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23rd August 2012, 01:57 PM  #26871 
diyAudio Moderator

Vague and poorly worded question, easy to do with a couple of symmetry assumptions (Laplace will work, there's easier ways) once that was teased out of Ed, no real motivation. Time waster if the point of the "exercise" isn't given and the goalposts are likely to be moved.
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You might be screaming "No, no, no" and all they hear is "Who wants cake?" Let me tell you something: They all do. They all want cake. Wilford Brimley 
23rd August 2012, 01:58 PM  #26872 
diyAudio Member
Join Date: Apr 2002
Location: Prague

Well John, I am 57 and might be considered rusty as well. But, my everyday experience dictates me to be able to learn continuously (hw, sw, cad etc.) and not to forget basics. Forgetting basics means one is unable to assess technical/engineering issues properly.

23rd August 2012, 02:13 PM  #26873 
diyAudio Member

Brad,
Have you ever had any engineering use for the average value of a half sine? 
23rd August 2012, 02:20 PM  #26874 
diyAudio Member


23rd August 2012, 02:27 PM  #26875  
diyAudio Member

Quote:
Yes, the mean average deviation portion of computing amplifier efficency by class A/B, H, G etc. It's contained as an appendix in my class H ADSL driver paper (ISSCC). It has been referenced in several text books since. PMA  some of us just don't want to play.
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23rd August 2012, 02:29 PM  #26876 
diyAudio Member

Back to the problem.
Pavel showed a nice simulation and answer that the power dissipation is the same in both resistors. Now for some that doesn't seem reasonable as one is in series with a capacitor and the other is directly across the power source. So lets look at what happens. As the current starts from zero and rises to one amp the capacitor charges and the voltage across the capacitor rises. This drops the voltage across R2 but since we have a current source the voltage rises across R1. When the current source goes from 1 amp to 1 amp there is still a charge on the capacitor and this must discharge through R2 so the initial surge current is a bit greater than from just the current source. As the capacitor charges in the other direction we again see more energy into R1. Now the next issue was can the circuit be simplified? The answer is to treat the current source and R1 as a Thevenin equivalent. The attached image shows what happens. For those who don't understand the transform, compare the open circuit voltage and the short circuit current, they remain the same. So the two circuits are equivalents and make it clear that the power in R1 must equal that of R2. Now that we have discussed power in an RC circuit and reduced the problem to one, What happens if we calculate the power in R2 by using a Fourier series to represent the square wave voltage source? 
23rd August 2012, 02:36 PM  #26877  
diyAudio Member

Quote:
That makes sense on both counts. SY, A big part of engineering is defining the problem. If you found it confusing you just could have asked. Pavel, I am afraid I will want to do vectors and the unit circle bit as some here don't seem to know it. But I will keep it to one post. You can take a nap then! Dave, I actually expected someone to give the answer of 10 watts as capacitors do not dissipate power. But most of the PM'd answers were 2.5 watts! (And I will not admit to having made the same mistake myself once or twice!) ES 

23rd August 2012, 02:37 PM  #26878 
diyAudio Member

Scott, is there a downloadable version?

23rd August 2012, 02:37 PM  #26879 
diyAudio Moderator

I found it vague and pointless. Your most recent "problem" even more so: "Solve this using an inappropriate tool."
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You might be screaming "No, no, no" and all they hear is "Who wants cake?" Let me tell you something: They all do. They all want cake. Wilford Brimley 
23rd August 2012, 03:46 PM  #26880  
diyAudio Member

Quote:
The use of loop equations is the first step in analyzing any circuit. Although on a very simple circuit many can isolate the issue and use an easier method of solution, but then when confronted with a large complex circuit having the skills to dissect it and solve it are needed. Now today most folks will use a simulation program that figures out the loops and does the calculations automatically. But I think most of us even here recognize understanding how the simulation works even if we could not do the entire job by ourselves is a great benefit. Now the questions were dead bang simple and what you get out of the answers is for some a refresher, for others a new experience and for some it casts new light on old problems. You certainly don't have to pay any attention to it. However as you chide John for not doing double blind tests, I reserve the right to chide you for not even attempting the simplest question. I picked that one because I figured almost every one knows that when the reactance equals the resistance that is the half power point. ES 

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