Mean Power is not RMS Power for a sinusiodal voltage.
Dr. David Kirkby, G8WRB. 26/3/2005


This was done in an attmept to settle an argument about RMS power, that appeared on the  'amps' mailing list in March 2005. Note that arugments every seem to get settled there (espeically the parasitic oscillations), but this might be a bit more scientific.

Definition of RMS

This is the definition of the RMS value of a repetative waveform . You integreate over the time period T

rms = (1/T∫_0^T  f (t)^2  dt)^(1/2) 

where T is the period of the waveform . This is valid for any repetative waveform .

Definition of mean value of a continusous repetative waveform
mean= 1 T 0 T   f(t)  dt

Finding the RMS value of a sine wave.
    

First find the RMS value of a sine wave, with a peak of 1.
The period T of a sine wave is 360° or, expressed more mathematically, it is 2π radians.

Here's a plot of a sine wave over 2π radians

In[37]:=

volts = Plot [ Sin [ t ] , { t , 0 , 2 Pi } , PlotStyle Green ]

[Graphics:HTMLFiles/rmspower_4.gif]

Out[37]=

Graphics

You can see, that is a complete cycle.
    
    Now the rms value of this will be found. we define the voltage as sin(t).
    

In[38]:=

voltage [ t_ ] := Sin [ t ]

In[39]:=

rms = Sqrt [ 1 / ( 2 Pi ) * Integrate [ voltage [ t ] ^ 2 , { t , 0 , 2 Pi } ] ]

Out[39]=

1 2

Convert the above to a floating point number.
    

In[40]:=

N [ % ]

Out[40]=

0.7071067811865475

In[73]:=


So no great surprise there.

Finding the mean of a sine wave

In[41]:=

Sqrt [ 1 / ( 2 Pi ) * Integrate [ voltage [ t ] , { t , 0 , 2 Pi } ] ]

Out[41]=

0

In[30]:=

So the mean value of voltage on a sine wave is 0, the rms is 0.707 times the peak.
    

Instantanous power.

In[73]:=

The power developed in a resistor will not be constant throughout the whole cycle , but will be the product of two sine waves - the current waveform and the voltage waveform . So it is is a sin ( t ) 2 function .

We will define the current to be sinusoidal too, just like for a voltage.
    

In[42]:=

current [ t_ ] := Sin [ t ] ;


Now here's a plot of the instantanous power.

In[43]:=

watts = Plot [ voltage [ t ] * current [ t ] , { t , 0 , 2 π } , PlotStyle Red ]

[Graphics:HTMLFiles/rmspower_5.gif]

Out[43]=

Graphics


Note, that is not the same as a rectified sine wave, which which is found by plotting the absolute value of a sine wave, as below.

In[44]:=

rectified = Plot [ Abs [ Sin [ t ] ] , { t , 0 , 2 Pi } , PlotStyle Blue ]

[Graphics:HTMLFiles/rmspower_6.gif]

Out[44]=

Graphics

Finding the RMS value of power.

To find the rms value of the power, we note that power is the product of the voltage and current. and go back to our definitinon of RMS
rms= 1 T 0 T f ( t ) 2 dt 2
    

In[45]:=

rmsPower = Sqrt [ ( 1 / ( 2 Pi ) ) * Integrate [ ( voltage [ t ] * current [ t ] ) ^ 2 , { t , 0 , 2 Pi } ] ]

Out[45]=

3 2 2


Now to find the numerical value of that.

In[46]:=

N [ % ]

Out[46]=

0.6123724356957945

In[47]:=

Finally, here is a graph showing the voltage (green), the power (red) and that of a rectified sine wave (blue)

In[48]:=

Show [ { volts , watts , rectified } ]

[Graphics:HTMLFiles/rmspower_7.gif]

Out[48]=

Graphics


Note, since the power waveform actually repeats every π radians, we could integrate over π rather than 2π in the case of the power, but not in the case of the voltage or current, as they have a period of 2π radians.


    Conclusions

  
  The RMS vale of a sine wave is 0.707 x the peak value. Hence the mean voltage of a sine wave is 0.
The instantaneous power, which is the product of voltage and current, looks similar(ish) to a rectified sine wave, but it is a sine wave squared. The RMS value of the power waveform can be found, and the answer is 0.612372 times the peak of the voltage waveforum. The mean value of power of is 0.5*Vpeak*Ipeak, or Vrms*Irms.

The most important conclusion is that the mean power and rms power are not the same for a sinusoidal voltage applied to a resistor. The RMS power is larger by a factor
3 2 = 1.224744871391589


I'm sure this will never settile an arugument on 'amps', but I tried.

73 de G8WRB. 26/3/2005


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