# RMS value of a DC offset saw tooth wave



## rg1 (Jun 2, 2017)

I have calculated RMS values of a saw tooth waveform in two ways.

1. One is taking  take square of whole wave, integrate and take mean and then root of it.

2. Two split the wave into two parts; one DC and the AC parts. Take RMS of both separately. Then sqrt the squares of both.

Theoretically I find both ways are correct. Then why answers not matching. Can some one help me where am I wrong.

View attachment 9715


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## rg1 (Jun 15, 2017)

Is there something wrong in this question?


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## cos90 (Jun 27, 2017)

You are a dangerous man, professor. These kinds of questions are a distraction from the workman attitude you need to use for the PE.

I used to enjoy thinking about stuff when I was in school, but for my PE exam I am not entertaining it.

After all that I will say I think your problem is the DC component of the sawtooth is (1.2+.75)/2 = 0.975


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## cos90 (Jun 29, 2017)

rg1, here is the rest of the problem, please never post any problems like this again  :beerchug:  I have many pages of work to get to this point and have wasted much time on this problem.







I actually think I have seen this problem before in Camera's Practice Problems. If you are using that book I suggest throwing it in a trash barrel, lighting it on fire, shooting the barrel with a cannon, rolling it off a cliff, hitting it with a car, etc. Then focus on practicing PE style problems, which I have noticed you have an above average handle on, but I think it's worth repeating you should focus on the test you are taking and not academic style problems.


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## rg1 (Jun 29, 2017)

cos90 said:


> rg1, here is the rest of the problem, please never post any problems like this again  :beerchug:  I have many pages of work to get to this point and have wasted much time on this problem.
> 
> 
> 
> ...


I found the problem  in CI-4, Q-2. It asks for rms value of this wave. He takes rms of constant part =.75 and varying saw tooth (all positive) wave as .26. So far so good  but the total he takes as .75+.26=1.01, which I felt was wrong because rms values are not added like this.. When I started doing it from basics the rms value of the wave comes as .9836 which is correct from your answer too. But I remembered another way of combining waves which is under root of squares which brought me to a third answer .7937. Now what is right? It is not that such question can not be asked in PE but we are missing something. However if it is taking disproportionate time and efforts we can leave it. View attachment 9809


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## cos90 (Jun 29, 2017)

Oh, it comes from CI, so it's not a waste of time. My post is how you would calculate method 2(II) on your sheet for an arbitrary offset and peak to peak voltage.

Summary: 

DC component =(1.2+.75)/2= .975

AC component = (1.2-.75)/sqrt(12) = 0.1299

Combined = sqrt( .975^2 + .1299^2) = .9836

It seems that the method given in CI Exam 4-2 is wrong. The worst case would be when the DC &amp; AC component are equal in magnitude.

I do think a question like this is highly unlikely on the PE Exam. It takes too long to solve.


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## BigWheel (Jul 1, 2017)

cos90 said:


> I do think a question like this is highly unlikely on the PE Exam. It takes too long to solve.


Agreed.

I predict both of you will pass your PE Exams, and will have plenty of time to play with this later.


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## rg1 (Jul 1, 2017)

BigWheel said:


> Agreed.
> 
> I predict both of you will pass your PE Exams, and will have plenty of time to play with this later.


Thanks @BigWheel Ameen, Amen, Tathaastu.


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## BigWheel (Jul 1, 2017)

rg1 said:


> Thanks @BigWheel Ameen, Amen, Tathaastu.


I can tell English is a second language for you, but the mathematics language is universal. You post good work on here, @rg1...I have no doubts about your passing.

@cos90 is also mathematically strong.

Both of you guys will do well.

Just don't make the mistake of overthinking things. The actual PE exam questions have been vetted through-and-through before they appear in the actual exam, so if you miss the question, well, you missed it fair-and-square. In other words, you should not have to "figure out what they mean;" it will be obvious to the casual observer what they're asking for.

You will see questions that are obviously "test questions;" these are questions that they are "testing" for viability, but won't be actually included in the final scoring. I recognized one right away because the grammar, spelling, and setup was so utterly confusing that I couldn't figure out what it was EXACTLY they were looking for. 

Honestly, the questions are worded so carefully that there will be _no_ room for interpretation. You either know the answer or you don't. 

This is why I know you both will pass. Both of you find mistakes in your reference material and are able to deduce why it's wrong, what should have been asked based on what the "right" answer is, and can rationalize both _why_ the question and answer conflict.

I would wish you both good luck, but neither of you need it. Instead, I'll wish you both a foregone conclusion of "CONGRATULATIONS!"


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## cos90 (Jul 2, 2017)

Thank you, BigWheel. I will certainly be trying my hardest to not have to repeat this process.


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## rg1 (Jul 2, 2017)

Thanks @BigWheel .


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