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Hello, I have a question above from random material and tried to solve it.

Here, I assume signal power to be S1(dB) and noise power to be N1(dB) at input, and S2(dB) and N2(dB) at output.

So below is what I started from "S1-N1=S2-N2"

Step 1 :

S1-N1=S2-(2*N1)

N2 becomes (2*N1) because question said noise power at output is twice as much as input.

Step 2 :

S1-N1=(S1-10)-(2*N1)

S2 now becomes (S1-10) because question also said there is attenuation of the channel which is 10dB, and it also says "The channel does not attenuate the noise", which means that it only attenuate signal power instead. So I subtract 10 from S1 and set it to S2.

Step 3 :

Then I get N1=-10, then we know N2=-20 (because Question said N2 is twice as N1)

Step 4 :

Now, question want me to determine SNR in DB at output. So, here I used SNR(dB) formula, which is "SNR(dB)=10log(SNR) where base is 10".

10log(-20) = 13.01 dB

And I guess I am done.

So my question is, is this even correct way to solve this question? And did I get the answer correct?

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  • 1
    \$\begingroup\$ "N2 becomes (2*N1) because question said noise power at output is twice as much as input." might want to think about what doubling power looks like when figuring in dB. \$\endgroup\$ – Chris Stratton Jun 21 '17 at 6:30
  • \$\begingroup\$ I agree with Chris. That sentence does not tell you that N2[dB] = 2 * N1[dB]. Instead, N2[W] = 2 * N1[W]. At least this is what I understand. \$\endgroup\$ – Rohat Kılıç Jun 21 '17 at 6:45
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If you are new to these types of problems, I would recommend converting all original dB terms to straight power ratios and then converting back to dB form in the end.

Start with the original SNR of 25 dB. To convert this to a power ratio, use 10 (25/10) which is 316.2. This means the signal contains 316.2 times more power than the noise.

Convert the 10 dB power loss figure to a power ratio the same way to find it is 10.

Now you can walk through the gains and losses as straight multiplication. The signal goes down by a factor of 10 so it is now 31.62. The noise doubles so it goes up to 2 (it was the 1 denominator of the 316.2).

So we have a new SNR of 31.62/2 or 15.81. Convert this back to dB by the formula 10*log 10 (power ratio) and you find that the new SNR is ~ 12 dB.

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