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Further on ISO invariance
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Jun 3, 2016 11:33:51   #
Apaflo Loc: Anchorage, Alaska
 
selmslie wrote:
Apaflo wrote:
There are not multiple definitions of dB, just one. The relation between amplitude and power is very simple: P = IE. And that is the "difference" shown in the chart at that URL. ...

I guess that't the only definition you understand. Too bad!

Hilarious. There is only one definition. I do understand it.
selmslie wrote:
The table is clear. A power ratio of 10 is represented by 10 dB (the classic definition) but the amplitude ratio of 10 is 20 dB. That's two definitions.[/quote
No it is just two very different values to start with. If either component of "amplitude" goes up by a factor of 10, the other necessarily does too. E = IR is the formula. Plug that into P = IE and wholla, you get a power ratio. It is the same change in power, whether you measure power or instead just measure either current of voltage. Just different ways to calculate exactly the same thing.
[quote=selmslie]Even if you can't understand the difference it will be clear to anyone else who looks at that table and reads the article.
The article wrote:
A change in power by a factor of 10 corresponds to a 10 dB change in level. At the half power point an audio circuit or an antenna exhibits an attenuation of approximately 3 dB. A change in voltage by a factor of 10 results in a change in power by a factor of 100, which corresponds to a 20 dB change in level.

Maybe the rest of us are smarter than you are.
The table is clear. A power ratio of 10 is represe... (show quote)

The above example from the Wiki article demonstrates something you are missing. I've actually measured both power and voltage, as well as current, in high power antenna systems. If you measure voltage and get a 3 dB change, rest assured that measuring power will also show a 3 dB change. They are exactly the same. The ratio of power levels will not be the same as the ratio of voltage levels... which is why two different calculations are needed. And the relationship is in the formula P = IE. The 3 dB is exactly the same 3 dB.

I've designed a fair number of antenna systems. I built and measured some multiple times that many. I'd done trouble shooting measurements of some multiple of that. Not exactly uncommon for someone with 4 decades of experience in the communications industry...

What kind of engineering do you have experience with?

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Jun 3, 2016 13:30:15   #
selmslie Loc: Fernandina Beach, FL, USA
 
Apaflo wrote:
Hilarious. There is only one definition. I do understand it.

Maybe it's time for you to reacquaint yourself Ohm's law: I=V/R

In P=VI, replace I with V/R and you get P=(V^2)/R? Did you ever see that one? It explains why dB that applies to power (P) is different than the one that applied to amplitude (E).
Apaflo wrote:
If you measure voltage and get a 3 dB change, rest assured that measuring power will also show a 3 dB change.

If you double the voltage on a fixed resistance you will also double the current and thereby quadruple the power. That's why a 10 dB change in voltage results in a 20 dB change in power.

This is is grade school physics and Algebra I. Too bad it is over your head.

You might want to look into some courses so you can earn your high school GED. You will never get a complete education looking stuff up on Google.

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Jun 3, 2016 14:10:44   #
BebuLamar
 
Selmie & Apaflo!
You 2 are the more knowledgeable guys here and I have a question about dB.
When you're measuring power ratio by simply measuring the voltage then a 10:1 change in voltage equals 20dB because it would cause a change of 100:1 in power. However when measuring the SNR of an imaging sensor I wouldn't know for sure which one we should use though and also since I don't think any of us actually make the measurement ourselves so how do we know which scale a lab like DxO uses?

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Jun 3, 2016 15:33:58   #
selmslie Loc: Fernandina Beach, FL, USA
 
BebuLamar wrote:
... When you're measuring power ratio by simply measuring the voltage then a 10:1 change in voltage equals 20dB because it would cause a change of 100:1 in power. However when measuring the SNR of an imaging sensor I wouldn't know for sure which one we should use though and also since I don't think any of us actually make the measurement ourselves so how do we know which scale a lab like DxO uses?

That's the whole point. There is ambiguity about which version of dB should be used to measure of S/N. Should we use the power ratio or the amplitude ratio?

dB is a logarithmic representation for a raw arithmetic ratio. It's just easier to represent S/N using a logarithmic scale rather than pure ratios.

For the power ratio a 10:1 S/N ratio equates to 10 dB and a 100:1 S/N ratio is 20 dB . For the amplitude ratio a 10:1 S/N ratio equates to 3.162 dB and a 100:1 S/N ratio is 10 dB. So there really are two ratios, take your choice.

Nevertheless, since DxOMark is actually making the test and representing it however they want to, it does not really make any difference. When they say has a D810 has about a 40 dB S/N ratio at ISO 400, does that mean that the arithmetic S/N ratio is 100:1 or 10000:1? Do we care?

What is important is that they show a threshold around 20 dB below which they feel that the noise will be easily seen and a level above about 38 dB where it is nearly impossible to see noise. Their graph make it easier to compare one camera to another. They are surely better than Popular Photography's version of "extremely low, low ... unacceptable".

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Jun 3, 2016 15:37:12   #
Apaflo Loc: Anchorage, Alaska
 
BebuLamar wrote:
Selmie & Apaflo!
You 2 are the more knowledgeable guys here and I have a question about dB.
When you're measuring power ratio by simply measuring the voltage then a 10:1 change in voltage equals 20dB because it would cause a change of 100:1 in power. However when measuring the SNR of an imaging sensor I wouldn't know for sure which one we should use though and also since I don't think any of us actually make the measurement ourselves so how do we know which scale a lab like DxO uses?

You've got the relationship between voltage and power down pat. It's all the same dB (which in fact is power), but a voltage ratio has to be squared because it isn't just the voltage but also the current that changes.

When looking at an ISO vs exposure graph that measures exposure in dB it is almost always, but not necessarily, the result of a voltage measurement. Sensor designers might well have equipment to measure the power from the sensor and almost certainly actually do measure the power and use those figures. The reason for that is because a power meter will necessarily have a fixed impedance, while the input impedance of the ISO amplifier might (probably) will vary enough to make a bridging voltage measurement unreliable. (In typical real systems nobody measures the voltage across the input to a device; instead at least a 6 dB pad is put between the point of measurement and the device input, to swamp any impedance changes across whatever frequency range is being measured.)

The point is that it makes no difference how it was measured, with a voltmeter or with a power meter, the results will be the same if the proper technique is used. The amplitude (current or voltage) has to be squared, while a power measurement does not. The resulting dB value is exactly the same either way.

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Jun 3, 2016 16:05:01   #
selmslie Loc: Fernandina Beach, FL, USA
 
Apaflo wrote:
... When looking at an ISO vs exposure graph that measures exposure in dB it is almost always, but not necessarily, the result of a voltage measurement. ...

I have never seen an ISO vs exposure graph where exposure is expressed in dB. Exposure is always* expressed in EV stop or steps, which are already logarithmic. Only ratios (like S/N) are expressed in dB.
Apaflo wrote:
... Sensor designers might well have equipment to measure the power from the sensor and almost certainly actually do measure the power and use those figures. ...

You can directly measure the voltage but not the power. To measure the voltage you have to have to apply it to a high resistance (otherwise you drain the voltage).

To measure DC power you need to measure both the voltage and the current or use a fixed resistance and the square of the voltage you have applied to it.

Impedance relates to alternating current. Sound amplifiers use alternating current since sound is produced by alternating current.

Cameras use direct current. Sensors operate more like capacitors. You can partially or fully charge them and then measure their voltage to determine how much charge they received. Then you dump the charge after reading it to reset for the next image.

* I hesitate to say "always" since you will now probably start a Google search for some obscure source that expresses exposure in dB. Happy hunting.

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Jun 3, 2016 16:06:47   #
Apaflo Loc: Anchorage, Alaska
 
selmslie wrote:
For the power ratio a 10:1 S/N ratio equates to 10 dB and a 100:1 S/N ratio is 20 dB . For the amplitude ratio a 10:1 S/N ratio equates to 3.162 dB and a 100:1 S/N ratio is 10 dB. So there really are two ratios, take your choice.

Total nonsense.

If we change the voltage in a circuit from say 1 Volt to 10 Volts, there is a 10:1 change ratio for the voltage. That is a 20 dB change.

If we measure the power change when that voltage was changed, it will be a 100:1 change ratio. That is a 20 dB change.

We can easily calculate that too. If the impedance of the circuit is 1000 ohm that means that, given E= IR then 0.001 * 1000 = 1 Volt. At the original voltage we will have 0.001 Amps of current. And the power in the circuit is P = IE, so we have initially 1 * 0.001 = 0.001 Watt. If we increase the voltage from 1 volt to 10 volts we have a 10:1 amplitude change. The nice little chart showed, or we can calculate using 20log(10), that is a 20 dB change. The same exact dB change happens if we measure power. The power initially was 0.001 Watt, but when the voltage is increased to 10 volts the current changes from 0.001 to 0.010 Amps. The power is now 0.01 * 10 = 0.1 Watts (P = IE). The power ratio is 100:1 (0.1 to 0.001). That calculates to 20 dB. Exactly the same as the 20 dB calculated using only the voltage (or if we calculate using only the current).

Same circuit, same change, same dB.

20 dB is 20 dB is still just and only 20 dB. Just one kind of dB.

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Jun 3, 2016 16:48:10   #
Apaflo Loc: Anchorage, Alaska
 
selmslie wrote:
I have never seen an ISO vs exposure graph where exposure is expressed in dB. Exposure is always* expressed in EV stop or steps, which are already logarithmic. Only ratios (like S/N) are expressed in dB.

Then you clearly have indeed seen just such a chart. For linear data:

1 EV = 1 f/stop = 0.301 density units = 6.02 dB

Any chart labeled with one of those units is a chart that is exactly the same as one labeled with any other of those units with different scaling.
selmslie wrote:
You can directly measure the voltage but not the power. To measure the voltage you have to have to apply it to a high resistance (otherwise you drain the voltage).

Look, lets not get too far off the track with this silliness. You can measure power directly. And the statement that "you have to apply it to a high resistance" is ignorant.

If you want a tutorial on how to accurately make power and voltage measurements please start a separate thread in the Chit-Chat section where that belongs. I'll write you book if you'd like! Not here though.

selmslie wrote:
Cameras use direct current.

More ignorance. The signal being read from a sensor is not DC.

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Jun 3, 2016 17:32:52   #
selmslie Loc: Fernandina Beach, FL, USA
 
Apaflo wrote:
More ignorance. The signal being read from a sensor is not DC.

Really? Then it's A/C? You have to be kidding! It has to be one or the other.

So far you have demonstrated that you don't understand the algebra, only the arithmetic. That puts you somewhere below 7th grade level

You don't know the difference between resistance (D/C) and impedance (primarily an A/C concept). That is possibly elementary school general science but maybe high school physics.

You didn't know the difference between voltage and power until I cleared it up for you.

You don't understand the difference between a power ratio and a voltage ratio so you don't realize that there are two different ways to look at dB.

You don't realize that the sensor stores a charge as a result of collecting photons that is eventually measured as a voltage.

Apparently you are extrapolating any casual knowledge of amplification that you may have collected by putting together sound equipment over to the world of photography. There are some terms that they share but they don't mean the same thing.

You never acknowledge that there is a limit to what you know and understand - you pretend that you know it all and never ask any questions.

You don't know how to apply what you claim to know to the process of photography and thereby cannot produce a single visual demonstration of your claims.

Your qualification for participating in this discussion is non-existent.

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Jun 3, 2016 17:56:25   #
Apaflo Loc: Anchorage, Alaska
 
selmslie wrote:
Really? Then it's A/C? You have to be kidding! It has to be one or the other.

It's AC. No question about it.
selmslie wrote:
You don't know the difference between resistance (D/C) and impedance (primarily an A/C concept).

Actually, the difference is imaginary! Well, that is true in theory. The imaginary component is reactance.

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Jun 3, 2016 18:24:17   #
selmslie Loc: Fernandina Beach, FL, USA
 
Apaflo wrote:
It's AC. No question about it.

You are truly delusional!
Apaflo wrote:
Actually, the difference is imaginary! Well, that is true in theory. The imaginary component is reactance.

Is reactance another term you do not understand? Clearly, you are in over your head.

It's time for you to stop pretending that you are omniscient and infallible. Give us all a break!

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Jun 3, 2016 18:48:32   #
BebuLamar
 
AC has a frequency. What do you think this frequency may be?

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Jun 3, 2016 19:03:33   #
Apaflo Loc: Anchorage, Alaska
 
selmslie wrote:
Apaflo wrote:
It's AC. No question about it.

You are truly delusional!

If the voltage at the output of the sensor were DC... then every single sensor location recorded exactly the same voltage. But of course it could be that every other sensor location recorded maximum and minimum voltages. That would produce a varying voltage from 0 to 1 volt at whatever the rate the sensor data is clocked out at.

It's an AC voltage! Very basic engineering.

selmslie wrote:
Apaflo wrote:
Actually, the difference is imaginary! Well, that is true in theory. The imaginary component is reactance.

Is reactance another term you do not understand? Clearly, you are in over your head.

From your cite (emphasis added):

"... a uniform notion of reactance as the imaginary part of impedance ...

...

X is the reactance, ... It is the imaginary part of the impedance ..."

You should have read the material you cited. This is common knowledge/terminology for any EE.

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Jun 3, 2016 20:03:11   #
selmslie Loc: Fernandina Beach, FL, USA
 
Apaflo wrote:
It's an AC voltage! Very basic engineering. ...This is common knowledge/terminology for any EE. ...

You know absolutely nothing about engineering, EE or otherwise, and apparently very little about electronics either.

You have revealed yourself to be a fraud.

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Jun 3, 2016 20:49:45   #
Apaflo Loc: Anchorage, Alaska
 
selmslie wrote:
You know absolutely nothing about engineering, EE or otherwise, and apparently very little about electronics either.

You have revealed yourself to be a fraud.

Just like I didn't know anything about reactance and imaginary...

The fact is that DC is something that is very hard to find in the wild. The terminals of a battery have a DC voltage, as long as there is no load! Don't think so? I've put a spectrum analyzer on the -48 VDC feed in a communications center, and you can trust that the amount of alternating voltage, even just a few feet away from the battery plant, was amazing to even a bunch of crusty old comm types. The problem was pretty simple, our radio and carrier equipment was in the same room and used the same battery plant as a major telephone switching system. Big problem...

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