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I have this simple circuit: https://groups.io/g/LTspice/files/Temp/MEAS_RMS_error.asc
.meas rms command gives wrong values when injecting DC+AC signal, so must have to calculate the average of the signal Vm to calculate the difference between signal and Vm to calculate RMS
In the example there is a square waveform 0-1V which RMS value is sqrt((0.5^2*n/n))=0.5, not 0.707!!!!!!!!!!!!!
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Vt1 has the same rms value as a 0.707V peak value sinewave.
LTspice uses peak values.
Le 15/07/2024 à 10:58, Javier Lopez a
écrit :
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I have this simple circuit: https://groups.io/g/LTspice/files/Temp/MEAS_RMS_error.asc
.meas rms command gives wrong values when injecting DC+AC
signal, so must have to calculate the average of the signal Vm
to calculate the difference between signal and Vm to calculate
RMS
In the example there is a square waveform 0-1V which RMS value
is sqrt((0.5^2*n/n))=0.5, not 0.707!!!!!!!!!!!!!
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Javier,
The RMS value of a pulse wave is sqrt(duty cycle) which equals sqrt(0.5) for a square wave which equals 0.707.
All for now
Sent: Monday, July 15, 2024 at 4:58 AM
From: "Javier Lopez via groups.io" <jlopez2022@...>
To: LTspice@groups.io
Subject: [LTspice] .MEAS RMS error
I have this simple circuit: https://groups.io/g/LTspice/files/Temp/MEAS_RMS_error.asc
.meas rms command gives wrong values when injecting DC+AC signal, so must have to calculate the average of the signal Vm to calculate the difference between signal and Vm to calculate RMS
In the example there is a square waveform 0-1V which RMS value is sqrt((0.5^2*n/n))=0.5, not 0.707!!!!!!!!!!!!!
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Actually, the rms value of a 50% duty ratio square wave is 0.5
times the peak amplitude.
The rms value of a sinewave is 0.707 times the peak amplitude.
In order to show the same rms value, the sinewave must have a max
amplitude of 0.707.
Le 15/07/2024 à 12:27, eewiz via
groups.io a écrit :
переключити цитоване повідомлення
Показати цитований текст
Javier,
The RMS value of a pulse wave is sqrt(duty cycle) which
equals sqrt(0.5) for a square wave which equals 0.707.
All for now
I have this simple circuit: https://groups.io/g/LTspice/files/Temp/MEAS_RMS_error.asc
.meas rms command gives wrong values when injecting
DC+AC signal, so must have to calculate the average of
the signal Vm to calculate the difference between
signal and Vm to calculate RMS
In the example there is a square waveform 0-1V which
RMS value is sqrt((0.5^2*n/n))=0.5, not
0.707!!!!!!!!!!!!!
|
There are no sinewaves here. We shouldn't talk about sinewaves.
Out of curiosity, what is going on in the uploaded schematic, why are there two sources? The two pulse voltage sources are identical. Why are there two, or is it just by coincidence and you left one in for redundancy?
Javier wrote, ".meas rms command gives wrong values when injecting DC+AC signal". Why did you write that? Did you come to this question assuming that the .MEAS command gives wrong values? Or does your circuit attempt to show that it is wrong?
Hint: The .MEAS command is correct.
Did you believe that RMS should remove the DC portion of a signal, before computing RMS? That would be incorrect. Remember, the RMS value of a 5 V DC signal plus a 0 V AC signal is 5 V. RMS is often described as the voltage that gives the same amount of heating, and both DC and AC components must be counted.
The mean average of V(t1)^2 between 0 and 0.5 ms is 1.0. The mean average of V(t1)^2 between 0.5 and 1.0 ms is 0.0. The mean average over the interval 0 to 1 ms is 0.5, and that exact pattern repeats indefinitely. The square root of 0.5 is 0.707107 volts. LTspice's answer from .MEAS is 0.707107. It is the correct answer.
The RMS value of V(t1) or V(t2) is not 0.5. I don't know why you think it should have been 0.5.
Now, if your PULSE voltage alternated between -0.5 and +0.5 V, so that its peak-to-peak amplitude was the same but without the DC offset, then that signal's RMS voltage would be 0.5 V. But that is a different signal, and if you examined the heat from the resistor, it would indeed be less.
LTspice is doing the right thing. Your interpretation or your understanding of RMS is in error.
Andy
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The process of computing an RMS value assumes that you know that the order of operations is essential. The correct order of operations is:
- Square one period of the signal. (N.B. All values of the squared signal are positive and >= 0)
- Compute the mean of the squared signal over one period (N.B. this actually preserves any DC component)
- Take the square root of the mean value computed in step 2. (N.B. this is also a positive value)
Any other sequence of steps will produce garbage.
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Le 15/07/2024 à 13:31, Andy I a écrit :
There are no sinewaves here. We shouldn't talk about sinewaves.
That's because of the 0.707 factor mentioned by the OP, which is pertinent to sinewaves.
Out of curiosity, what is going on in the uploaded schematic, why are there two sources? The two pulse voltage sources are identical. Why are there two, or is it just by coincidence and you left one in for redundancy?
Javier wrote, ".meas rms command gives wrong values when injecting DC+AC signal". Why did you write that? Did you come to this question assuming that the .MEAS command gives wrong values? Or does your circuit attempt to show that it is wrong?
Hint: The .MEAS command is correct.
Did you believe that RMS should remove the DC portion of a signal, before computing RMS? That would be incorrect. Remember, the RMS value of a 5 V DC signal plus a 0 V AC signal is 5 V. RMS is often described as the voltage that gives the same amount of heating, and both DC and AC components must be counted.
The mean average of V(t1)^2 between 0 and 0.5 ms is 1.0. The mean average of V(t1)^2 between 0.5 and 1.0 ms is 0.0. The mean average over the interval 0 to 1 ms is 0.5, and that exact pattern repeats indefinitely. The square root of 0.5 is 0.707107 volts. LTspice's answer from .MEAS is 0.707107. It is the correct answer.
The RMS value of V(t1) or V(t2) is not 0.5. I don't know why you think it should have been 0.5.
Now, if your PULSE voltage alternated between -0.5 and +0.5 V, so that its peak-to-peak amplitude was the same but without the DC offset, then that signal's RMS voltage would be 0.5 V. But that is a different signal, and if you examined the heat from the resistor, it would indeed be less.
LTspice is doing the right thing. Your interpretation or your understanding of RMS is in error.
Andy
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Javier,
Asking for the RMS value of a pulse wave is simply not a correct question to ask.
Although mathematically correct the answer is meaningless.
RMS applies only to waveforms that do not spend an equal amount time at each point value.
A sine wave spends less time at it's 5% value then it does at it's 95% value.
Since energy is power * time, the RMS value of a sine wave is usefull.
For a pulse wave, RMS provides a result that does not reflect the heating value.
A unipolar square wave will fully heat a resistor half the time and provide no heat for the other half.
The energy contained in a pulse wave is related to the average value of that wave not the RMS value.
The average value of a pulse wave is it's on-time divided by the period of one cycle.
On-time divided by period is a pulse wave's duty cycle.
The energy of a unipolar pulse wave is directly related to it's duty cycle not it's RMS value.
All for now.
Sent: Monday, July 15, 2024 at 11:29 AM
From: "Jerry Lee Marcel via groups.io" <jerryleemarcel@...>
To: LTspice@groups.io
Subject: Re: [LTspice] .MEAS RMS error
Le 15/07/2024 à 13:31, Andy I a écrit :
There are no sinewaves here. We shouldn't talk about sinewaves.
That's because of the 0.707 factor mentioned by the OP, which is pertinent to sinewaves.
Out of curiosity, what is going on in the uploaded schematic, why are there two sources? The two pulse voltage sources are identical. Why are there two, or is it just by coincidence and you left one in for redundancy?
Javier wrote, ".meas rms command gives wrong values when injecting DC+AC signal". Why did you write that? Did you come to this question assuming that the .MEAS command gives wrong values? Or does your circuit attempt to show that it is wrong?
Hint: The .MEAS command is correct.
Did you believe that RMS should remove the DC portion of a signal, before computing RMS? That would be incorrect. Remember, the RMS value of a 5 V DC signal plus a 0 V AC signal is 5 V. RMS is often described as the voltage that gives the same amount of heating, and both DC and AC components must be counted.
The mean average of V(t1)^2 between 0 and 0.5 ms is 1.0. The mean average of V(t1)^2 between 0.5 and 1.0 ms is 0.0. The mean average over the interval 0 to 1 ms is 0.5, and that exact pattern repeats indefinitely. The square root of 0.5 is 0.707107 volts. LTspice's answer from .MEAS is 0.707107. It is the correct answer.
The RMS value of V(t1) or V(2) is not 0.5. I don't know why you think it should have been 0.5.
Now, if your PULSE voltage alternated between -0.5 and +0.5 V, so that its peak-to-peak amplitude was the same but without the DC offset, then that signal's RMS voltage would be 0.5 V. But that is a different signal, and if you examined the heat from the resistor, it would indeed be less.
LTspice is doing the right thing. Your interpretation or your understanding of RMS is in error.
Andy
|
eewiz wrote:
. "Asking for the RMS value of a pulse wave is simply not a correct question to ask.
Although mathematically correct the answer is meaningless.
RMS applies only to waveforms that do not spend an equal amount time at each point value."
I have to disagree.
The RMS value is not meaningless. For example, it is useful for computing heat dissipated in a resistive load. And who knows what else.
It applies to ANY waveforms of ANY shape.
"For a pulse wave, RMS provides a result that does not reflect the heating value."
That is not true.
"The energy contained in a pulse wave is related to the average value of that wave not the RMS value."
Again, incorrect.
"The energy of a unipolar pulse wave is directly related to it's duty cycle not it's RMS value."
The energy is related to its duty cycle, AND to its RMS value, and the RMS value depends on its duty cycle.
Andy
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Very strange assertion. I'd like to see a proof.
Le 15/07/2024 à 19:28, eewiz via
groups.io a écrit :
переключити цитоване повідомлення
Показати цитований текст
Javier,
Asking for the RMS value of a pulse wave is simply not a
correct question to ask.
Although mathematically correct the answer is meaningless.
RMS applies only to waveforms that do not spend an equal
amount time at each point value.
A sine wave spends less time at it's 5% value then it does
at it's 95% value.
Since energy is power * time, the RMS value of a sine wave
is usefull.
For a pulse wave, RMS provides a result that does not
reflect the heating value.
A unipolar square wave will fully heat a resistor half the
time and provide no heat for the other half.
The energy contained in a pulse wave is related to the
average value of that wave not the RMS value.
The average value of a pulse wave is it's on-time divided
by the period of one cycle.
On-time divided by period is a pulse wave's duty cycle.
The energy of a unipolar pulse wave is directly related
to it's duty cycle not it's RMS value.
All for now.
Le 15/07/2024 à 13:31, Andy
I a écrit :
There are no sinewaves here. We shouldn't
talk about sinewaves.
That's because of the 0.707 factor mentioned by the OP,
which is pertinent to sinewaves.
Out of curiosity, what is going on in the uploaded
schematic, why are there two sources? The two pulse
voltage sources are identical. Why are there two, or
is it just by coincidence and you left one in for
redundancy?
Javier wrote, ".meas
rms command gives wrong values when injecting DC+AC
signal". Why did you write that? Did you come
to this question assuming that the .MEAS command gives
wrong values? Or does your circuit attempt to show
that it is wrong?
Hint: The .MEAS command is correct.
Did you believe that RMS should remove the DC portion
of a signal, before computing RMS? That would be
incorrect. Remember, the RMS value of a 5 V DC signal
plus a 0 V AC signal is 5 V. RMS is often described as
the voltage that gives the same amount of heating, and
both DC and AC components must be counted.
The mean average of V(t1)^2 between 0 and 0.5 ms is
1.0. The mean average of V(t1)^2 between 0.5 and 1.0
ms is 0.0. The mean average over the interval 0 to 1
ms is 0.5, and that exact pattern repeats
indefinitely. The square root of 0.5 is 0.707107
volts. LTspice's answer from .MEAS is 0.707107. It is
the correct answer.
The RMS value of V(t1) or V(2) is not 0.5. I don't
know why you think it should have been 0.5.
Now, if your PULSE voltage alternated between -0.5 and
+0.5 V, so that its peak-to-peak amplitude was the
same but without the DC offset, then that signal's RMS
voltage would be 0.5 V. But that is a different
signal, and if you examined the heat from the
resistor, it would indeed be less.
LTspice is doing the right thing. Your interpretation
or your understanding of RMS is in error.
Andy
|
O.K. I have prepaired my meal of crow and am now prepared to eat it.
Andy is correct. I was wrong.
My error was not realizing that an RMS voltage value will produce an RMS current value.
Assuming a 1 volt peak square wave driving a 1 Ohm load.
Related to duty cycle; 1A * 1V * 0.5 time + 0V * 0A * 0.5 time = 0.5 W
as well as, the RMS calculation; 0.7071V * 0.7071A = 0.5W.
Using the RMS value of a pulse wave provides a correct power result.
So, if an RMS voltage value of a pulse wave is available, use it.
It's just not necessary to spend effort to determine the RMS value of a pulse wave where one is not readily available.
Peak value times duty cycle, which is often easier to determine, provides the same result.
Sent: Monday, July 15, 2024 at 4:43 PM
From: "Jerry Lee Marcel via groups.io" <jerryleemarcel@...>
To: LTspice@groups.io
Subject: Re: [LTspice] .MEAS RMS error
Very strange assertion. I'd like to see a proof.
Le 15/07/2024 à 19:28, eewiz via groups.io a écrit :
Javier,
Asking for the RMS value of a pulse wave is simply not a correct question to ask.
Although mathematically correct the answer is meaningless.
RMS applies only to waveforms that do not spend an equal amount time at each point value.
A sine wave spends less time at it's 5% value then it does at it's 95% value.
Since energy is power * time, the RMS value of a sine wave is usefull.
For a pulse wave, RMS provides a result that does not reflect the heating value.
A unipolar square wave will fully heat a resistor half the time and provide no heat for the other half.
The energy contained in a pulse wave is related to the average value of that wave not the RMS value.
The average value of a pulse wave is it's on-time divided by the period of one cycle.
On-time divided by period is a pulse wave's duty cycle.
The energy of a unipolar pulse wave is directly related to it's duty cycle not it's RMS value.
All for now.
Le 15/07/2024 à 13:31, Andy I a écrit :
There are no sinewaves here. We shouldn't talk about sinewaves.
That's because of the 0.707 factor mentioned by the OP, which is pertinent to sinewaves.
Out of curiosity, what is going on in the uploaded schematic, why are there two sources? The two pulse voltage sources are identical. Why are there two, or is it just by coincidence and you left one in for redundancy?
Javier wrote, ".meas rms command gives wrong values when injecting DC+AC signal". Why did you write that? Did you come to this question assuming that the .MEAS command gives wrong values? Or does your circuit attempt to show that it is wrong?
Hint: The .MEAS command is correct.
Did you believe that RMS should remove the DC portion of a signal, before computing RMS? That would be incorrect. Remember, the RMS value of a 5 V DC signal plus a 0 V AC signal is 5 V. RMS is often described as the voltage that gives the same amount of heating, and both DC and AC components must be counted.
The mean average of V(t1)^2 between 0 and 0.5 ms is 1.0. The mean average of V(t1)^2 between 0.5 and 1.0 ms is 0.0. The mean average over the interval 0 to 1 ms is 0.5, and that exact pattern repeats indefinitely. The square root of 0.5 is 0.707107 volts. LTspice's answer from .MEAS is 0.707107. It is the correct answer.
The RMS value of V(t1) or V(2) is not 0.5. I don't know why you think it should have been 0.5.
Now, if your PULSE voltage alternated between -0.5 and +0.5 V, so that its peak-to-peak amplitude was the same but without the DC offset, then that signal's RMS voltage would be 0.5 V. But that is a different signal, and if you examined the heat from the resistor, it would indeed be less.
LTspice is doing the right thing. Your interpretation or your understanding of RMS is in error.
Andy
|
Really strange how lengthy such a discussion about nothing can be. Do you know that there is a great source of knowledge called internet? https://en.wikipedia.org/wiki/Root_mean_square
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Pukses, as wll as sinewaves, are particular cases.
The rms formula is valid for any waveform.
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Javier Lopez
eewiz