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Given an IIR system with the system function,

$$ H(z) = \frac{1}{2 + 1.4z^{-1} + 1.8z^{-2}} $$

I need to draw the lattice structure realization.

I have drawn the lattice structure for $\frac{1}{1 + 0.7z^{-1} + 0.9z^{-2}}$, which is shown in the figure below.

enter image description here

Now since $H(z) = \frac{1}{2(1 + 0.7z^{-1} + 0.9z^{-2})} $, should I multiply the signal $x(n)$ by $2$ or $1/2$ before feeding to the system? My intuition is that I should multiply with $\frac 1 2$, but then since $2$ appears as the multiple in the denominator polynomial, I got confused.

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As mentioned by LamebrainEddy, you should find the difference equation of your system. To do so, recall that:

$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{2 + 1.4z^{-1} + 1.8z^{-2}}$

So:

$Y(z)(2 + 1.4z^{-1} + 1.8z^{-2}) = X(z)$

$2Y(z) + 1.4z^{-1}Y(z) + 1.8z^{-2}Y(z) = X(z)$

And if you anti-transform you get:

$2 y[n]+1.4 y[n-1]+1.8 y[n-2]$ = x[n]

Finally, solving for $y[n]$:

$y[n]=-0.7y[n-1]-0.9y[n-2]+0.5x[n]$

From this equation you can notice that x[n] is scaled by 0.5

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  • $\begingroup$ Is there a method for reducing a difference equation directly to the Lattice structure? Only method I know is to recursively reduce the polynomial in the denominator and finding the coefficients. I mean whenever I'm given a difference equation like this, I first take Z transform and convert it into H(z) form and proceed thereafter. Am I missing something? $\endgroup$ – bikalpa Sep 13 '19 at 2:49
  • $\begingroup$ But after getting the difference equation I think the answer to my question becomes apparent. Since $x(n)$ has been scaled by $\frac 1 2$ in the difference equation, I should multiply $x(n)$ by $\frac 1 2$ before feeding into the structure I drew in the question, right? Correct me if I'm wrong. $\endgroup$ – bikalpa Sep 13 '19 at 2:54
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    $\begingroup$ That's right, it is scaled by $\frac{1}{2}$. $\endgroup$ – Axel Mancino Sep 13 '19 at 12:56
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Rearrange into a difference equation form and your answer will become clearer!

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