decibel

fiberman

If db is a unit of change, typically -3db equals half the power in dbm. Then why when you connect a -3db atenuator to an optical Fibre do you only get a reduction in 3dbm, ie -15dbm with a -3db atenuator added will produce a resultant power of -18dbm???

biko

From Blogs forum...

icdg

it doesn't really belong here either though....before I send it to Engineering anyone have any better thoughts

Roxoid

fiberman wrote: »

If db is a unit of change, typically -3db equals half the power in dbm. Then why when you connect a -3db atenuator to an optical Fibre do you only get a reduction in 3dbm, ie -15dbm with a -3db atenuator added will produce a resultant power of -18dbm???

dBm is the decibel referenced to 1 milliwatt. So still part of the logarithmic decibel scale & obviously -15 - 3 = -18.

Antenna

fiberman wrote: »

If db is a unit of change, typically -3db equals half the power in dbm. Then why when you connect a -3db atenuator to an optical Fibre do you only get a reduction in 3dbm, ie -15dbm with a -3db atenuator added will produce a resultant power of -18dbm???

-3dB change is half the power, regardless of dBm dBW etc
+3dB is double the power

-18dBm is indeed half the power of -15dBm (were you wrongly thinking -30dBm would be half the power of -15dBm !? )

read

http://en.wikipedia.org/wiki/DBm

or

http://en.wikipedia.org/wiki/Decibel

amongst many uses, dB/decibel is of course encountered in TV signal reception and distribution

fiberman

Cheer thanks for that. it does appear to me that there is a direct proportional relationship between db and dbm because the attenuation is always directly proportional, ie 1db per attenuator always results in 1dbm reduction in power.

Kensington

fiberman wrote: »

Cheer thanks for that. it does appear to me that there is a direct proportional relationship between db and dbm because the attenuation is always directly proportional, ie 1db per attenuator always results in 1dbm reduction in power.

Look at this way (converting back to linear):
-15dBm... which is ~31.6uW
-3dB..... subtracting 3dB from -15dBm...

---

-18dBm... which is ~15.8uW (i.e. half the power)

fiberman

so if you are adding db can you simply add + without having to use log scale? because when checking inserion and span loos I have 0.5 + 0.5 + 3.5 - what will the resultant power loss be in db? - Because when I use a db calculator I get 6.5 db, but how can these values be added together 0.5+0.5+3.5=4.5??

Roxoid

it does appear to me that there is a direct proportional relationship between db and dbm because the attenuation is always directly proportional, ie 1db per attenuator always results in 1dbm reduction in power.

The decibel is a separate concept from the actual unit of measurement. The logarithmic scale is intended to make loss & gain calculations easier, by allowing addition & subtraction, rather than multiplication or division e.g. for power, -3 dBm is the same as dividing the actual milliwatts figure by 2.

fiberman

perfect but when calculating loss we have insertion loss and span loss which would be 0.5db + 0.5db + 3.5db = 4.5db. but when I use a db calculator the figure is 6.5db. So applying this to power would be incorrect as a loss form -5dbm to -9,5dbm (in the first instance) and -5dbm to 11.5dbm (in the second instance) - obviously the second instance is wrong but i used a db calculator??

Roxoid

Working in decibels, you are already using a log scale. Just add or subtract the figures, assuming the same reference is used.

fiberman

Great stuff, thanks a lot. So how do you add channels together - with a calculator or manually 1+1 assuming the same reference? say you had ch1 -6db ch2 -8db ch3 -7db
What would be the composite power be in both db and dbm?

Kensington

fiberman wrote: »

perfect but when calculating loss we have insertion loss and span loss which would be 0.5db + 0.5db + 3.5db = 4.5db. but when I use a db calculator the figure is 6.5db. So applying this to power would be incorrect as a loss form -5dbm to -9,5dbm (in the first instance) and -5dbm to 11.5dbm (in the second instance) - obviously the second instance is wrong but i used a db calculator??

Is your loss expressed or being treated in dB or dBm?
dB is a ratio, whereas dBm is an absolute power value.

If it is in dB, then your loss is 0.5dB + 0.5dB + 3.5dB = 4.5dB, a 4.5dB loss.
If it is in dBm, then your power loss is 1.12mW + 1.12mW + 2.24mW = 4.48mW, which is a ~6.5dBm loss.

Roxoid

fiberman wrote: »

So how do you add channels together - with a calculator or manually 1+1 assuming the same reference? say you had ch1 -6db ch2 -8db ch3 -7db
What would be the composite power be in both db and dbm?

dB isn't a measure of anything. It's a ratio, meaningless without a reference unit.

I'm no engineer btw, so don't know what is "composite power".

fiberman

basically:

I measure the signal with a power meter at A side in dbm, say -5dbm
I then add an attenuator -2db
I have a span loss of -7db
I have connector loss of 0.5db + 0.5db
When I go to B side my power is absolutely relative with a result of -15dbm
I get this by adding the loss together and subtracting it form the signal

This works as a calculation, but the reference is the same?

Roxoid

I should have put it that dB can only be used to express gain or loss, rather than it doesn't measure "anything".

Kensington

fiberman wrote: »

basically:

I measure the signal with a power meter at A side in dbm, say -5dbm
I then add an attenuator -2db
I have a span loss of -7db
I have connector loss of 0.5db + 0.5db
When I go to B side my power is absolutely relative with a result of -15dbm
I get this by adding the loss together and subtracting it form the signal

This works as a calculation, but the reference is the same?

Not sure I understand what you mean by "reference is the same"...?

fiberman

My mistake.

is it okay to calculate the loss at detailed below?

I measure the signal with a power meter at A side in dbm, say -5dbm
I then add an attenuator -2db
I have a span loss of -7db
I have connector loss of 0.5db + 0.5db
When I go to B side my power is absolutely relative with a result of -15dbm
I get this by adding the loss together and subtracting it form the signal

Roxoid

fiberman wrote: »

is it okay to calculate the loss at detailed below?

I measure the signal with a power meter at A side in dbm, say -5dbm
I then add an attenuator -2db
I have a span loss of -7db
I have connector loss of 0.5db + 0.5db
When I go to B side my power is absolutely relative with a result of -15dbm
I get this by adding the loss together and subtracting it form the signal

Yes, you just add together the dB losses or gains. Why wouldn't it be "absolutely relative", if everything is decibels referenced to 1 milliwatt?

fiberman

thanks again!