HWK #1 Solutions - Mallard ECE 290: Computer Engineering I...

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Mallard ECE 290: Computer Engineering I - Spring 2007 - HWK #1 Sol. .. https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=mding3_. .. 1 of 4 4/29/2007 4:23 PM HWK #1 Solutions Problem 1.1 (a) i. 17 10 = (1,2,2) 3 = 1x9 + 2x3 + 2x1 So to weigh the 17-gram object, we will need one 9-gram weight, two 3-gram weights, and two 1-gram weights . Similarly, 31 10 = (1,0,1,1) 3 = 1x27 + 0x9 + 1x3 + 1x1 So to weigh the 17-gram object, we will need one 27-gram weight, one 3-gram weight, and one 1-gram weights . ii. This works because the weights (and the number of weights we have) can be used as a basis for base 3 arithmetic. (b) We only need one weight each of 1, 3, 9, and 27 grams (i.e. we do not need to have two of each). i. 17 10 = (1,-1,0,-1) 3 = 1x27 + 0x9 - 1x3 - 1x1 So to weigh the 17-gram object, we will need one 27-gram weight, one 3-gram weight, and one 1-gram weight . We will place the 27-gram weight on the opposite side of the scale, and place the 3- and 1-gram weights on the same side of the scale as the object we are weighing. Systematically, 17 10 = (1,2,2) 3 = (1,3,-1) 3 = (2,0,-1) 3 = (1,-1,0,-1) 3 . Now, as we have seen before, 31 10 = (1,0,1,1) 3 . Here we need make no change in order to accommodate our new set of weights. ii. The reason is that we can effectively have a -1 coefficient in our base 3 representation (practically, we would accomplish this by putting the weight on the same side of the scale as the object whose weight we are trying to measure). 2x3 k = 3x3 k - 3 k = 1x3 k+1 - 1x3 k . Problem 1.2 (a) For a 4-bit standard binary counter, there is 1 bit flip between 0000 and 0001. 2 bit flips between 0001 and 0010. 1 bit flips between 0010 and 0011. 3 bit flips between 0011 and 0100. ... To count through all 16 numbers, we will have a total of 1+2+1+3+1+2+1+4 + 1+2+1+3+1+2+1+4 = 30 bit flips. Therefore, the average power consumed when counting from one number to the next is (30/16) x C . For a 4-bit Gray code counter, there is 1 bit flip between all successive numbers. Therefore, the average power consumed when counting from one number to the next is
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This note was uploaded on 06/21/2011 for the course ECE 290 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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HWK #1 Solutions - Mallard ECE 290: Computer Engineering I...

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