Problem 4 (5 points) The truth table for a 5-to3 priority encoder is given below: 14 0 0 12 0 0 02 X 0 0 0 0 0 0 0 11 0

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Problem 4 5 Points The Truth Table For A 5 To3 Priority Encoder Is Given Below 14 0 0 12 0 0 02 X 0 0 0 0 0 0 0 11 0 1 (173.64 KiB) Viewed 40 times

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Problem 4 (5 points) The truth table for a 5-to3 priority encoder is given below: 14 0 0 12 0 0 02 X 0 0 0 0 0 0 0 11 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 0 0 0 IO 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 01 X 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 0 13 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 1 1 1 1 0 0 00 X 0 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 0 Note that 02 follows 14, while 01 and O0 are always low when 14 is high. Hence, to find expressions for 01 and 00 you can construct a four variable k-map using 13, 12, 11, and IO. Then, whatever expression you get and it with not 14. 1. Find the Sum-Of-Product expression for 00, implement it using 2-input Nand gates and count the number of transistors assuming input are available as active high or low. 2. Find the Product-of-Sum expression of 01, implement it using 2-input Nor gates and count the number of transistors assuming inputs are available as active high or low.