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I and my friend have different answer for this block diagram

enter image description here

Mine: $$y[n] = -\frac23x[n] + x[n-1] -\frac12[n-2]$$

Hence

$$H(z) = \frac{1}{-\frac23 + z^{-1} -\frac12z^{-2}}$$

My friend: $$q[n]=x[n]-\frac12q[n-1]$$ $$y[n] = -\frac23x[n]+q[n-1]$$ $$Y(z) = \frac23X(z)+z^{-1}Q(z)=\frac23X(z)+z^{-1}\frac{X(z)}{1+\frac12z^{-1}}$$

Hence

$$H(z)=\frac{Y(z)}{X(z)}=\frac23+\frac{z^{-1}}{1+\frac12z^{-1}} = \frac{4z^{-1}+2}{3+\frac32z^{-1}}$$

I don't know how can he come to that result but I'm not sure about mine either

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  • $\begingroup$ Your formula does not make sense, please correct it (what is $\frac12 [n-2]$?) How can there be a term $n-2$ if you only have one delay element? Furthermore, your difference equation doesn't match your transfer function. $\endgroup$
    – Matt L.
    Commented Dec 18, 2015 at 8:19

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As pointed out in the other answers, your friend's approach is correct. However, his end result for $H(z)$ isn't.

With $q[n]$ the input signal of the delay element, you get $q[n-1]$ at its output, and the given difference equation for $q[n]$ follows easily. The output $y[n]$ can then be written in terms of $x[n]$ and $q[n-1]$. In the $\mathcal{Z}$-transform domain you get algebraic equations in terms of $X(z)$, $Q(z)$, and $Y(z)$, from which you can eliminate $Q(z)$, leaving you with an expression for $H(z)=Y(z)/X(z)$:

$$H(z)=\frac{Y(z)}{X(z)}=-\frac23+\frac{z^{-1}}{1+\frac12 z^{-1}}=-\frac23\frac{1-z^{-1}}{1+\frac12 z^{-1}}\tag{1}$$

So it was basically just a sign error that messed up the end result.

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Your friend is correct. You have a feedback loop inside the block diagram so you need to introduce another state variable q[n], which in this case is the signal right after the first summing mode. You could also choose the one after the delay block but the result would be the same.

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Aukxn: Your friend represented the output of the first adder as $q[n]$ and then he wrote his $q[n]$ and $y[n]$ equations. The center part of his $Y(z)$ equation is the z-transform of his $y[n]$ equation. In that center part he has a $Q(z)$ term. He replaced that $Q(z)$ term with the z-transform of his $q[n]$ equation in order to produce the right side of his $Y(z)$ equation. Next he wrote $H(z)=\frac{Y(z)}{X(z)}$ as the sum of two ratios. Finally he put the numerators of the two ratios over a common denominator to produce the right side of his $H(z)$ equation. I'm willing to bet a pint of pale ale that your friend's $H(z)$ expression is correct.

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    $\begingroup$ Would you like to pay via PayPal? $$H(z)=-\frac23\frac{1-z^{-1}}{1+\frac12 z^{-1}}$$ $\endgroup$
    – Matt L.
    Commented Dec 18, 2015 at 13:57
  • $\begingroup$ Ah, yes. Friend dropped a minus sign in his Y(z) equation. Good catch Matt L. The score is now Lyons 2, Matt L. 1. $\endgroup$ Commented Dec 19, 2015 at 9:45
  • $\begingroup$ Does your score count the number of errors? Where is mine? $\endgroup$
    – Matt L.
    Commented Dec 20, 2015 at 19:31

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