----- Original Message ----- From: "Samudra Haque" [email protected] To: "Amsat-bb" [email protected] 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