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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 $\omega_{3dB}$.

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    $\begingroup$ 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. $\endgroup$ – Marcus Müller Jan 16 '16 at 12:48
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    $\begingroup$ "can someone provide me with an answer?": Probably, yes, but you have not asked a question! $\endgroup$ – Marcus Müller Jan 16 '16 at 12:50
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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.

enter image description here

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.

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  • $\begingroup$ 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... $\endgroup$ – Slev1n Jan 17 '16 at 10:25
  • $\begingroup$ @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. $\endgroup$ – Matt L. Jan 17 '16 at 13:43

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