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Homework Help: Digital control: Z transform

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    The Plots bellow show two discrete time signals, in the plots only the initial proportion is shown. Then the signals remain constant and equal to the last value shown. Derive the Z transform of each Signal.
    image1.JPG image2.JPG

    The attempt at a solution
    I can get part A, giving an answer of (z9+z8+z7+z6+z5+z4+z3+z2+z+1)/z9

    I apply the same concept to the second image but im getting lost and not getting near the correct answer of (1/z9)x((z10+1)/z-1)

    Im very new to digital control, so ive potentially missed a key detail out of the question, so if i have let me know!

    Any tips about how to go about this would be greatly appreciated
  2. jcsd
  3. Apr 8, 2015 #2
    You may not have covered this yet but have you gotten to the point where you are using the transform tables?

    Can you think of a way to build the signals as a combinations of other easier signals???

    It helps if you can start to think of ways to approach these questions which may not be immediately clear.
    On one midterm we were asked to find the convolution of 2 square pulses. Most people ended up taking 2 pages of math to get there, where a few of us used the answer from the previous question (what is the fourter xform of the square wave) and got to the answer in about 3 lines :D
  4. Apr 8, 2015 #3
    You mean that table above ^^?

    I have but i am struggling to apply it to this scenario and I have found little in the way of help from online notes
  5. Apr 8, 2015 #4
  6. Apr 8, 2015 #5


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    They both look like step responses to me......
  7. Apr 8, 2015 #6
    So can you solve the question using the known step transform result, and the "addition" property? (will need the delayed unit step though)
  8. Apr 8, 2015 #7


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    yes, but I'm not going to do the problem for him
  9. Apr 8, 2015 #8
    Haha I'm dum and I failed to notice that you weren't him :D
  10. Apr 8, 2015 #9


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    haha, it happens...

    I am curious
    why do you say this is the correct answer?
  11. Apr 9, 2015 #10
    Ok thanks, i'll give that a go now and let you know how it goes!

    I know that's the answer as it is a question off a pass paper that our lecturer gave us along with a markscheme
  12. Apr 9, 2015 #11
    Ok I am literally getting nowhere with this, any other pointers?
  13. Apr 9, 2015 #12
    If you just forget the Z-transform for a bit, try to imagine how you can "build" the sequences shown in your graphs as the sum of a unit step u[n] and a time-shifted unit step u[n - m].

    If you can do that, then the rest is just a matter of checking the Z-transform tables you've already been given.
  14. Apr 9, 2015 #13
    Are you sure its (1/z^9)x((z^10+1)/z-1) and not (1/z^9)x((z^10-1)/z-1)

    I get to the Z^10 - 1 answer 2 different ways.
    I'll describe the process and let you do the maths (I'd suggest doing it both ways for practice)

    First way: You have a series of dirac's shifted. So using the dirac z Transform 1 and the shift property you end up getting 1+z^-1....+z^-9 (where you were in the first post)
    From there you use the geometric summation formula and rearrange ending up (unless I made a maths error) with (1/z^9)x((z^10-1)/z-1)

    Second way you as Miles said notice that the series is a combination of 2 step functions (which ones though???)

    then you use the transform for the unit step, shift property and linearity property and you'll end up with X[z] = A + B rearrange and you'll end up at the same result

    (sorry I suck at latex lol)
  15. Apr 9, 2015 #14
    Are you looking at (a) or (b)? It's correct for (b).
  16. Apr 9, 2015 #15
    attention to detail has never been my strong point.... I was looking at (a) =D
  17. Apr 9, 2015 #16
    Ok thanks for that, I am starting to get there now i managed to get the answer you go of (1/z^9)x((z^10-1)/z-1) for part A, however its the fact that the step goes up to 2, in part B thats throwing me, heres what I have done so far, am I on the right track? (Its a picture as it takes me forever to type in the equations :P )
  18. Apr 9, 2015 #17
    I remember having issues with the infinite sums like what you're trying to do in part 2.

    Can question 2 be put in terms as a linear combination of 2 other functions?

    If you can get used to using the known transforms and properties early it'll make your like a lot easier in the long run :)
  19. Apr 9, 2015 #18


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    one thing to note, I assumed that before time(0) the signal was not zero, so both a and b are comprised of a single step
  20. Apr 9, 2015 #19
    What do you mean?
  21. Apr 9, 2015 #20
    Are there 2 simple functions that you can add together to get the function in the question? (Hint both functions will be the same, one will just be shifted)
  22. Apr 9, 2015 #21
    Ok I think I may have done it, how does this look:

    function of initial line calculation:

    Then function of the second line:

    then the addition of the two to give a final answer:

    Is my working correct here? Or have i just fluked my way to the right answer haha
  23. Apr 9, 2015 #22
    Seems fine, although your notation is a bit odd in places, e.g. ##\mathcal{Z}\{u[n]\}## is for the Z-transform of u[n].

    There's an easier (in my opinion) way to go about it, though. How would the sequence u[n] + u[n - 10] look, and how would you find its Z-transform using tables?
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