Relation between Eb/N0 and SNR

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If you are looking for a function to generate a signal with given SNR, please see this latest post.

I have seen a lot of questions being posted in several forums on the relationship between Eb/N0 and SNR. Like, whether SNR and Eb/No are same or not. In fact , the relationship is very easier to understand.

Lets start with the basic equation and try to verify its authenticity.

$$\frac{S}{R_{b}}=E_{b}\;\;\;\;\; \rightarrow (1)$$

where,
\(R_b\) = bit rate in bits/second
\(E_b\) = Energy per bit in Joules/bit
\(S\) = Total Signal power in Watts

We all know from fundamental physics that Power = Energy/Time. Using SI units, lets verify the above equation.

$$ \frac{S}{R_{b}}=E_{b}$$

$$ \frac{Watts}{\frac{bits}{second}} = \frac{Joules}{bit}$$

$$ \Rightarrow Watts = \frac{Joules}{second}$$

This verifies the power, energy relationship between \(S\) and \(E_b\). Now, introducing the noise power \( N_{0}\) in equation (1)

$$ \Rightarrow \frac{Eb}{N0} = \frac{S}{ \left( Rb*N0 \right )}$$

$$\Rightarrow SNR = \frac{Rb*Eb}{N0}\;\;\;\;\; \rightarrow (2)$$

This equation implies that the SNR will be more than \( \frac{Eb}{N_{0}}\) by a factor of \( R_{b}\) (if Rb > 1 bit/second)

Increasing the data rate will increase the SNR, however , increasing \(R_b\) will also cause more noise and noise term also increases ( due to ISI – intersymbol interference , since more bits are packed closer and sent through the channel).

So we cannot increase SNR by simply increasing \(R_b\). We must strike a compromise between the data rate and the amount of noise our receiver can handle.

Note:

Actually,

$$ \frac{S}{N} = \frac{\left( R_b*E_b \right)}{B*N_0}$$

where B is the noise bandwidth. Here, I have taken an unity noise bandwidth (ie. B=1) for simplification.

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See also:

[1] Colored Noise Generation in Matlab
[2] Sampling Theorem – Baseband Sampling
[3] Sampling Theorem – Bandpass or Intermediate or Under Sampling
[4] Window Functions – An Analysis
[5] FFT and Spectral Leakage
[6] Raised Cosine Filter
[7] Moving Average Filter ( MA filter )

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  • Jayme Milanezi Jr.

    suppose I have a digitalized signal ‘Sig’ with Fs snapshots over a time
    symbol T. Suppose that this signal is free of noise. Let us now imagine another
    vector with Fs data, which is my noise vector ‘Noi’, with std. ‘NL’, so that Noi =
    NL*randn(1,Fs). Consider time symbol T = 10e-6 s = 1 us. How can I
    calculate Eb/N0 is this case for Signal = SIG+NOI? I’m working with
    FSK-4, and I’m not sure about the Eb/N0 with regards to ‘NL’. Thanks in
    advance.

  • Khusbu Mehar

    Can anyone prove and explain? SNR= (Eb*C) / (No*BW)

    where C is Shanon’s Capacity & BW is bandwidth

  • Yes it is valid for bandpass signals too. Only difference is -> Substitute W by noise equivalent bandwidth. If you modulate with a carrier, then you have to take the bandwidth (difference between Max modulated frequency and min modulated frequency) into consideration.

  • Is the relation (S/N=(Eb/No)*(R/W))valid for band pass signals?
    For Example BPSK signal modulated at carrier frequency 8KHz, bit rate is 1KHz. The signal is sampled at 96KHz.
    Can any body suggest what values of R and W should be used to calculate S/N for given values of Eb/N0?

  • Yes.. Tuseef. You are right

    S/N = (R/W)*Eb/N0

    In the above post W is taken as 1 to simplify the explanation

  • Tauseef ahmad

    as much i know about this relation is that Eb/No = S/N(W/R)
    where w is the bandwidth and R is the data rate and S as we know is the signal power

    we use E/No to study the DCS. that which system is working best on the basis of E/No