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Re: preamps--mast mount vs. in shack



Hi Jon,

Your method is correct for calculating the noise figure of the amplifier and
the noise figure at the antenna feed does increase in proportion to the loss
in the coax preceeding the amplifier. However the noise figure numbers
assume that all components are at 290K and the antenna temperature at radio
frequencies is much different. The antenna temperature depends on noise
picked up from the sky at which the antenna is pointed. SInce all components
are not at the same temperature, the noise figure must be converted to noise
temperature referenced to the output of the antenna and then all noise
contributions must be added to determine the total noise level.

The effect of noise figure on signal to noise ratio (SNR) is easier to
calculate by converting to noise temperarture. Assuming a 50-ohm system, the
noise figure is the amount of additional noise created by a two port device
with a 50 ohm resistor on the input of the device compared to the thermal
noise generated by the 50-ohm resistor alone at 290 K. If the noise figure
of the amplifier is 0.5 dB, the input referred noise level has increased by
a factor of 1.122 and the noise contributed by the amplifier is 0.122 * 290
K = 35 K. The same method can be used to calculate the additional noise
added by a lossy piece of coaxial cable or the noise figure at the antenna
can be calculated and then converted to noise temperature.

The noise temperature of the coax was calculated using the 1.2 dB of
attenuation given by Mark for 75 feet of Belden 9913 at 150 MHz. To get the
attenuation at 450 MHz, I multiplied by the square root of 450/150. This is
a good approximation for coaxial cable loss below 1000 MHz.

Noise above 20 MHz, when ionospheric propogation is not present, is mainly
of galactic origin and decreases with frequency at a rate of about 20 dB per
decade. A rule of thumb is that the average noise level present when you
point a low gain antenna at the sky is 20 dB above thermal noise (2900 K) at
20 MHz and decreases to 20 dB below thermal noise (2.9 K) at 2000 MHz.
Between 2 and 10 GHz the the sky temperature is about 2.7 K due to residual
radiation from the "big bang."

When the antenna temperature is higher than 290 K (below 200 MHz), reducing
the receiver noise figure has an effect on SNR that is smaller than
indicated by the difference in noise figures. When the antenna temperature
is lower than 290 K, the effect on SNR is larger than indicated by the
difference in noise figures.

The noise level on 2 meters is actually higher than I indicated above due to
man-made noise. CCIR Report 258-3 shows noise levels of about 1,400, 3,600
and 11,000 K in rural, residential and business locations, respectively,
using omnidirectional antennas. Some of this noise will get into the
receiver through antenna sidelobes that may only be 10 dB down on small
yagis.

73,

John
KD6OZH

----- Original Message -----
From: "Jon Ogden" <na9d@mindspring.com>
To: "John Stephensen, KD6OZH" <kd6ozh@AMSAT.Org>; "AMSAT-BB"
<AMSAT-BB@AMSAT.Org>; "Mark L. Hammond" <hammond@surrealnet.net>
Sent: Wednesday, 29 November 2000 05:40 UTC
Subject: Re: [amsat-bb] 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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