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*Subject*: Re: [amsat-bb] preamps--mast mount vs. in shack*From*: Jon Ogden <na9d@xxxxxxxxxxxxxx>*Date*: Tue, 28 Nov 2000 23:40:49 -0600*In-Reply-To*: <00d401c059ad$5ba92580$a4b2173f@house>*User-Agent*: Microsoft-Outlook-Express-Macintosh-Edition/5.02.2022

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

**Follow-Ups**:**Re: preamps--mast mount vs. in shack***From:*Jens Schmidt

**Re: preamps--mast mount vs. in shack***From:*John Stephensen, KD6OZH

**References**:**Re: preamps--mast mount vs. in shack***From:*John Stephensen, KD6OZH

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