Multiple Access Technique (SATELLITE)

[jeejou]

1. Consider a transponder with a 56 MHz bandwidth and an available single carrier C/No value of 97.5 dB Hz. Using a single channel per carrier – voice activated system (assume ∝ = 0.35) how many channels of QPSK modulated PCM can be sustained? Allow 3 dB loss for intermodulation and 10% for guard bands. Ignore companding.

For the same transponder, consider the case of two video carriers, each frequency modulated with an NTSC video signal. Calculate the video SNR. Allow 2.0 dB for intermodulation loss in C/No below the single carrier value.




Can someone tell me how to solve this question? What multiple access scheme should I use? Where can I find the formula for this question? Please help. I am clueless.

 

[KataKoniK]

Hello,

To solve this question, you will need to use the following formula:

C/No = 10log10(∑Ci/2kB) - I - E - α

Where:
C/No = Carrier-to-Noise ratio
Ci = Individual carrier power
kB = Boltzmann's constant (1.38 x 10^-23 J/K)
I = Intermodulation loss
E = Guard band loss
α = System loss factor (in this case, 0.35)

For the first part of the question, you will need to calculate the total bandwidth required for a single channel. This can be done by adding the 3 dB loss for intermodulation and 10% for guard bands to the 56 MHz bandwidth. This gives a total bandwidth of 62.08 MHz.

Next, you will need to calculate the total power required for a single channel. This can be done by using the formula:

Pt = (C/No + I + E + α) x 2kB x B

Where:
Pt = Total power required
B = Total bandwidth (62.08 MHz in this case)

Using the given C/No value of 97.5 dB Hz, we can calculate Pt to be 1.58 x 10^-16 W.

Now, to calculate the number of channels that can be sustained, we need to divide the total available power (1.58 x 10^-16 W) by the power required per channel (Pt). This gives us a total of 7.78 x 10^6 channels.

As for the multiple access scheme, you can use Frequency Division Multiple Access (FDMA) for this system.

For the second part of the question, where two video carriers are considered, you will need to use the following formula to calculate the video SNR:

SNR = 10log10((C/No + I + E + α) x 2kB x B/V)

Where:
SNR = Signal-to-Noise ratio
V = Video bandwidth (6 MHz for NTSC)

Using the given C/No value of 97.5 dB Hz, and taking into account the 2.0 dB intermodulation loss, the SNR for the two video carriers can be calculated to be 36.4 dB.

I hope this helps you solve the question. Good luck!

 

[piercas]

I can help you understand the concept and provide guidance on how to approach this problem. Firstly, the multiple access technique being discussed here is the satellite communication system, which allows multiple users to share the same frequency band for transmission. This technique is used in satellite communication to efficiently use the limited bandwidth available for communication.

Now, let's break down the given information and understand the problem. We have a transponder with a bandwidth of 56 MHz and a single carrier C/No value of 97.5 dB Hz. This means that the transponder can handle a total of 97.5 dB of noise power per Hz of bandwidth. The value of α is given as 0.35, which represents the voice activity factor, indicating that only 35% of the time the channel will be in use for voice communication.

To determine the number of channels that can be sustained, we need to use the formula for the channel capacity of a satellite communication system, which is given as:

C = W log2 (1+ C/No)

Where C is the channel capacity, W is the bandwidth, and C/No is the carrier-to-noise ratio. In our case, we have W = 56 MHz and C/No = 97.5 dB Hz. Substituting these values in the formula, we get:

C = 56 x 10^6 log2 (1+ 10^(97.5/10))

C = 56 x 10^6 log2 (1+ 3.98 x 10^9)

C = 56 x 10^6 log2 (3.98 x 10^9)

C = 56 x 10^6 x 31.94

C = 1.79 x 10^9 bps

Now, we need to consider the loss due to intermodulation and guard bands. Intermodulation occurs when multiple signals are transmitted on the same frequency band, resulting in distortion and noise. For this problem, we are assuming a 3 dB loss due to intermodulation, which means that the effective C/No value will be 94.5 dB Hz (97.5 dB Hz - 3 dB). We also need to allocate 10% of the bandwidth for guard bands, which means we can only use 90% of the total bandwidth. Therefore, the effective bandwidth will be 0.9 x 56 MHz =