RMS of a triangle wave???

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bwts
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RMS of a triangle wave???

Post by bwts » Tue Dec 09, 2003 4:41 am

This is buggin' me coz I have a feeling that the answer is simple. I have an AC triangle wave alternating between +5 and -5 but no time scale (ie no frequency!) given, so what is its RMS value?<p>B)
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MrAl
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Re: RMS of a triangle wave???

Post by MrAl » Tue Dec 09, 2003 5:55 am

Hello Mr Bwtz,<p>To get the rms of a triangle a shortcut is
to divide the peak by the square root of 3.<p>Vrms=Vp/sqrt(3)=Vp/1.732
where
Vp=Vpp/2<p>As a side note, "rms" stands for
"The ROOT of the MEAN of the SQUARE"
where
the Square is simply the waveform squared,
the Mean is simply the average,
and the Root is simply the square root.<p>Taken together,
[1]
First you form the square of the wave, which
results in a second wave.
[2]
Then you calculate the mean of that resulting wave found in step 1.
[3]
Lastly, you take the square root of the mean
value calculated in step 2.<p>Normally when you calculate the mean of the
square of the wave you have the square of the
wave in the form of an equation and you use
integration to get the mean of one period, but
you can also use small steps in the time
variable to go through the wave and add up all
the results of the squaring, then divide the sum
up by the length of time of one period, then
take the square root and you have the answer to
a good degree of accuracy. This method is good
for any wave that is periodic, not only
those of a specific form such as triangle.<p>
As a quick example, lets use the equation
y=x to represent a triangle wave. To make
this appear periodic, we'll use x values from
0 to 1 and assume that the wave repeats
(even though in this form it really doesnt).<p>Taking the square of the wave is easy:
The wave squared is y^2 = x^2 or x*x<p>Taking the integral of x^2:
integ(x^2, from x=0 to 1)=0.3333333
Then divide by the period (1):
0.3333333/1=0.3333333<p>Taking the square root of 0.3333333:
sqrt(0.3333333)=0.57735<p>So the rms value of this wave (a triangle)
is 1/sqrt(3), which is the Vp/1.732 as above.<p>
Take care,
Al
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Re: RMS of a triangle wave???

Post by rosborne » Tue Dec 09, 2003 10:19 am

Mr. AI is correct, although I can not help myself but add the following. A triangle wave is often refered to as a sawtooth wave and consists of two triangles. And to change his notation from x and y to t and y. The integration takes place over the period (Tau) over the two parts of the triagle defined as
y = t from t = 0 to t = Tau/2 and
y = -t from t = Tau/2 to t = Tau. Square 'em and
integrate t^2 and (-t)^2 from 0 to Tau/2 and Tau/2 to Tau respectively.
The result is t^3/3 evaluated from 0 to Tau/2 +
t^3/3 evaluted from Tau/2 to Tau which equals
Tau/6 + Tau/6 = Tau/3.
This result is averaged over the period Tau resulting in 1/3, take the sqrt() and Mr. AI's correct answer pops out.
The slope of Mr. AI's triangle is 1 that is not necessarily the case for all triangle waves. I think the general solution for a sawtooth wave rms value is
m/sqrt(3) where m is the slope of the triangles.
-Rick

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Re: RMS of a triangle wave???

Post by bwts » Wed Dec 10, 2003 5:25 am

Thanx this has freed some extra space in my thought processes (not that things like this keep me awake at night u understand ;) )<p>B)
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MrAl
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Re: RMS of a triangle wave???

Post by MrAl » Wed Dec 10, 2003 8:29 am

Hello again,<p>Sorry rosborne, but Vp/1.732 is correct for
triangles and sawtooth waves.
Perhaps i misunderstood the point you were
trying to make?<p>The only difference for waves that appear
above the time axis and also touch it
(such as a sawtooth rising from 0 to 5v
and back to zero) is that Vp is taken as
the height of the triangle, not the Vpp/2 .<p>For a triangle sawtooth that rides above
zero, there is also a trick:<p>Vrms=sqrt[Vrmst^2+Vp*Vd]
where
Vrmst is the rms value of the sawtooth as if
it was sitting on the time axis (0vdc)
and
Vp is the max height of the triangle
and
Vd is the min height of the triangle.<p>That trick was derived from knowing that the
total rms value of a wave is equal to the
square root of the sum of the squares of the
individual component rms values.<p>Also, for waves that have a duty cycle where
part of the wave is at zero for some time,
you can multiply the triangle rms value (taken
alone for it's time duration) by the square root
of the duty cycle to get the true rms value.
For example a triangle that is 'on' for only
half the time would have it's rms value (taking
the triangle alone) multiplied by sqrt(0.5) to
get the true rms of the entire wave.<p>BACK TO THE ORIGINAL QUESTION<p>In the simple case where the triangle runs
equally plus and minus above and below the
time axis (as was the original question) the
rms value is simply the peak value divided by
the square root of three, where the peak value
is obtained by dividing the peak to peak value
by two.<p>Take care,
Al
LEDs vs Bulbs, LEDs are winning.

rosborne
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Re: RMS of a triangle wave???

Post by rosborne » Sat Dec 13, 2003 3:06 pm

Mr. AI said,
>Sorry rosborne, but Vp/1.732 is correct for
>triangles and sawtooth waves.
Perhaps i misunderstood the point you were
trying to make?

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Re: RMS of a triangle wave???

Post by rosborne » Sat Dec 13, 2003 3:20 pm

Whoops, and<p>Mr AI, I have no idea what you could be sorry about. I thought that what I wrote complemented and expanded on what you wrote. I even came up with the exact same answer Vp/sqrt(3) = Vp/1.732 although I see that I forgot the Vp.<p>As far as what happens when there is a DC component I think you can use superposition and just add the DC value. But maybe I'm mistaken. And, of course, i agree that the triagle or sawtooth is sitting on the 0V line.<p>The duty cycle stuff lost me. I'll have to look at it more closely later. Looks like a trick I'd like to know.<p>Sorry if miscommunicated or irritated,
Rick

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Re: RMS of a triangle wave???

Post by rshayes » Sun Dec 14, 2003 3:00 am

The RMS voltage can be viewed as the DC voltage which, when applied to a resistor, will produce the same heating effecting as a complex waveform applied to the same resistor. Some RMS voltmeters actually use thermal conversion to measure the RMS value.<p>The power is proportional to the square of the applied DC voltage. If the waveform is only applied one half of the time, the heating effect is one half. Producing one half the heating effect means dividing the DC voltage by the square root of two instead of two. This is the basis for the RMS voltage being proportional to the square root of the duty cycle.<p>In general, adding two waveforms using the root-sum-of-squares of the RMS value is not valid. Consider two waveforms, each consisting of 1 volt DC, and thus 1 volt RMS. Adding these with a root-sum-of-squares would give a value of 1.414 volts RMS. The actual sum is a DC level of 2 volts, which is also 2 volts RMS.<p>If the waveforms do not overlap, then the root-sum-of-squares addition is probably valid. This is a special case however. Consider two 1 volt pulses of 50% duty cycle. The RMS value of each pulse is .707 volts. If the two pulses do not overlap, they provide a constant 1 volt level. This is the same answer that would be given by a root-sum-of-squares addition. If the pulses coincide, they form a 2 volt pulse of 50% duty cycle. This would have an RMS value of 1.414 volts, which is not the result of a root-sum-of-squares operation.

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MrAl
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Re: RMS of a triangle wave???

Post by MrAl » Sun Dec 14, 2003 8:24 am

Hello again Rick and hello stephen,<p>Rick:
I thought you were trying to say that the RMS value was
equal to the slope m divided by the square root of three
because you said that you thought that was the general
solution to a sawtooth. I liked your idea about
integrating from -T/2 to T/2 though.
The duty cycle shortcut was derived directly from
the integral equation for calculating rms.<p>stephen:
I liked to hear your ideas about the reasoning behind some
of the shortcuts used in calculating rms values, but what on
earth would make you want to add two separate waveforms by
using the root-sum-of-squares method?<p>
Take care,
Al
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rshayes
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Re: RMS of a triangle wave???

Post by rshayes » Sun Dec 14, 2003 11:01 am

Sometimes a complex waveform can be considered as the sum of several simpler waveforms. If there was a consistent rule for summing RMS values, this would make calculating the RMS value easier. One example would be adding a DC offset to a waveform whose RMS value is known. A root-sum-of-squares calculation is simpler than calculating the RMS value by integration. Unfortunately, the root-sum-of-squares is not always valid.<p>An example of a composite waveform would be the current in a deflection yoke in a television set or computer monitor. This is a sawtooth during the trace period and a sine wave during the retrace period. Both these waveforms have known RMS values. The overall RMS value can be computed by using these previously known results, weighted by the duty cycle of each waveform. This is easier than calculating the integral of the squared waveform.<p>The RMS voltages of uncorrelated noise sources do add in a root-sum-of-squares fashion. If the noise sources are correlated, the RMS voltage of the combined waveforms is higher, depending on the degree of correlation.

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MrAl
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Re: RMS of a triangle wave???

Post by MrAl » Mon Dec 15, 2003 4:36 am

Hello stephen,<p>Sorry if i didnt make it clear, but i had said:<p>"That trick was derived from knowing that the
total rms value of a wave is equal to the
square root of the sum of the squares of the
individual component rms values."<p>I assumed that most readers would understand that
'component' means "Fourier component", which is
NOT the same as two separate waveforms. You got
me wondering why you would add two dc values to
try to somehow disprove this important
conclusion, which cannot be disproved by adding
two dc voltages or currents as you did.<p>In any case, let me restate this so it becomes
more clear...<p>"The rms value of a waveform is equal to the
square root of the sum of the squares of the
individual Fourier component rms values."<p>Does that make this more clear to you now?<p>I was going to recommend using the integral
anyway because of the misunderstandings that
sometimes come up when trying to use shortcuts
like this.<p>
Take care,
Al
LEDs vs Bulbs, LEDs are winning.

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Re: RMS of a triangle wave???

Post by rosborne » Mon Dec 15, 2003 6:11 am

AI, <p>You are right about the slope thing, that was a half baked idea that occured to me while I was defining the shape of the triangle. I forgot that the slope of the triangle wave has an impact on the period of the waveform. I came up with an idea that you rightly called me on because it was BS. Thanks, once again I learn something that I already thought I knew, only better. . . if that makes sense.<p>You wrote Vrms=sqrt[Vrmst^2+Vp*Vd]
I don't get it. Did you mean
Vrms=sqrt(Vrmst^2+(Vp-Vd)^2) or
Vrms=squt(Vrmst^2) + abs(Vp-Vd)? Which I think are two ways of saying about the same thing, but definitly does not say the same thing as your eq'n.<p>-Rick

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MrAl
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Re: RMS of a triangle wave???

Post by MrAl » Mon Dec 15, 2003 6:49 am

Hi again rosborne,<p>In order to find a formula that is easy to apply i thought
it would be easiest to pick the two points (on the triangle
riding on the dc level) to be the very top most point
(peak) and call this Vp, and the point at the 'bottom' of
the triangle, which is above zero when the triangle rides
on a dc level, and call this point Vd.
I'll use the number 1.732 as the square root of 3.<p>Taking these two points, Vp and Vd, lets form a simple
expression for what we'll call the 'zero ac' level:
Vz=(Vp-Vd)/2<p>Now form an expression for the rms value of the triangle
alone at the 'zero ac' level:<p>Vrmst=Vz/1.732
Notice this is the same as for a regular triangle from
the "Vp/1.732" formula, so this part is simple.<p>Now the dc level we use here is:
Vdc=Vd+Vz
Notice this calculation is simple too.<p>Now to get the rms value of these two together
we can take the square root of the sum of the
squares of the individual component rms values:
Vrms=sqrt(Vrmst^2+Vdc^2)<p>Notice this is still a fairly simple calculation,
but this isnt the formula i presented earlier.
The earlier formula:
Vrms=sqrt[Vrmst^2+Vp*Vd]
is the same however, just in a slightly different
form, and you have to take care not to confuse
Vd in this formula with Vdc of the second formula.
You still have to calculate Vrmst though, as:
Vrmst=(Vp-Vd)/1.732 noting that this value
is different then the other one because we will
assume a different dc level in order to simplify
the end formula.<p>If you dont like the earlier formula for some
reason, then you can always do the second one
but you have to calculate three additional things:<p>Vz=(Vp-Vd)/2
Vrmst=Vz/1.732
Vdc=Vd+Vz
then finally:
Vrms=sqrt(Vrmst^2+Vdc^2)<p>This was supposed to make it easier to calculate the rms
value of a triangle riding on a dc level,
that's all :-)<p>
Take care,
Al
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Re: RMS of a triangle wave???

Post by rshayes » Fri Dec 19, 2003 5:18 am

Most of the times that I have had to calculate RMS values have been in power supply design. Some swotching power supplies generate ripple currents that are trapezoidal pulses of variable duty cycle. These can be viewed as sawtooth waveforms added to square pulses. It is very convenient if the overall RMS value can be calculated from the values of two or more simpler waveforms. The dead time and the variable duty cycle make the fourier series for the waveforms very complicated.<p>Actually, one of the worst calculations is finding the RMS ripple current in the filter capacitor of a linear power supply. The input voltage is a sine wave, but the current tends to flow in short pulses with high peak values. The conduction angle depends on the load current, which may vary as the capacitor voltage varies. The easiest way that I have found to do this calculation is pure brute force- Use a spread sheet to calculate the current at 50 or 100 points and then square the values and sum them numerically.<p>When you square a fourier series, you get the squares of the individual terms as well as a large number of cross products, such as sine two omega tee times sine three omega tee. These cross terms are not zero, but, over a cycle, they integrate to zero. This leaves only the square terms, and the root-sum-of-squares would be valid for this case.<p>Adding a DC component to a waveform can be done using root-sum-of-squares if the waveform does not have a DC component. As an example, consider a 1 volt RMS sine wave with a 1 volt DC offset. The RMS value would be 1.414 volts. Now take this waveform and offset it by 1 volt. A root-sum-of-squares calculation using the previous RMS value would give a value of 1.732 volts RMS. However, this waveform can also be considered as a waveform with a 1 volt sine wave with a 2 volt offset. The RMS value of this waveform is actually 2.236 volts RMS rather than 1.732 volts.

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MrAl
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Re: RMS of a triangle wave???

Post by MrAl » Fri Dec 19, 2003 9:31 am

Hello again stephen,<p>I think i agree with everything you said
except the last paragraph. Im sorry to
say this but your argument isnt valid.<p>The reason for this is that the moment you
say "Fourier component" you automatically
preclude the use of more than one dc value.
You cant break up a dc value into more then
one part like that because there is only one
dc component in a Fourier expansion.<p>For another example, you cant break up 1vdc
into 0.2+0.2+0.2+0.2+0.2 because the Fourier
dc component is 1vdc and only this single value.
I was trying to point this out when you broke
up the dc value in another post apparently to
try to disprove something maybe?<p>The same holds true for any of the harmonics
i think. Each component gets summed before
you can calculate anything else.<p>The correct rms value for a pure sine with a
peak equal to the sqrt(2) riding on a 2vdc
level is sqrt(5).
The correct rms value for a pure sine with a
peak equal to the sqrt(2) riding on a 1vdc
level is sqrt(2).
In both these cases, if you take the dc component
to be a single component (just as you do with
any of the harmonic components) you always get the
correct answer if you use the sum of the sqrt
of the component rms values method.<p>Take care,
Al
LEDs vs Bulbs, LEDs are winning.

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