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Homework Help: Help with this Summation?

  1. Dec 15, 2011 #1
    It's been a while since I took Calc 2 and I am in a Linear Systems and Signals class right now. I'm looking at a solution on how to obtain a zero state response of a discrete time signal, but performing the summation confuses me. Can someone explain the steps they did?

    This is part A

    http://imageshack.us/photo/my-images/684/partal.jpg/

    This is part B

    http://imageshack.us/photo/my-images/11/part2sy.jpg/

    In Part A, where did the +1 come from?
    Also, how would you simplify the first fraction appearing after the equal sign of Ystep?

    In Part B, where did the -1 's come from? Originally, h[m] was just (6/5)[itex]\sum[/itex](-1/2)m + (24/5)Ʃ(1/3)m

    I have a Final next week on this stuff so, any help would be appreciated.
     
    Last edited: Dec 15, 2011
  2. jcsd
  3. Dec 15, 2011 #2

    SammyS

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    Part A: Where did the 1 come from?
    [itex]\displaystyle \sum_{m=0}^{k}(-2)^m=(-2)^0+\sum_{m=1}^{k}(-2)^m[/itex]

    and of course, (-2)0=1​

    As for simplifying the fraction, they have done it quite well.
    [itex]\displaystyle \frac{1-(-2)^{k+1}}{1-(-2)}=\frac{1-(-2)(-2)^{k}}{3}[/itex]

    [itex]\displaystyle =\frac{1}{3}+\frac{2}{3}(-2)^{k}[/itex]​
     
  4. Dec 15, 2011 #3

    SammyS

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    I assume the sum in the original form of h[m] started with m=1. The -1s come from the fact that in Part B, the sums start with m=0.

    BTW: You appear to still be very rusty regarding working with summations, considering that your final is next week.

    Good Luck !
     
  5. Dec 16, 2011 #4
    For Part A, the index for the final answer is m = 0 though. You have it at m=1 for (-2)m
     
  6. Dec 16, 2011 #5

    SammyS

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    Yes. I left it for you to identify what is what & for you to do a little algebra.

    [itex]\displaystyle \underbrace{\sum_{m=0}^{k}(-2)^m}_{\text{from 2nd line}}=(-2)^0+\underbrace{\sum_{m=1}^{k}(-2)^m)}_{\text{from 1st line}}[/itex]
     
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