Showing Time Delays in Inverse Z Tranform Results

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NoamVH
NoamVH 2015 年 11 月 24 日
Hello,
I need some help.
Say we have the next function:
F(z) = z / (z - 0.5)
The inverse Z function to this function will result as:
f(k) = 0.5^(k)
And Matlab agrees with me:
>> syms z k
>> F = z / (z - 0.5);
>> f = iztrans(F, k);
>> f = vpa(f, 2)
f =
0.5^k
If we'll divide F(z) by z we'll have the next function:
F(z) = 1 / (z - 0.5)
Therefore, we'll expect to have a time delay in the inverse Z transform function:
f(k) = 0.5^(k - 1)
However, this time Matlab won't cooperate:
>> F = 1 / (z - 0.5);
>> f = iztrans(F, k);
>> f = vpa(f, 2)
f =
2.0*0.5^k - 2.0*kroneckerDelta(k, 0)
The result we get is (of course) correct, but very messy and not very nice to look at, this is a big problem with more complex inverse Z tranforms, the way I'd like to see the result is in the same way f(k) is shown. How can I do that?
Also, the unit step function is for some reason shown as such in Matlab after a Z transform:
1 - kroneckerDelta(k, 0)
This is also a big part of the unreadable mess I get when using Matlab for inverse Z tranforms, is there anything I can do?
Thank-you,
NoamVH.

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