# oversampled coefficient for existing exponential smoothing

Say I have an exponential smoothing for certain $\Delta t$, $t_{i+1} = t_i + \Delta t$. In this sampling, I choose a particular $\alpha$ to filter signal $z_i$ like $$v_1 = z_0 \\ v_{i+1} = \alpha\:z_i + (1-\alpha)\:v_i$$

and we have some sort of agreement as to what $\alpha$ to use.

Now, someone comes along and tells me that I am able to get, say, five times more samples in the same $\Delta t$ period. I need to modify the smoothing factor $\alpha$. It seems to me, intuitively, that the new factor, say $\alpha'$, that has the same discount effect for previous values but with five times more iterations should be $$(1-\alpha')^5 = 1-\alpha$$

Is this correct? Is there any principled approaches to justify this?

Note: The formula above seems sensible in the sense that the original asymptotic variance (for a constant model) $$\mathbb{V}ar[v] \to \frac{\alpha}{2-\alpha} \sigma^2$$ becomes $$\mathbb{V}ar[v'] \to \frac{\alpha'}{2-\alpha'} \sigma^2$$ which, under the equation above, is actually smaller (I checked this). This makes sense as more data would give more precision for $v_t$ as estimator of a constant $z_t = k + \epsilon$, where $\epsilon \sim \mathcal{N}(0,\sigma)$.

I think it's instructive to take a look at the impulse responses, taking absolute time into account. I'm assuming the following recursion:

$$y_n=\alpha x_n+(1-\alpha)y_{n-1}\tag{1}$$

The impulse response corresponding to the recursive system described by (1) is

$$h_n=\alpha(1-\alpha)^n$$

Assume now that we have a discrete-time system with sample interval $T$, and another one with sample interval $T^{\prime}=T/L$. Let $\beta=1-\alpha$, and let $\alpha^{\prime}$ and $\beta^{\prime}$ be the constants of the system with sample interval $T^{\prime}$. In order for the two systems to have the same time constant and the same scaling we require (considering a point in time $t_n=nT=nLT^{\prime}$)

$$\alpha\beta^{n}=\alpha^{\prime}{\beta^{\prime}}^{nL}\tag{2}$$

From (2) we see that

$$\alpha=\alpha^{\prime},\quad \beta=\beta^{\prime L}\quad \text{(i.e. \beta^{\prime}\neq 1-\alpha^{\prime})}$$

In this case the impulse response of the system with the higher sampling rate is simply an interpolated version of the other system's impulse response. Note that the condition on $\beta^{\prime}$ is the one that you came up with intuitively, and it is also the more important of the two because it affects the time constant of the system. The choice $\alpha=\alpha^{\prime}$ only affects the scaling and in this sense is secondary. Choosing $\alpha^{\prime}=1-\beta^{\prime}$ is of course also possible and just changes the scaling.

• View at the impulse response, great idea! – carlosayam May 29 '14 at 0:59