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Sorry my poor knowledges about advanced math.

Consider this post that talks about biquad filters.

At the beginning of the post the author shows the filter's transfer equation is

$$ H(z) = \frac{Y(z)}{X(z)} = \frac {b_0 + b_1 z^{-1} + b_2 z^{-2}} {1 + a_1 z^{-1} + a_2 z^{-2}} \tag{1} $$

a couple of lines below that, the author shows an equation, in terms of discrete values, that can be used easily to create a program to process discrete signals in an array...

$$ y[n] = b_0 x[n] + b_1 x[n-1] + b_2 x[n-2] - a_1 y[n-1] - a_2 y[n-2] \tag{2} $$

what magic is involved to convert one equation into the other?

Duck
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1 Answers1

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For integer $m$, $z^{-m}$ is the operator that takes $x_r$ and $y_r$ to $x_{r-m}$ and $y_{r-m}$, respectively. So (multiplying by the denominator of the RHS) $$ (1+a_1z^{-1}+a_2z^{-2}) Y(z) \Longrightarrow y_n +a_1 y_{n-1} + a_2 y_{n-1} $$ and similarly multiplying the numerator of the RHS by the denominator of the LHS.

This yields precisely the equation for $y_n$, after moving all but the $y_n$ term to the right of the equal sign.

Mark Fischler
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