# Lattice structure realization of an all-pole system

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.

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.

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

• 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? Commented Sep 13, 2019 at 2:49
• 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. Commented Sep 13, 2019 at 2:54
• That's right, it is scaled by $\frac{1}{2}$. Commented Sep 13, 2019 at 12:56

Rearrange into a difference equation form and your answer will become clearer!