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WI11-exam2-sol

WI11-exam2-sol - UNIVERSITY OF CALIFORNIA-DAVIS DEPARTMENT...

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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 W INTER 2011 E XAM II S TUDENT I NFORMATION I NSTRUCTIONS The exam is closed book and notes. A single double-sided cheat sheet is allowed. Print your name and your ID number. There are four problems in the exam. Solve all of them and show your work. If you need more space for your solution, use the back of the sheets. E XAM G RADE Name Hussain Al-Asaad ID Number xxx-xx-xxxx Problem Maximum Points Student Score 1 25 25 2 25 25 3 25 25 4 25 25 Total 100 100
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UC Davis 2 Hussain Al-Asaad 1. A RITHMETIC C IRCUITS (15 + 10 = 25 POINTS ) Let X be a 3-bit unsigned number represented by X 2 X 1 X 0 , Y be a 4-bit unsigned number repre- sented by Y 3 Y 2 Y 1 Y 0 , Z be a 5-bit unsigned number represented by Z 4 Z 3 Z 2 Z 1 Z 0 , and let W be a 9-bit unsigned number represented by W 8 W 7 W 6 W 5 W 4 W 3 W 2 W 1 W 0 . 1.1 Design a gate-level circuit C that computes the function . Assume that the input combinations ( X 2 X 1 X 0 = 110) and ( X 2 X 1 X 0 = 111) will never appear at the inputs of C . Minimize the number of gates in your design. Assume that the literals and their complements are available. 1.2 Suppose that we have an 8-bit adder (shown below), show how we can implement the function . W 25 X 2 = X W X 2 X 1 X 0 W 8 W 7 W 6 W 5 W 4 W 3 W 2 W 1 W 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 1 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 1 1 0 x x x x x x x x x 1 1 1 x x x x x x x x x From a K-map we get: We choose , since it yields less number of gates. So, we have Based on the above, we need three gates to implement C . W
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WI11-exam2-sol - UNIVERSITY OF CALIFORNIA-DAVIS DEPARTMENT...

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