

Re: Path loss and signal calc...

The equation of losses in free space is:

losses= ((4*pi*d)/lambda)^2  (it's elevated to square)

pi: 3.14159....
lambda: wavelenght

In a simplier equation:

losses (dB)= 32.45 + 20 log f (MHz) + 20 log d (Km)

This give the total losses in dB if you put frequency in Megahertz and
distance in Kilometers.

Then, to calculate the receive power, you have:

P_rx(dBm)= P_tx (dBm) + G_tx (dB) + G_rx (dB) - losses (dB)

(All this without consider other losses than it produces distance)

You can join P_tx + G_tx  as EIRP (PIRE in spanish).

To calculate the minimum recepcion power of you receiver from the SINAD,
you have:

i.e: Minimun voltage on you receiver 0.2 uv
     Imput Impedance 50 Ohms
    (consider that antena impedance and receiver impedance are matched)

P_rx_min=(V*V)/Z= (0.2*10^-12)/50=8*10^-16 watts

in dBm: P_rx_min (dBm)=-121 dBm

If the P_rx is greater than your P_rx_min you can receive something, if
it is not greater you must increase your Pr_rx adding preamps, better
antenna, etc ....

I hope this can help to your questions.

About losses in a ionospheric propagation is difficult to explain in a
message because are included too many factors.


Anthony Bombardiere escribió:

> Hi all,
>   AF9Y calculated the path loss for the Mars probe at -179dbm.
> If the average receiver has a minimum disernable signal of -140dbm.
> How can one hear a signal thats -39db below the receivers noise floor?
>  I would assume that  antenna gain and any pre-amplification
> is taken into account, and the fact that Mike,(AF9Y) used his DSP
> program to graphically display this signal that could not be heard
> by the human ear , but how can one receive a signal that is far
> below that of the receivers noise floor?
>  Is the calculation for this problem as simple as taking the
> preamp gain and antenna gain and adding them to the path loss
> of the signal? If so, assuming a total gain in the system of
> 50db then....
>  -179dbm + 25db(pre-amp) + 25db(antenna gain) = -129dbm ?????
>  Also, if the sensititvity of a receiver is said to be .2 micro
> volts at (10db s+n/n), that would mean that the receiver would
> produce a signal of 10db above the noise for an input of .2 uv.
>  How can one relate this mesurment with  signal strengths
> that are expressed in dbm or decibles as they relate to microvolts?
> That is, how does a signal that is said to be -140dbm relate to
> the measument of .2uv at (10db s+n/n)
>  Its obvious that I'm not sure of my math skills, but I had to
> ask...
>  Thanks , AB2CJ

Raul Gomez Fernandez .-.-.-.-.-.-.-.-. EB4GZO
    Email_1= mailto:raul.gomez@coitt.es
    Email_2= mailto:rgf41@rcua.alcala.es
P.O.Box:55.180 - C.P.:28.080 - Madrid - Spain