ASSIGNMENT SUBMISSION &
FEEDBACK
School: Architecture, Computing & Engineering
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BENG HONS Electrical and Electronic
Engineering
Embedded Systems and IC Design
EE3003
Andrew Chanerley
Assessment Deadline: 08/May/2013
Assessment :
Component Number: 001
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I confirm that no part of this assignment, except where clearly quoted and referenced, has been copied from material belonging to any other person e.g. from a book, handout, another student. I am aware that it is a breach of UEL regulations to copy the work of another without clear acknowledgement and that attempting to do
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Once processed at 120 ns, Cout1 is produced without being included in the second block, making all the numbers produced by the blocks after it, wrong as well. Figure 2.2 – Block diagram of the state of the 4-bit ripple adder at time 130 ns after system start
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That is, of course, until time 129.99 ns,where the second block now “sees” the new value of Cout1 and so processes it, changing the final result into 20 (10100B) as seen in figures 1.1 and 2.2 (time 130 ns), which is also an erroneous answer.
This is due to the same reason as in figure 2.1. The Cout2 of the second block does not exist at the time the third and the rest of the blocks are processing the applied inputs (time 129.99 ns).
This phenomenon also occurs for the third block at time 140 ns, until the final result is obtained at
150 ns, where the correct answer 24 (11000B) is finally obtained.
This is why it is called a ripple adder, as at each stage the carry out of the previous block is fed in, after a delay into the next block, and so the carry “ripples” from the first to the last block of the ripple adder, in stages/iterations.
These carries are stored as temporary values in the code and so are not included in the waveforms produced. 3. Modify the test bench code to include the value of the carry-in (ci) to be a logic ‘1’ after 10ns.
This task was accomplished by adding the command
“ci 1001B +
means by the fourth unit it will take 80% of the time of the second and the eighth unit will take
Overflow occurs when the two numbers of similar signs are added together and a result with an opposite sign is produced.
the operands are not the same in the operation. An attacker could exploit the weakness of this
3.1 Using the program shown in Figure 3.30 explain what the output will be at Line A.
It will increase or decrease each division to the choosing time which will change the
by 1, and then the new value is used in the expression in which it appears. For example,
This produces a 106% error causing a very large range of possible values causing our results to be very imprecise.
Off course I had a challenge in the processing time, because of interruption from my system and my function in fact not well processing for on my coding system either way. The steps that must be performed the program has stopped solve my solutions; this was one of the challenge I had in the unit. But overall it is amazing unit of the beginning of
If the response time for this experiment by saying the word” now” falls between 100 and 200 milliseconds, I would conclude that
After running a process flow [see Exhibit 2], it becomes apparent that a main bottleneck exists at the
The theoretical flow time of each path is determined by adding the work contents of the activities along that path. Thus the theoretical flow time of path 1 is 100 minutes and that of path 2 is 127 minutes. Path 2 is then the critical path and the theoretical flow time for the Deluxe model is 127 minutes.
Now we can apply Little’s Law to calculate the throughput time which is equal to the manufacturing lead time in this case.
This is a pattern that we also can see. Starting form value 0,04 to 0,10 then results do not have as big of a difference as 0,02 and 0,04
This equated to the total handling time (TH) for all
This could be prevented if the person who is counting the oscillations also times them themselves.