+ Post New Thread
Results 1 to 9 of 9

24th December 2010, 21:24 #1
The difference between RMS and average value
Yesterday I was in class and we were talking about signals and then question was raised about RMS value. What I have read about RMS value is "It is the value of Sine Wave which will produce same heat as DC" but I was confused when one of student said that Average Value of Sine Wave is equal to DC.
Can you please help me what is the difference between RMS and Average.

24th December 2010, 21:24

24th December 2010, 22:34 #2
 Join Date
 Sep 2006
 Posts
 168
 Helped
 52 / 52
 Points
 3,629
 Level
 14
Re: RMS vs Average Value
The average voltage of a sine wave is equal 0V.
The average voltage is simply mathematical average nothing more.
RMS represent DC equivalent for power calculation.
For example if I have 230V DC and I plug this voltage across light bulb.
So to get the same amount of light with AC voltage you need connect 325Vp (peak) sin wave. Vrms = Vp/√2 = 325/1.41 = 230V
But to mess up a little bit the average AC power is equal DC power

24th December 2010, 22:42 #3
Re: RMS vs Average Value
Average Value is the DC value ok
is integrating the signal for one period and divide it by the period
avg = (1/T) integration(sin(t))dt [from 0 >>> T]
T is the period
for sin wave the integrating is Zero because the positive part will be eliminated with the negative part
and also for square wave will be zero for the same reason
but for
pulse wave for example it have an average and called the dc component in that wave
any wave can be represented or( reformed) as sum of orthogonal signals like sin and cos signals as fourier said
so if we tried to represent the pulse wave it will have a DC component that is the average of the signal
the average value for the Dc is the DC level itself
but for RMS value
first is too important most instruments used it to indicate to AC values
and it equal the value of the DC value which WiLL GENERATE THE SAME HEAT POWER IN LOAD (resistor for example)
I = square root( lim [t>> infinity] (1/T) integrating( (sin(t)^2 dt) [wehere integrating boundries from T/2 to T/2]) ) T is the period
for volts and currents don't forget to emerge the peak value in integrations
hope to be useful

24th December 2010, 22:42

25th December 2010, 01:10 #4
Awards:
 Join Date
 Jul 2009
 Location
 Aberdyfi, West Wales, UK
 Posts
 10,798
 Helped
 3513 / 3513
 Points
 65,725
 Level
 62
Re: RMS vs Average Value
Look at it this way:
If you take the average of a sine wave the result is zero because it's positive and negative swings are equal. Adding them together cancels them out. The voltage/current is clearly not always zero!
If you square a negative number it becomes positive, so imagine you square samples of a sine wave, all the samples will now be positive.
Take the average of all those squared samples, because they are all positive you get a positive result.
Because the values were initially squared, the average is now squared so take the square root of it to recover it's real value.
So the squaring process is a workaround to make sure the values are always positive.
Brian.

25th December 2010, 01:10

25th December 2010, 02:29 #5
 Join Date
 Apr 2010
 Posts
 132
 Helped
 88 / 88
 Points
 2,377
 Level
 11
Re: RMS vs Average Value
Average Value:
It is simple mathematical average value as said above.
RMS Value:
It is used to determine the average power in an alternating current. Since the voltage in an A/C system oscillates between + and , the Average value is zero. The RMS or "nominal" voltage is related to the average of the absolute value (or average vector magnitude) of the current, and is about 70% of the peak value.
As you have asked for the sine wave case. It needs a lil bit more explanation.
There are 3 different ways to quantify the magnitude of a sine wave.
1Peak voltage 2 peakpeak voltage 3RMS voltage
1Peak voltage:
Peak voltage tells you how far the voltage swings, either positive or negative, from the point of reference(In case of sine wave this point of reference is the DC level). Peak voltage is only a moderately useful way of measuring voltage when trying to express the amount of work that will be done when driving a specified load. As AC voltage(assume sine wave) is constantly changing and is at or near the highest and lowest points in the cycle for only a tiny fraction of the cycle, the peak voltage is not a good way to determine how much work can be done by an AC power source.
2 PeakPeak voltage:
As you can guess from the name,it tells you how far voltage swings from one peak value to the next.It is rarely used for measuring voltages in case of sine wave. It is probably more useful in the case of a nonsymmetrical wave form.
3 RMS voltage:
RMS voltage is absolutely the most common way to measure/quantify AC voltage. It is also the most useful since it will give you the ability to exactly(more or less) predict how much work will be done by an AC voltage(source)unlike the case of peak voltage. The RMS value(voltage) of a pure sine wave is approximately .707*peak voltage. All voltmeters generally gives the RMS voltage of the wave form.
From where the name RMS came: If you try to find out the average value of a sine wave,it will be equal to ZERO(DC) because of equal positive and negative half of sine wave in one complete cycle but this ZERO value is not correct(since when an AC voltage/current wave osciallte it perform some work and avrage work cant be ZERO). To get the right average value,
S:sine wave is first squared (so that both halves of sine wave become +ve),
M:after this average is being taken just like normal averaging way
R:In the end to remove the Effect of square which has been done in first step, Square root is being taken to get actual Average value(amount of work being done)
If you read it in reverse order you will get 'RMS'(Root mean square) name.
If you are wondering about the number 0.707,this is how it comes.
If you take one cycle of Sine wave with a peak power of '1volt' and measures the instantaneous voltage at regular time intervals on this sine wave. Then 'squares' all of the voltages at each point and adds the squared values together and then calculates the average (mean) from the squared values and finally in the end calculate the square root of the average (mean) value you will get the value of 1(peak value)*0.707 (1volt* 1/square root of 2).
If you have sinewave with a peak value other than 1volt simply replace 1 with that peak value in above equation and you will get true average value of sine wave.
Hope it helps.
/SC
2 members found this post helpful.

25th December 2010, 09:16 #6
 Join Date
 Oct 2010
 Location
 Earth
 Posts
 709
 Helped
 196 / 196
 Points
 4,515
 Level
 15
Re: RMS vs Average Value
As Mahmoud alluded to above, taking the mathematical average of a 50% duty cycle square wave would net you zero and as betwixt states "the squaring process is a workaround".
We know that voltage and current even in a periodic alternating waveform that would average to zero has the ability to do useful work just as DC does so the mathematical construct of square root of the mean (average) of the squares of all points in the waveform (RMS) was invented.
Another term that is sometimes used to describe the power in an alternating waveform is "area under the curve" in reference to the fact that the power in an alternating waveform is due to the excursion of the voltage away from zero.

25th December 2010, 19:34 #7
Re: RMS vs Average Value
Guys my question was very simple that "What voltage value of Sine wave equals DC voltage".

25th December 2010, 19:34

25th December 2010, 19:53 #8
 Join Date
 Jan 2008
 Location
 Bochum, Germany
 Posts
 40,147
 Helped
 12264 / 12264
 Points
 232,490
 Level
 100
Re: RMS vs Average Value
Guys my question was very simple that "What voltage value of Sine wave equals DC voltage".
But in a short, your original assumption about rms value has been correct.

26th December 2010, 00:53 #9
Re: RMS vs Average Value
Thanks for helping me
+ Post New Thread
Please login