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SashulF [63]
4 years ago
12

In previous counters that you have created, you set the upper limit as the value just past the last digit you wanted displayed.

That is not the case with this design. Why is how you set the range different for this design?
Computers and Technology
1 answer:
koban [17]4 years ago
4 0

Answer:

The Synchronous load is synchronous and the Asynchronous load input is delayed with the counter pulse by one when it counts up.

Explanation:

The Synchronous load is synchronous and when it goes through the input load then the count from the Q outputs will stay the same. The Asynchromous load input is bit different and it delays the counters pulse by one when it counts up. it adds 1 to the pulse when it counts down. This means that it will either add or subtract 1 to the restart/maximum binary count.

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kolezko [41]

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3 years ago
Suppose that the five measured SampleRTT values are 106 ms, 120 ms, 140 ms, 90 ms, and 115 ms.
jeyben [28]

Answer:

a) 100.75,   103.15,    107.756,    105.536,   106.72

b) 5.06,    8,       14.061,       14.43,    12.89

Explanation:

Measured sample RTT values : 106 ms, 120 ms, 140 ms,  90 ms,  115 ms

First estimated RTT value = 100 ms

∝ = 0.125

β = 0.25

First DevRTT value = 5 ms

<u>A)   Compute the EstimatedRTT after each of these SampleRTT values is obtained,</u>

Formula for estimated RTT = ( 1 - ∝ ) . estimated RTT + ∝.sample RTT

B). <u>Compute  the DevRTT after each sample is obtained,</u>

Formula for DevRTT = ( 1 - β ) . DevRTT + β. | sample RTT - Estimated RTT |

attached below is a detailed solution

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Oxana [17]

Answer:

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Explanation:

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