Find Eb/No : signal to noise ratio

AI Thread Summary
The discussion revolves around calculating the E_b/N_0 for a digital communication system with a given received signal level and noise temperature. The solution provided indicates that E_b/N_0 equals 12 dBW, derived from the received signal level of -151 dBW, effective noise temperature of 1500K, and a transmission rate of 2400 bps. The user attempts to apply formulas related to capacity and thermal noise but struggles with the number of variables. The forum rules emphasize that while hints and guidance can be offered, users must complete their homework independently. The thread highlights the importance of understanding the underlying principles rather than seeking direct answers.
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Homework Statement


If the received signal level for a particular digital system is -151dBw and the receiver system effective noise temperatyre is 1500K, what is E_b/N_o for a link transmitting 2400bps?

solution says:
(Eb/N0) = –151 dBW – 10 log 2400 – 10 log 1500 + 228.6 dBW = 12 dBW

My attempt:
Im trying to use C=Blog_2(1+SNR) and N=kTB
But I have 3 variables and only 2 equations. B's will be canceled but N and SNR(E_b/N_o) are unknown
 
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Thermal noise in Watt= kTb
Thermal noise in dBW(N)=10logk +10logT +10logb
N=-228.6+10log1500+10log2400=-163.037dBW
Eb/N0=S/kTb
(Eb/No)dB=S dBW -N dBW=-151 -(-163.037)
=12.03 dBW
 
Rajeswari Rajasingh said:
Thermal noise in Watt= kTb
Thermal noise in dBW(N)=10logk +10logT +10logb
N=-228.6+10log1500+10log2400=-163.037dBW
Eb/N0=S/kTb
(Eb/No)dB=S dBW -N dBW=-151 -(-163.037)
=12.03 dBW
Welcome to the PF. :smile:

The rules do not allow us to provide solutions to homework problems. We can give hints, ask questions, find mistakes, etc. But the student must do the bulk of the work on their homework assignments. Please keep that in mind in your future posts here.

But since this thread is about 8 years old, and this student likely has their PhD by now, I'll allow your post to remain. :smile:
 
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