RMS Derivation of Piecewise Sinusiodal Function

In summary, the conversation discusses the calculation of the root mean square of a function v(t) under different conditions. The integral form of the calculation is provided, along with a simplified version. However, there is a discrepancy in the result when evaluated at theta = pi, leading to the possibility of incorrect problem definition or integral calculation. Further investigation is suggested, with additional calculations at different values of theta provided. There is also a discussion on the expected behavior of the RMS value of an alternating signal, with the expectation that it should always be an increasing function.
  • #1
j0shha1nes
2
0
Calculate the root, mean, square of v(t) where
v(t) = 0 0 < wt <= theta
v(t) = Vm*sin(wt) theta < wt <= pi()
v(t) = 0 pi() < wt <= pi()+theta
v(t) = Vm*sin(wt) pi()+theta < wt <= 2*pi()

In integral form, I am confident that it looks like this.

Vrms = SQRT(1/T*INTEGRAL(Vm^2*sin^2(wt)dt)) from theta/w to pi/w AND pi+theta/w to 2*pi/w

I believe it integrates like this. My thought was that I could integrate only one of the ranges and then double my answer due to symmetry in the sinusoidal function.

Vrms = Vm*SQRT(w/2*pi*(t/2-sin(2wt)/4w)) evaluated from theta/w to pi/w

After substitions and simplification (which I have been over about 10-15 times) I get this.

Vrms = Vm - Vm*SQRT(1/pi*(theta-sin(theta)*cos(theta)))

I know this is wrong because when I evaluate this equation at theta = pi, the value is zero. The correct value is Vrms = Vm/SQRT(2) when theta = pi. I have either defined the problem wrong somehow or I have performed the integral wrong.

Many thanks for any help.
 
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  • #2
j0shha1nes said:
I know this is wrong because when I evaluate this equation at theta = pi, the value is zero. The correct value is Vrms = Vm/SQRT(2) when theta = pi. I have either defined the problem wrong somehow or I have performed the integral wrong.

Many thanks for any help.

It appears to me that when theta = pi you should get zero. You have extended the domain of the V(t) = 0 function to the entire interval. Vrms = Vm/SQRT(2) should be the result for theta = 0. Your result does not appear to be giving that, but I'll let you check it. Check also at theta = pi/2. At that point your integral should be half the result for theta = 0 (assuming I'm right above), so Vrms = Vm/2.
 
  • #3
Thanks for the reply.

Not sure what you mean when you say that the domain of v(t) = 0 has been extended ?to the entire interval? (confusing part between the question marks). v(t) equal zero only for a fraction of the first half cycle (and second half cycle) of the T period (0 to 2pi radians).

I will check some of the other quadrants that you mentioned. I have not done that yet. Here they are.

Vrms(theta=pi/2) = Vm*(1-1/sqrt(2))
Vrms(theta=3pi/2) = Vm*(1/sqrt(2)-1)

One thing that I keep running into that I do not understand is that the RMS value over the entire period of an alternating signal should not be zero. I expect it to be an increasing function, not linearly, but always increasing as theta varied from 0 to 2pi radians. I would expect this even when there are equal positive and negative area(s) under the curve. Do you agree?
 

1. What is RMS derivation of piecewise sinusoidal function?

The RMS derivation of piecewise sinusoidal function is a mathematical method used to calculate the root mean square (RMS) value of a piecewise sinusoidal function. This value represents the effective or average value of the function over a given time interval.

2. Why is RMS derivation of piecewise sinusoidal function important?

RMS derivation of piecewise sinusoidal function is important because it allows us to accurately determine the power or energy in a system that is represented by a piecewise sinusoidal function. It is commonly used in electrical engineering and physics to calculate the power dissipated in a circuit or the energy carried by a wave.

3. How is RMS derivation of piecewise sinusoidal function calculated?

RMS derivation of piecewise sinusoidal function involves squaring the amplitude of the function at each point, taking the average of these squared values, and then finding the square root of the average. This can be represented mathematically as: RMS = sqrt(1/T * integral from 0 to T of [f(x)^2] dx). T represents the time interval over which the function is defined.

4. What are the assumptions made in RMS derivation of piecewise sinusoidal function?

The main assumptions made in RMS derivation of piecewise sinusoidal function are that the function is continuous and has a finite period. Additionally, the function is assumed to be a pure sinusoid and not a combination of multiple frequencies.

5. How is RMS derivation of piecewise sinusoidal function applied in real-world situations?

RMS derivation of piecewise sinusoidal function is commonly used in various fields such as electrical engineering, physics, and signal processing. It is used to calculate the power dissipated in electrical circuits, the energy carried by sound or radio waves, and the strength of seismic waves in geology. It is also used in data analysis to determine the average value of a signal over a given time period.

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