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*Subject*: Re: [amsat-bb] preamps--mast mount vs. in shack*From*: Jens Schmidt <j.schmidt@xxxxxxxxxxxxxxx>*Date*: Thu, 30 Nov 2000 01:28:41 +1300

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 > > ---- > Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. > To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org ---- Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

**Follow-Ups**:**Re: preamps--mast mount vs. in shack***From:*Jon Ogden

**References**:**Re: preamps--mast mount vs. in shack***From:*Jon Ogden

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