Q15 n 16 of Assignment3 sol clarification -reg

[murthy...@gmail.com]

Dear sir/madam,
1.
Q15:If the Z-transform of a sequence isX(z)=0.5zz−2it is given that the region of convergence of X(z) includes the unit circle. The value of x[0] is and solution shown as ANS:B;
if we use, initial Value theorem, x(0) = 0.5;i.e. ANS:A

so pls check once again

2.
Q16:
For a rational and stable system function, which of the following is correct
and the solution says the ANS:D, i.e.ROC must include unit circle;

but for any stability conditions using Z-transform, most imp condition is that, all the poles must lie inside the unit circle.even though some times ROC may not include UNIT circle. so ANS:A is more appropriate than ANS:D.

So pls look into them and do the needful..!
Pls help me to understand BETTER..!

THANKS in advance..MURTHY

[shivama...@gmail.com]

Regarding Q15:
If you look closely the pole is located at Z=2. The ROC includes the unit circle, this means the system function corresponds to the non-causal sequence given by x[n] = 0.5(2^n) u[−n − 1]. so at n=0 signal has the value 0. Also, the initial value theorem hold only for causal sequences but the sequence here take values for the negative time axis so one can not apply initial value theorem . This is one of the questions where one can see the importance of specifying the ROC of a z-transform along with the expression.

Regarding Q16:
Consider the system function given by h[n]=0.5(2^n) u[−n − 1]. Now the Z-transform of this sequence has a pole at z=2 but it is an absolutely summable(and hence stable) sequence(you can check it!!!) so your claim is incomplete. here in this question we asked what are the condition for a rational Z-transform to to be stable so the necessary and sufficient condition is to have an ROC must include the unit circle.

Hope you are satisfied,

Thanks
TA