Let's say I have a true signal $x$ which is corrupted by noise $n$ such that:

$y = x + n$

I am attempting to use a Weiner filter to estimate the true signal $x$. This estimator is of the form:

$\hat{x} = Gy$

Now let's say I keep the original equation and variables and add a known input bias $c$

$y + c = x + c + n$

The Weiner filter estimator would now take on the form

$\widehat{x+c} = G(y + c)$

The issue with this form is that I am now estimating $x + c$ instead of just $x$. Assuming I know the input bias value $c$, is it "okay" to simply break up the estimated output like the following?

$\widehat{x} = G(y + c) - c$

My confusion primarily comes about breaking up the output of an estimator. I assume because the Weiner filter is linear, this is okay.


1 Answer 1


There is nothing wrong about subtracting the bias. For the Weiner filter, the operation is a matrix multiply, there are no non-linear operations taking place which is reason everything works out. There are two possible ways to do it:

  1. Subtract the bias before the filter. In this case you'll have a Weiner filter, $G_1$, which will produce $\hat{x}$ given $x+n$.

\begin{align} \hat{x}&=G_1(y-c) \end{align}

  1. Subtract the bias after the filter. In this case you'll have a Weiner filter, $G_2$, which will produce $\widehat{x+c}$ given $x+c+n$ (this is the case you presented).

\begin{align} \widehat{x+c}&=G_2(y+c)\\ \hat{x}&=\widehat{x+c}-c \end{align}

Note: Be careful with the notation of $G$. The Weiner filter is constructed using the data so the filter constructed using the $x+n$ data will be different from the filter constructed using the $x+c+n$ data.

  • $\begingroup$ Let's say I take approach 1 where I subtract the bias before the filter and generate G1. Is it possible to reuse filter G1 to estimate (x + c) given measurement (y + c)? $\endgroup$
    – Izzo
    Mar 23, 2021 at 1:21
  • $\begingroup$ @Izzo since you say you know $c$, I'd say that is the better way to do this. Approach #2 "kind of" estimates both $x$ and $c$, well their sum at least. Any estimation will have some error so you'd be introducing un-needed error by including a known factor $c$ in the estimate. To your point, if you want to produce $x+c$, then use $G_1$ to produce $\hat{x}$ and then simply add $c$ as you say. $\endgroup$
    – Engineer
    Mar 23, 2021 at 11:01

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