db Question
 

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

db Question
 

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
--

db Question
 

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

db Question
 

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

db Question
 

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

db Question
 

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

db Question
 

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!

db Question
 

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

db Question
 

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

db Question
 

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

db Question
 

Hm. I was taught that the egotistical SOB named it after himself, no honor
involved.

Jim



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

db Question
 

RST Engineering wrote:
Hm. I was taught that the egotistical SOB named it after himself, no honor
involved.

Jim




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




For a (fairly) comprehensive write up on the "decibel" go to
https://ewhdbks.mugu.navy.mil/decibel.htm .

Also:

Alexander Graham Bell
From Wikipedia, the free encyclopedia


Bell and decibel

The bell is a unit of measurement invented by Bell Labs and named after
Bell. The bell was too large for everyday use, so the decibel (dB),
equal to 0.1 B, became more commonly used. The dB is commonly used as a
unit for measuring sound intensity.

Dave WD9BDZ

db Question
 

If Roy's answer works for you, skip mine. I think an answer of mine on this
must be in the archives.

Rule number one:
The Decibel is a measure of the ratio of two power levels ONLY, ALWAYS and
NEVER anything else.
Rule Number two. We like to violate rules and drive newbies crazy.


However, I'll go back to Rule Number One first.


The so called "Voltage dB", "Current dB" and "Power dB' are all figments of
your (well some people's)imagination. I used to have fun with new Engineers
tripping them up on this. "Why are Voltage dBs twice as big as power
dB's?". This is a false-reasoned question.

While, it is a fact that there is only one "kind" of dB, there are, however,
three ways to _calculate_ them depending on what quantities you have at
hand. Because Power is Volts times Amps, there is something special
happening when we do the math. Lets call this V x A for this discussion.

For a simple case of doubling the power - or a power ratio of two to one:
In order to get twice the power, we need to change the Voltage AND Current
enough so that VxA is twice what we started.
1 Volt and 1 Amp gives 1 Watt
For 2 Watts we'll need:
1.414 Volts and 1.414 Amps 1.414 x 1.414 = 2

Looking, we see that the voltage did NOT double. It went up less that
twice.

However, we can calculate the decibel difference from the voltage difference
OR the power difference. Using the "Power dB" formula for dB we get the
correct 3.whatever dB and using the "Voltage dB" formula we get the VERY
SAME amount of 3.whatever dB.

So the dB are not different, only the starting quantities and, therefore the
formulas.


Rule Number 2:
We can, and, we do use the voltage formula ( if it's there, use it... just
like the mountain climber's answer to why does he climb the
mountain...because it's there) in cases where we are not actually doing
power ratio comparisons. Why? Because it is easy to compare large
differences this way..that is, in the log format. Frequently, voltage gain
in OP amp circuits is shown in dB even though there is practically an
INFINITE power gain. This is because the input resistance is very high and
the output resistance is very low and ignoring the resistance and ONLY
looking at the voltage gives erroneous dB numbers. HOWEVER, those in the
field know and understand this and all is well in the universe, so to speak.
By measuring the voltage gain, and using the dB formula, it is easier to
show 50 dB "Voltage" gain than showing a gain of 100,000. Silly Engineers!

73, Steve, K9DCI



"dansawyeror" wrote in message
...
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

db Question
 

"Richard Harrison" wrote in message
...
In the telephone industry, the most common reference level is one
milliwatt.

Ditto the RF industry.

Signal loss is common.

Pretty ubiquitous... something about the three laws of thermodynamics... ;-)

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,

The cable TV industry likes to use dBuV, for whatever odd reason.