Yes I am surprised no one actually gave any response. The thing is i am not sure if Z^2 gives delta(n+2), since my prof never gave such an example. I hope you can take a look at my work later today and let me know what you think.Hi,
I am in a bit of a hurry today i can look this over later today if no one else does first.
But, in the mean time why dont you apply the Z transform to your results and see if you get the original question expression?
If you take the Z transform of an expression y and you get Y, then when you take the inverse transform of Y you should get y back again. Likewise if you take the inverse transform of Y and get y then when you take the Z transform of y you should get Y back again.
This is really the best way to handle this anyway so you can always have a way to check your results yourself in the future.
Hello again,Yes I am surprised no one actually gave any response. The thing is i am not sure if Z^2 gives delta(n+2), since my prof never gave such an example. I hope you can take a look at my work later today and let me know what you think.
I know my handwriting is small, so I rewrote it in a bigger font, under the sun. Is it okay or do you want me to type it?Hi,
Sorry but maybe you better type it out so i can read it ok. That second set is almost worse than the first unfortunately
Sorry but we have to have something that is not too hard to read or else it makes it much more difficult to help.
Sure take your time, but take into consideration the ROC given in the question in the beginning of the threadHi,
That looks better, and yes one of the problems was the tiny 'font' in the previous set.
Give me some time to take a look.
Hello there,Hello,
@MrAl ,
This seems to be an other version of the book:
https://archive.org/details/discrete-time-signal-processing/mode/2up
There are the same tables in chapter 4:
View attachment 206535
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Bertus
I actually have no problem. I just want someone to verify if my answers are correct or notHello again,
Thanks for the table that's a pretty good one actually.
What did you do to get the inverse transform of z^2 ?
I ask because if we try to use entry #4 in the table which is z^(-m) we would have to make m=-2 and that would seem to imply that the result would be delta[n+2].
Oh that is what you got ok
Is that the part you had the problem with for that question number?
Using that table this seems pretty simple.
What is amazing is that Wolfram can not find the solution. It does not know what to do with z^2.