I'm adding this as a corollary to Olin's post, in case an exact formula for attenuation is needed.

That basic IIR is derived from the exponential moving average, and its impulse response is:

$$h[n]=\alpha x[n]+(1-\alpha)x[n-1]$$

which translates into this transfer function:

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

This can be used to determine analytically the attenuation at a particular frequency, by simply substituting $z^{-1}=\text{e}^{-j\Omega}$:

\begin{align*} H(j\Omega)=\frac{\alpha}{1-(1-\alpha)\text{e}^{-j\Omega}} \\ |H(j\Omega)|=\frac{\alpha}{\sqrt{2(\alpha-1)(\cos(\Omega)-1)+\alpha^2)}}\tag{1} \end{align*}

The -3 dB point can also be calculated from $(1)$:

\begin{align*}|H(\Omega)|^2=\frac12=\frac{\alpha^2}{2(\alpha-1)(\cos(\Omega)-1)+\alpha^2}\quad\Rightarrow \\ 2(\alpha-1)(\cos(\Omega)-1)+\alpha^2=2\alpha^2\quad\Rightarrow \\ \cos(\Omega)-1=\frac{\alpha^2}{2(\alpha-1)}\quad\Rightarrow \\ \Omega=\arccos\left(1+\frac{\alpha^2}{2(\alpha-1)}\right)\end{align*}

This can be easily tested with any simulator. As an example, for $\alpha=0.4$, the attenuation at $0.37\frac{f_s}{2}$ and the -3 dB point are:

\begin{align*} |H(0.37\pi)|=\frac{0.4}{\sqrt(2(0.4-1)(\cos(0.37\pi)-1)+0.4^2)}=0.4255746548210215 \\ f=\frac{1}{\pi}\arccos\left(\frac{0.4^2}{2(0.4-1)}+1\right)=0.1662579714811903 \end{align*}

(where I used a normalized [0...Nyquist])

posted about 4 years ago

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4y ago by Mithrandir24601‭

a concerned citizen‭

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