Hello:

When a table gives the attenuation (at some freq.) for a length of a
particular coax in db,
are they referring to db in voltage or db in power ?

Is is correct to ask a question as simply as the following:

If the attenuation is given as, e.g., 2 db, what Percentage therefore of a
received signal is "lost"
going thru the coax length ?

Thanks,
Bob

On Fri, 10 Feb 2006 18:46:13 -0500, "Robert11"
wrote:

Hello:

When a table gives the attenuation (at some freq.) for a length of a
particular coax in db,
are they referring to db in voltage or db in power ?

Is is correct to ask a question as simply as the following:

If the attenuation is given as, e.g., 2 db, what Percentage therefore of a
received signal is "lost"
going thru the coax length ?

Decibels are a means of expressing a power ratio when used properly.
Of course, the power ratio is implied by a voltage or current ratio
when the impedance is the same in the two cases being compared. For
that reason, you can say that the power ratio in db is 10*log(P1/P2)
or 20*log(V1/V2) provided Z remains constant.

Nevertheless, voltage or current ratios are often compared using dB
where the Z is not the same, and stricly speaking, the power ratio
rquires an adjustment for the changed impedance.

If the loss is 2dB, the the ratio of power out to power in is
10^(-2/10) or 63%. If the impedance is the same, the ratio of the
voltages will be the square root of 63% or 79%.

Owen
--

Robert11 wrote:
Hello:

When a table gives the attenuation (at some freq.) for a length of a
particular coax in db,
are they referring to db in voltage or db in power ?

Both. When voltage and power ratios are measured across the same single
impedance, the ratio of voltages in dB is the same as the ratio of power
in dB. That is, dB in voltage equals dB in power if only one impedance
is involved. That's why there are different formulas for dB voltage and
power ratios, to make this happen. And it's the case if a coax cable is
terminated in its characteristic impedance so the input and output
impedances are the same. It's under this condition that cable loss is
specified, so the dB loss represents the loss of both voltage and current.

But when you compare the power at the output of a coax cable with the
power at the input and the cable isn't terminated in its characteristic
impedance, output and input impedances can be very different. The power
loss will be somewhat higher than the specification (probably not much,
when the matched loss is only 2 dB), but you can have much less or more
voltage at the output than the input. So the power and voltage loss in
dB can be very different, and you can actually have voltage gain --
although it can be argued that defining dB for a ratio of voltages
across two different impedances is a bit shaky and perhaps not too
meaningful.

Is is correct to ask a question as simply as the following:

If the attenuation is given as, e.g., 2 db, what Percentage therefore of a
received signal is "lost"
going thru the coax length ?

100 * (1 - 10^(-dB/10)) ~ 37% is the fraction of power lost.

100 * (1 - 10^(-dB/20)) ~ 21% is the fraction of voltage lost.

These assume that the coax is terminated with its characteristic impedance.

And you don't need to put "lost" in quotation marks. It is truly lost as
a signal, having been turned into heat.

Roy Lewallen, W7EL

There are three terms, idb, pdb, and vdb. They are not the same. idb and pdb
have the same value: 10*log10 (i/i0), while vdb is 20*log10(v/v0).

Dan


Roy Lewallen wrote:
Robert11 wrote:

Hello:

When a table gives the attenuation (at some freq.) for a length of a
particular coax in db,
are they referring to db in voltage or db in power ?


Both. When voltage and power ratios are measured across the same single
impedance, the ratio of voltages in dB is the same as the ratio of power
in dB. That is, dB in voltage equals dB in power if only one impedance
is involved. That's why there are different formulas for dB voltage and
power ratios, to make this happen. And it's the case if a coax cable is
terminated in its characteristic impedance so the input and output
impedances are the same. It's under this condition that cable loss is
specified, so the dB loss represents the loss of both voltage and current.

But when you compare the power at the output of a coax cable with the
power at the input and the cable isn't terminated in its characteristic
impedance, output and input impedances can be very different. The power
loss will be somewhat higher than the specification (probably not much,
when the matched loss is only 2 dB), but you can have much less or more
voltage at the output than the input. So the power and voltage loss in
dB can be very different, and you can actually have voltage gain --
although it can be argued that defining dB for a ratio of voltages
across two different impedances is a bit shaky and perhaps not too
meaningful.


Is is correct to ask a question as simply as the following:

If the attenuation is given as, e.g., 2 db, what Percentage therefore
of a received signal is "lost"
going thru the coax length ?


100 * (1 - 10^(-dB/10)) ~ 37% is the fraction of power lost.

100 * (1 - 10^(-dB/20)) ~ 21% is the fraction of voltage lost.

These assume that the coax is terminated with its characteristic impedance.

And you don't need to put "lost" in quotation marks. It is truly lost as
a signal, having been turned into heat.

Roy Lewallen, W7EL

Although the other answers given are correct as best as I can remember from
school (long time ago), for most practical cb or ham radio purposes just
remember that every 3db at a given impedance will 1/2 the power, from radio
to antenna, or antenna to radio.

RoD


"Robert11" wrote in message
...
Hello:

When a table gives the attenuation (at some freq.) for a length of a
particular coax in db,
are they referring to db in voltage or db in power ?

Is is correct to ask a question as simply as the following:

If the attenuation is given as, e.g., 2 db, what Percentage therefore of a
received signal is "lost"
going thru the coax length ?

Thanks,
Bob

dansawyeror wrote:
There are three terms, idb, pdb, and vdb. They are not the same. idb and
pdb have the same value: 10*log10 (i/i0), while vdb is 20*log10(v/v0).


Not quite sure what all these are. But to express a voltage or current
ratio in dB, you use:

dB = 20 * log(V2/V1)
dB = 20 * log(I2/I1)

and to express a power ratio in dB, you use:

dB = 10 * log(P2/P1)

If the voltages or currents are measured at the same impedance as the
power, the numerical dB values will be the same. For example, let's
suppose we have a 100 ohm resistor with one amp flowing through it. Then
there's 100 volts across it, and it's dissipating 100 watts.

Double the current to 2 amps. The dB increase in current is

20 * log(2/1) = 6.02 dB

The voltage has also doubled to 200 volts, so the dB increase in voltage is

20 * log(200/100) = 6.02 dB

The power dissipation is now 400 watts, so the dB increase in power is

10 * log(400/100) = 6.02 dB

Cool, huh?

Roy Lewallen, W7EL

In article ,
Roy Lewallen wrote:

The power dissipation is now 400 watts, so the dB increase in power is

10 * log(400/100) = 6.02 dB

Cool, huh?

Not unless it's a pretty big resistor mounted on a very large
heatsink!

--
Dave Platt AE6EO
Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

dansawyeror wrote:
There are three terms, idb, pdb, and vdb. They are not the same. idb
and pdb have the same value: 10*log10 (i/i0), while vdb is
20*log10(v/v0).


"A decibel is a decibel is a decibel". There is only one kind of
decibel, and by definition it is based on a power ratio.

Decibels are always used for comparing power levels. If you're comparing
current or voltage levels, you first have to convert them into a power
ratio, and that is the reason why you have to use a different formula.
Your starting-point was different, but the definition of the decibel
remains the same.


There is a similar-looking term, sometimes written as either "dBV",
"dBmV" or "dBuV", but that has a different meaning. Anything tagged on
after the "dB" indicates a standard reference level. For example, "dBV"
means "a power ratio in decibels, relative to the power level of a 1V
rms signal measured at the same point in the circuit."


--
73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Ian White GM3SEK wrote:
dansawyeror wrote:

There are three terms, idb, pdb, and vdb. They are not the same. idb
and pdb have the same value: 10*log10 (i/i0), while vdb is
20*log10(v/v0).


"A decibel is a decibel is a decibel". There is only one kind of
decibel, and by definition it is based on a power ratio.

Decibels are always used for comparing power levels. If you're comparing
current or voltage levels, you first have to convert them into a power
ratio, and that is the reason why you have to use a different formula.
Your starting-point was different, but the definition of the decibel
remains the same.


There is a similar-looking term, sometimes written as either "dBV",
"dBmV" or "dBuV", but that has a different meaning. Anything tagged on
after the "dB" indicates a standard reference level. For example, "dBV"
means "a power ratio in decibels, relative to the power level of a 1V
rms signal measured at the same point in the circuit."




Actually what I was taught was the standard unit is the BEL, named in
honor of Alexander Graham Bell. This is a large value hence the DECIBEL
or 1/10 of a BEL is generally used.

Dave WD9BDZ

Ian, GM3SEK wrote:
"Decibels are always used for comparing power levels."

True.

The decibel or dB is a power ratio. The dB alone says nothing about the
quantity of power, only the ratio between powers. Thus, the dB is useful
to express the gain of a linear amplifier or the loss of a resistance
pad.

The dB can be used to express a power level by reference to some known
quantity of power. In this case, the ratio is said to be so many dB
above or below the reference level.

In the telephone industry, the most common reference level is one
milliwatt. Signal loss is common. Cables fave loss per foot or mile.
Attenuation is a division process. Amplification is a multiplication
process. It is convenient to express the amount of power at a particular
point in a system as being so many dB above or below a reference power
of one milliwatt because adding and subtracting decibels gives the same
answers as multiplying and dividing powers. Decibels above or below one
milliwatt is usually abbreviated as + or - dBm.

DBa and dBrn indicate the interfering effect of noise in a
communications channel. DBx is used to indicate crosstalk coupling into
telephone circuits.

VU is a volume unit. Reference is to one milliwatt, same as the dBm, but
the ballistics of the meter are specified so that a standard response is
made to varying signal level.

Other dB units are easily contrived. There are dBw (dB referenced to one
watt), dBk (dB referenced to one kilowatt), dBRAP (dB above reference
acoustical power which is defined at 10 ro the minus 16 watts), etc,
etc, etc.

Best regards, Richard Harrison, KB5WZI