Thursday 19 January 2017

signal - Solving Difference Equation Using Z-Transform


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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 ?



Answer



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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