I've just one question :

How can I write a model like y = x + w,( with w a white gaussian noise) with a fixed SNR and a noise variance equal to 1. What coefficient may I have before x ?

Thanks !

  • $\begingroup$ What is the amplitude/power of $x$? A standard model is that $y$ is a sample of the output of a (possibly matched) filter, $w$ is a standard Gaussian random variable, and $x$ is a sample of the signal output of the filter. If $|x| = A$, then the SNR is $A$. If your model for $x$ is a unit-amplitude output, then you want to write $y = Ax+w$. Warning: there are different meanings ascribed to SNR depending on the application. Some people like to write $P_e = Q(SNR)$, others $P_e = Q\left(\sqrt{SNR}\right)$, still others $P_e = Q\left(\sqrt{2\cdot SNR}\right)$. All get the same $P_e$ .... $\endgroup$ Commented Aug 26, 2014 at 14:17
  • $\begingroup$ ... because they are using different definitions of SNR. $\endgroup$ Commented Aug 26, 2014 at 14:18

1 Answer 1


If you define the SNR as the ratio of the signal power and the noise power in dB, you have

$$SNR_{dB}=10\log \left(\frac{P_s}{P_w}\right)\tag{1}$$

where $P_s$ is the power of the desired signal and $P_w$ is the noise poiwer. If the noise $w$ has a mean of zero, then $P_w=\sigma^2_w=1$. From (1) (with $P_w=1)$ you get the desired value of $P_s$ for a given value of $SNR_{dB}$:


In order to normalize the signal $x$ such that it has the desired power $P_s$ you first need to know its power $P_x$. Dividing $x$ by $\sqrt{P_x}$ will give you a unity power signal, which can then be multiplied by $\sqrt{P_s}$ in order to obtain the desired SNR:


  • $\begingroup$ Hey! Just another question. How does it work when you work with complex variable ? A coefficient of 2 ? $\endgroup$
    – Henry
    Commented Aug 29, 2014 at 8:50
  • $\begingroup$ @Henry: complex variables will change nothing. You just compute the powers (which are real-valued) and you do the same as before. For computing the powers you need to use the squared magnitude instead of simple squares. $\endgroup$
    – Matt L.
    Commented Aug 29, 2014 at 16:13

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