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Can someone explain how the output of this control system is derived?

Engineering Asked on September 1, 2021

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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

One Answer

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: $$tilde{y}(n) = b_0net(n) - b_0a_1x(n-1) - b_0a_2x(n-2)+b_1net(n-1) - b_1a_1x(n-2) - b_1a_2x(n-3)+b_2net(n-2) - b_2a_1x(n-3) - b_2a_2x(n-4)$$ $$ = b_0net(n)+b_1net(n-1)+b_2net(n-2) - a_1left(b_0x(n-1) + b_1x(n-2) + b_2x(n-3)right) - a_2left(b_0x(n-2) + b_1x(n-3) + b_2x(n-4)right)$$ 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: $$tilde{y}(n) = b_0net(n)+b_1net(n-1)+b_2net(n-2) - a_1tilde{y}(n-1) - a_2tilde{y}(n-2)$$

Correct answer by Petrus1904 on September 1, 2021

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