Hi everyone, I would like to get some clarification please.
I have seen dynamic range described using various units, as a number of EV, stops, dB, bits.
Firstly, are all those units legitimate units of dynamic range?
Also, are there any conversion factors?
- 12 EV of dynamic range
- 12 stops of dynamic range
- 12 dB of dynamic range
- 12 bits of dynamic range
- 12=12=12=12?
Finally, does the appropriate unit vary with what is being measured?
- the live scene in front of me has 12 (unit) of dynamic range
- the sensor has 12 (unit) of dynamic range
- the image has 12 (unit) of dynamic range
- the PC monitor has 12 (unit) of dynamic range
As an aside, some PC monitors or video projectors seem to use a contrast ratio (like 5000:1) when they are describing what I would call the dynamic range. Can one convert from contrast ratio to EV / stops / dB / bits?
Thanks for any responses.
I see what you're getting at. Dynamic range has no dimensions. As others have said here, it is the ratio of two quantities. In photography, those quantities have the same dimensions (like volts), so the dimensions cancel when the ratio is created, and you are left with a dimensionless number. That number can be expressed in many ways. Say it's 5000, which is the example you used in your original post.
log2(5000) = 12.29. Since each stop is a doubling of light, that's 12.29 stops, or that many EV. Since one bit position is also a doubling, you could call that 12.29 bits.
If we're talking voltage, 20*log10(5000) = 74 decibels (dB). See how the dB number is different? That's because dB is based on log base 10, not log base 2.
Jim
This might be a silly question, but what is the motivation for the 20x multiplier in the dB scale?
d stands for deci-, factor of 10x.
2x is because log(V^2) = 2*log(V)
Iliah is correct. Let me unpack that for you, Jonas.
One Bel is one factor of ten in power. The relationship between power, P, and current, I, and voltage, V, is as follows:
P = E * I
Now, let's assume that the voltage is measured across a resistance of R. Ohm's law states that:
E = I * R
Let's solve that for I:
I = E / R
Now let's plus that into the power equation:
P = E * E / I, or P = E^2 / I
That's where the V^2 that Iliah is talking about comes from. Some engineers, like me, use E to indicate a voltage. Some, like Iliah, use V. Po-tay-to, po-tah-to.
So 1 Bel is a factor of ten in power (log10(10) = 1), and a factor of 3.162 in voltage (2*log10(3.162) = 1.
A decibel is one tenth of a Bel, so, if we have a power ratio of x, it's
10 * log10(x) decibels
If we have a voltage ratio of y, it's
20 * log10

decibels
If we raise the voltage to 3.162 times what it was before, the power goes up by:
20*log10(3.162) = 10 dB
If we raise the power to 10 times what it was before, the relationship is:
10*log10(10) = 10 dB
Jim