What is the difference between the RMS and Average voltage of a full-wave rectified sine wave?
usen...@hotmail.com
full-wave rectified sine wave?
Obviously, the average voltage of a sine wave is 0, but I have always
assumed that the RMS and average voltage of a full-wave rectified sine
wave is .707 of the
peak of the waveform. But today, I plotted several points between 0 and
180 degrees
in .25 degree increments in a spreadsheet and calculated the Sine of
each sample, assuming a peak voltage of 1. Then added all measurements
up and divided by the total of measurements, which I would assume give
me .707, but instead my result was .633 The reason I tried this
experiment is that I am planning an FPGA design that measures the RMS
voltage of a waveform, by sampling a waveform with an external A/D and
integrating the measurements (sum and divide) Perhaps I am missing
something. Here is a portion of my spreadsheet
phase angle
in (degrees) sin(angle)
column a column b
0 x
.25 x
.50 x
.75 x
.
90 1
.
180 0
RMS voltage = sum of column b results /# samples
Sent via Deja.com http://www.deja.com/
Before you buy.
Bob Penoyer
>What is the difference between the RMS and Average voltage of a
>full-wave rectified sine wave?
>
>Obviously, the average voltage of a sine wave is 0, but I have always
>assumed that the RMS and average voltage of a full-wave rectified sine
>wave is .707 of the
>peak of the waveform. But today, I plotted several points between 0 and
>180 degrees
> in .25 degree increments in a spreadsheet and calculated the Sine of
>each sample, assuming a peak voltage of 1. Then added all measurements
>up and divided by the total of measurements, which I would assume give
>me .707, but instead my result was .633
Congratulations for taking the bull by the horns and pursuing your
reasoning with real effort.
When a sine curve has a peak amplitude of 1 and the period has angular
measure (instead of time, for example) then the area under one
alternation of the sine curve is exactly 2. Your estimate of area
should be close to 2 since you used 0.25-degree increments.
Since the area of one alternation is 2 and one alternation occupies a
"width" of pi radians, then the average amplitude is 2 / pi, or about
0.637.
If you use a DC voltmeter to observe a full-wave rectified sine wave
with a peak amplitude of 1 through an appropriate low-pass filter, the
meter will read 0.637 volts. (The low-pass filter acts as an
averager.)
If a sine curve is squared, the area under one period is exactly pi/2.
Since the period is pi, dividing by pi yields an average value of
exactly 1/2. If you square your sine samples and average them, you
should get a result of very nearly 0.5. But RMS stands for ROOT mean
square. 0.5 is the mean of the square. The last step is to take the
(square) root. The square root of 1/2 is about 0.707. Thus, the RMS
value of a sine wave is about 0.707 times the peak value.
The RMS value can be treated like a DC value when doing calculations
dealing with power.
Viktor Velisavljev
<usen...@hotmail.com> wrote in message news:85r8kj$k0c$1...@nnrp1.deja.com...
> What is the difference between the RMS and Average voltage of a
> full-wave rectified sine wave?
> Obviously, the average voltage of a sine wave is 0, but I have always
> assumed that the RMS and average voltage of a full-wave rectified sine
> wave is .707 of the peak of the waveform.
This is true for RMS, but not for average.
>But today, I plotted several points between 0 and 180 degrees
> in .25 degree increments in a spreadsheet and calculated the Sine of
> each sample, assuming a peak voltage of 1. Then added all measurements
> up and divided by the total of measurements, which I would assume give
> me .707, but instead my result was .633
You calculated average of full-wave rectified sine wave. Average is what you
get if you add all samples and divide the result by the total number of
samples.
RMS stands for root mean square and that tells you how to calculate RMS. To
calculate RMS numerically you should first calculate square of each sample
(second power of each sample), then add them all, then divide the result by
total number of samples and finally find square root of that. The last
operation ( square root ) can be tricky to implement digitally.
Best regards, Jelena
Rolie Baldock
The FSQRT in the maths co-processor calculates the square root no
sweat. If you want to do it on an MCU such as the HC11 there are a
couple of options. You can use Gordon Doughman's floating point math
package, or if integer arithmetic is adequate then I suggest you
remember that the square root is the antilog of half the log of the
original number. This is neatly done using logs to the base 2. I
developed a routine to do this somewhere in a HC11 program, If you
relly want to get a copy you will have to email me and I will try to
fish it out of my MCU asm file sub directory.
Rolie Baldock. email: <berd_ka...@techemail.com>
Phil Harrison
<berd_ka...@techemail.com> writes
http://www.embedded.com/98/9802fe2.htm
Also see here for a simplified routine that works well:
http://www.embedded.com/98/9805pt.htm
--
Phil Harrison
Robert Macy
The average of a sine wave is 2/pi.
The rms of a sine wave is 1/sqrt(2)
So you got close to the right answer of 0.6366198
You got a 0.6% error. Probably from your accuracy as you added.
Your accumulator only has so many digits and as you add tiny values to it,
it will round off and round off, always causing you to get less than you
should.
Studying a book like Introduction to Numerical Analysis by Carl-Erik Froberg
published by Addison Wesley in 1965 will show how to predict and quantify
this phenomenon.
One simple solution is to do interim additions which always straddle your
min max capability. Just check your accumulator to make certain that the
number you add to it is large enough to make a difference or if too small go
add it to a different accumulator until the contents of that accumulator is
large enough to make a difference then go add that to your first accumulator
and so on. You should be able to reduce your error to acceptible limits
with only two accumulators (you may need 3+ ?)
That is the key, though acceptible limits. I'd arbitrarily set my limits to
1/4 the measurement sampling accuracy to be sure that the accumulator error
did not dominate.
- Robert -
Anthony Wong wrote in message <38C44D29...@cfhc.caritas.edu.hk>...
>hi
>
>usen...@hotmail.com wrote:
>
>> What is the difference between the RMS and Average voltage of a
>> full-wave rectified sine wave?
>>
>> Obviously, the average voltage of a sine wave is 0, but I have always
>> assumed that the RMS and average voltage of a full-wave rectified sine
>> wave is .707 of the
>> peak of the waveform. But today, I plotted several points between 0 and
>> 180 degrees
>> in .25 degree increments in a spreadsheet and calculated the Sine of
>> each sample, assuming a peak voltage of 1. Then added all measurements
>> up and divided by the total of measurements, which I would assume give
>> me .707, but instead my result was .633 The reason I tried this
>> experiment is that I am planning an FPGA design that measures the RMS
>> voltage of a waveform, by sampling a waveform with an external A/D and
>> integrating the measurements (sum and divide) Perhaps I am missing
>> something. Here is a portion of my spreadsheet
>>
>> phase angle
>> in (degrees) sin(angle)
>>
>> column a column b
>>
>> 0 x
>> .25 x
>> .50 x
>> .75 x
>> .
>> 90 1
>> .
>> 180 0
>>
>> RMS voltage = sum of column b results /# samples
>>
>> Sent via Deja.com http://www.deja.com/
>> Before you buy.
>
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Robert Macy
Forgot to mention that an arbitrary waveform you can get way off by simpling
averaging the waveform. Picture a spike signal. It is possible to start
getting magnitudes off there. For square wave to sine wave, you'll only be
off around 10%
But you really should try squaring the value, summing it, averaging the
total, then simply leave it alone for the average power.
Anthony Wong
Don Klipstein
: usen...@hotmail.com wrote:
:
: > What is the difference between the RMS and Average voltage of a
: > full-wave rectified sine wave?
: >
: > Obviously, the average voltage of a sine wave is 0, but I have always
: > assumed that the RMS and average voltage of a full-wave rectified sine
: > wave is .707 of the
: > peak of the waveform. But today, I plotted several points between 0 and
: > 180 degrees
: > in .25 degree increments in a spreadsheet and calculated the Sine of
: > each sample, assuming a peak voltage of 1. Then added all measurements
: > up and divided by the total of measurements, which I would assume give
: > me .707, but instead my result was .633 The reason I tried this
: > experiment is that I am planning an FPGA design that measures the RMS
: > voltage of a waveform, by sampling a waveform with an external A/D and
: > integrating the measurements (sum and divide) Perhaps I am missing
: > something.
Average is average absolute value which for a sinewave is peak times
2/pi. This is peak times .6366.
RMS is square root of average square, which for a sinewave is peak times
the square root of 1/2. This is peak times .7071.
Since average power dissipation in a resistor is determined by the
average square of any rapidly cyclicly varying voltage, RMS voltage is the
effective voltage which determines the average power dissipation in a
resistor.
- Don Klipstein (d...@misty.com)
John Woodgate
<aw...@cfhc.caritas.edu.hk> inimitably wrote:
>Obviously, the average voltage of a sine wave is 0, but I have always
>> assumed that the RMS and average voltage of a full-wave rectified sine
>> wave is .707 of the
>> peak of the waveform. But today, I plotted several points between 0 and
>> 180 degrees
>> in .25 degree increments in a spreadsheet and calculated the Sine of
>> each sample, assuming a peak voltage of 1. Then added all measurements
>> up and divided by the total of measurements, which I would assume give
>> me .707, but instead my result was .633 The reason I tried this
>> experiment is that I am planning an FPGA design that measures the RMS
>> voltage of a waveform, by sampling a waveform with an external A/D and
>> integrating the measurements (sum and divide) Perhaps I am missing
>> something.
I noticed some very erudite responses that may do nothing to dispel your
confusion. To get the rms value, you need to square the sines, add up
the squared values and take the square root of the result. Your
calculation found the average value, or d.c. component, correctly, so
you should get the correct rms value by using the right algorithm.
--
Regards, John Woodgate, OOO - Own Opinions Only. Phone +44 (0)1268 747839
Fax +44 (0)1268 777124. http://www.jmwa.demon.co.uk
I wanted to make a fully-automated nuclear-powered trawler,
but it went into spontaneous fishing.
Rick Daniel
phase locked to the signal youare trying to get the rms value of. you
need to start and stop your calculations on an exact number of
waveforms.
>hi
>
>usen...@hotmail.com wrote:
>
>> What is the difference between the RMS and Average voltage of a
>> full-wave rectified sine wave?
>>
>> Obviously, the average voltage of a sine wave is 0, but I have always
>> assumed that the RMS and average voltage of a full-wave rectified sine
>> wave is .707 of the
>> peak of the waveform. But today, I plotted several points between 0 and
>> 180 degrees
>> in .25 degree increments in a spreadsheet and calculated the Sine of
>> each sample, assuming a peak voltage of 1. Then added all measurements
>> up and divided by the total of measurements, which I would assume give
>> me .707, but instead my result was .633 The reason I tried this
>> experiment is that I am planning an FPGA design that measures the RMS
>> voltage of a waveform, by sampling a waveform with an external A/D and
>> integrating the measurements (sum and divide) Perhaps I am missing
dwight_...@my-deja.com
Anthony Wong <aw...@cfhc.caritas.edu.hk> wrote:
> hi
>
> usen...@hotmail.com wrote:
>
> > What is the difference between the RMS and Average voltage of a
> > full-wave rectified sine wave?
> >
Hi
I think everyone else has been dancing around the question
without really answering it. RMS stands for "root mean square".
What the measurement is is ( I have quoted an important man :)
the value of DC voltage that will create the same power to be
dissipated in a resistor. Since the power in a resistor is
P= E^2 / R, the power that is integrated over any function
will be proportional the square root of the sum of the square
of the voltages. What you are doing is just the average voltage
and not the average power effective voltage. For a sine wave,
it is simply 0.707 times the peak as others have mentioned.
For a more comples wave forms, you have to integrate the
square of the voltage of a cycle and then square root it
to get the RMS or what I call the same power effect.
Does that make anymore sense or am I just confusing
things more.
Dwight
Win Hill
> Anthony Wong <aw...@cfhc.caritas.edu.hk> wrote:
>> usen...@hotmail.com wrote:
>>
>>> What is the difference between the RMS and Average voltage of a
>>> full-wave rectified sine wave?
>
> For a sine wave, it is simply 0.707 times the peak as others have
> things more. Dwight
The ratio of a sine wave's RMS value to its rectified average value,
obtained by integrating over a half cycle, is pi/2^3/2 = 1.11, as we
point out in AoE page 455. The average value (90% of rms, and 63.6%
of the amplitude) is what you get at the output of a low-pass filter,
or what's shown on a meter movement. Generally ac voltmeters are
calibrated to read RMS voltage, even though they are responding to
the ac average.
- Win
Winfield Hill
Rowland Institute for Science
100 Edwin Land Blvd
Cambridge, MA 02142-1297
Arthur Jernberg
wave versus the peak voltage levels. That is, when a voltage is applied to a
motor to cause the motor to produce work the RMS readings are more closely
related to the effective horsepower produced by the motor. I know, probably
adding more pluthera to the pile.
Win Hill <wh...@mediaone.net> wrote in message
news:38C7EDE4...@mediaone.net...
Mark Kinsler
>I was taught that RMS reflected the effective working ability of the sine
>wave versus the peak voltage levels. That is, when a voltage is applied to a
>motor to cause the motor to produce work the RMS readings are more closely
>related to the effective horsepower produced by the motor. I know, probably
>adding more pluthera to the pile.
This is my favorite way of looking at it. It's easiest from a historical
perspective:
It's 1910. Your house has been wired for electric lighting, and is
supplied by 120v DC. Your living room is lit with several 100 watt lamps,
which obediently allow a current of 100w/120v=0.833 amperes through them.
Now, the power company has announced that it's going to convert from DC
power to AC power. What are you, the homeowner, supposed to do about
this?
Well, ideally nothing. You're expecting your lights to be as bright as
they always were. To do this, the power company has to produce AC that'll
make the lamps glow as brightly as they did at 120v DC. It turns out that
this can be done by supplying a voltage of 120 x sqrt(2), or about 169
volts peak. This is necessary because the AC waveform spends a lot of
time at low voltages, down there near plus-or-minus ten volts or fifteen
volts or 50 volts. So we make up for that by having it spend a certain
amount of time at voltages _higher_ than 120 volts.
>> The ratio of a sine wave's RMS value to its rectified average value,
>> obtained by integrating over a half cycle, is pi/2^3/2 = 1.11, as we
>> point out in AoE page 455. The average value (90% of rms, and 63.6%
>> of the amplitude) is what you get at the output of a low-pass filter,
>> or what's shown on a meter movement. Generally ac voltmeters are
>> calibrated to read RMS voltage, even though they are responding to
>> the ac average.
>>
>> - Win
But I didn't know that, either. Or if I did, I'd forgotten. Thanks.
Mark Kinsler
--
............................................................................
114 Columbia Ave. Athens, Ohio USA 45701 voice740.594.3737 fax740.592.3059
Home of the "How Things Work" engineering program for adults and kids.
See http://www.frognet.net/~kinsler
Steve
>
> I was taught that RMS reflected the effective working ability of the sine
> wave versus the peak voltage levels. That is, when a voltage is applied to a
> motor to cause the motor to produce work the RMS readings are more closely
> related to the effective horsepower produced by the motor. I know, probably
> adding more pluthera to the pile.
The squaring and square-rooting done is to reverse compensate for the
equivalency being made with the power delivered by AC and DC. This is
done since Power = I*V = V^2/R = I^2*R , so that power is always
proportional to the square of current or voltage. A sinewave of 10V rms
delivers the same power to a resistor as 10VDC. If you work it backwards
with calculus over time this will become obvious.
-Steve
--
-Steve Walz rst...@armory.com ftp://ftp.armory.com:/pub/user/rstevew
-Electronics Site!! 1000 Files/50 Dirs!! http://www.armory.com/~rstevew
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