# (Eb/N0) Overall calculation

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#### Bram Jester

##### Newbie level 4 Hi Guys,

Just wondering if someone could help me out:

I have the following:

I get 7.19dB

I think my problem is that I have to convert dB to something, but I don't know how do I convert the dB so that I can use it in my equation?

Cheers,

#### afz23

##### Full Member level 3 Use this rule,
10*logX=value in dB,
Where log is to the base 10.

Find X for the values in dB,and substitute in the formula,you will get the answer,
Don't forget to reconvert the answer in dB .

#### vfone Actually you have to convert dB to linear using: linear = 10^(dB/10)
So, 18.4dB = 69.18 linear
and 11.8dB = 15.13 linear
we get (69.18*15.13) / (69.18+15.13) = 12.41 linear
in dB = 10*LOG(12.41) = 10.9 dB

• Bram Jester

### Bram Jester

Points: 2

#### Bram Jester

##### Newbie level 4 I do have another question. Still trying to get my head around Eb/N0.

Calculating Eb/N0 with a C/NO of 73.68dB, a bandwidth (fb) of 75Mbit/s. Is giving me -5.08dB

Is the formula Eb/N0= C/N0-10 LOG fb

Is this right? What am I missing?

#### vfone I think you have to put the channel bandwidth in your equation, which is not the same thing as bit rate.

Eb/No(dB) = C/No(dB) - 10*LOG[Bit_Rate(bps)/BW(Hz)]

http://www.sss-mag.com/ebn0.html

#### Bram Jester

##### Newbie level 4 Thank mate.
I think that formula is actually:
C/N = Eb/No - 10LOG (B/fb)

I figured out what I did wrong to get my negative answer. I made a mistake early on in my Link Budget analysis that gave me an incorrect C/No.

The correct C/No was 106.1dB which in turn gave be an Eb/No of 27.35dB (and an Eb/No overall of 10.57)

Thanks again.

#### vfone Is somehow related to the difference between S/No and C/No.
Very often the published equations looks like:

Eb/No(dB) = S/No(dB) + 10*LOG[BW(Hz)/Bit_Rate(bps)]
Eb/No(dB) = C/No(dB) - 10*LOG[Bit_Rate(bps)/BW(Hz)]

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