# Exponential weighted moving average time constant

I think it is an easy question but I am stuck.

I want to derive that $$\alpha = -e^{\frac{-T}{\tau}}\,.$$

Can someone provide me with an answer?

[EDIT] Sorry guys, this was really a senseless question. Here are some more informations. The differential equation of the EWMA: $$y(n)=\alpha\cdot x(n)+(1-\alpha)y(n-1)$$ and the corresponding frequency magnitude response: $$H_E(z)=\frac{\alpha}{1-(1-\alpha)z^{-1}}$$ and in the frequency domain for $\omega_s=1 \rightarrow T=1$ : $$H_E(\omega)=\frac{\alpha}{\sqrt{1-2(1-\alpha)\cos(\omega)+(1-\alpha)^2}}$$ $$\tau=\frac{1}{2\pi\omega_{3dB}}$$

Hence $$H_E(\omega_{3dB})=\frac{\alpha}{\sqrt{1-2(1-\alpha)\cos(\omega_{3dB})+(1-\alpha)^2}}\overset!=\frac{1}{\sqrt{2}}\text.$$

I'm looking for $\alpha$, given an $\omega_{3dB}$.

• You will have to explain what $\alpha$ should be. Also, if you ask me, a moving average can't be exponentially weighted -- otherwise, it wouldn't really be an average (by the usual definition of average), but just a low pass filter. – Marcus Müller Jan 16 '16 at 12:48
• "can someone provide me with an answer?": Probably, yes, but you have not asked a question! – Marcus Müller Jan 16 '16 at 12:50
• (This should have been a comment if I had enough reputation.) In the comment to Matt's comprehensive answer, as well as in the original question, @Slev1n has asked about the relation suggesting $\alpha = -e^{-\frac{T}{\tau}}$, while Matt has arrived at the formula $1 - e^{-\frac{T}{\tau}}.$ While it is impossible to have $\alpha = -e^{-\frac{T}{\tau}}$ as $\alpha$ is assumed to be positive, it is possible to have $\alpha = e^{-\frac{T}{\tau}}$ with another interpretation of $\alpha$. In some sources, $\alpha$ is the coefficient before the previous output rather than the current input, so \$\alp – Yury Kartynnik yesterday

If I understood you correctly, you want to compute the value of $$\alpha$$ that results in a specified 3dB cut-off frequency for an exponentially weighted moving average filter. If you square your last equation, you get

$$\frac{\alpha^2}{1-2(1-\alpha)\cos(\omega_c)+(1-\alpha)^2}=\frac12\tag{1}$$

which can be rearranged into the following quadratic equation:

$$\alpha^2+2\alpha(1-\cos(\omega_c))-2(1-\cos(\omega_c))=0\tag{2}$$

with the positive solution

$$\alpha = \cos(\omega_c)-1+\sqrt{\cos^2(\omega_c)-4\cos(\omega_c)+3}\tag{3}$$

So, e.g., for a desired cut-off frequency $$\omega_c=0.2\pi$$, you obtain from $$(3)$$ a value of $$\alpha=0.455886780102867$$. The figure below shows the magnitude of the frequency response of the resulting exponentially weighted moving average filter, from which you can see that the desired cut-off frequency is achieved.

EDIT: The formula for $$\alpha$$ in your question should actually be

$$\alpha=1-e^{-T/\tau},\qquad \tau=1/\Omega_c\tag{4}$$

Note that unlike $$\omega_c$$ in Eq. $$(3)$$, $$\Omega_c$$ in Eq. $$(4)$$ is not normalized by the sampling frequency. So we have $$-T/\tau=-\Omega_cT=-\omega_c$$.

Eq. $$(4)$$ is an approximation, and it comes from applying the impulse invariant transformation to the continuous-time transfer function

$$H(s)=\frac{1}{1+s\tau}\tag{5}$$

which has a 3dB cut-off frequency $$\omega_c=1/\tau$$. Applying the impulse invariant transformation to $$(5)$$ gives

$$H(z)=\frac{T}{\tau}\frac{1}{1-e^{-T/\tau}z^{-1}}\tag{6}$$

Comparing the denominator of $$(6)$$ with the denominator of the discrete-time transfer function of an EWMA filter

$$H(z)=\frac{\alpha}{1-(1-\alpha)z^{-1}}\tag{7}$$

results in the given formula. Note, however, that this is only an approximation. Especially for cut-off frequencies close to Nyquist, the error of formula $$(4)$$ becomes relatively large.

• Hey Matt, thanks for your help. Your formula is correct, but I also found the relation: $$\alpha = -e^{\frac{-T}{\tau}}\,.$$ And now I am wondering how to get from yours to mine... – Slev1n Jan 17 '16 at 10:25
• @Slev1n: I've updated my answer to explain where that approximation comes from. Note that it is not exact, unlike the formula in my answer. – Matt L. Jan 17 '16 at 13:43