The idea here is that C(s) and D[z] have the same response at the sample times to a unit step Solve for D[z] from L−1(C(s)s1​)∣t=kT​=z−1(D[z]z−1z​) D[z]=zz−1​z(L−1(C(s)s1​)∣t=kT​)