Adding noise given by Noise Factor

> 
> Well, so long as you don't conclude this adds in an automatic 3 dB loss 
> when computing your link margin.  Because it doesn't.  You aren't
> actually dissipating half of the received power in the antenna, and
> half in the LNA.  Ideally antennas do not dissipate RF power.
> The antenna model I described above is a valid circuit model for
> a receiving antenna, but it is non-physcial in the sense that the 
> voltage source internal this model is chugging out twice the RF power 
> than actually exists physically.
> 
> Steve

If I may,,, this brings us to a question I've had for a while...

the assumption for part of this discussion was that the LNA was NOISLESS i.e.
NF = 0 dB.  In this case the SNR is unchanged by the LNA. OK so far.

Also it is implied  but hasn't been stated, the antenna Temp = 273K

But in order for the  LNA to be matched and NF = 0 dB the 50 Ohm R inside  the LNA must be at Temp =0 K.

So OK, this works if the LNA is physically  at 0 K, it has NF= 0 dB and the SNR is not reduced.  OK so far.

But what about the more typical case when the LNA is at 273K.
Then it seems we do get a 3 dB  SNR reduction and therefore the NF = 3 dB.

It seems that if we assume a matched system, and the physical temp of the LNA = 273K, then the __BEST___ noise figure you could hope to get would be 3 dB.

But you can buy off the shelf LNA with NF=1 dB and they do not require cooling.

So something doesn't fit.   

Does it mean that roomtemp LNAs with <3 dB NF ___CANNOT___ be matched?

    

On 2016-02-17 6:30, makolber@yahoo.com wrote:
> [...]
> 
> If I may,,, this brings us to a question I've had for a while...
> 
> the assumption for part of this discussion was that the LNA was NOISLESS i.e.
> NF = 0 dB.  In this case the SNR is unchanged by the LNA. OK so far.
> 
> Also it is implied  but hasn't been stated, the antenna Temp = 273K
> 
> But in order for the  LNA to be matched and NF = 0 dB the 50 Ohm R inside  the LNA must be at Temp =0 K.

For a physical system probably yes.

> So OK, this works if the LNA is physically  at 0 K, it has NF= 0 dB and the SNR is not reduced.  OK so far.
> 
> But what about the more typical case when the LNA is at 273K.
> Then it seems we do get a 3 dB  SNR reduction and therefore the NF = 3 dB.

No

> It seems that if we assume a matched system, and the physical temp of the LNA = 273K, then the __BEST___ noise figure you could hope to get would be 3 dB.

No

> But you can buy off the shelf LNA with NF=1 dB and they do not require cooling.
> 
> So something doesn't fit.   
> 
> Does it mean that roomtemp LNAs with <3 dB NF ___CANNOT___ be matched?

The issue is that you assume resistive matching (which is the most
obvious way to do it and good for wideband matching but the worst for
noise).

But you can generate input impedances in other ways. For example, a
Common Gate stage has input resistance of 1/gm (which is 50 Ohm for
gm=20mS). The noise factor is approximately F = 1 + gamma (gamma is MOS
noise factor). You can use boosting to reduce this to F = 1+gamma/(A+1).
The gain can be made from a capacitive divider (in a differential
design) and hence it's noise-less and you get F=1+gamma/2.

If linearity/wideband match is not an issue you can use inductive
degeneration for best possible NF. The input impedance is then:

Zin = 1/(jCgs omega) + jL omega + L omegaT

which looks purely resistive (L omegaT = 50) at resonance w0=1/sqrt(L
Cgs). The higher Q, the better NF (but less bandwidth and linearity).

Hence NF=1dB (or even <1dB) at 400K are physically realizable.


Peter

Peter Mairhofer  <63832452@gmx.net> wrote:

>On 2016-02-17 6:30, makolber@yahoo.com wrote:

>> the assumption for part of this discussion was that the LNA was NOISLESS i.e.
>> NF = 0 dB.  In this case the SNR is unchanged by the LNA. OK so far.

>> Also it is implied  but hasn't been stated, the antenna Temp = 273K

>> But in order for the  LNA to be matched and NF = 0 dB the 50 Ohm R
>inside  the LNA must be at Temp =0 K.

>For a physical system probably yes.

>> So OK, this works if the LNA is physically  at 0 K, it has NF= 0 dB
>and the SNR is not reduced.  OK so far.

>> But what about the more typical case when the LNA is at 273K.
>> Then it seems we do get a 3 dB  SNR reduction and therefore the NF = 3 dB.

>No

>> It seems that if we assume a matched system, and the physical temp of
>> the LNA = 273K, then the __BEST___ noise figure you could hope to get
>> would be 3 dB.

>No

>> But you can buy off the shelf LNA with NF=1 dB and they do not 
>> require cooling.

>> So something doesn't fit.   

>> Does it mean that roomtemp LNAs with <3 dB NF ___CANNOT___ be matched?

>The issue is that you assume resistive matching (which is the most
>obvious way to do it and good for wideband matching but the worst for
>noise).

>But you can generate input impedances in other ways. For example, a
>Common Gate stage has input resistance of 1/gm (which is 50 Ohm for
>gm=20mS). The noise factor is approximately F = 1 + gamma (gamma is MOS
>noise factor). You can use boosting to reduce this to F = 1+gamma/(A+1).
>The gain can be made from a capacitive divider (in a differential
>design) and hence it's noise-less and you get F=1+gamma/2.

>If linearity/wideband match is not an issue you can use inductive
>degeneration for best possible NF. The input impedance is then:

>Zin = 1/(jCgs omega) + jL omega + L omegaT

>which looks purely resistive (L omegaT = 50) at resonance w0=1/sqrt(L
>Cgs). The higher Q, the better NF (but less bandwidth and linearity).

>Hence NF=1dB (or even <1dB) at 400K are physically realizable.

Peter - thanks.

Steve

<makolber@yahoo.com> wrote:

>If I may,,, this brings us to a question I've had for a while...

>the assumption for part of this discussion was that the LNA was NOISLESS i.e.
>NF = 0 dB.  In this case the SNR is unchanged by the LNA. OK so far.

>Also it is implied  but hasn't been stated, the antenna Temp = 273K

>But in order for the  LNA to be matched and NF = 0 dB the 50 Ohm R
>inside  the LNA must be at Temp =0 K.

>So OK, this works if the LNA is physically  at 0 K, it has NF= 0 dB and
>the SNR is not reduced.  OK so far.

>But what about the more typical case when the LNA is at 273K.
>Then it seems we do get a 3 dB  SNR reduction and therefore the NF = 3 dB.

>It seems that if we assume a matched system, and the physical temp of
>the LNA = 273K, then the __BEST___ noise figure you could hope to get
>would be 3 dB.

>But you can buy off the shelf LNA with NF=1 dB and they do not require cooling.

>So something doesn't fit.   

>Does it mean that roomtemp LNAs with <3 dB NF ___CANNOT___ be matched?

Peter provided the detailed answer, but to address one of the
meta-issues here:

A model can be valid in a simulation under defined conditions, without 
the internal pieces of the model making physical sense individually.  

I mentioned one example of this in my previous post: the antenna
receiving a RF signal can be usefully modeled as a voltage source
in series with a resistor.  But the power output of that voltage
source, viewed in isolation, is twice the power of the physical 
RF signal.  This voltage source is non-physical.

Similarly the LNA can be modeled as embodying an input impedance,
an equivalent input noise source, and and ideal voltage amplifier.
But you can't take the resistor in that model and start attributing
thermal noise to it -- that is already accounted for by the equivalent 
input noise.  The resistor in the LNA model is non-physical.

The whole purpose of defining the noise figure is so that you
can abstract the noise density into a single number; if you
tear the model apart and attribute noise to its components, you have
defeated that purpose.

(In physics, a common saying is "All models are wrong; some models
are useful".  This applies in engineering as well.)

Steve

> 
> A model can be valid in a simulation under defined conditions, without 
> the internal pieces of the model making physical sense individually.  
> 
> I mentioned one example of this in my previous post: the antenna
> receiving a RF signal can be usefully modeled as a voltage source
> in series with a resistor.  But the power output of that voltage
> source, viewed in isolation, is twice the power of the physical 
> RF signal.  This voltage source is non-physical.
> 
I think the antenna model is more physical than you think.
An antnna absorbs power from the passing EM wave which creates a current distribution in the elements.

If the terminals are open circuit, a voltage is developed, but ALL the power is re-radiated. (we're taking ideal of course)

If the terminals are loaded, 1/2 the power is delivered to the load and 1/2 is still re-radiated.

It seems to fit the model perfectly.

The source resistance in the radiation resistance of the antenna transformed by however the load is connected to the antenna.  The radiation resistance __is__ physical, it is all the "stuff" that the antenna "sees".

Mark









> Similarly the LNA can be modeled as embodying an input impedance,
> an equivalent input noise source, and and ideal voltage amplifier.
> But you can't take the resistor in that model and start attributing
> thermal noise to it -- that is already accounted for by the equivalent 
> input noise.  The resistor in the LNA model is non-physical.
> 
> The whole purpose of defining the noise figure is so that you
> can abstract the noise density into a single number; if you
> tear the model apart and attribute noise to its components, you have
> defeated that purpose.
> 
> (In physics, a common saying is "All models are wrong; some models
> are useful".  This applies in engineering as well.)
> 
> Steve

>An antnna absorbs power from the passing EM wave which creates a current
>distribution in the elements.

>If the terminals are open circuit, a voltage is developed, but ALL the
>power is re-radiated. (we're taking ideal of course)

>If the terminals are loaded, 1/2 the power is delivered to the load and
>1/2 is still re-radiated.

I have never encountered this information before -- do you have
a reference handy?

Steve

On Thu, 18 Feb 2016 07:25:13 -0800 (PST), makolber@yahoo.com wrote:

>
>> 
>> A model can be valid in a simulation under defined conditions, without 
>> the internal pieces of the model making physical sense individually.  
>> 
>> I mentioned one example of this in my previous post: the antenna
>> receiving a RF signal can be usefully modeled as a voltage source
>> in series with a resistor.  But the power output of that voltage
>> source, viewed in isolation, is twice the power of the physical 
>> RF signal.  This voltage source is non-physical.
>> 
>I think the antenna model is more physical than you think.
>An antnna absorbs power from the passing EM wave which creates a current distribution in the elements.
>
>If the terminals are open circuit, a voltage is developed, but ALL the power is re-radiated. (we're taking ideal of course)
>
>If the terminals are loaded, 1/2 the power is delivered to the load and 1/2 is still re-radiated.
>
>It seems to fit the model perfectly.
>
>The source resistance in the radiation resistance of the antenna transformed by however the load is connected to the antenna.  The radiation resistance __is__ physical, it is all the "stuff" that the antenna "sees".
>
>Mark

There are multiple sources of resistance (or impedance, whatever), wrt
the antenna.   There is natural resistance of the electrons moving
around in the material of the antenna, and there is the impedance to
the interface to free space (which affects the ability to both collect
and re-radiate power).

The details seem to get pretty crazy, so I've always tried to just use
off-the shelf stuff and keep to the manufacturer's gain and impedance
specs.  These days with active loading, etc., it seems to get crazier,
which appears to be apropos to the OP's issues.


Eric Jacobsen
Anchor Hill Communications
http://www.anchorhill.com

On 2016-02-16 14:41, Steve Pope wrote:
> Peter Mairhofer  <63832452@gmx.net> wrote:
> 
>> On 2016-02-16 12:00, Steve Pope wrote:
> 
>>> Surely power = V^2 / R, not V^2 / (2*R) !!
> 
>> Actually, this is what I meant in <n9vt8g$8s1$2@news.albasani.net>,
>> however, got it wrong for some reason.
> 
> Great, so we have resolved the 3 dB discrepancy between formulae.
> 
>> So to summarize:
> 
>> In case of matching, Pin = Vsig^2/(2*R)
> 
> Yes, if the antenna is being modeled as a voltage source with a series 
> resistance R, and Vsig is that voltage source, then this is then the power 
> going into the input of the LNA.  (But see below.)

Unfortunately I need to come back to this again. It seems I still had a
mistake in my code (drawing from rand rather than randn) giving me
coincidentially the correct result.

In case of matching Pin = Vsig^2/(4*R) !!

Because:

Vsig^2 ------|  R  |---Vx^2----|   R    |---- GND

Vx^2 = Vsig^2 * (R/(R+R))^2 = Vsig^2 / 4

Power at the second resistance: Psig = Vx^2 / R = Vsig^2 / (4R)

This is then consistent with the fact that the noise power is just kTB
(i.e., no factor of 4 or resistance).

>> But then the same is true for noise: Vn^2 = 4kTBR*R/(2*R) = 2kTR

No, this is wrong.

The noise power is then exactly kTR:

Pnoise = 4kTBR * (R/(2R))^2 / R = kTB

> [...]
>> If no matching is assumed (LNTA has infinite input impedance),
>> Pin=Vsig^2/R, Vn^2=4kTR

This is nonsense. (1) If LNTA has infinite input impedance there is zero
power (but 100% voltage transfer).

Similarly, if LNTA has zero input impedance (2), the calculation does
not make any sense because there is no input impedance and hence no
signal at the LNTA (*).

Also, assuming a zero input impedance (3) is meaningless because it is
inconsistent with the initial assumptions (source noise).

So given Rs != 0, some kind of LNTA input impedance (load impedance)
must be assumed. Now we could choose Rs != RL (no matching) but this
gives expressions with both Rs and RL. So assuming matching (Rs = RL =
R) the expressions are nice:

Psig = Vsig^2 / (4R)
Pnoise = kTB

(*): Mathematically (2) still works out. Then Psig = Vsig^2/R and
Pnoise=4kTB resulting in the same SNR. But in this setup we would "read"
the signal at the source assuming an ideal voltage source with Rs=0 and
we are back to (3).

>> In my opinion, the former should be favored because in an
>> actual system there would be a match between antenna and LNA
>> (and both; noise and signal would be halfed).

This reinforces my opinion: The one calculation which makes sense is
assuming matching and calculating Psig = Vsig^2 / (4R), Pnoise = kTB .

> Well, so long as you don't conclude this adds in an automatic 3 dB loss 
> when computing your link margin.  Because it doesn't.  You aren't
> actually dissipating half of the received power in the antenna, and
> half in the LNA.

Although not clearly stated I was talking about signal *voltage* but not
signal *power*. This is exactly the point about matching: While the
amplitude is halfed, the signal power transfer is 100% (and hence no 3dB
loss) - again reinforcing the matching approach from above.

> Ideally antennas do not dissipate RF power.
> The antenna model I described above is a valid circuit model for
> a receiving antenna, but it is non-physcial in the sense that the 
> voltage source internal this model is chugging out twice the RF power 
> than actually exists physically.

Sure but this is resolved once we think about signal power instead of
voltages.

Peter

Peter Mairhofer  <63832452@gmx.net> wrote:

>On 2016-02-16 14:41, Steve Pope wrote:

>> Peter Mairhofer  <63832452@gmx.net> wrote:
>> 
>>> On 2016-02-16 12:00, Steve Pope wrote:
>> 
>>>> Surely power = V^2 / R, not V^2 / (2*R) !!
>> 
>>> Actually, this is what I meant in <n9vt8g$8s1$2@news.albasani.net>,
>>> however, got it wrong for some reason.
>> 
>> Great, so we have resolved the 3 dB discrepancy between formulae.
>> 
>>> So to summarize:
>> 
>>> In case of matching, Pin = Vsig^2/(2*R)
>> 
>> Yes, if the antenna is being modeled as a voltage source with a series 
>> resistance R, and Vsig is that voltage source, then this is then the power 
>> going into the input of the LNA.  (But see below.)
>
>Unfortunately I need to come back to this again. It seems I still had a
>mistake in my code (drawing from rand rather than randn) giving me
>coincidentially the correct result.

>In case of matching Pin = Vsig^2/(4*R) !!
>
>Because:
>
>Vsig^2 ------|  R  |---Vx^2----|   R    |---- GND
>
>Vx^2 = Vsig^2 * (R/(R+R))^2 = Vsig^2 / 4
>
>Power at the second resistance: Psig = Vx^2 / R = Vsig^2 / (4R)
>
>This is then consistent with the fact that the noise power is just kTB
>(i.e., no factor of 4 or resistance).

My reaction to this is that, you certainly want to model the 
matching case, before modeling less ideal cases.  So let's
assume we're modeling the matching case.  

You also might want to have a model for the antenna in the
presence of the received signal, connected to a second model for the LNA.

At this point, under the matching assumption, you can choose lots of
possible model pairs, and so long as combined model gives you the
right answer, it is a valid model.

And again under the assumption that there is matching, there's no rule 
that says you even have to include the resistance R anywhere
in these models.  You could have a voltage source for the signal, a 
voltage source for the noise, and an ideal voltage amplifier and nothing
else.

(In fact, this is what I said to do in my first post to the thread --
but without going through the above assumptions and chain of reasoning).

So when you say the power going into the LNA is Vsig^2 / (4*R),
for some Vsig, I still believe the Vsig is non-physical, it is necessary
to make the chosen model work, with the values R in place, but other 
models also work.

>>> But then the same is true for noise: Vn^2 = 4kTBR*R/(2*R) = 2kTR
>
>No, this is wrong.
>
>The noise power is then exactly kTR:
>
>Pnoise = 4kTBR * (R/(2R))^2 / R = kTB
>
>> [...]
>>> If no matching is assumed (LNTA has infinite input impedance),
>>> Pin=Vsig^2/R, Vn^2=4kTR
>
>This is nonsense. (1) If LNTA has infinite input impedance there is zero
>power (but 100% voltage transfer).
>
>Similarly, if LNTA has zero input impedance (2), the calculation does
>not make any sense because there is no input impedance and hence no
>signal at the LNTA (*).
>
>Also, assuming a zero input impedance (3) is meaningless because it is
>inconsistent with the initial assumptions (source noise).
>
>So given Rs != 0, some kind of LNTA input impedance (load impedance)
>must be assumed. 

Not true, if you are instead assuming matching, such that you don't
even need to accurately model mismatched cases.

>Now we could choose Rs != RL (no matching) but this
>gives expressions with both Rs and RL. So assuming matching (Rs = RL =
>R) the expressions are nice:
>
>Psig = Vsig^2 / (4R)
>Pnoise = kTB
>
>(*): Mathematically (2) still works out. Then Psig = Vsig^2/R and
>Pnoise=4kTB resulting in the same SNR. But in this setup we would "read"
>the signal at the source assuming an ideal voltage source with Rs=0 and
>we are back to (3).
>
>>> In my opinion, the former should be favored because in an
>>> actual system there would be a match between antenna and LNA
>>> (and both; noise and signal would be halfed).
>
>This reinforces my opinion: The one calculation which makes sense is
>assuming matching and calculating Psig = Vsig^2 / (4R), Pnoise = kTB .

>> Well, so long as you don't conclude this adds in an automatic 3 dB loss 
>> when computing your link margin.  Because it doesn't.  You aren't
>> actually dissipating half of the received power in the antenna, and
>> half in the LNA.

>Although not clearly stated I was talking about signal *voltage* but not
>signal *power*. This is exactly the point about matching: While the
>amplitude is halfed, the signal power transfer is 100% (and hence no 3dB
>loss) - again reinforcing the matching approach from above.

>> Ideally antennas do not dissipate RF power.
>> The antenna model I described above is a valid circuit model for
>> a receiving antenna, but it is non-physcial in the sense that the 
>> voltage source internal this model is chugging out twice the RF power 
>> than actually exists physically.
>
>Sure but this is resolved once we think about signal power instead of
>voltages.

I don't know why this needs to be resolved.  It is what it is.

I'm still curious about your statement about a receiving antenna, 
matched to a receiver, re-radiating half the intercepted power 
from the EM field.

When I visualize this, it seems it violates the antenna 
recipricocity theorem as it would suggest driving the antenna
with a given power would create twice that power in the EM
field.  Something it not right with the idea, and I'd like
to see more of a description, perhaps a literature reference.

Steve

> 
> When I visualize this, it seems it violates the antenna 
> recipricocity theorem as it would suggest driving the antenna
> with a given power would create twice that power in the EM
> field.  Something it not right with the idea, and I'd like
> to see more of a description, perhaps a literature reference.
> 
> Steve


that was my statement. (not Peters)
I spent a little time doing a literature search but didn't find anything worth quoting..

but I'll give you these thoughts....

every conductor with accelerated charges (which I will now refer to as current) 
radiates a field.

If you place an antenna in an existing field, there will be a current induced into the conductors.

THe current distribution will be changed depending upon how the antenna terminals are terminated.  The three corner cases are open circuit, short circuit and matched.  Since the current distribution changes, the re-radiated field changes.  You can think of it in terms of conservation of energy as well, the power delivered to the LNA must be taken from the EM field and hence the field is changed by the presence of the loaded antenna.  And the EM field is  changed differently if the antenna is shorted or open instead or matched.

Another thought is the operation of a Yagi antenna.  The reflector and directors obviously re-radiate a field in that case.  The spacing and length of these are adjusted such that the rre-radiated field re-enforces the field at the "driven element".


Due to conservation of energy (ignoring losses) a shorted or open antenna must re-radiate ALL the energy it absorbs.    A matched antenna delivers some power to the load, therefore the amount re-radiated must be less. 


Mark