Having a signal X(n) with variance "var" and a known filter H(n), what will be the variance of the filtered signal Y(n), where Y(n) = X(n) * H(n)?
For a FIR digital filter :
var(Y) = Sum[ Sum[ Ci * Cj * Covar(i,j), {i,1,2,...,n} ], {j,1,2,...,n} ]
where n is the number of coefficients of the FIR, Ci the i-th coefficient, and COV the covariance matrix defined as: Covar(i,j) = , ( is mean value).
note.1. If the stochastic components of the signal (i.e. noise) is stationary then ==
note 2. If the input signal X is also standardized (i.e. =0 and var(X)=1) then Covar(i,j) is the autocorrelation function ACF(|i-j|) which depends only on the absolute value of the difference of indices i and j.
Dear Abdelhak,
You can return to the definition of the variance to calculate it for any sequence.
If we have a y[n] sequence with sufficient length N we can calculate the the average value as y[n] av = Sum y[n] from n=0 to N,
The standard deviation sigma y square = Sum [ y[n] - y[n]av]^2 /N FOR n=0 to N.
y[n] is calculated by convolution of h[n] and x[n]. where x[n] is the random sequence with the required variance.
This is a primitive method but works for all cases.
wish you success
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