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Find the Initial and final Halves, x(z)=1+z−11−0.25z−2
1 Answer
written 2.3 years ago by |
Solution:
Initial Value,
x(0)=ltz→∞x(z)=ltz→∞1+z−11−0.25z−2
=1+∞−11−0.250−2=1+01−0=1
x(0)=1
Final Value,
x(∞)=ltz→1(1−z−1)×(z)=ltz→1(1−z−1)(1+z−1)1−0.25z−2
=ltz→11−z−21−0.25z−2
=ltz→1z2−1z2−25
=1−11−.25=0