For Roy Lewallen et al: Re Older Post On My db Question
 

Hi Roy,

Thanks for reply re my older post on db question. Very clear explanations
by all.
Will put this as a new thread, though.

New at this, and relize I'm not thinking about this the correct way,
probably.

Relative to a receiving antenna's signal: what does the receiver actually
respond to; power or voltage at its input ?

Bob

--------------------------------------------------------------


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

For Roy Lewallen et al: Re Older Post On My db Question
 

Robert11 wrote:

Hi Roy,

Thanks for reply re my older post on db question. Very clear explanations
by all.
Will put this as a new thread, though.

New at this, and relize I'm not thinking about this the correct way,
probably.

Relative to a receiving antenna's signal: what does the receiver actually
respond to; power or voltage at its input ?

Bob

--------------------------------------------------------------

(snip)

Boy, that almost demands a philosophical answer.

But only almost.

Power. You can (and most really low noise receivers do) transform the
impedance at the antenna into some ideal source impedance for the RF
preamp (or mixer if you're operating in a band where atmospheric noise
exceeds thermal noise by a good amount). Assuming an ideal* impedance
transformation each time you can go from almost any source impedance,
and the noise figure (which is a power ratio) will remain the same.

* Almost never a match. Low noise amplifiers usually do _not_ want a
matched source for best noise figure, although there are ways to force
it to happen.

--

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

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

For Roy Lewallen et al: Re Older Post On My db Question
 

Again, BOTH, tho, usually, sensitivity is expressed in Micro-Volts, but
the dB range also holds. And, to further confuse facts, sensitivity ,
at VHF/UHF/Microwave, uses Noise (figure/factor), which refers to a
Perfect Reciever (no internal noise), vs. your front end rf stage, or
at your antenna's output, before coax loss's and other things (like
BACKGROUND Received NOISE, ect.) Jim NN7K


Robert11 wrote:
Hi Roy,

Thanks for reply re my older post on db question. Very clear explanations
by all.
Will put this as a new thread, though.

New at this, and relize I'm not thinking about this the correct way,
probably.

Relative to a receiving antenna's signal: what does the receiver actually
respond to; power or voltage at its input ?

Bob

--------------------------------------------------------------



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

For Roy Lewallen et al: Re Older Post On My db Question
 

Relative to a receiving antenna's signal: what does the receiver
actually
respond to; power or voltage at its input ?

Bob

--------------------------------------------------------------

The S-meter is actually a wattmeter. S9 = 50 pico-watts into a
50-ohms receiver input impedance.

It indicates signal strength and strength is watts.

A typical meter is scaled from -54 dB to +50 dB relative to S9.

That's why S9 is roughly half way along the scale - about 50 dB on
each side of it.

-54 dB corresponds roughly to receiver internal noise level.

+50 dB corresponds to receiver overload level.
----
Reg, G4FGQ

For Roy Lewallen et al: Re Older Post On My db Question
 

Robert11 wrote:
Hi Roy,

Thanks for reply re my older post on db question. Very clear explanations
by all.
Will put this as a new thread, though.

New at this, and relize I'm not thinking about this the correct way,
probably.

Relative to a receiving antenna's signal: what does the receiver actually
respond to; power or voltage at its input ?

The short answer is power. You can't have one without the other, but you
can choose to have various voltages for the same amount of power. A
receiver has a fixed source impedance, and this determines the fixed
relationship between power and voltage at its input. You can transform
the impedance with various passive circuits, resulting in various other
impedances with correspondingly different voltages. But this won't
change the amount of power available from the antenna.

I'm avoiding discussion of the effects of impedance matching in order to
keep the answer brief and hopefully clear.

Roy Lewallen, W7EL

For Roy Lewallen et al: Re Older Post On My db Question
 

It is power which leaves the transmitter.

Power is what is received by the receiver.

So the S-meter is a power meter.

It is a great pity that the usual S-meter is scaled partly in S-units
and partly in decibels relative to S9.

But given that 1 S-unit = 6 dB, the modern meter scale fits very
nicely between receiver internal noise level and the receiver overload
point.

And, for example, it's so much easier to report signal strength as S5
rather than 0.047 micro-microwatts.

In the same vein, we could, of course, report a signal strength of 20
dB over S9 as S12 and be done with decibels.
----
Reg, G4FGQ.

For Roy Lewallen et al: Re Older Post On My db Question
 

Bob,

To more directly answer your question...

One way to look at it is, that it is power which does the deed. To be
fair, however, you can't have power without the voltage and visa-versa.
This is when we are talking about receivers not OP amps. When a receiver
receives a signal, there is a voltage present and current flowing at its
input. They both occur. They both have to. Together these represent some
amount of power. The receiver's input stage needs some RF energy to tickle
its input so it can amplify it for the next stage. It is this power which
is the energy. Because both of these quantities are there, we can use
either to talk about the sensitivity of said receiver. Although not
"normal" we could just as easily use current as a metric for receiver
sensitivity.

Voltage is directly measurable so it is easy to use as a measure. It has
been used over the centuries (not really) as a standard way of measuring
what is happening in a circuit.

Power, on the other hand, cannot be directly measured it has to be
calculated. Power, we know, is Volts times Amps or E x I. There are other
ways to figure out power which I won't go into here, but this is how we must
be satisfied doing it.

Why, you may ask then, do we use these power numbers instead of just voltage
AND to make matters worse, why do we express them in some obscure thing like
dBm? Good question! So we Engineers can keep you Hams in a steady state of
confusion and therefore keep us in high esteem and power... Not really, but
I'll stop here to see if this helped any.

73, Steve, K9DCi




You can't have
"Robert11" wrote in message
...
Hi Roy,

Thanks for reply re my older post on db question. Very clear explanations
by all.
Will put this as a new thread, though.

New at this, and relize I'm not thinking about this the correct way,
probably.

Relative to a receiving antenna's signal: what does the receiver actually

respond to; power or voltage at its input ?

Bob

--------------------------------------------------------------


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

For Roy Lewallen et al: Re Older Post On My db Question
 

Technically, is it not energy that leaves the transmitter and is
received by the receiver?

Actual integration being performed by later receiver stages and/or the
human ear/brain?


Chuck, NT3G

Reg Edwards wrote:
It is power which leaves the transmitter.

Power is what is received by the receiver.

So the S-meter is a power meter.

It is a great pity that the usual S-meter is scaled partly in S-units
and partly in decibels relative to S9.

But given that 1 S-unit = 6 dB, the modern meter scale fits very
nicely between receiver internal noise level and the receiver overload
point.

And, for example, it's so much easier to report signal strength as S5
rather than 0.047 micro-microwatts.

In the same vein, we could, of course, report a signal strength of 20
dB over S9 as S12 and be done with decibels.
----
Reg, G4FGQ.

For Roy Lewallen et al: Re Older Post On My db Question
 

Steve Nosko wrote:
. . .
Power, on the other hand, cannot be directly measured it has to be
calculated. Power, we know, is Volts times Amps or E x I. There are other
ways to figure out power which I won't go into here, but this is how we must
be satisfied doing it.
. . .

Power meters are in wide use and directly measure power by detecting the
amount of heat it produces.

Roy Lewallen, W7EL

For Roy Lewallen et al: Re Older Post On My db Question
 

chuck wrote:
Technically, is it not energy that leaves the transmitter and is
received by the receiver?

Technically, RF energy passing a point/plane during a unit
of time is RF power (joules/sec). We can't have one without
the other.
--
73, Cecil http://www.qsl.net/w5dxp

For Roy Lewallen et al: Re Older Post On My db Question
 

OK, I'm a little confused. Well, maybe more than a little.

Starting with energy as the "ability to do work" and power as the rate
at which energy is "transformed" into work, things quickly get muddy.

Energy passing through an imaginary surface (or point or plane) would
not actually do any work in passing through, and in fact would retain
its full potential to do work after having passed through.

What then is power density? Is it the amount of work that the energy
passing through a unit area of the surface "could have done" had it been
actually and fully "captured" at that surface? There is no real power at
that surface, is there?

While power (and work) absolutely require energy, it strikes me as
metaphysical whether all energy ultimately does do work and produce
power. I don't think physics requires that, and it seems that lot of
radiated energy is not obviously being transformed into work. So is
energy without power really impossible, Cecil?

Been away from this for a longer bit than I'm comfortable mentioning.

Chuck. NT3G

Cecil Moore wrote:
chuck wrote:

Technically, is it not energy that leaves the transmitter and is
received by the receiver?


Technically, RF energy passing a point/plane during a unit
of time is RF power (joules/sec). We can't have one without
the other.

For Roy Lewallen et al: Re Older Post On My db Question
 

chuck wrote:
OK, I'm a little confused. Well, maybe more than a little.

Starting with energy as the "ability to do work" and power as the rate
at which energy is "transformed" into work, things quickly get muddy.

Power is the rate at which energy is transferred or used. Period.

Energy passing through an imaginary surface (or point or plane) would
not actually do any work in passing through, and in fact would retain
its full potential to do work after having passed through.

Yes.

What then is power density? Is it the amount of work that the energy
passing through a unit area of the surface "could have done" had it been
actually and fully "captured" at that surface?

No. Power is not an "amount of work", nor is power density. Power is the
rate at which power is being transferred, so it can tell you only the
rate at which work can be done, not the amount. The rate at which energy
is being passed through a given cross sectional area of a surface is the
power density. If you integrate the power going across the boundary for
some period of time, you then know the amount of energy which has
passed, and therefore the amount of work which can be done.

There is no real power at
that surface, is there?

There is real energy passing that surface, and the rate at which it's
passing is the power at that surface. So yes, there is.

While power (and work) absolutely require energy, it strikes me as
metaphysical whether all energy ultimately does do work and produce
power. I don't think physics requires that, and it seems that lot of
radiated energy is not obviously being transformed into work. So is
energy without power really impossible, Cecil?

Cecil loves metaphysical arguments, so this is an ideal question for him.

Been away from this for a longer bit than I'm comfortable mentioning.

Try finding a basic physics textbook at your local library. Most high
school level texts should cover the topic adequately. For a more
mathematical and quantitative treatment, a freshman level college text
would be fine. A couple of the more popular ones are Resnick & Halliday,
and Weidner & Sells. In the older editions of both, at least, electrical
phenomena are covered in the second volume. However, power, work, and
energy aren't restricted to electricity so are covered in general terms
in the first volume.

Roy Lewallen, W7EL

For Roy Lewallen et al: Re Older Post On My db Question
 

Correction:

Roy Lewallen wrote:
. . .
No. Power is not an "amount of work", nor is power density. Power is the
rate at which power is being transferred, so it can tell you only the
rate at which work can be done, not the amount. . .

I of course meant ". . .Power is the rate at which *energy* is being
transferred. . ."

Thanks, Owen!

Roy Lewallen, W7EL

For Roy Lewallen et al: Re Older Post On My db Question
 

chuck wrote:
So is energy without power really impossible, Cecil?

Roy has already answered the basic question from an engineering
viewpoint. Physicists often consider "power" to have a different
definition than the engineering definition.

From the IEEE Dictionary: "power - The rate of generating,
transferring, or using energy." (agrees with Roy)

From "University Physics" by Young and Freedman: "power is the
time rate at which work is done." i.e. only the "using energy"
portion of the engineering definition.

A certain physicist I know will argue that no work is being done
in a lossless transmission line so there is no power there. He will
say the existence of 100 watts at 1000 points along the line means
there must be 100,000 watts in the line. He will say that a Bird
wattmeter doesn't measure watts. He will say that a power
generating plant doesn't generate power and a transmission line
doesn't transfer power. He will also say that reflected power
doesn't exist because it is not doing any work. He will say that
for an EM wave in free space, ExH has the dimensions of watts
but it isn't power because no work is being done.

As Roy indicated, engineers have a wider definition of "power".
Energy without power is certainly possible, e.g. a DC battery
with zero current. However, for a constant steady-state power
level associated with an EM wave, energy and power are inseparable.

I like to use a one-second long lossless transmission line in
some of my examples because it is impossible to hide the joules.
A one-second long line with 200 watts forward power and 100 watts
reflected power contains 300 joules that have been generated but
have not reached the load. Since EM wave energy cannot stand still
(or slosh around side to side) it is only logical to assume that
200 of those joules are in the forward wave and 100 of those joules
are in the reflected wave both traveling at the speed of light.
--
73, Cecil http://www.qsl.net/w5dxp

For Roy Lewallen et al: Re Older Post On My db Question
 

chuck wrote:

Energy passing through an imaginary surface (or point or plane) would
not actually do any work in passing through, and in fact would retain
its full potential to do work after having passed through.

What then is power density?

The full name is power *flux* density, implying the rate at which energy
*flows through* unit area of a defined reference plane. SI units are
watts per square metre.

Is it the amount of work that the energy passing through a unit area
of the surface "could have done" had it been actually and fully
"captured" at that surface?

Yes, that is the implication - except that it's the *rate* of energy
capture, ie the amount that could be captured from unit area in unit
time.

This is only a concept, because it isn't physically possible to
intercept the power flux through unit area of an EM wavefront - your
wave-catcher would disturb the wavefront around its edges, and the
shadow behind it would be filled in by diffraction. However, very
similar concepts apply to power flux in a transmission line - and in
that case you really *could* capture the exact steady-state power flux
at any point, by cutting the line and substituting a dummy load of the
correct impedance.


There is no real power at that surface, is there?

That rather depends on your personal definitions of the words "real",
"at" and possibly "is" :-) I think you'd caught it correctly in the
previous paragraph... but if you squeeze too hard, the waves will slip
through your fingers.




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

For Roy Lewallen et al: Re Older Post On My db Question
 

Thank you all for your assistance. It will take a little time (and a
little work) for the difference in physicist/engineer definitions of
power to sink in.

Chuck, NT3G

Ian White GM3SEK wrote:
chuck wrote:


Energy passing through an imaginary surface (or point or plane) would
not actually do any work in passing through, and in fact would retain
its full potential to do work after having passed through.

What then is power density?


The full name is power *flux* density, implying the rate at which energy
*flows through* unit area of a defined reference plane. SI units are
watts per square metre.

Is it the amount of work that the energy passing through a unit area
of the surface "could have done" had it been actually and fully
"captured" at that surface?


Yes, that is the implication - except that it's the *rate* of energy
capture, ie the amount that could be captured from unit area in unit time.

This is only a concept, because it isn't physically possible to
intercept the power flux through unit area of an EM wavefront - your
wave-catcher would disturb the wavefront around its edges, and the
shadow behind it would be filled in by diffraction. However, very
similar concepts apply to power flux in a transmission line - and in
that case you really *could* capture the exact steady-state power flux
at any point, by cutting the line and substituting a dummy load of the
correct impedance.


There is no real power at that surface, is there?

That rather depends on your personal definitions of the words "real",
"at" and possibly "is" :-) I think you'd caught it correctly in the
previous paragraph... but if you squeeze too hard, the waves will slip
through your fingers.

For Roy Lewallen et al: Re Older Post On My db Question
 

I think the differences in this discussion go back to very fundamental
definitions of power and energy (and power flux, which might not be the same
as power). One set of terminology came from classical Thermodynamics and
the other set from more general principles. More fun may be had dealing with
the Second Law (which arrived from two very different viewpoints) and, for
example, quantum effects in the same discussion. The discussion can quickly
hinge on precise meanings of common words, such as "power."

Mixing terminology produces such wonderful concepts as measuring 100 watts
flowing at 100 points in a transmission line and concluding that we have
found 10,000 watts.

Bill
W2WO

For Roy Lewallen et al: Re Older Post On My db Question
 

Why unnecessarily complicate matters with words.

Watts = Amps times Volts.

.. . . . and that's all there is to it.
----
Reg.

For Roy Lewallen et al: Re Older Post On My db Question
 

Reg Edwards wrote:
Watts = Amps times Volts.

Only for DC or in-phase AC/RF.

Volt-Amps = Amps times Volts = SQRT(watts^2 + vars^2)
Watts = Amps times Volts times cos(A) = real power
Vars = Amps times Volts times sin(A) = reactive power
Reference: "Alternating Current Circuits", Kerchner/Corcoran,
3rd edition, (C) 1938, 1943, 1951.
--
73, Cecil http://www.qsl.net/w5dxp