This question is in continuation to the one asked earlier Help in understanding the formula of Signal-to-Noise-Ratio (SNR) - Part 1

Consider a model $$y[n] = x[n] + w[n]$$

Signal means the desired information which in this case is x and y is the output noisy observation.

Question A

Case 1: When noise is AWGN $ \sim C\mathcal N(0,2\sigma^2_w)$; $y$ and $x$ is also complex valued i.e., $w \sim CN(0,2\sigma^2_w), x \sim CN(0,2\sigma^2_x)$

Then would SNR = $\frac{E[x^2]}{2\sigma^2_w} = \frac{2\sigma^2_x}{2\sigma^2_w} $ ?

Case 2: AWGN $ \sim \mathcal N(0,\sigma^2_w)$; $y$ and $x$ are real valued. Then would SNR = $\frac{E[x^2]}{\sigma^2_w}$?

These topics are not clearly mentioned in the Answers given for the other Question. The Answers explain what is SNR but my confusion remains that if the signal and the noise are complex valued, then would there be any change in the SNR due to the real and imaginary component? In the denominator of the formula would there be a 2? That is why I have asked this here.

Question B: https://en.wikipedia.org/wiki/Signal-to-noise_ratio shows 2 formulae for SNR. In many research articles, the formula of SNR is not used, instead EbN0 is used. I am using Least squares estimator to estimate channel coefficients and then trying to plot the mean square error (in db) of the channel estimates with varying SNR. NExt, I want to plot the bit-error rate (using simulation) for bpsk and say QAM modulation. I used SNR for the plot of mean square error. So, should I use SNR for Bit error rate? OR is there no such rule that both the plots should be using SNR and not EbNo formula?


1 Answer 1


Question A: For the considered system model $$ y[n] = x[n] + w[n], $$ as already pointed out in your previous question, it holds: $$ E\{ y[n]y[n]^* \} = E\{ (x[n]+w[n])(x[n]+w[n])^* \} \\= E\{ x[n]x[n]^* \} + E\{ x[n] \}\underbrace{E\{ w[n]^* \}}_{= 0} + \underbrace{E\{ w[n] \}}_{= 0}E\{ x[n]^* \} + E\{ w[n]w[n]^* \} \\= E\{ x[n]x[n]^* \} + E\{ w[n]w[n]^* \}.$$ As a result, your $SNR$ will always be$$ SNR = \frac{E\{ x[n]x[n]^* \}}{E\{ w[n]w[n]^*\}}. $$ So for both of your cases the $SNR$ you specify is correct. It just depends on your definition of $x[n]$ and $w[n]$.

Question B: For your application I would recommend to use $\frac{E_b}{N_0}$, because the energy per bit depends on the modulation scheme (BPSK, QAM etc.). Therefore comparing the $BER$ in terms of $\frac{E_b}{N_0}$ will be more fair.

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  • $\begingroup$ Thank you so much for your answer. Just to confirm, if my understanding is correct or not, the SNR for any kind of input signal - real valued or complex, for the model $y = x + w$ (y is the output and x is the desired signal (input)) is $\frac{\sigma^2_x}{\sigma^2_w}$ ? $\endgroup$
    – Ria George
    Jul 12, 2017 at 21:48
  • $\begingroup$ Yes, that is correct, if $E\{xx^*\} = \sigma_x^2$ and $E\{ww^*\} = \sigma_w^2$. $\endgroup$
    – Enzo
    Jul 13, 2017 at 12:09
  • $\begingroup$ Thank you, lastly here the operator $\ast$ denotes the conjugate or the complex conjugate? $\endgroup$
    – Ria George
    Jul 14, 2017 at 20:11
  • 1
    $\begingroup$ Here $( \cdot )^*$ denotes the complex conjugate. If $x[n]$ and $w[n]$ are real, then you can ignore it. For a more general real or complex random vector $x[n]$ you can generalize the result to $E\{ x[n] x[n]^H \}$, where $( \cdot )^H$ denotes the complex conjugate transpose. $\endgroup$
    – Enzo
    Jul 16, 2017 at 12:33

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