(15)n.u(n)−C(15)n−1u(n−1)=δ(n)
(15)n[u(n)−(15)−1C u(n−1)]=δ(n)
[u(n)−(15)−1C u(n−1)]=5n δ(n)
[u(n)−(15)−1C u(n−1)]=δ(n)
u(n)−δ(n)=(15)−1C u(n−1)
u(n−1)=(15)−1C u(n−1)
1=(15)−1.C
C=15=0.2
Alternate Solution:
H(z)=11−15z−1 .....(i)
H(z)−Cz−1H(z)=1
H(z)=11−Cz−1 .....(ii)
Comparing equation (i) and (ii),
C=15=0.2