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I understand equation (1), but I really don't see the connection between equation (2) and the picture.

For starters, shouldn't the $b_0$ constant be multiplied by

$$net(n)-(a_1 \cdot net(n - 1) + a_2 \cdot net(n-2))$$

Edit: Honestly even an "I'm not sure" would be helpful


1 Answer 1


Well as a fresh start: net(n) is the value before the node, this can be seen by equation (1). The value after the node is the following (lets call this $x$ for simplicity): $$x(n) = net(n) - a_1x(n-1) - a_2x(n-2)$$ Therefore: $$\tilde{y}(n) = b_0x(n)+b_1x(n-1)+b_2x(n-2)$$ Now substituting the first equation into the second one yields the following: $$\tilde{y}(n) = b_0net(n) - b_0a_1x(n-1) - b_0a_2x(n-2)+b_1net(n-1) - b_1a_1x(n-2) - b_1a_2x(n-3)+b_2net(n-2) - b_2a_1x(n-3) - b_2a_2x(n-4)$$ $$ = b_0net(n)+b_1net(n-1)+b_2net(n-2) - a_1\left(b_0x(n-1) + b_1x(n-2) + b_2x(n-3)\right) - a_2\left(b_0x(n-2) + b_1x(n-3) + b_2x(n-4)\right)$$ From here, it can be observed the parts multiplied with $a_1$ and $a_2$ are actually equal to a shifted version of the original equation of $\tilde{y}(n)$! by replacing them with the these shifted outputs, your equation in question will show up: $$\tilde{y}(n) = b_0net(n)+b_1net(n-1)+b_2net(n-2) - a_1\tilde{y}(n-1) - a_2\tilde{y}(n-2)$$

  • 1
    $\begingroup$ An easier way of showing this I think would be by using $z$ such that $x(n+k)=z^k\,x(n)$. Such that $x(n)\,(1+a_1\,z^{-1}+a_2\,z^{-2})=net(n)$ and $\tilde{y}(n)=x(n)\,(b_0+b_1\,z^{-1}+b_2\,z^{-2})$, then solve the first equation for $x(n)$ and substitute that into the second equation. $\endgroup$
    – fibonatic
    Jul 19, 2020 at 14:49

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