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Rewriting x(k) So i can calc a Z Transform

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Let x(k)=
    0 k<0
    0.6^k 0<=k<=21
    0 k>21

    Calculate X(z)
    2. Relevant equations

    Z transform

    3. The attempt at a solution

    Ok so i think calculating the Z transform of this wont be too difficult for me at all. I just have a hard time with this type of notation. So what i would like to some help on is how to rewrite the equation x(k).

    I understand that x(k)=0.6^k and for K is between 0 and 21. Any values outside of that x(k) = zero..... So how do I assemble X(k) in a form that I can do a z transform to it? It seems like it should be easy but I just dont see it. Something with a summation maybe and a limit of 0 -> 21?

    Thanks for any comments guys!
     
  2. jcsd
  3. Dec 9, 2011 #2

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    Hi Evo8! :smile:

    X(z) is the z-transform.

    It is given by:
    [tex]X(z)\overset{\mathrm{def}}{~=~}\mathcal{Z}\{x(k)\}\overset{\mathrm{def}}{~=~}\sum_{k=-\infty}^{\infty} x(k)z^{-k}[/tex]
     
    Last edited: Dec 9, 2011
  4. Dec 9, 2011 #3
    Hi I like Serena.

    I understand the z transform formula. What I don't understand is what my x(k) actually is to do the z transform too.

    Thanks for your help!
     
  5. Dec 9, 2011 #4

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    I'm not sure I understand your question.

    You're meant to substitute your definition of x(k) in the summation...
    Then adjust the bounds, since it's only non-zero in part of its range.
     
  6. Dec 10, 2011 #5
    Yes but my problem is I dont really understand what my x(k) is in this case.

    x(k)=0.6^k with the bounds 0->21 How do I apply this? I dont know why this is hard for me to see. Im sure its simple. I just dont fully understand the notation I guess.

    Would it just be

    X(z)=(0.6^k) (z^-1)?
     
  7. Dec 10, 2011 #6

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    There is a summation over k:
    [tex]X(z)=... + x(-1)z^{-(-1)} + x(0)z^{-0} + x(1)z^{-1} + ...[/tex]

    And
    [tex] ..., \quad x(-1)=0, \quad x(0)=0.6^0, \quad x(1)=0.6^1, \quad ..., \quad x(21)=0.6^{21}, \quad x(22)=0, \quad ...[/tex]
     
  8. Dec 10, 2011 #7
    Ok I think I understand. I think my issue was more understaning how the summation works. Im still on the edge of understanding. Either way this is what I have come up with after following a similar example from my book.

    x(k)=[itex]\sum[/itex](0.6^k) 0->21

    X(z)= [itex]\sum[/itex](0.6^k z^-k) 0->21

    X(z)=[itex]\sum[/itex](0.6^k (1/z^k)) 0->21

    X(z)=[itex]\sum[/itex]((0.6/z)^k) 0->21

    X(z)=1/(1-(0.6/z))

    X(z)=z/0.4

    Am I on the right track? I used the generalized geometric series to get this form. Well I tried to anyway.

    Thanks again for your help!
     
  9. Dec 10, 2011 #8

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    Let's see...

    No, x(k) itself is not a summation, but a function with a value that depends on the value of k.


    Looking good, except for that last step.
    That is the sum of a geometric series summing to infinity.
    However, you should only sum to k=21.


    Geometric series is the way to go.
    Just get the right formula for it. ;)
     
  10. Dec 10, 2011 #9
    Ok so up until

    [itex]\sum(\frac{0.6}{z})^{k}[/itex]

    Im still on the right track?

    The formula for the geometric series that I have is

    [itex]\sum z^{k}[/itex]=[itex]\frac{z^{m}}{1-z}[/itex],m≥0 and lzl<1

    My m in this case would equal 0 because my limits are 0-21. so the z in the numerator becomes 1 and the z in the denominator is my z which equals (0.6/z). This was my reasoning anyway. Do I just need to have notation instructing to stop the sum at 21?

    Like lzl≤21?
     
  11. Dec 10, 2011 #10

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    Yep!


    I don't know that formula. :confused:
    Where did you get it?

    Here's the formula I mean:
    http://en.wikipedia.org/wiki/Geometric_series#Formula
     
  12. Dec 11, 2011 #11
    I got that formula from my DSP book. Ill take a look at the one you have posted from wikipedia and see if i can understand how to apply it.

    Thanks again.
     
  13. Dec 12, 2011 #12
    Ok so I looked back and Im unsure if im using the geometric series correctly.

    This is what I have for a definition in my book.

    geometricseriesJPG.jpg

    However I made a stupid mistake on my Algebra....

    Up to this point im ok.. X(z) = [itex]\sum(\frac{0.6}{z})^{k}[/itex]

    Now when I apply the geometric series I get X(z)=[itex]\frac{1}{1-(\frac{0.6}{z})}[/itex]

    Now when I simplify it should be X(z)=[itex]\frac{z}{z-0.6}[/itex]

    This looks ok to me i think. Especially since there is an example that is very similar in my book that I followed. The only problem is I dont have the limit of 21 defined anywhere. The example in the book is 0->∞ My problem above is 0->21... Where does this upper limit of 21 come into play?

    Here is the example for your reference.
    ROC1.jpg
    ROC2.jpg

    The X[itex]_{c}[/itex](z) in this example is what I believe to be very similar to my problem above.


    In addition I have to find the region of convergence of my X(z). I am a bit confused on what the ROC actually is but im still reading that section over in my book. I may have a few questions on this shortly. Not sure if its worth starting another thread for though.

    Thanks,


    Edit: If my X(z) is correct would my ROC simply be 21>lzl.0.6 ?
     
    Last edited: Dec 12, 2011
  14. Dec 12, 2011 #13

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    Sorry, but your X(z) is not correct yet.
    As you already surmised the upper limit of 21 should go somewhere.
    All the formulas that you have shown sum up to infinity.

    To sum a series [itex]r^k[/itex] up to some limit n, you need:
    [tex]\sum_{k=0}^n r^k = {1 - r^{n+1} \over 1 - r}[/tex]

    Alternatively you can write your sum as:
    [tex]\sum_{k=0}^n r^k = \sum_{k=0}^\infty r^k - \sum_{k=n+1}^\infty r^k[/tex]



    When you sum up to infinity, there is a chance that the sum does not exist, for instance because it tends to infinity.
    The region of convergence is where the sum exists.
    But let's discuss that when you have your X(z).
     
  15. Dec 12, 2011 #14
    Hmmm.....ok. I think I understand what you mean.

    Going by the first formula you posted I would get

    [itex]\frac{1-(\frac{0.6}{z})^{22}}{1-(\frac{0.6}{z})}[/itex]

    If I understand this correctly I see now what you mean by the series I had first written was to ∞ not to 21...

    However I need to do something else here to simplify I assume? Or is this the final form? I dont see anything else I can do algebraically except for writing out the entire series. I may be missing something though as my algebra skills are sub par in my opinion.
     
  16. Dec 12, 2011 #15

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    Nope. That's it! :)
    As far as I'm concerned that's the final form.
    There's no useful algebraic simplification.

    As for the region of convergence.
    For which values of z is this formula well defined?
     
  17. Dec 12, 2011 #16
    I think it would be 0≤z≤21 right?

    Or would it be 0.6≤z≤21 ....? If I were to guess I feel like my first answer is correct. Im still a little confused on this though.
     
  18. Dec 12, 2011 #17

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    What happens if you substitute z=0?
    Does it exist?

    What if z=22?

    Btw, note that z can be any complex number, not just a real number.
     
  19. Dec 12, 2011 #18
    Well when you substitute z=o it does not exist. Thats for sure. So good point there.

    If you substitute z=22 it does indeed exist. So maybe my second answer is more correct? Or should it be 0.6≤z≤22?

    Or am I still way off track here?
     
  20. Dec 12, 2011 #19

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    You're still a bit off track. ;)

    What if z=0.6?
    What if z=23?
    What if z=0.1?
    What if z=i?
     
  21. Dec 12, 2011 #20
    Well I tried all of those values and

    z=0.6 comes up undefined of course so thats out
    z=23 comes out at just over 1. 1.02679....
    z=0.1 is some large number.
    z=i comes up as a complex number with real and non real parts. Im not sure how to understand this.

    I can see that the lower limit (i think) has to larger then 0.6 to keep the expression form being 0. So z>0.6 .... ?

    As for an upper limit it seems like 23 and higher would work no problem but were looking for the limit of the convergence right so it would be ≤ 21?

    So 0.6< z ≤21?
     
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