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Re: preamps--mast mount vs. in shack



on 11/29/00 6:28 AM, Jens Schmidt at j.schmidt@paradise.net.nz wrote:

> 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

Well, increased thermal noise will cause an increase in the overall noise
temperature of the system and will therefore cause a degradation in Noise
Figure.  However, the same principle of putting your "low noise/High Gain"
stage before the lossy coax still follows.  All of my examples were shown at
room temp, but the principles still hold for higher temps, etc.

73,

Jon
NA9D

> 
> 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
>> 
>> ----
>> Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA.
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> 


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