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dBm dBm/Hz what is the relationship

I understand that when we measure noise floor, we use dBm/Hz, that is,
the noise power per 1Hz bandwidth. The ideal noise floor is -174
dBm/Hz.

However, when I measure a signal power at spectrum analyzer, I would
get, for example, -30dBm. What is the bandwidth of my measurement? Is
the bandwidth that we use in measuring the signal is less than 1Hz?

Thanks,
feilip
 
T

Tim Wescott

Jan 1, 1970
0
I understand that when we measure noise floor, we use dBm/Hz, that is,
the noise power per 1Hz bandwidth. The ideal noise floor is -174
dBm/Hz.

However, when I measure a signal power at spectrum analyzer, I would
get, for example, -30dBm. What is the bandwidth of my measurement? Is
the bandwidth that we use in measuring the signal is less than 1Hz?

Thanks,
feilip
The bandwidth of your measurement is the "noise bandwidth" of the filter
you're using in your spectrum analyzer. This should be specified in the
analyzer documentation.

-174dBm is about 2pW (yes, pico watts). -174dBm means that you see 2pW
per Hz of bandwidth. For voice communications with a 2500Hz wide filter
you'd expect that -174dBm/Hz would give you about 5nW, or -106dBm.

Does that help?

P.S. There is no 'ideal' noise floor. The ideal noise floor looking
into a resistor at room temperature (300K or so) is, indeed -174dBm/Hz.
But if you point a microwave antenna at the sky you'll find that most
points have a noise temperature much lower than 300K -- and if you
happen to point it at the sun you'll find a higher noise temperature.

Microwave guys often talk about their circuits in terms of "noise
temperature" rather than "noise figure" for just this reason.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google? See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html
 
J

Joel Kolstad

Jan 1, 1970
0
However, when I measure a signal power at spectrum analyzer, I would
get, for example, -30dBm. What is the bandwidth of my measurement?

The spectrum analyzer's manual will tell you this; usually it's some constant
multiplied by the IF bandwidth set you've set the analyzer to use (the
constant is a function of the shape of the IF filter).
Is
the bandwidth that we use in measuring the signal is less than 1Hz?

Generally, no! In many cases use spectrum analyzer power measurements when
they've set their IF bandwidth width width enough to capacity the vast
majority of the energy in the signal they're observing. That is, they're
attempting to find the *total power* in the signal (or the channel, for
channelized systems). But of course there are plenty of other uses too...
 
G

Genome

Jan 1, 1970
0
I understand that when we measure noise floor, we use dBm/Hz, that is,
the noise power per 1Hz bandwidth. The ideal noise floor is -174
dBm/Hz.

However, when I measure a signal power at spectrum analyzer, I would
get, for example, -30dBm. What is the bandwidth of my measurement? Is
the bandwidth that we use in measuring the signal is less than 1Hz?

Thanks,
feilip

dB measurements are made relative to something else.

Wot you do is divide what you have measured by what is the reference and
then take the log of the answer and you get your answer in dB.

Then you have to say what your answer was reffered to and make sure that any
other bits you used in the referenced are included in the thing you measured
and what it was referenced to.

So.... if your reference is the weight of an Elephant and you were looking
to measure a pig then you would take the weight of your pig and divide it by
the weight of your elephant and take the log of it and express your answer
as dB/Welph.

Now, things get complicated when you are looking at how much your pig weighs
and the frequency it farts at when compared to your reference Elephant and
the frequency it farts at so then you end up with units like dB/WElphHz.

You will only realise the worth of these units when you have to work with
fly farts per weight of a fly referenced to elephant farts per weight of an
elephant or maybe not but most times you can write 2 dBWF/WElphHz rather
than 0.000000000000001299876 or
399691589691256120.

Most times when you are comparing stuff in their original units then you
have to multiply the dB answer by 10. However, if the units are different
then you have to use 20 instead because E=mC^2 and the squared term is a
multiply by two in log terms.

See...

DNA
 
A

Andrew Holme

Jan 1, 1970
0
I understand that when we measure noise floor, we use dBm/Hz, that is,
the noise power per 1Hz bandwidth. The ideal noise floor is -174
dBm/Hz.

However, when I measure a signal power at spectrum analyzer, I would
get, for example, -30dBm. What is the bandwidth of my measurement? Is
the bandwidth that we use in measuring the signal is less than 1Hz?

When measuring a discrete tone such as an unmodulated carrier, you don't
need to state the bandwidth. If you change resolution bandwidth, the noise
floor will change, but the tone is always -30 dBm.

Noise power is proportional to the bandwidth you measure it in, so you must
either state both, or quote dBm/Hz.
 
A

Andrew Holme

Jan 1, 1970
0
Andrew Holme said:
Clarification:

When measuring a discrete tone such as an unmodulated carrier, you don't
need to state the bandwidth. If you change resolution bandwidth, the
noise floor will change, but the tone is always -30 dBm.
SO IT IS ENOUGH JUST TO SAY THE SIGNAL LEVEL WAS -30 dBm.

Noise power is proportional to the bandwidth you measure it in, so you
must either state both, or quote dBm/Hz.
WHEN YOU ARE TALKING ABOUT NOISE POWER.
 
I understand that when we measure noise floor, we use dBm/Hz, that is,
the noise power per 1Hz bandwidth. The ideal noise floor is -174
dBm/Hz.

However, when I measure a signal power at spectrum analyzer, I would
get, for example, -30dBm. What is the bandwidth of my measurement? Is
the bandwidth that we use in measuring the signal is less than 1Hz?

Thanks,
feilip

It's the ratio of the vbw/rbw...

You'll find that changing the resolution bandwidth(RBW) and video
bandwidth (VBW)setting on the spectrum analyzer with give you slightly
different readings.


RBW filter: The resolution bandwidth filter of a
spectrum analyzer. This is the filter whose selectivity
determines the analyzer's ability to resolve
(indicate separately) closely spaced signals.


VBW Filter: The Video Bandwidth filter, a low-pass
filter that smoothes the output of the detected IF
signal, or the log of that detected signal.
 
R

rick H

Jan 1, 1970
0
Tim Wescott said:
-174dBm is about 2pW (yes, pico watts). -174dBm means that you see 2pW
per Hz of bandwidth.


Hello Tim. -174dBm is not 2pW. It's 10^(-174/10)/1000 = 4e-21 W. This
is the product of Boltzmann's constant and the ambient temperature
in Kelvin (1.38e-23 * 300 ~ 4e-21 W).

For voice communications with a 2500Hz wide filter
you'd expect that -174dBm/Hz would give you about 5nW, or -106dBm.


-174dBm/Hz in a 2500Hz BW is not 5nW, and 5nW is not -106 dBm (-it's
-53 dBm).

-174 dBm in 2500 Hz = -174 + 10*log10(2500) = -140dBm.

or, put another way:
10*log10 (k * T * BW / 1mW) = 10*log10(1.38e-23 * 300 * 2500 / 1e-3)
= -140 dBm.


Cheers,
 
B

Bo

Jan 1, 1970
0
rick H said:
Hello Tim. -174dBm is not 2pW. It's 10^(-174/10)/1000 = 4e-21 W. This
is the product of Boltzmann's constant and the ambient temperature
in Kelvin (1.38e-23 * 300 ~ 4e-21 W).




-174dBm/Hz in a 2500Hz BW is not 5nW, and 5nW is not -106 dBm (-it's
-53 dBm).

-174 dBm in 2500 Hz = -174 + 10*log10(2500) = -140dBm.

or, put another way:
10*log10 (k * T * BW / 1mW) = 10*log10(1.38e-23 * 300 * 2500 / 1e-3)
= -140 dBm.


Cheers,

I've always been confused on how one knows whether a dBx unit is 10
log(variable/x) or 20 log(variable/x). Any help? I do realize the 20 has
reference to V^2/R--but sometimes its not clear if the (dB) is V/V or P/P...

Bo
 
J

Jeroen Belleman

Jan 1, 1970
0
Bo said:
I've always been confused on how one knows whether a dBx unit is 10
log(variable/x) or 20 log(variable/x). Any help? I do realize the 20 has
reference to V^2/R--but sometimes its not clear if the (dB) is V/V or P/P...

Well, dB is a ratio of signal *power*, so if it's 10 log P/P, it's
20 log V/V because power is proportional to V^2.

Jeroen Belleman
 
T

Tim Wescott

Jan 1, 1970
0
Bo said:
I've always been confused on how one knows whether a dBx unit is 10
log(variable/x) or 20 log(variable/x). Any help? I do realize the 20 has
reference to V^2/R--but sometimes its not clear if the (dB) is V/V or P/P...

Bo
A dB is _always_ a power ratio. You can bend the definition dB to use
it with voltages, or currents, or anonymous numbers in a DSP system, but
to do so in a sensible way you have to assume that ultimately you're
driving some linear dissipative thing, which leads to the V^2 bit.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google? See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html
 
T

Tom Bruhns

Jan 1, 1970
0
need to state the bandwidth. If you change resolution bandwidth, the noise
floor will change, but the tone is always -30 dBm.

Noise power is proportional to the bandwidth you measure it in, so you must
either state both, or quote dBm/Hz.

Of course, the power per unit bandwidth is useful for things other
than noise. It's useful for any signal with significant bandwidth. A
few examples: phase noise is almost always spec'd in terms of dBc/Hz
at a particular offset: dB relative to the "carrier" or main signal,
per Hz bandwidth, at a frequency offset from the carrier. Audio, or
the sidebands from modulation, or video may all be usefully
characterized in terms of power per unit bandwidth, or dBm/Hz. In
doing measurements of RF noise emitted from a piece of equipment, for
qualification testing, it makes a lot of sense to talk in terms of
both single-tone power (dBm) and power per unit bandwidth (dBm/Hz, or
dBm/MHz). Some standards specify how you are to measure the noise,
and how you are to express it. (Some poor ones are less specific...)

Cheers,
Tom
 
T

Tom Bruhns

Jan 1, 1970
0
is the product of Boltzmann's constant and the ambient temperature
in Kelvin (1.38e-23 * 300 ~ 4e-21 W).

-53 dBm).

-174 dBm in 2500 Hz = -174 + 10*log10(2500) = -140dBm.

or, put another way:
10*log10 (k * T * BW / 1mW) = 10*log10(1.38e-23 * 300 * 2500 / 1e-3)
= -140 dBm.

Cheers,

Thanks for posting that correction, Rick. You did it more gracefully
than I probably would have. ;-)

I sometimes like to ponder that 100 watts (+50dBm) radiated from a
simple antenna on, say, 20MHz, can be heard half-way around the world,
about 12000 miles or 20000km away, when "skip" conditions are right.
The received signal level may well be -130dBm or more. That's a path
loss of 180dB or so. By comparison, at 21MHz, coaxial line with
copper conductors, 6" diameter (one of the largest commonly available
commercial sizes, and quite expensive), impedance optimized for lowest
loss, will have about 1dB/mile loss, or 12000dB loss (give or take a
little) in that length.

Cheers,
Tom
 
R

Richard Fry

Jan 1, 1970
0
I've always been confused on how one knows whether a dBx unit
is 10 log(variable/x) or 20 log(variable/x). Any help? I do realize
the 20 has reference to V^2/R--but sometimes its not clear if the (dB) is
V/V or P/P...
____________

Some of it probably is related to the following.

Through convention most readers seeing the term "dBm" know that it means
decibels with respect to one milliwatt. But the real unit of measure (the
watt) is not stated. Better to write the complete forms of these terms: in
this case, dBmW.

An example in the voltage realm is "dBu." Almost always the meaning is
dBuV, and it would eliminate assumption to write it that way. Some even
use dBu to express radiated field strength, for which the correct form needs
to include the distance across which that voltage is developed, such as in
dBuV/m.

And of course, there is the problem of upper case vs lower case following
the rules for SI units. Even some technically sharp people write things
like MW, Mw or mw when it is clear from the context that they mean mW.

Pardon the rant.
 
T

Tom Bruhns

Jan 1, 1970
0
____________

Some of it probably is related to the following.

Through convention most readers seeing the term "dBm" know that it means
decibels with respect to one milliwatt. But the real unit of measure (the
watt) is not stated. Better to write the complete forms of these terms: in
this case, dBmW.

An example in the voltage realm is "dBu." Almost always the meaning is
dBuV, and it would eliminate assumption to write it that way. Some even
use dBu to express radiated field strength, for which the correct form needs
to include the distance across which that voltage is developed, such as in
dBuV/m.

And of course, there is the problem of upper case vs lower case following
the rules for SI units. Even some technically sharp people write things
like MW, Mw or mw when it is clear from the context that they mean mW.

Pardon the rant.


Oh, no! dBu is dB referenced to 0.775 volts ... (1mW into 600
ohms...commonly used for audio work.)

(In case you missed it, that's a tongue-in-cheek seconding of your
rant...)

And of course it people are talking about MW, I assume they mean
something like the Naval Transmitting Station across the way,
nominally 1MW at about 20kHz, +/-, and not the signal coming out of my
bench signal generator.

Fortunately, when I'm not writing papers that will get wide
distribution, but am only communicating with folk around here, we're
all comfortable enough with what we mean that we can use a certain
amount of shorthand and rightfully assume that we understand each
other. But someone communicating through something like a newsgroup
should understand that if there's any room for misinterpretation, you
can count on being misinterpreted. I've used pronouns referring to
something in the previous sentence and been told that a reader thought
it referred to something posted by someone else in an entirely
different post that I hadn't even copied in my reply!

Cheers,
Tom
 
J

jasen

Jan 1, 1970
0
I sometimes like to ponder that 100 watts (+50dBm) radiated from a
simple antenna on, say, 20MHz, can be heard half-way around the world,
about 12000 miles or 20000km away,

somewhat closer straight-line distance,
when "skip" conditions are right.
The received signal level may well be -130dBm or more. That's a path
loss of 180dB or so. By comparison, at 21MHz, coaxial line with
copper conductors, 6" diameter (one of the largest commonly available
commercial sizes, and quite expensive), impedance optimized for lowest
loss, will have about 1dB/mile loss, or 12000dB loss (give or take a
little) in that length.

radiated power is reduced according to an inverse square law,
cable losses are inverse exponential, at a long enough distance radiated
power beats conducted,
 
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