# How can this problem be solved with partial fractions? (Zero order holder with g(s) block function)

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:

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}).$$