1 Star2 Stars3 Stars4 Stars5 Stars (7 votes, average: 4.57 out of 5)
Loading...

 

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.

1 Star2 Stars3 Stars4 Stars5 Stars (7 votes, average: 4.57 out of 5)
Loading...

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 )

Recommended Books:

More Recommended Books at our Integrated Book Store

  • 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

  • http://www.blogger.com/profile/14472055926898539663 mathuranathan

    Yes.. Tuseef. You are right

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

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

  • http://www.blogger.com/profile/00915057980146702637 avais

    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?

  • http://www.blogger.com/profile/14472055926898539663 mathuranathan

    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.

  • Khusbu Mehar

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

    where C is Shanon’s Capacity & BW is bandwidth