[Solved] A Satellite Communication Link Has Uplink C/No Of 50 DB-Hz A

Home » C/no Db-hz » [Solved] A Satellite Communication Link Has Uplink C/No Of 50 DB-Hz A

Maybe your like

Concept:

In satellite communication:

\(\frac{1}{{{\left( \frac{C}{N} \right)}_{total}}}=\frac{1}{{{\left( \frac{C}{N} \right)}_{uplink}}}~+~\frac{1}{~{{\left( \frac{C}{N} \right)}_{downlink}}}\)

Where C/N is the carrier to noise ratio;

\(\Rightarrow {{\left( \frac{C}{N} \right)}_{Total}}=\frac{{{\left( \frac{C}{N} \right)}_{U}}.{{\left( \frac{C}{N} \right)}_{D}}}{{{\left( \frac{C}{N} \right)}_{U}}+{{\left( \frac{C}{N} \right)}_{D}}}\)

Calculation:

Given:

\({{\left( \frac{C}{N} \right)}_{U}}=50~dB=10\log {{\left( \frac{C}{N} \right)}_{U}}\)

\({{\left( \frac{C}{N} \right)}_{U}}={{10}^{5}}\)

Similarly,

\({{\left( \frac{C}{N} \right)}_{D}}47dB=10\log {{\left( \frac{C}{N} \right)}_{D}}\)

\( {{\left( \frac{C}{N} \right)}_{D}}={{10}^{4.7}}\)

\(10\log {\left( {\frac{C}{{{N_0}}}} \right)_{OVERALL}} = 10\log \left( {\frac{{{{10}^{9.7}}}}{{{{10}^5} + {{10}^{4.7}}}}} \right) = 45.23\;dB\) Download Solution PDF Share on Whatsapp

Tag » C/no Db-hz