0
$\begingroup$

I have two signals: $x(n)$ is a white-noise signal with a given variance, $y(n)$ is the sum of white noise plus a sum of sinusoids. $$ x(n) = v_1(n) $$ $$ y(n) = \sum_{i=1}^{q} [ a_i \cos(\omega_i n) + b_i \sin(\omega_i n) ] + v_2(n) $$ $$ v_{1,2}(n) \sim W.N.(0,\sigma^2) $$

And that is how I generate the two signals on Matlab. Now I'd like to normalise $y(n)$ so that its power is the same as $x(n)$, how can I do that?

$\endgroup$
0
$\begingroup$

Multiply y(n) by std(x)/std(y)

$\endgroup$
  • $\begingroup$ To normalize power, shouldn't it be the ratio of variances? $\endgroup$ – Omegaman Jan 9 '16 at 3:47
  • $\begingroup$ You would think so, but no. $\endgroup$ – CMDoolittle Jan 9 '16 at 8:14
  • $\begingroup$ Right, units... It would be $sqrt(\text{var}(x)\cdot y(n)^2 / \text{var}(y))$, which simplifies to std(x)/std(y)... $\endgroup$ – Omegaman Jan 10 '16 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.