probset3_solutions - Department of Electrical and Computer...

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Department of Electrical and Computer Engineering The University of Texas at Austin EE 306, Fall 2009 Problem Set 3 Solutions Due: October 12, before class Yale N. Patt, Instructor TAs: Aater Suleman, Chang Joo Lee, Ameya Chaudhari, Antonius Keddis, Arvind Chandrababu, Bhargavi Narayanasetty, Eshar Ben-dor, Faruk Guvenilir, Marc Kellermann, RJ Harden 1. Moved from problem set 2. We want to make a state machine for the scoreboard of the Texas vs. Oklahoma Football game. The following information is required to determine the state of the game: 1. Score: 0 to 99 points for each team 2. Down: 1, 2, 3, or 4 3. Yards to gain: 0 to 99 4. Quarter: 1, 2, 3, 4 5. Yardline: any number from Home 0 to Home 49, Visitor 0 to Visitor 49, 50 6. Possesion: Home, Visitor 7. Time remaining: any number from 0:00 to 15:00, where m:s (minutes, seconds) (a) What is the minimum number of bits that we need to use to store the state required? (100*100)*4*100*4*101*2*901 = 2912032000000. 2^41 < 2912032000000 < 2^42 so we need 42 bits (b) Suppose we make a separate logic circuit for each of the seven elements on the scoreboard, how many bits would it then take to store the state of the scoreboard? 1. 7 x 2 bits 2. 2 bits 3. 7 bits 4. 2 bits 5. 7 bits 6. 1 bit 7. 4 bits for minutes 6 bits for seconds Total 43 bits (c) Why might part b be a better way to specify the state? The assignments in (b) are easier to decode 2. Moved from problem set 2. A logic circuit consisting of 6 gated D latches and 1 inverter is shown below: Figure 1 Figure 2
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Let the state of the circuit be defined by the state of the 6 D latches. Assume initially the state is 000000 and clk starts at time t = 0. Question: What is the state after 50 cyles. How many cycles does it take for a specific state to show up again? Every 6 clock cycles a pattern repeates (shown below). Because 50 = 6*8+2 after 50 cycles the state will be the same as after 2 cycles. It will be in state 111000 after 50 cycles 3. (3.31) Moved from problem set 2. If a particular memory has 8 byte addressability and a 4 bit address space, how many bytes of memory does that computer have? Number of byes = address space x adressability. 2^4 x 2^3 = 2^7 = 128 bytes
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This note was uploaded on 01/21/2010 for the course EE 306 taught by Professor Ambler during the Spring '07 term at University of Texas at Austin.

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probset3_solutions - Department of Electrical and Computer...

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