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

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/

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

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

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

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.

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

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.

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

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