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In the complex frequency domain a step input can be represented as R(s) and we have seen that it adds a pole at s 0. The transfer function of a zero order hold block is 1-e-Ts Which despite having a free s in the denominator, I promise you does not adod a pole at s = 0, Prove this fact.

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damain t- u(L-T) L-T L:T -TS Ct) t then

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