 2. ALU D ESIGN The aim of this problem is to design a bit slice of the ALU shown below. The ALU performs one of the eight operations (summarized in the table below) on the two data inputs A = A n – 1 ...A 0 and B = B n – 1 ...B 0 according to the values of the selection input S = S 2 S 1 S 0 . The block diagram of slice i is shown above. There are two outputs for each slice. For slice i , the outputs are F i and C i + 1 . We can rewrite the function table above for slice i as shown below. For the case of A plus B , we have: For the case of B minus A , we have: B A = B + (– A ). In order to achieve this result, we need to get the 2’s com- plement of A and add it to B . This can be done by inverting the A i bits and adding 1 to the result. Adding 1 to the result is equivalent to setting C 0 = 1. Therefore, we have: C 0 = 1 The rest of the table can be determined in a similar way. Based on the table above, we can determine the equations of F i , C i + 1 , and C 0 . Y
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This note was uploaded on 02/21/2012 for the course EEC 180A taught by Professor Redinbo during the Spring '08 term at UC Davis.

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WI12-pp2-sol - UNIVERSITY OF CALIFORNIA-DAVIS DEPARTMENT OF...

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