Satellite Link Design: Find Tx for Satellite & Ground Sides

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SUMMARY

The discussion focuses on calculating the transmission power (Tx) for both satellite and ground sides using Shannon's equation. The given parameters are a bit rate of 9600 bits/second and a channel bandwidth of 10 MHz. The user attempted to compute the signal-to-noise ratio (S/N) but arrived at an incorrect value of -31.77 dB. A key conclusion is that the provided bit rate is incompatible with the specified bandwidth, indicating a fundamental error in the initial assumptions.

PREREQUISITES
  • Understanding of Shannon's theorem and its application in communication systems
  • Familiarity with signal-to-noise ratio (S/N) calculations
  • Knowledge of Effective Isotropic Radiated Power (EIRP) concepts
  • Basic principles of satellite communication systems
NEXT STEPS
  • Review Shannon's equation and its implications for bandwidth and bit rate
  • Learn about the relationship between bit rate and channel bandwidth in communication systems
  • Study Effective Isotropic Radiated Power (EIRP) calculations for satellite links
  • Explore common pitfalls in signal-to-noise ratio computations
USEFUL FOR

Students and professionals in telecommunications, satellite communication engineers, and anyone involved in link budget analysis for satellite systems.

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Homework Statement



Given:
Bit rate = C = 9600 bits/second
Channel Bandwidth = B = 10 MHz
Find Tx for satellite side and ground side

Homework Equations



Shannon's equation: C = B*log {1+(S/N)} (log is to the base 2)

The Attempt at a Solution



Well this is how I decided to approach the problem
Use Shannon's equation> calculate (S/N)> account for all the losses> calculate EIRP> calculate Tx
But I seem to be going wrong in the beginning itself
After computation, I calculated (S/N) to be -31.77 dB or -1.77 dBm (which is definitely wrong!)

Could you please point out the mistake? Is it in my computation or my interpretation of the Shannon's equation? Or is my entire approach gravely wrong?
 
Last edited:
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Bit rate = C = 9600 bits/second
Channel Bandwidth = B = 10 MHz

IMHO the given bitrate is not correct. 9,6 kbps carrier can't take 10Mhz of bandwidth.
 

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