Suppose that you are adapting $w$ to minimize $\text{E}(y[n]-w[n]*u[n])^2$ where
$$y[n]=h[n]*u[n]+\nu[n]$$
$y[n]$ and $u[n]$ are known and $\nu[n]$ is an additive noise component.
With a long enough FIR filter you can model any linear system $h$ with any accuracy. You can opt for a shorter filter length in expense of limited modelling accuracy. A key parameter is the sampling rate of the signals. If the signals are over sampled unnecessarily then the required length for $w$ increases.
Suppose that we settle for a certain modelling accuracy and we want to find $w$. If all the samples of $u[n]$ are available and $h$ remains constant you can find an estimate of $h$ namely $w$ using a least square estimation. This estimation will have a limited accuracy. But it is the best you can do. This error is referred to as the Least square error which here I (a bit wrongly) consider the same as Mean Square Error (MSE error).
Since $h$ may change, we need to continuously estimate $h$, and that is why we are using the adaptive filters. Let's say that we use the LMS adaptation to find $w$.
Now on top of the MSE error we have extra error owing to the way that the LMS adaptation is performed. This error is called misadjustmentis and it is caused by the limited window where $u[n]$ is used in the adaptation. If the LMS step size $\mu$ approaches zero we are including all sample of $u[n]$ in estimation, but then there is no adaptation. It can be proved (Take a look at any adaptive filters book) that misadjustment is proportional to
- The filter length
- The step size
The other issue is the tracking behavior of the filter. Since $h$ is changing we like to adapt the filter as fast as possible. To increase the adaption speed we can increase $\mu$. To keep the filter stable we should have
$\mu<\frac{1}{\lambda_{\text{max}}}$
where $\lambda_i$ are the eigenvalues of the covariance matrix of $u[n]$.
The situation is worse if the sample of $u[n]$ are correlated (apparantly your scenario). In this case the covariance matrix has different eigen values and the adaptive filter converges with different modes where the time constant $\tau_i$ for each mode will be
$$\tau_i=\frac{1}{2 \mu \lambda_i}>\frac{\lambda_{\text{max}}}{2 \lambda_i}$$
The ratio of largest to smallest eigenvalue $\frac{\lambda_{\text{max}}}{\lambda_{\text{min}}}$ increases as the samples of $u[n]$ become more correlated and the LMS filter becomes slower. Unfortunately the speed cannot be increased idefinitely by increasing $\mu$. Since $\mu$ is limited from above by $\frac{1}{\lambda_{\text{max}}}$.
Back to your case:
It seems that your data is highly correlated that is why removing 14/15 of samples does not change the performance that much.
Throwing away samples of $u[n]$ can help in your case for the following reasons:
- You are decreasing the length of the part that you are adapting. This can decrease the misadjustment.
- You are whitening your data by down sampling it. This decreases the eigen ratio and improves the tracking behavior of the filter.
- Another issue with highly correlated $u[n]$ is the potential divergence of filter at frequencies where $u[n]$ has almost no component. If $u(e^{j\omega_i})=0$ then $h(e^{j\omega_i})$ can go to infinity if the filter length is infinite. The shorter the filter the least likely that this occurs. This problem however can be solved by introducing leakage.
The negative impact of throwing way samples of $u[n]$ is that your are imposing a certain assumption on $w$. By using only $[u(k), u(k-15), u(k-30), u(k-45), u(k-60)]$ you are assuming that $w$ takes a form like:
$$w=[w_0~...~0~w_{15}~0~...0~w_{30}~...~w_{60}]$$
This limits the ability of $w$ to model an arbitrary $h$.