[amsat-bb] Re: requesting help on a RF link solution (imaginary ka-bandlink!)

[i8cvs]

----- Original Message -----
From: "Samudra Haque" <samudra.haque at gmail.com>
To: "Amsat-bb" <amsat-bb at amsat.org>
Sent: Tuesday, October 27, 2009 11:03 AM
Subject: [amsat-bb] requesting help on a RF link solution (imaginary
ka-bandlink!)

> Hi, amsat-bb
>
> CQ any satellite link budget expert !
>
> I'm trying to do a calculation on my own based upon published specs
> for the NASA MRO Ka-band experiment, but am getting some unexpected
> results for a Ka-band simplex link with Temp=3000K (hypothetical),
> operating with a Signal to Noise ratio (unitless) figure of 1.171
> (representing 4.5 dB eb/no with a data rate of 1 Gbps and a bandwidth
> of 2.4x10^9 Hz)
>
> Question : is 1 gbps not 1x10^9 bps ?
>
> Question : if both antennas are 3m parabolic (both are the same type)
> with 56.4 dBi boresight gain, what would you think the furthest
> distance the link can perform with SNR of 1.171. I have actually used
> a padding of 3 dB Eb/No in my link budget, so am not worried about any
> further signal loss at first (ok, I should be ..) For the exercise, I
> am choosing a 10 Watt estimated output on an arbitrary basis.
>
> So:
>
> P_t = 10W
>
> G_t = 56.4 dBi = G_r , can we assume the same gain for TX and RX on a
> parabolic dish ?
>
> T = 3000K at receiver
>
> SNR = 1.171 required
>
> f=32.2 GHz
>
> B = 2.4E9 Hz, (bpsk, ldpc code 0.5)
>
> DR = 1E9 bps
>
> So, I am puzzled why this link budget says the range with these
> parameters is equal to 4.644 x 10^9 Km -- that seems to be a long
> distance ! What am I not able to conceptualize.
>
 > BTW, I know if I send this out, the answer will come to me soon
> thereafter, but for education, I would like to know where the problem
> in my understanding lies !
>
> Samudra N3RDX

Hi Samudra, N3RDX

If I well understand your question is to know what is the maximum
free space distance at which you can get a S/N ratio of 4.5 dB using
two identical  transmitting and receiving systems having the following
characteristics:

1) Antenna gain for TX and RX = 56.4 dBi at 32.2 GHz

2) Frequency = 32.2 GHz

3) Overall receiving system noise temperature: T = 3000 kelvin

4) Bandwidth of receiving system = 2.4 x 10^9 Hz

5) TX power 10 W

6) Required Signal to Noise ratio S/N at the unknown distance = 4.5 dB

With the above data we first calculate the receiver noise floor Pn = KTB
where:

K = Boltzmann constant = 1.38 x 10^ -23 (Joule/kelvin)
T = Overall System Noise Temperature = 3000 kelvin
B = Bandwidth of receiving system = 2.4 x 10^9 Hz 

Working out the numbers we get the following RX noise floor 

Pn = (1.38 x 10^ -23) x (3000) x (2.4 x 10^9) = 1 x 10^-10 watt

and 10 x  [ log   (1 x 10^-10)] = - 100 dBW or  - 70 dBm
                       10

                                       Link budged calculation

TX power = 10 W =..................+ 40 dBm
TX antenna gain ........................+ 56.4 dBi
                                                     ------------
Transmitted EIRP......................+ 96.4 dBm

Free space attenuation for
61.000 km at 32.2 GHz............- 218.3 dB
                                                     ------------
Received power over isotropic
ant. at 61.000 km distance........- 121.9 dBm

RX  antenna gain.......................+56.4 dB
                                                     -------------
Received power at RX input... - 65.5  dBm
Receiver noise floor.................- 70.0  dBm
                                                    --------------
Received S/N Ratio................. + 4.5 dB

Conclusion:

Using two boresight identical parabolic dishes having each a gain of
56.4 dBi at 32.2 GHz one transmitting with 10 watt and the other one
receiving with a receiving system having a noise temperature of 3000 kelvin
into a bandwidth = 2.4 x 10^9 Hz the free space distance  at which the
signal is received with a  S/N ratio  = + 4.5 dB  is only 61.000 km so that
your hypothetical system is not suitable for the NASA MRO Ka-band
experiment because the distance Earth to Mars is about 1 AU i.e.
1 Astronomical Unit  corresponding to 149 Million/ km 

73" de

i8CVS Domenico