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Re: preamps--mast mount vs. in shack



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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