Accurate Measurements on pulsed system's harder than you think.

[digitalindustry]

I am trying to show you that things are opposite to what they should be.
Look at the scope shots i posted in my last post. The first two show what the current trace should look like across the CVR. But in the scope shot from the DUT,the current wave form is opposite to what it should be. This is showing the bulbs resistance is decreasing during the pulse--not increasing as it should be. As i have been trying to say,the incandescent bulb is showing/doing opposite to what it should be. Measurements being taken from something that dose not act as it should can only lead to measurement error.

Now you have to figure out why the current trace is opposite to what it should be-->then we can start talking about correct measurements.

Brad

I entertained this thought in TK measurements.

he stated the resistance of the bulb at the pulse , and calculated that as the over all resistance, however then i thought that might have been accounted for by the duty cycle.

i.e no pulse = no resistance.

EM

you have to admit this is a strange and curious effect, why not lets learn from it if you totally understand it, granted you know more about this than i do, i'll happily tuck away my human pride to learn some more here.

you have to admit when TM posted this you didn't say, 'oh that's simple it's due to the pulse duty cycle average and bulb filament'

if you had said that sentence we could all agree you totally understood it?

[TinselKoala]

You all are just confusing yourselves with more Red Herrings and misunderstandings. STOP already!

1. By multiplying the values for Current and Voltage at each instant of time, you generate an "instantaneous power curve" that represents the actual power at each instant in your sample set. Phase Angle _does not enter_ into this process! The IP curve is correct and gives the actual power at each instant, no matter the phase angle between V and I.

2. When you tell the scope to compute CH1 x CH2 it is following this process. For each timeslice in its memory it is taking a value from one channel and a value from the other channel, multiplying them, and plotting the resulting value at that timeslice on the Math trace to generate a point on the Instantaneous Power Curve. Then it moves on to the next pair of values for the next timeslice, and repeats until it is done.

3. When the scope computes the "average" or mean of a signal it is doing the following: for each timeslice in the sample set, it is adding the value to the sum of the preceding timeslice values, then is dividing the result by the  number of timeslices in the sample set.

So for a constant voltage, it is doing this to find the average voltage over, say, twelve samples:
6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 = 72.0
72.0 / 12 = 6.0
and you get a valid average (mean) value of 6.0 volts across that 12 timeslices.

But for a pulse train with a duty cycle of 1/3 High and 2/3 Low (i.e. 33 percent or 0.33 High), the scope is doing this:
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 6.0 + 6.0 + 6.0 + 6.0 = 24.0
24 / 12 = 2.0
and 2.0 is the valid average (mean) value for this pulse train of one period in 12 samples.
NOTE that this result is the SAME as what you get by taking the Peak Value (or the "average high" during the pulse ON timeslices) of 6.0 and multiplying it by the Duty Cycle of 0.33.

4. So, let's say you have a set of voltage and current values at a series of timeslices and you want to find the IP curve and the average power. Do you find the average voltage, the average current, then multiply those values to give an average Power? Or do you find the instantaneous VxI at each timeslice to generate an IP value, then average those IP values over the full sample set as in Part 3 above? Let's see what happens if we have a pulse train with ON values of V = 3.0, I = 3.0 and a Duty Cycle of 1/3 ON (33 percent, 0.33 ON) , with a sample size of 12 timeslices.

The first way:
Average voltage is 0+0+0+0+0+0+0+0+3+3+3+3= 12, and 12/12 is 1, so average voltage is 1 volt. This is correct, it is the average voltage.
Average current is 0+0+0+0+0+0+0+0+3+3+3+3= 12, and 12/12 is 1, so average current is 1 amp. This is correct, it is the average current.
Now multiply those averages together and you get 1 Watt "average". This is clearly wrong by inspection, since you can easily see that you have 9 watts during 1/3 of the timeslices and 0 Watts during 2/3 of the timeslices.

The second, correct way:
First we multiply each V and I pair to generate the IP curve values at each timeslice:
0,0,0,0,0,0,0,0,9,9,9,9 and taking the average of those we get
0+0+0+0+0+0+0+0+9+9+9+9= 36, and 36/12 = 3 Watts average.
Also we can see that a constant 9 watts during the ON time x 0.33 duty cycle = 3 Watts average over the full 12-slice (100 percent) period. 

5. The error in the first method basically boils down to using the Duty Cycle of 33 percent ON twice in the calculation by computing the separate averages before the multiplication, when it should only be used once, by finding the average after multiplication.

6. The behaviour of a single bulb in a simple circuit can't be directly compared to the behaviour of the bulb in the Bedini SSG circuit because there is a lot of other stuff going on in the Bedini circuit that determines the shape of the pulse displayed on the scope. Those differences are irrelevant when computing the power, though, since the scope's timeslices are very fine indeed and can easily take the pulse shape into account when computing the IP curve values and the subsequent average power.

[digitalindustry]

EM TM TK and other not to be off topic :

but could we consult with any possible people living on this new planet that was just discovered in our solar system :

https://voat.co/v/Contact/comments/719767

: D

and just think EM was all worried about world war 3 ha ha. (jokes)

we're going to need an accurate pi measurement to get this far out ha ha

: D

[TinselKoala]

Sure, ask them. Be sure to let me know when you get an answer.

[EMJunkie]

You all are just confusing yourselves with more Red Herrings and misunderstandings. STOP already!

1. By multiplying the values for Current and Voltage at each instant of time, you generate an "instantaneous power curve" that represents the actual power at each instant in your sample set. Phase Angle _does not enter_ into this process! The IP curve is correct and gives the actual power at each instant, no matter the phase angle between V and I.

2. When you tell the scope to compute CH1 x CH2 it is following this process. For each timeslice in its memory it is taking a value from one channel and a value from the other channel, multiplying them, and plotting the resulting value at that timeslice on the Math trace to generate a point on the Instantaneous Power Curve. Then it moves on to the next pair of values for the next timeslice, and repeats until it is done.

3. When the scope computes the "average" or mean of a signal it is doing the following: for each timeslice in the sample set, it is adding the value to the sum of the preceding timeslice values, then is dividing the result by the  number of timeslices in the sample set.

So for a constant voltage, it is doing this to find the average voltage over, say, twelve samples:
6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 + 6.0 = 72.0
72.0 / 12 = 6.0
and you get a valid average (mean) value of 6.0 volts across that 12 timeslices.

But for a pulse train with a duty cycle of 1/3 High and 2/3 Low (i.e. 33 percent or 0.33 High), the scope is doing this:
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 6.0 + 6.0 + 6.0 + 6.0 = 24.0
24 / 12 = 2.0
and 2.0 is the valid average (mean) value for this pulse train of one period in 12 samples.
NOTE that this result is the SAME as what you get by taking the Peak Value (or the "average high" during the pulse ON timeslices) of 6.0 and multiplying it by the Duty Cycle of 0.33.

4. So, let's say you have a set of voltage and current values at a series of timeslices and you want to find the IP curve and the average power. Do you find the average voltage, the average current, then multiply those values to give an average Power? Or do you find the instantaneous VxI at each timeslice to generate an IP value, then average those IP values over the full sample set as in Part 3 above? Let's see what happens if we have a pulse train with ON values of V = 3.0, I = 3.0 and a Duty Cycle of 1/3 ON (33 percent, 0.33 ON) , with a sample size of 12 timeslices.

The first way:
Average voltage is 0+0+0+0+0+0+0+0+3+3+3+3= 12, and 12/12 is 1, so average voltage is 1 volt. This is correct, it is the average voltage.
Average current is 0+0+0+0+0+0+0+0+3+3+3+3= 12, and 12/12 is 1, so average current is 1 amp. This is correct, it is the average current.
Now multiply those averages together and you get 1 Watt "average". This is clearly wrong by inspection, since you can easily see that you have 9 watts during 1/3 of the timeslices and 0 Watts during 2/3 of the timeslices.

The second, correct way:
First we multiply each V and I pair to generate the IP curve values at each timeslice:
0,0,0,0,0,0,0,0,9,9,9,9 and taking the average of those we get
0+0+0+0+0+0+0+0+9+9+9+9= 36, and 36/12 = 3 Watts average.
Also we can see that a constant 9 watts during the ON time x 0.33 duty cycle = 3 Watts average over the full 12-slice (100 percent) period. 

5. The error in the first method basically boils down to using the Duty Cycle of 33 percent ON twice in the calculation by computing the separate averages before the multiplication, when it should only be used once, by finding the average after multiplication.

6. The behaviour of a single bulb in a simple circuit can't be directly compared to the behaviour of the bulb in the Bedini SSG circuit because there is a lot of other stuff going on in the Bedini circuit that determines the shape of the pulse displayed on the scope. Those differences are irrelevant when computing the power, though, since the scope's timeslices are very fine indeed and can easily take the pulse shape into account when computing the IP curve values and the subsequent average power.

Brad, TK and others following - I agree with TK's over all Power Conclusion. We cant be mixing Oranges and Apples if it is just Apples we want to observe.

TK, I think it needs to be said, do you think anything unusual is going on in this circuit?

   Chris Sykes
       hyiq.org