rms value of the halfwave rectified sinusoidal waveby yy112 Tags: halfwave, rectified, sinusoidal, wave 

#1
Nov2711, 07:31 PM

P: 7

1. The problem statement, all variables and given/known data
I know the rms of a half wave is half the peak value. But the peak value is not given to me. Instead, the V(t) function of 4cos(20pi(x)) is given. Also the period T = 100ms 2. Relevant equations vrms = vpeak/2 But the peak is not given! 3. The attempt at a solution Im not sure exactly how to integrate it in order to obtain the peak value and then divide it by 2 in order to get the rms. Much help is appreciated! 



#2
Nov2711, 07:59 PM

Emeritus
Sci Advisor
PF Gold
P: 5,198

Do you really think that you haven't been given the peak value of this cosine wave? Consider this: a cosine function oscillates between +1 and 1. Therefore, a cosine function that has been multiplied by 4 will oscillate between +__ and __. Therefore, the peak value (a.k.a. amplitude) is __. You fill in the blanks. 



#3
Nov2711, 08:09 PM

P: 7

Sorry. You are correct. it is a cosine function. The question states that it is a halfwave rectified sinusoidal function.
I would think that the peak value is 4. So the Vrms value is 4/2 = 2. But the question wants to see me integrate the function in some sort to get the Vrms value. 



#4
Nov2711, 08:12 PM

P: 7

rms value of the halfwave rectified sinusoidal wave
The integration needs to be in a form of cos^2(x)=1/2 or 1+cos2x




#5
Nov2711, 08:33 PM

Mentor
P: 11,446

First you should explain your understanding of what the rms value is. How does one calculate the rms value of a given periodic function?




#6
Nov2711, 08:40 PM

Emeritus
Sci Advisor
PF Gold
P: 5,198

Yeah, gneill is right. The problem is asking you to use integration to derive the result that the rms value of a halfwave rectified sinusoidal function is half of its peak value (rather than just assuming it's true). In order to do this, you need to know what the definition of the rms value of a function is. Hint: it stands for "root mean square."




#7
Nov2711, 08:41 PM

P: 7

Vrms = sqrt ( 1/T * integral from 0T of v^2(t) dt)
Thats the equation i would use. Would it be appropriate to apply it to this question? 



#8
Nov2711, 09:10 PM

Emeritus
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PF Gold
P: 5,198





#9
Nov2711, 09:21 PM

P: 7

That makes perfect sense. Now i understand it. But if i had a function such as v(t)=15+10cos(20πt) would i also apply the same equation or would i have to do something with the angular frequency?




#10
Nov2711, 09:51 PM

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P: 11,446

For your example the cosine function is periodic over an angle of [itex]2 \pi[/itex]. Thus your rms calculation becomes [tex] RMS = \frac{1}{2 \pi}\int_0^{2 \pi} (15 + 10 cos(\theta))^2 d \theta [/tex] 



#11
Nov2711, 09:59 PM

P: 7

I understand that. I would get an RMS of 275. Would that be correct?




#12
Nov2711, 10:06 PM

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P: 11,446




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