Z-transform of unit step sequence
Solution :
The definition of unit step sequence is \[u_n=1 if n=0,1,2,3,….0 if n=-1,-2,-3,…\] We know that Z-transform of a signal yn is given as \[Z\left(y_n\right)=\sum_{n=0}^{\infty}{y_nz^{-n}}\] Therefore Z-transform of unit step sequence is \[Z\left(u_n\right)=\sum_{n=0}^{\infty}{u_nz^{-n}}\] using definition of unit step sequence \[Z\left(u_n\right)=1+z^{-1}+z^{-2}+z^{-3}+\ldots.\] This is an infinite series whose constant ratio is z^−1. Hence the sum of the first N terms is \[S_N=1+z^{-1}+z^{-2}+z^{-3}+\ldots.+z^{-(N-1)}\] \[S_N=\frac{1-z^{-N}}{1-z^{-1}}\] If |z^-1|<1 then SN converges to 1/(1-z^-1) as N tends to infinity. Therefore Z-transform of unit step sequence is \[Z\left(u_n\right)=\frac{1}{1-z^{-1}} \]