dBm question

I have a professor who claims that it's not OK to add two signals expressed
in dbM, because that would amount to multiplication in the linear domain,
which yields units of "mW squared".  Using the same train of thought he then
claimed that it *is* OK to subtract the same two signals (both expressed in
dbM) because that would amount to division in the linear domain, which
yields a unitless result.

I claim that you can't do either...that you must first express both signals
in the linear doamin, then operate on them (add or subtract), and then take
the result and transform it back to logarithmic result.

Am I wrong?

"Frank Camsen" <fcamsen@yahoo.com> wrote in message
news:bemb6v$g9d$1@home.itg.ti.com...
> I have a professor who claims that it's not OK to add two signals
expressed
> in dbM, because that would amount to multiplication in the linear domain,
> which yields units of "mW squared".  Using the same train of thought he
then
> claimed that it *is* OK to subtract the same two signals (both expressed
in
> dbM) because that would amount to division in the linear domain, which
> yields a unitless result.
>
> I claim that you can't do either...that you must first express both
signals
> in the linear doamin, then operate on them (add or subtract), and then
take
> the result and transform it back to logarithmic result.

It depends on what you want to do.  If you want the ratio of two powers
expressed in dBm, it's easier and perfectly correct to subtract them and get
a result in dB.  This is common practice in electronics engineering.

Hello Frank,

Your prof's objection is flawed since dBm is already unitless.

dBm = 10*log(power/(reference power))

As you can see the argument to the log function is unitless as is the
argument to any transcendential function.

The "m" in dBm implies the reference power is one milliwatt.  Likewise you
will sometimes see dBW where the reference is one watt.

So go and add or subtract dBm as necessary.

Clay




"Frank Camsen" <fcamsen@yahoo.com> wrote in message
news:bemb6v$g9d$1@home.itg.ti.com...
> I have a professor who claims that it's not OK to add two signals
expressed
> in dbM, because that would amount to multiplication in the linear domain,
> which yields units of "mW squared".  Using the same train of thought he
then
> claimed that it *is* OK to subtract the same two signals (both expressed
in
> dbM) because that would amount to division in the linear domain, which
> yields a unitless result.
>
> I claim that you can't do either...that you must first express both
signals
> in the linear doamin, then operate on them (add or subtract), and then
take
> the result and transform it back to logarithmic result.
>
> Am I wrong?
>
>

"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message
news:foyPa.9641$ru2.1013141@news20.bellglobal.com...
> "Frank Camsen" <fcamsen@yahoo.com> wrote in message
> news:bemb6v$g9d$1@home.itg.ti.com...
> > I have a professor who claims that it's not OK to add two signals
> expressed
> > in dbM, because that would amount to multiplication in the linear
domain,
> > which yields units of "mW squared".  Using the same train of thought he
> then
> > claimed that it *is* OK to subtract the same two signals (both expressed
> in
> > dbM) because that would amount to division in the linear domain, which
> > yields a unitless result.
> >
> > I claim that you can't do either...that you must first express both
> signals
> > in the linear doamin, then operate on them (add or subtract), and then
> take
> > the result and transform it back to logarithmic result.
>
> It depends on what you want to do.  If you want the ratio of two powers
> expressed in dBm, it's easier and perfectly correct to subtract them and
get
> a result in dB.  This is common practice in electronics engineering.
>
>

Right.  The result of subtracting or adding two power signals expressed in
dBm is - as you say - the *ratio* of the two signal power levels, expressed
in dBm.  But it is *not* the power in dBm of the difference or sum of the
two signal powers.  I can't just say, for instance, that I've got two
signals, one of 13 dBm and one of 20 dBm, and that the power of the combined
signal is then just 33 dBm.

This is a computer science class on wireless and cellular communications,
and we're speaking in this particular case of a mobile phone "seeing" two
radio signals from two different base stations.  The professor is a CS prof
trying to arm-wave this idea over the class (which is about 50/50 CS and EE
sutdents) - I think he's on the right track in cautioning students to not
simply add or subtract signals that are represented in dBm, but I think that
his reasoning is a bit fishy.

Hello Frank,

When it comes to adding powers in the case of RF interference, you have to
add in the linear domain


Power P = P1 + P2


So using the dBm = 10 log (P/pref)

we find  (the pref cancels out in the calculation)

X = 10 log (  10^(X1/10) + 10^(X2/10) )


where X is the resulting level in dBm and X1 and X2 are the individual
levels in dBm.


For example X1 = 13dBm and X2 = 20 dBm

X = 20.79 dBm


A simple rule of thumb is when the louder signal exceeds the quieter one by
more than 10 dB, the answer is basically the louder one's level. When the
signals are equal, the level goes up by only 3.01dB,

I hope this helps.

Clay





"Frank Camsen" <fcamsen@yahoo.com> wrote in message
news:bemi8d$s83$1@home.itg.ti.com...
> "Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message
> news:foyPa.9641$ru2.1013141@news20.bellglobal.com...
> > "Frank Camsen" <fcamsen@yahoo.com> wrote in message
> > news:bemb6v$g9d$1@home.itg.ti.com...
> > > I have a professor who claims that it's not OK to add two signals
> > expressed
> > > in dbM, because that would amount to multiplication in the linear
> domain,
> > > which yields units of "mW squared".  Using the same train of thought
he
> > then
> > > claimed that it *is* OK to subtract the same two signals (both
expressed
> > in
> > > dbM) because that would amount to division in the linear domain, which
> > > yields a unitless result.
> > >
> > > I claim that you can't do either...that you must first express both
> > signals
> > > in the linear doamin, then operate on them (add or subtract), and then
> > take
> > > the result and transform it back to logarithmic result.
> >
> > It depends on what you want to do.  If you want the ratio of two powers
> > expressed in dBm, it's easier and perfectly correct to subtract them and
> get
> > a result in dB.  This is common practice in electronics engineering.
> >
> >
>
> Right.  The result of subtracting or adding two power signals expressed in
> dBm is - as you say - the *ratio* of the two signal power levels,
expressed
> in dBm.  But it is *not* the power in dBm of the difference or sum of the
> two signal powers.  I can't just say, for instance, that I've got two
> signals, one of 13 dBm and one of 20 dBm, and that the power of the
combined
> signal is then just 33 dBm.
>
> This is a computer science class on wireless and cellular communications,
> and we're speaking in this particular case of a mobile phone "seeing" two
> radio signals from two different base stations.  The professor is a CS
prof
> trying to arm-wave this idea over the class (which is about 50/50 CS and
EE
> sutdents) - I think he's on the right track in cautioning students to not
> simply add or subtract signals that are represented in dBm, but I think
that
> his reasoning is a bit fishy.
>
>

Frank Camsen wrote:
> 
> I have a professor who claims that it's not OK to add two signals expressed
> in dbM, because that would amount to multiplication in the linear domain,
> which yields units of "mW squared".  Using the same train of thought he then
> claimed that it *is* OK to subtract the same two signals (both expressed in
> dbM) because that would amount to division in the linear domain, which
> yields a unitless result.
> 
> I claim that you can't do either...that you must first express both signals
> in the linear doamin, then operate on them (add or subtract), and then take
> the result and transform it back to logarithmic result.
> 
> Am I wrong?

I'm confused. Probably. Maybe your professor is too, or maybe you or he
isn't expressing things clearly. Are you adding the signals, or adding
their measures? Does addition open the possibility of constructive or
destructive interference?

The sum of uncorrelated signals of equal amplitude is 3 dB more than
either one of them. As the difference in level becomes less, the
difference between the sum and the larger rapidly decreases.

The decibel, being a ratio of powers, is dimensionless. If your
professor really believes otherwise, shame on him.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

(snip)

> Right.  The result of subtracting or adding two power signals expressed in
> dBm is - as you say - the *ratio* of the two signal power levels,
expressed
> in dBm.  But it is *not* the power in dBm of the difference or sum of the
> two signal powers.  I can't just say, for instance, that I've got two
> signals, one of 13 dBm and one of 20 dBm, and that the power of the
combined
> signal is then just 33 dBm.

The result of subtracting two dBm numbers will be in dB, not dBm.

You could also add a dBm and a dB with the result in dBm.

I can't think of many cases where you would multiply powers, so adding two
dBm would be unusual.

> This is a computer science class on wireless and cellular communications,
> and we're speaking in this particular case of a mobile phone "seeing" two
> radio signals from two different base stations.  The professor is a CS
prof
> trying to arm-wave this idea over the class (which is about 50/50 CS and
EE
> sutdents) - I think he's on the right track in cautioning students to not
> simply add or subtract signals that are represented in dBm, but I think
that
> his reasoning is a bit fishy.

In any case were power ratios make sense you can subtract dB or dBm.

You can add or subtract dB to dB or any dBx where x is a known reference
power unit.

Note, though, that you sometimes see dBu (that should be a mu), which is
referenced to 1 microvolt.  As a power unit this requires a source impedance
to be specified.

-- glen

> "Frank Camsen" <fcamsen@yahoo.com> wrote in message
> news:bemb6v$g9d$1@home.itg.ti.com...
> > I have a professor who claims that it's not OK to add two signals
> expressed
> > in dbM, because that would amount to multiplication in the linear
domain,
> > which yields units of "mW squared".  Using the same train of thought he
> then
> > claimed that it *is* OK to subtract the same two signals (both expressed
> in
> > dbM) because that would amount to division in the linear domain, which
> > yields a unitless result.
> >
> > I claim that you can't do either...that you must first express both
> signals
> > in the linear doamin, then operate on them (add or subtract), and then
> take
> > the result and transform it back to logarithmic result.
> >
> > Am I wrong?

Clay is correct about these being ratios and, thus, dimensionless.  So the
reasoning that one would get mW^2 by adding isn't correct as such.
Most often we are talking about sinusoids or noise.  So, dB or dBm aside,
it's *not* OK to add two signals according to their amplitudes alone.  If
they are sinusoids of the same frequency, we also need to know their phase.

If you have two sinusoids *in phase* and are of different amplitudes.  It's
pretty easy to do the math and create examples:

Assume we will talk in volts and milliwatts.
signal 1: 1 vrms across 600ohms would dissipate 1.6666mW.
signal 2: 2 vrms across 600ohms would dissipate 6.6666mW.
Signal 1 expressed in dBm is 10*log10(1.6666)=2.2dBm
Signal 2 expressed in dBm is 8.2dBm
If we assume the signals are in phase then the sum of the voltages is 3 vrms
and the power is (1+2)^2/600 = 9/600 = 15mw which is 11.76dBm.

Had you added dBm, you would have gotten 10.4dBm which would equate to a
ratio of 10.96 or 10.96mW.  This would result from a composite voltage  of
2.56vrms - which is obviously not correct.  So, you can't add dBm unless
there are other constraints / conditions.

You can use the same example and subtract 1vrms from 3vrms and show it's
different than subtracting 2.2dBm from 11.76dBm.

So, in neither case can you add *or* subtract dBm.

Here's an example of when you *can* subtract dBm:

Signal 1 is measured at the input to a filter as 10dBm.
Signal 2 is measured at the same time at the output of the filter as 7dBm.
Q: What is the attenuation of the filter?
A: 10dBm-7dBm = 3dB (not dBm)

How does this work?  (We note that the output may not be in phase with the
input here)
First, we note that the power is only known into a known load because we are
generally measuring voltage and not power - so using dBm is often a
"conversion" from "volts across 600 ohms" to "milliwatts" even though the
impedance isn't always exactly 600 ohms.  (Just a little practical
information..... ).  It's really a measure of voltage in practice.

The description of the situation is saying:
There is apparently 10mW at the input which is equivalent to 2.45 vrms
There is apparently 5mW at the output which is equivalent to 1.734 vrms.
The ratio of powers is 2 and the ratio of voltages is sqrt(2) which is
quickly recognized (or calculated) as 3dB.

Note that we are neither adding nor subtracting voltages here.  We are
comparing their amplitudes by adding or subtracting dB or dBm.

What happened to the "m" in dB???
The input is expressed as a ratio relative to 1 mW.
The output is expressed as a ratio relative to 1 mW.
The attenuation is expressed as a ratio of the input relative to the output.
The input could have been expressed as:
10*log10(Vin^2/600ohms//0.7746v^2/600ohms) which is a ratio of powers with
1mW in the demoninator.  It simplifies to:
= 10*log10(Vin^2/0.6v^2) which is a ratio of volts squared.
The output could be expressed the same way:
= 10*log10(Vout^2/0.6v^2)
The ratio of the input to the output can be expressed as:
10*log10[(Vin^2/0.6v^2)/(Vout^2/0.6v^2)]
= 10*log10[(Vin^2)/(Vout^2)]= 20*log10(Vin/Vout)

So, we went from a specific ratio using 1mW as a reference to a ratio of
arbitrary voltages or powers.  However, note that we weren't needing to be
concerned about phase in this case - just the ratios.

It's probably best to get completely away from dBm in order to examine this
question.  After all, dBm is commonly just another way to say dBv (with a
suitable conversion factor) - that is the measure is still generally a
voltage measure and not really power.  That said, dB is either a power ratio
measure or a volts^2 ratio measure - the only difference having to do with
assumptions about impedance.

Let's stick with a volts^2 measure and examine the question about adding or
subtracting dB.

dB = 10*log10(v1^/v2^2) = 20*log(v1/v2)

Let's use dBv for convenience, meaning that v2=1=v2^2 so that
dBv = 20*log10(v1)

But this is too convenient and may lead to a misunderstanding so let's also
recognize that if we use an arbitrary reference for v2, then the expression
becomes:

dB1=20*log(v1/v2)=20*log10(v1)-20*log10(v2)
dB3=20*log10(v3)-20*log10(v2)

So: dB1+dB2=20*log(v1) + 20*log10(v3) - 40*log10(v2)
= 20*log10(v1*v3/v2^2)                               (A)
where what we really expected to get was:
 20*log10((v1+v3)/v2)                                (B)
And:
dB1-dB3=20*log(v1) - 20*log10(v3)
= 20*log10(v1/v3)                                    (C)

So, it does appear that the reference cancels out if you're subtracting
*and* the result achieved is the ratio of the two voltages - not their
difference.
So equation A isn't what you wanted and equation C might be what you wanted
sort of but isn't what you expected!  It's a measure of the ratio and not of
the difference.  It is expressed in dB and you can't know the difference
without knowing one of them in absolute terms.

Probably more than you wanted....

While, like Clay, I don't agree with the assertion, it's interesting to note
that the professor's observation that you get mW^2 if you add, does sort of
come out of this:  the term -40*log10(v2) is -20*log10(v2^2) or something
like (-10*log10(mW^2) if an impedance is assumed.  You get the ratio of a
product of powers to a power squared (which is also a kind of product).  I
don't know what good that is ....... so have to agree with the professor in
principle - as you did.

Fred

On Fri, 11 Jul 2003 11:51:38 -0400, "Clay S. Turner"
<physicsNOOOOSPPPPAMMMM@bellsouth.net> wrote:

>Hello Frank,
>
>When it comes to adding powers in the case of RF interference, you have to
>add in the linear domain
>
>
>Power P = P1 + P2
>
>
>So using the dBm = 10 log (P/pref)
>
>we find  (the pref cancels out in the calculation)

Since Pref cancels when you add two terms that expressed in dBm the
result is just dB.  The milliwatt reference is no longer valid.

>X = 10 log (  10^(X1/10) + 10^(X2/10) )

>where X is the resulting level in dBm and X1 and X2 are the individual
>levels in dBm.

Yes, this way preserves the mW reference so that the result is really
in dBm.   Note that the resulting quanitity is different, too, so the
interpretations have to be carefully made.  I think the Prof. is
merely touching on the point that log quantities with a unit
reference, e.g., dBm, dBW, etc., have to be treated differently than
those without.   e.g., when managing gain stages it's perfectly fine
to add all of the dB gain values to get the total gain of the system.

>For example X1 = 13dBm and X2 = 20 dBm
>
>X = 20.79 dBm
>
>
>A simple rule of thumb is when the louder signal exceeds the quieter one by
>more than 10 dB, the answer is basically the louder one's level. When the
>signals are equal, the level goes up by only 3.01dB,
>
>I hope this helps.
>
>Clay
>
>
>
>
>
>"Frank Camsen" <fcamsen@yahoo.com> wrote in message
>news:bemi8d$s83$1@home.itg.ti.com...
>> "Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message
>> news:foyPa.9641$ru2.1013141@news20.bellglobal.com...
>> > "Frank Camsen" <fcamsen@yahoo.com> wrote in message
>> > news:bemb6v$g9d$1@home.itg.ti.com...
>> > > I have a professor who claims that it's not OK to add two signals
>> > expressed
>> > > in dbM, because that would amount to multiplication in the linear
>> domain,
>> > > which yields units of "mW squared".  Using the same train of thought
>he
>> > then
>> > > claimed that it *is* OK to subtract the same two signals (both
>expressed
>> > in
>> > > dbM) because that would amount to division in the linear domain, which
>> > > yields a unitless result.
>> > >
>> > > I claim that you can't do either...that you must first express both
>> > signals
>> > > in the linear doamin, then operate on them (add or subtract), and then
>> > take
>> > > the result and transform it back to logarithmic result.
>> >
>> > It depends on what you want to do.  If you want the ratio of two powers
>> > expressed in dBm, it's easier and perfectly correct to subtract them and
>> get
>> > a result in dB.  This is common practice in electronics engineering.
>> >
>> >
>>
>> Right.  The result of subtracting or adding two power signals expressed in
>> dBm is - as you say - the *ratio* of the two signal power levels,
>expressed
>> in dBm.  But it is *not* the power in dBm of the difference or sum of the
>> two signal powers.  I can't just say, for instance, that I've got two
>> signals, one of 13 dBm and one of 20 dBm, and that the power of the
>combined
>> signal is then just 33 dBm.
>>
>> This is a computer science class on wireless and cellular communications,
>> and we're speaking in this particular case of a mobile phone "seeing" two
>> radio signals from two different base stations.  The professor is a CS
>prof
>> trying to arm-wave this idea over the class (which is about 50/50 CS and
>EE
>> sutdents) - I think he's on the right track in cautioning students to not
>> simply add or subtract signals that are represented in dBm, but I think
>that
>> his reasoning is a bit fishy.
>>
>>
>
>
>

Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org

Bob Bruhns wrote:

<Tim, the term "dBm" means "decibels, referenced against 1 milliwatt".
<Although one usually measures voltage applied to a load, dBm is an 
<expression of power level.  DBm is often misunderstood because AC 
<voltmeters often have "dBm" calibrations referenced to 600 ohm
systems,
<and people incorrectly think dBm means a voltage level, independent
of
<impedance.
<
<  You must define the load resistance (R ), and measure the RMS
voltage
<applied to that load (E),               L
<and then calculate:
<
<                            /  2        \
<power (dBm) = 10 *  log    |  E  * 1000  |
<                       (10)| ----------- |
<                            \    R      /
<                                  L
<
<
<For a 600 ohm impedance, and 1.22766 volts RMS,
<
<              /         2        \                   /       \
<10 *  log    | (1.22766)  * 1000  |  =  10 * log    | 2.51192 |
<         (10)| ------------------ |             (10) \       /
<              \        600       /                  
<
<
< = 10 * (0.40) = 4 dBm (a standard "0 VU" level in professional
audio).
<
<Bob Bruhns, WA3WDR, bbruhns@li.net


A little nit-picking. This isn't quite true.  dBm is not a power.  It
is a dimensionless number that is the ratio of two powers where the
reference is 1mw.  The general equation for dB when using voltages is

           /    \             /    \
          | E(2) |           |  R1  |
   20 log | ---- |  + 10 log | ---- |
          | E(1) |           |  R2  |
           \    /             \    /


You can ignore the impedance ratio ONLY if the load and the reference
impedance are the same.

What this says is that the power in the 600 ohm load is a level that
is 4 dB above a milli-watt in 600 ohms.  We often say (I do it too)
that we have a power level of 4 dBm.  This is OK if we, and everyone
who sees it, understands what this really represents.

(Quickly, without calculating, if the impedance remains constant, and
I increase the voltage by 6 dB how much of an increase in power (*in
dB) will I see?)
Maurice Givens