# Linear Prediction Filter

• May 6th 2009, 09:24 PM
krispiekream
Linear Prediction Filter
• May 7th 2009, 07:04 AM
cl85
Let $\displaystyle \hat{s} = h_1u(n-1) + h_2u(n-2) + h_3u(n-3)$ where $\displaystyle h_i, i=1,2,3$ are the coefficients to be determined.

Then apply the orthogonality principle
$\displaystyle E[(s(n)-\hat{s})u(n-i)] = 0, i = 1,2,3$
Solve for the three coefficients using the 3 equations.

After computing the coefficients, compute
$\displaystyle E[(s(n)-\hat{s})^2]$
for the MMSE.
• May 7th 2009, 04:16 PM
krispiekream
what is the the orthogonality principle?
I couldnt get the 3 equations you mentioned. please give me more details?
• May 7th 2009, 07:54 PM
cl85
You didn't learn about it in class? Do you have a formula involving matrix inversion?

In its simplest form, the Orthogonality Principle states that "the error vector of the optimum estimator is orthogonal to any other possible estimator."

I assumed that you have had some linear algebra and understood the concept of inner product. If two vectors are orthogonal, then its inner product is zero, ie $\displaystyle <\bar{x} , \bar{y}> = 0$.

Here, the inner product of two random variables $\displaystyle X$ and $\displaystyle Y$ is defined to be $\displaystyle <X , Y> = E[XY]$.

Now let the optimum estimator be $\displaystyle \hat{s} = h_1u(n-1) + h_2u(n-2) + h_3u(n-3)$ where $\displaystyle h_i, i= 1, 2, 3$ are the coefficients to be determined. You can think of $\displaystyle u(n-1), u(n-2), u(n-3)$ as three linearly independent vectors(like the cartesian basis $\displaystyle \hat{i},\hat{j},\hat{k}$, although I should add that the u's are not orthogonal like the cartesian basis)

The error of the optimum estimator is thus $\displaystyle s(n) - \hat{s}$.

Now apply the orthogonality principle, and we have
$\displaystyle E[(s(n)-\hat{s}) A] = 0$ where $\displaystyle A$ is any possible estimator.

We have three unknowns to solve for so we need three linearly independent equations. They can be obtained by substituting $\displaystyle A$ with $\displaystyle u(n-1), u(n-2), u(n-3)$ (each of the u's is a possible estimator by itself). So the three equations are

$\displaystyle E[(s(n)-\hat{s})u(n-1)] = 0$
$\displaystyle E[(s(n)-\hat{s})u(n-2)] = 0$
$\displaystyle E[(s(n)-\hat{s})u(n-3)] = 0$

Substitute in the expression for $\displaystyle \hat{s}$, expand out, plug in the values, and solve for the 3 unknown coefficients and you're done.
• May 7th 2009, 08:00 PM
krispiekream