----- Original Message ----- From: "Samudra Haque" samudra.haque@gmail.com To: "Amsat-bb" amsat-bb@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