dBs

[Cursitor Doom]

I'm feeling cognitively-declined today, probably as a consequence of my
vast age and general ignorance of matters mathematical and everything else
in fact, with the sole exception of "fatuous conspiracy theories." Can
some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator in
line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.

CD.

[Jeroen Belleman]

Post by Cursitor Doom
I'm feeling cognitively-declined today, probably as a consequence of my
vast age and general ignorance of matters mathematical and everything else
in fact, with the sole exception of "fatuous conspiracy theories." Can
some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator in
line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
CD.

That would be -13 + 20 = +7dBm, provided that impedances
are matched everywhere.

Jeroen Belleman

[Cursitor Doom]

Post by Cursitor Doom
I'm feeling cognitively-declined today, probably as a consequence of my
vast age and general ignorance of matters mathematical and everything
else in fact, with the sole exception of "fatuous conspiracy theories."
Can some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator
in line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
CD.

That would be -13 + 20 = +7dBm, provided that impedances are matched
everywhere.

I was under the impression that one couldn't simply just add dBs to dBms?

[Grant Taylor]

Post by Cursitor Doom
I was under the impression that one couldn't simply just add dBs to dBms?

I've had very similar question in the laser power on fiber and was given
the following understanding:

dB is simply a ratio while
dBm is a ratio referenced against a known thing.

dBm is anchored while dB is floating.

It's sort of like saying someone raised the picture four feet (dB) after
the picture started two feet off the ground (dBm).

It's been years since I had it explained to me and I may be wrong and /
or the context it was explained to me is different than the context you
are asking in.

N.B. the measured aspect of dBm is somewhat of a nomenclature but I
believe the m references 1 mW of power.

Then there's the logarithmic aspect and decimal aspect making dB / dBm
somewhat more unintuitive to me.

[Phil Hobbs]

Post by Cursitor Doom

Post by Cursitor Doom
I'm feeling cognitively-declined today, probably as a consequence of my
vast age and general ignorance of matters mathematical and everything
else in fact, with the sole exception of "fatuous conspiracy theories."
Can some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator
in line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
CD.

That would be -13 + 20 = +7dBm, provided that impedances are matched
everywhere.

I was under the impression that one couldn't simply just add dBs to dBms?

To use logarithms sensibly, you have to start with a dimensionless number.
Vanilla decibels express power ratios. Watts divided by watts is
dimensionless.

Decorated decibels express absolute power values under various measurement
conditions. This works because the denominator is a constant power fixed by
convention. There are many kinds: dBm, dBW, dBa, dBc, dBm0, dBrnC0
(“dibrinco”), and so forth. (*)

When you subtract two decorated decibel values of the same type, you’re
implicitly dividing the underlying ratios.

Since the denominators are equal, they cancel, leaving the decibel ratio of
the numerators.

This of course is a dimensionless power ratio, expressed in vanilla
decibels.

Doing this backwards, if you want to apply a gain to a decorated-decibel
value, you add the gain in vanilla decibels.

If you add two decorated decibel values, the result is nothing useful,
because you get two copies of the denominator instead of one.



Cheers

Phil Hobbs

[Phil Hobbs]

Post by Phil Hobbs

Post by Cursitor Doom

Post by Cursitor Doom
I'm feeling cognitively-declined today, probably as a consequence of my
vast age and general ignorance of matters mathematical and everything
else in fact, with the sole exception of "fatuous conspiracy theories."
Can some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator
in line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
CD.

That would be -13 + 20 = +7dBm, provided that impedances are matched
everywhere.

I was under the impression that one couldn't simply just add dBs to dBms?

To use logarithms sensibly, you have to start with a dimensionless number.
Vanilla decibels express power ratios. Watts divided by watts is
dimensionless.
Decorated decibels express absolute power values under various measurement
conditions. This works because the denominator is a constant power fixed by
convention. There are many kinds: dBm, dBW, dBa, dBc, dBm0, dBrnC0
(“dibrinco”), and so forth. (*)
When you subtract two decorated decibel values of the same type, you’re
implicitly dividing the underlying ratios.
Since the denominators are equal, they cancel, leaving the decibel ratio of
the numerators.
This of course is a dimensionless power ratio, expressed in vanilla
decibels.
Doing this backwards, if you want to apply a gain to a decorated-decibel
value, you add the gain in vanilla decibels.
If you add two decorated decibel values, the result is nothing useful,
because you get two copies of the denominator instead of one.

Forgot the promised footnote.

(*) There are more complicated sorts of decorated decibels, such as the
other sort of dBc, i.e. modulation sideband power in dB with respect to the
carrier, and all sorts of other stuff that just muddies the water at this
level, e.g. dBV, dBm/Hz, and such truck.

[Jeroen Belleman]

Post by Cursitor Doom

Post by Cursitor Doom
I'm feeling cognitively-declined today, probably as a consequence of my
vast age and general ignorance of matters mathematical and everything
else in fact, with the sole exception of "fatuous conspiracy theories."
Can some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator
in line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
CD.

That would be -13 + 20 = +7dBm, provided that impedances are matched
everywhere.

I was under the impression that one couldn't simply just add dBs to dBms?

You can. That's what decibels were invented for.

Let's spell it out then. You know 0 dBm is 1 mW. So -13 dBm is
10^(-13/10) times 1 mW, or 50 uW.

A 20dB attenuator divides power by a factor of 10^(20/10), that
is, a factor of 100. So before the attenuator, you had 5 mW.

5mW is 10*log(5) is +7 dBm.

Jeroen Belleman

[Cursitor Doom]

Post by Jeroen Belleman

Post by Cursitor Doom

Post by Cursitor Doom
I'm feeling cognitively-declined today, probably as a consequence of
my vast age and general ignorance of matters mathematical and
everything else in fact, with the sole exception of "fatuous
conspiracy theories."
Can some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator
in line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
CD.

That would be -13 + 20 = +7dBm, provided that impedances are matched
everywhere.

I was under the impression that one couldn't simply just add dBs to dBms?

You can. That's what decibels were invented for.
Let's spell it out then. You know 0 dBm is 1 mW. So -13 dBm is
10^(-13/10) times 1 mW, or 50 uW.
A 20dB attenuator divides power by a factor of 10^(20/10), that is, a
factor of 100. So before the attenuator, you had 5 mW.
5mW is 10*log(5) is +7 dBm.
Jeroen Belleman

Oh I know you're figures are correct, Jeroen. But to check them I had to
use look-up tables off the net:

-13dBm = 0.05mW
20dB = 100X
0.05X100 = 5mW
5mW = =7dBm

Sometimes you can just straight add-up dBs and other times you can't and I
can never remember when it's appropriate and when it's not. To be safe, I
revert to the method I showed above. It's longer, but at least I know I
can rely on the result. Whoever invent dBs "to make things simpler" needs
to have their grave desecrated and their name effaced from history IMO.

[Jeroen Belleman]

Post by Cursitor Doom

Post by Jeroen Belleman

Post by Cursitor Doom

Post by Cursitor Doom
I'm feeling cognitively-declined today, probably as a consequence of
my vast age and general ignorance of matters mathematical and
everything else in fact, with the sole exception of "fatuous
conspiracy theories."
Can some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator
in line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
CD.

That would be -13 + 20 = +7dBm, provided that impedances are matched
everywhere.

I was under the impression that one couldn't simply just add dBs to dBms?

You can. That's what decibels were invented for.
Let's spell it out then. You know 0 dBm is 1 mW. So -13 dBm is
10^(-13/10) times 1 mW, or 50 uW.
A 20dB attenuator divides power by a factor of 10^(20/10), that is, a
factor of 100. So before the attenuator, you had 5 mW.
5mW is 10*log(5) is +7 dBm.
Jeroen Belleman

Oh I know you're figures are correct, Jeroen. But to check them I had to
-13dBm = 0.05mW
20dB = 100X
0.05X100 = 5mW
5mW = =7dBm
Sometimes you can just straight add-up dBs and other times you can't and I
can never remember when it's appropriate and when it's not. To be safe, I
revert to the method I showed above. It's longer, but at least I know I
can rely on the result. Whoever invent dBs "to make things simpler" needs
to have their grave desecrated and their name effaced from history IMO.

It's not that bad! Adding dB values works fine in the context of
chains of gain and attenuation. It's easier to add decibels than
to multiply gain values. One gets used to it. I can usually do
it in my head. Anyone working in RF or in control systems gets
proficient at it very quickly.

I agree that there are situations where using decibels can get
confusing. For example, in light detectors, optical power gets
converted into current, so a 20dB change in optical power would
result in only 10dB change in the electrical signal power. But
let's not go there just yet.

Jeroen Belleman

[boB]

On Sun, 26 May 2024 23:42:15 +0200, Jeroen Belleman

Post by Jeroen Belleman

Post by Cursitor Doom

Post by Jeroen Belleman

Post by Cursitor Doom

Post by Cursitor Doom
I'm feeling cognitively-declined today, probably as a consequence of
my vast age and general ignorance of matters mathematical and
everything else in fact, with the sole exception of "fatuous
conspiracy theories."
Can some kind soul assist?
If my RF power meter is reading -13dbm when there's a 20dB attenuator
in line, what is the true power level, please?
I've got an exhaustive App Note from Rhode & Schwartz which claims to
cover everything about decibels, but, er, doesn't.
CD.

That would be -13 + 20 = +7dBm, provided that impedances are matched
everywhere.

I was under the impression that one couldn't simply just add dBs to dBms?

You can. That's what decibels were invented for.
Let's spell it out then. You know 0 dBm is 1 mW. So -13 dBm is
10^(-13/10) times 1 mW, or 50 uW.
A 20dB attenuator divides power by a factor of 10^(20/10), that is, a
factor of 100. So before the attenuator, you had 5 mW.
5mW is 10*log(5) is +7 dBm.
Jeroen Belleman

Oh I know you're figures are correct, Jeroen. But to check them I had to
-13dBm = 0.05mW
20dB = 100X
0.05X100 = 5mW
5mW = =7dBm
Sometimes you can just straight add-up dBs and other times you can't and I
can never remember when it's appropriate and when it's not. To be safe, I
revert to the method I showed above. It's longer, but at least I know I
can rely on the result. Whoever invent dBs "to make things simpler" needs
to have their grave desecrated and their name effaced from history IMO.

It was Bell Labs a long time ago I believe that came up with that
great system. Don't shoot them ! Well, they're no longer around,
unfortunately.

Yep, Jeroen is right.
boB

Post by Jeroen Belleman
It's not that bad! Adding dB values works fine in the context of
chains of gain and attenuation. It's easier to add decibels than
to multiply gain values. One gets used to it. I can usually do
it in my head. Anyone working in RF or in control systems gets
proficient at it very quickly.
I agree that there are situations where using decibels can get
confusing. For example, in light detectors, optical power gets
converted into current, so a 20dB change in optical power would
result in only 10dB change in the electrical signal power. But
let's not go there just yet.
Jeroen Belleman