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Re: preamps--mast mount vs. in shack



Hi Jon,
I've only just read through your message, but it makes sense to me, in
conclusion at least.
Now, this is the question I have, -
what effect does the _real_ temperature, (bright sunlight / clear night sky)
have on the noise figures, in respect of RF devices ?
You may be briefer, if you should wish -:)
73 Jens    ZL2TJT

Jon Ogden wrote:

> on 11/28/00 8:35 PM, John Stephensen, KD6OZH at kd6ozh@gte.net wrote:
>
> > At 145 MHz galactic noise is about 3 dB above thermal noise so the location
> > of the preamp will make about a 0.6 dB difference in the signal to noise
> > ratio.
> >
> > 580 K  Galactic noise
> > 35 K  0.5 NF Preamp
> > ---------
> > 615 K  total noise temperature
> >
> > 580 K  Galactic noise
> > 92 K  noise from coax.
> > 35 K  0.5 NF Preamp
> > ---------
> > 707 K  total noise temperature
> >
> > 707/615 = 1.15 = +0.6 dB
>
> Huh?  I am not sure where you calculated this from.  For one, where did you
> get the noise temp from the coax as 92K?  The noise temp and hence noise
> figure of coax depends entirely on its length.  Of course, you are talking
> about SNR versus Noise Figure.  And perhaps at 2m you are correct.  However,
> just for argument sake:
>
> I dug out my college notes on Noise Figure and I quote:
>
> The noise performance of a 2-port can be characterized by "effective input
> temperature" (Te) for a given source impedance at a given frequency.
>
> Te is the temperature at which, when assigned to the input termination,
> gives the same output noise power as that of the actual two port.
>
> A piece of coax is a two port network and a lossy two port at that.
>
> In general computation of Te for a 2-port requires info about the internal
> "physics" of the 2-port.  But for the special case of a passive matched
> attenuator, Te can be computed directly.
>
> Passive Matched attenuator:
>
>                     -------------
>                     |           |
>           -----o-----           -----o----
>          Zs         |    G<1    |        Zl
>           -----o-----           -----o----
>                     |           |
>                     -------------
>
> The physical temperature of the device, we shall call Tatt where "att" stand
> for attenuator and G stands for gain.
>
> Zl and Zs are load and source impedances respectively.
>
> The actual attenuator might be a resistive T or Pi network or it may be a
> section of lossy transmission line.
>
> We have matched attenuator system with 50 Ohm coax, a 50 Ohm preamp (Zs), a
> 50 Ohm RX (Zl).
>
> We also will assume that all components of the system are of the same
> physical temperature (Tatt) (Now this is an assumption since if the preamps
> are outside they may be hotter or cooler than inside, the temperature of the
> coax might vary along the length, etc.  But for illustration purposes, we
> can easily make an assumption that the entire system is in thermal
> equillibrium.).  Again, call this Tatt.
>
> Thus the noise power available at the output of the 2 port must equal the
> noise power available from the load.
>
> We want to find Te for the attenuator.  Thus:
>
> GkTatt + GkTe = kTatt
>   (1)     (2)    (3)
>
> where:
>
> (1) = Noise Power at the output due to the source
> (2) = Noise Power at the output due to the 2 port
> (3) = Noise power from the load
>
> This:  Te = Tatt((1-G)/G)  or define L (loss) = 1/G
>
> Then Te = Tatt(L-1) where L =) 1
>
> The Noise Factor then is:
>
> F = 1+ Te/To  = 1 + (Tatt/To)*(L-1)  where To is the operating noise
> temperature of the system
>
> If the physicall temperature of the attenuator (coax in our case) is To (ie:
> Tatt = To):
>
> Then F=L or NF=10*Log(L) -> This is just the attenuation in dB!
>
> So the Noise Figure of a length of coax is equal to its attenuation.
>
> Now when looking at a cascade of two ports shown below:
>
> ======= 1 ======= 2 ====== ....... = n =
>
> Where the numbers represent the "2 ports"
>
> Te = Te1 + Te2/G1 + Te3/(G1G2) + ..... + Ten/(G1G2...G(n-1))
>
> F = 1 + (F1-1) + (F2-1)/G1 + (F3-1)/(G1G2) + ..... (Fn-1)/(G1G2...G(n-1))
>
> Consider an example of a 3 dB attenuator with Tatt = To followed by an
> amplifier with Gain of 15 dB and a Te=100K.
>
> Find effective input temp and NF of the cascade:
>
> First, we must find Te for the attenuator:
>
> Te1 = To(L-1)  A 3 dB attenuator has loss L = 10^(3/10) = 2
> Therefore Te1 = 290(2-1) = 290  (To = 290 K approx room temp)
>
> The "Gain" of the attenuator:  G1 = 1/L = 1/2
>
> Therefore: Te = Te1+Te2/G1
>
> =  290 + 100/0.5 = 290+200 = 490K
>
> F = 1+Te/To  = 1 + (490/290) = 2.69
>
> NF = 10 Log(F) = 4.3 dB
>
> The noise figure of the amplifier by itself can be computed as:
>
> F = 1+100/290 = 1.34 = 1.28  or approx 1.3 dB!!!!
>
> So a 3 dB loss in front of the amplifier degraded the noise figure by 3 dB!
> Hmmm.......
>
> Therefore, we have a special case that for a room temperature attenuator
> ahead of a 2-port with Noise Figure = NF2:
>
> ---------- Atten in dB ------------ NF2 G2 ------
>
> NF = L(dB) + NF2
>
> That is, the loss in dB adds DIRECTLY to the Noise figure of the second
> stage!
>
> So, my original comment after that lengthy mathematical proof was correct.
> The loss of a length of coax adds directly to the noise figure of your
> system.  So if you have a 0.5 dB noise figure preamp but have a run of coax
> with 1.5 dB of loss, your new noise figure now becomes 2 dB.
>
> Now, let's turn the above problem around.  Let's put the preamplifier BEFORE
> the coax.
>
> Therefore:
>
> Te of the amplifier was shown to be 100K
> Te of the attenuator was shown to be 290K
> G of the amplifier is 15 dB
>
> Now: Te = Te1 + Te2/G
>
> Te = 100 + 290/15 = 119.333
>
> F= 1 + Te/To = 1+(119.333/290) = 1.411
>
> NF = 1.5 dB
>
> So by putting the preamplifier before the lossy coax, we have a difference
> of 4.3 - 1.5 dB = 2.8 dB
>
> Amazing!  It's nearly the same as the loss of the coax!  WOW!
>
> So, where you put your long length of coax is definitely important!  The
> question should be "Will it matter if my Noise Figure increases?"  At HF, no
> it wouldn't.  At 2 meters, it will have an impact and as you get higher in
> frequency it will matter even more.  This is the original point made about
> the SNR.
>
> Hope you all enjoyed my long winded proof!
>
> 73,
>
> Jon
> NA9D
>
> -------------------------------------
> Jon Ogden
> NA9D (ex: KE9NA)
>
> Member:  ARRL, AMSAT, DXCC, NRA
>
> http://www.qsl.net/ke9na
>
> My President is George W. Bush -> The legal winner in Florida
>
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