[Moon-net] additional cable loss in front of LNA

[Leif Asbrink]

Hi Bodo,

I do not think it is so easy. First of all, signal loss
and loss of S/N are two different things.

Look at it this way: 

The wave front that arrives at the antenna will give
a specific electromotoric force at the feedpoint.
You will maximize the power extracted by putting a
50 ohm load there.

Now, imagine a very long superconducting 50 ohm cable 
with a nominal loss of 0.6 dB. Also assume that the incoming
signal has a very short duration. Just a pulse:-)

In this case, the antenna will see the 50 ohm cable as a 50 ohm load
and all the available power will start to travel along the cable.
When it arrives at the preamplifier (which we for ease of understanding
assume to have an infinite SWR) all the power is reflected and will
start to travel backwards through the cable. When it reaches the antenna
all of it will be re-radiated because the antenna is well matched.

The voltage at the LNA will be twice the voltage one would have with 
a 50 ohm load - and that is the normal operating circumstances for this 
particular LNA. Actually the neutralized LNAs that I have been using 
in the past have an input impedance above 500 ohms so the infinite SWR 
is not unrealistic. It is not even uncommon with negative input
impedances that cause the reflected wave to be even stronger than the 
incident wave.

The signal loss due to the cable is just 0.6 dB. The infinite impedance
LNA does not receive any power - it is just a device that amplifies the
voltage. If you move the LNA up to the feed point your signals would 
grow by 0.6 dB.

Another thing is that by moving the LNA to the feed point you could
get a lower noise floor and therefore a better S/N. Whether this would
happen depends on the antenna noise temperature.

The noise floor will be proportional to the total system noise.
0.5 dB of your preamp is 35 degrees Kelvin.
On 144 MHz your system temperature without cable losses
would be something like 280 K. The cable with 0.6dB loss 
and 300K will not change the system temperature much.

On 1296 your system temperature might be 50K and then a red hot
cable (300K) would matter even though the attenuation is only 0.6 dB.


With formulae you can see the situation as two cascaded amplifiers.

1) The cable. G1=0.871 (linear power gain) T1=300/0.871-300=44K
2) The LNA.   T2=35K G2=100 (G2 does not matter)

When series connected this formula applies: Tn= T1 + T2/G1
The result comes out as 44 + 40.2 = 84.2 K (NF=1.1dB)

On 144 your S/N loss would be (280+35)/(280+84.2)=0.865=0.63dB
On 1296 it could be (15+35)/(15+84.2)=0.504=3dB

In other words, on 144 the signal would drop by 0.6 dB while
the noise floor would not be affected. Under normal circumstances your
sky temperature will be high above 200K and then the noise floor 
would also drop for an even more insignificant influence on the S/N
due to the cable loss.

On 1296, signals will drop by 0.6 dB due to the cable but the noise
floor will increase by 2.4 dB for a total loss of 3dB.

The VSWR of your LNA does not enter the picture.
(Note that the LNA looks into 50 ohms in both cases.)

It is quite another thing when we want to transfer power through
the cable. Then the text-book formulae apply.

73

Leif / SM5BSZ




> my question:
> There are in antenna books and elsewhere
> tables showing the additional cable loss in respect
> to cable loss and SWR ratio.
> 
> Lets take following case:
> My LNA is situated a few meters behind my 
> perfectly (nearly) matched antenna. The SWR of
> the LNA is about 7:1 (return loss ca. 2-3dB) which
> is highly common with MGF1302 preamps.
> The loss of the cable between antenna and LNA
> is 0,6dB.
> According to a.m. table my additional cable loss would be
> ca. 1,3dB.
> Since any loss in front of the LNA must be added
> to the noise figure, the NF will be finally instead of 
> lets say 0,5dB
> then 0,5dB plus 1,3dB = 1,8dB.. Is this right???
> 
> Reply by the experts (to moon-net)
> would be appreciated!
> 
> 73 de Bodo/DL2FCN
> 
> 
> 
> 
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