EE450Fall2010HW1_Solution

EE450Fall2010HW1_Solution - EE450 Fall 2009 Homework 1...

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EE450 Fall 2009 Homework 1 ( solution ) Problem 4 (10 points, 6 points part a and 4 points part b) Tollbooths are 75 km apart, and the cars propagate at 100km/hr. A tollbooth services a car at a rate of one car every 12 seconds. a) There are ten cars. It takes 120 seconds, or 2 minutes, for the first tollbooth to service the 10 cars. Each of these cars has a propagation delay of 45 minutes (travel 75 km) before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 47 minutes. The whole process repeats itself for traveling between the second and third tollbooths. It also takes 2 minutes for the third tollbooth to service the 10 cars. Thus the total delay is 96 minutes. b) Delay between tollbooths is 8*12 seconds plus 45 minutes, i.e., 46 minutes and 36 seconds. The total delay is twice this amount plus 8*12 seconds, i.e., 94 minutes and 48 seconds. Problem 5
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This note was uploaded on 10/01/2010 for the course EE 450 taught by Professor Zahid during the Fall '06 term at USC.

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EE450Fall2010HW1_Solution - EE450 Fall 2009 Homework 1...

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