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im trying to figure out how to solve this problems for values a=2; actually im stuck, professor told me how to work it, but i cant figure out what to do exactly, we are using partial fractions inverse z transform to solve and this is the problem:

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it is asking for the transient response in unit step

I upload a example he sent to us with a=1, but i dont know how to work it properly to be true.

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(I tried to follow the things on the whiteboard but I couldn't get what is going on, but this is how you would get partial fractions once you find your system \$H(z)\$)

When you have a system with many poles such as $$ H(s) = \frac{s+d}{(s+a)(s+b)} $$ or a discrete one such as
$$ H(z) = \frac{z+d}{(z+a)(z+b)} $$ You can apply the Heaviside method to get the partial fraction expansion. With it, you want to change $$ H(z) = \frac{z+d}{(z+a)(z+b)} = \frac{d_1}{z+a}+\frac{d_2}{z+b}.$$

To do so, you multiply them by zeros which will cancel one of the poles, and evaluate them at \$z\$ value of the cancelled pole, to obtain those coefficients \$d_1, d_2\$. $$ \left[(z+a)H(z)\right]\big\rvert_{z=-a} = d_1,$$ $$ \left[(z+b)H(z)\right]\big\rvert_{z=-b} = d_2,$$ That way, you can find a partial fraction expansion of \$H(z)\$ $$ H(z) = \frac{d_1}{z+a}+\frac{d_2}{z+b}.$$

And since those terms are readily available in a Z transform table, you can convert them to obtain the time-domain of \$H(z)\$ $$ h(n) = \mathcal{Z}^{-1}(\frac{d_1}{z+a})+\mathcal{Z}^{-1}(\frac{d_2}{z+b}).$$

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  • \$\begingroup\$ Thank you so much for taking the time to answer, well, i think i got the partial fraction and evaluation part, but i cant figure out yet what to do with the chance of a=2, as i understand i have to multiply (1/2) in both side of this operation (a)/(s(s+a)=(1-e^at)/(z-1)(z-e^at), but i dont know what to do next with the problem, as i understand have to work with g(z), but dont know how, any advice? \$\endgroup\$ – Caesar Cruez Mar 25 at 1:21

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