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I was trying to solve a problem given in my textbook. I have attached the solution given in the book.

I thought some modifications where needed to it and tried to solve in a different way. I ended up getting absurd results. Where's the mistake in my approach ?

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Your approach is correct, but the problem is that the given response is not the response to a unit step, but to a scaled step. That's why you get \$\alpha y[-1]=-11/2\$ and \$1-\alpha y[-1]=8\$, which is incompatible. If you assume a scaled step \$ku[n]\$, you end up with the following \$\mathcal{Z}\$-transform of the output signal:

$$Y(z)=\frac{k-\alpha y[-1]+\alpha y[-1]z^{-1}}{(1-z^{-1})(1+\alpha z^{-1})}\tag{1}$$

Comparing \$(1)\$ to the \$\mathcal{Z}\$-transform of the given response you get

$$k-\alpha y[-1]=8,\quad \alpha y[-1]=-\frac{11}{2},\quad\alpha=-\frac12$$

from which you obtain

$$y[-1]=11\quad\text{and}\quad k=\frac{5}{2}$$

The values of \$\alpha\$ and \$y[-1]\$ do not change, but now the result is compatible with the given response.

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  • \$\begingroup\$ Thanks alot @MattL !! I tried everything but missed this subtle point. \$\endgroup\$ – Nikhil Kashyap Dec 30 '17 at 19:06
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\$u(n)\$ is the unit step sequence, and is zero for negative \$n\$, hence \$y(-1)=0\$.

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  • \$\begingroup\$ \$y[-1]\$ is the initial condition and is not represented by the given formula for \$y[n]\$, which is only valid for \$n\ge 0\$. If \$y[-1]\$ were zero, then the output would be different. \$\endgroup\$ – Matt L. Dec 30 '17 at 10:28
  • \$\begingroup\$ @MattL, can you suggest something ? \$\endgroup\$ – Nikhil Kashyap Dec 30 '17 at 12:10

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