# How do you "split apart" the output of a linear estimator?

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.

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.

• 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)?
– Izzo
Mar 23, 2021 at 1:21
• @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. Mar 23, 2021 at 11:01