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Given is this Wiener filter: enter image description here

From this we take \begin{equation} x[k]-a x[k-1]=v[k] \end{equation}

$v(k)$ is assumed to be a white gaussian noise.

In the textbook it is then stated that

The input $v[k]$ at time $k$ and the output $x[k − 1]$ at time $k − 1$ are statistically independent. Thus, $E\{v[k]x[k − 1]\} = 0$.

My intuition tells me thats wrong. The equation above can also be written as \begin{equation} x[k-1]=\frac{v[k]-x[k]}{-a} \end{equation}

in which I see that $x[k-1]$ depends on $v[k]$ and is therefor not independent of it.

Where is the mistake in my thoughts? I guess I look at it the wrong way. I'd like to build a better intuition for problems like this. They always mention it so easily in these textbooks, and I need hours to verify everything.

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Think it this way; assuming $x[-1] = 0$, then recusively compute the output $x[k]$ for $k \ge 0$ such as

$$\begin{align} x[0] &= v[0] \\ x[1] &= a x[0] + v[1] \\ x[2] &= a x[1] + v[2] = a^2 x[0] + a v[1] + v[2] = a^2 v[0] + a v[1] + v[2] \\ x[2] &= \left( a^2 v[0] + a v[1] \right) + v[2] \\ ... &= ... \\ x[k] &= \left( a^k v[0] + a^{k-1}v[1] + ... + a v[k-1] \right) + v[k] \\ x[k] &= a ~ x[k-1] + v[k] \\ \end{align} $$

As you can see at the bottom two lines, the output at time $k$ uses all input samples up to time $k-1$ and the new input sample $v[k]$. Since $v[k]$ is declared as an independent random process, then all of its random variables $v[0],v[1],v[2],...,v[k]$ are independent from each other. Then the current input $v[k]$ will be independent from the past output $x[k-1]$, as that output does not contain any contributions from the input sample $v[k]$;

So what's the illusion in this equation \begin{equation} x[k-1]=\frac{x[k]-v[k]}{a} \end{equation}

which seems to suggest a dependency between $x[k-1]$ and $v[k]$ ?

If you consider the causal solution of the difference equation as outlined recursively above, then it's seen that the current output sample $x[k]$ contains a contribution from current input sample $v[k]$ and plus all past input contributions from $v[0],v[1],...,v[k-1]$ but then the difference $x[k]-v[k]$ will cancel out the presence of the current input $v[k]$ at the right hand side of the equation. Hence the left hand side $x[k-1]$ will be actually depending on the past inputs alone and not on the current input (which apprently explicitly shows up there but is actually implicitly cancelled by the difference of $x[k]$ and $v[k]$).

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  • $\begingroup$ Thank you for the example. I get your point. It also makes sense when I speak it out. Old outputs do not depend on new inputs. Anyway my algebraic rearrangement looks correct. Is there something which should prevent me from doing this? I mean this was an easy example. But I guess it wont be the last time I did this. Also, do I assume correctly that my assumption would be correct if v itself was not declared as random process? $\endgroup$ – Mr.Sh4nnon Jun 29 '19 at 9:53
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    $\begingroup$ @Mr.Sh4nnon added an explanation $\endgroup$ – Fat32 Jun 29 '19 at 10:26
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    $\begingroup$ Good point! I get it know. Hopefully I'll see this in future problems directly :) $\endgroup$ – Mr.Sh4nnon Jun 29 '19 at 10:34
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    $\begingroup$ @Mr.Sh4nnon thanks. Always consider the solution of the difference equation for assessing the dependencies between various variables... $\endgroup$ – Fat32 Jun 29 '19 at 10:37
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The sequence $x[k]$ only depends on the current input $v[k]$ and on past inputs $v[k-l]$, $l>0$. Consequently, $x[k-1]$ only depends on past inputs $v[k-l]$, $l>0$. And since $E\{v[k]v[k-l]\}=0$ for $l>0$, you also have $E\{v[k]x[k-1]\}=0$.

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  • $\begingroup$ Matt and Fat32 answers were great but another way to think of the illusion is that the equality is true no matter what the two values in the numerator are, so the equality doesn't imply dependence. $\endgroup$ – mark leeds Jun 29 '19 at 15:33

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