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Rms value of the half-wave rectified sinusoidal wave

  • Thread starter yy112
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1. Homework Statement

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. Homework 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!
 

cepheid

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1. Homework Statement

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
Did you mean 4cos(20pi(t))? Also, note that this is not a half-wave rectified sinusoidal function, it's just a plain old sinusoidal function.

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. :wink:
 
7
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Sorry. You are correct. it is a cosine function. The question states that it is a half-wave 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.
 
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The integration needs to be in a form of cos^2(x)=1/2 or 1+cos2x
 

gneill

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

cepheid

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Yeah, gneill is right. The problem is asking you to use integration to derive the result that the rms value of a half-wave 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."
 
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Vrms = sqrt ( 1/T * integral from 0-T of v^2(t) dt)
Thats the equation i would use. Would it be appropriate to apply it to this question?
 

cepheid

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Vrms = sqrt ( 1/T * integral from 0-T of v^2(t) dt)
Thats the equation i would use. Would it be appropriate to apply it to this question?
Yup. [tex]\langle f \,\rangle \equiv \frac{1}{b-a}\int_a^b f(x)\,dx [/tex] is the definition for the average value of the function f(x) over the interval a < x < b. Therefore, what you have written there is [itex] \langle v^2(t) \rangle^{1/2} [/itex]. If you think about it, [itex] v^2(t) [/itex] is the square of the voltage, and hence [itex] \langle v^2 \rangle [/itex] is the "mean square" voltage (averaged over one period), and taking the square root of this gives you the "root mean square" voltage.
 
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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?
 

gneill

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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?
You use the function as given. That includes the "15" which is a DC component. Note that you can ignore the angular frequency if you can determine the angle over which a complete cycle occurs. In other words you don't have to integrate over time, but rather over a complete (angular) cycle of the function. The "period" then is the angular "distance" over which you integrate. For the common trig functions sine and cosine that angle is [itex]2 \pi[/itex].

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

gneill

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