{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HWK #1 Solutions

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

This preview shows pages 1–2. Sign up to view the full content.

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 (16/16) x C .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online