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From a diagram with input $x(n)$ a summer and three feedback delay taps I get the difference equation...

$$y(n) = x(n)+a_1y(n-1)+a_2y(n-2)+a_3y(n-3)$$

Then I am given values for ai coefficients.

Case1: $a_1=0,a_2=1,a_3=0$

When applying ai values I get

$y(n) = x(n)+y(n-2)$

$\mathcal{Z}$-transform

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

$$Y(z)-Y(z)z^{-2} = X(z)$$

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

$$\frac{Y(z)}{X(z)} = \frac{1}{1-z^{-2}}$$

I’m just wondering if this is correct. If not, I wouldn’t like to be given the answers I would just like to know what I’m doing wrong.

Thanks in advance to anyone who may help.

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Yes, that's correct.

Tip: If you write the difference equation in "standard" from

$$ \sum_i a_i y[n-i] = \sum_i b_i x[n-i] $$

it has a one to one correspondence with the z-transform.

$$H(z) = \frac{\sum_i b_i z^{-i}}{\sum_i a_i z^{-i}} $$

In your example $$y[n]-a_2 \cdot y[n-2] = x[n]$$ yields $$H(z) = \frac{1}{1-a_2 \cdot z^{-2}}$$

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  • $\begingroup$ Thanks for the edit lennon310 I will pay more attention to my formatting going forward. $\endgroup$
    – New2Dsp
    Aug 16 at 12:43
  • $\begingroup$ Thanks for the clarification Hilmar $\endgroup$
    – New2Dsp
    Aug 16 at 12:44

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