Discrete-Time Signals Properties

Discussion in 'Homework Help' started by jessica99, Mar 16, 2015.

  1. jessica99

    Thread Starter New Member

    Mar 16, 2015
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    Hi guys,
    I need help for this problem.



    I found that is linear, causal and not memoryless.
    But i didnt find this is stable? and time invariant?
    Can you explain why?
    Thanks alot.
     
    Last edited: Mar 17, 2015
  2. WBahn

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    Mar 31, 2012
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    How can we possibly explain why you found what you did since you give no information on how you found what you did? We aren't mind readers.

    Start by showing your work and how you came to the conclusions that you did. We can then look at your work and provide feedback?
     
  3. jessica99

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    Mar 16, 2015
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    Do you know about this subject?

    I can also show you causal and memoryless. But I didnt understand stable and time invariant for this question.
     
    Last edited: Mar 17, 2015
  4. Papabravo

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    Feb 24, 2006
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    I do not agree that the system is causal. Recheck the requirements of the definition of causal.
     
  5. jessica99

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    Mar 16, 2015
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    causal systems need to past values and current time values.
    this systems also dont need future values (x[n+1]). It need just past and current values. (x[n],...,x[3n-1])
    Is it true?
     
  6. Papabravo

    Expert

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    OK. You got the definition correct. Take an example, n = 7
    So we need the sum of
    x[7], x[8],...,x[20]
    Now tell me is x[20] in the future or the past with respect to x[7]?
     
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  7. jessica99

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    Mar 16, 2015
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    Hımm, thats correct. It is in future. Thanks for your help :)
     
  8. jessica99

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    Mar 16, 2015
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    I think also it is not stable. Because n-->infinite y[n]--> going to infinite.
    So it is not stable. But i am not sure :)
     
  9. Papabravo

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    In your proof for linearity: are you sure that you can pick the same value of n for both sequences?
    Suppose we have:
    a⋅y[5] + b⋅y[13]
    Is it still valid?
     
  10. Papabravo

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    I don't think so. Approaching infinity is different than being infinite.
    See if you can work out the argument.
     
  11. jessica99

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    Mar 16, 2015
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    I am pretty sure about linearity proof.
    If the x'=y' >> system is linear.
    summing operator's properties can show that.
    Its about the a and b. It is not about the y[n]. (Of course thats my opinion :) )
     
  12. jessica99

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    Mar 16, 2015
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    Hımm may be its correct. I dont know, I didnt understand stable and time invariant. :/
     
  13. Papabravo

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    I think stable just means the value of y[n] is bounded for all n.
    Ask yourself, "How many values am I adding together for a given value of n"
    Is the number of values for each n finite?
    Is each of the numbers I'm including in the sum finite?
    From these questions will come the answer.
     
    Last edited: Mar 16, 2015
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  14. Papabravo

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    My problem with your answer is moving the summation sign inside the brackets when the limits to the two sums may be different. I don't think you can do that.
    For example what is
    2⋅y[7] + 3⋅y[17] equal to??
    Is it even a member of the set of sequences defined by the definition?
     
    Last edited: Mar 16, 2015
  15. jessica99

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    Mar 16, 2015
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    I think you can not pick different value of n.
    You should pick same value of n. But you can pick different value of a and b.
     
  16. WBahn

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    I don't see you showing anything about causal or memoryless in your work.

    You are claiming it is causal but not showing your basis for that claim.

    What is required for the system to be causal? Keep in mind that it has to meet that requirement for ALL values of n.
     
  17. WBahn

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    What is the basis for saying that y[n] goes to infinity?

    For any finite value of n, is not the output a sum of a finite number of input values?

    What kind of stability is being asked for? BIBO stability?
     
  18. WBahn

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    Time invariant means that you CAN pick any value of time that you want. Basically, that which sample you call n=0 has no effect on the sequence of output values -- that when you reset your clock doesn't matter.

    Is that the case for this problem?
     
  19. jessica99

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    Mar 16, 2015
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    We already talked with Papabravo about causal. And I said my idea on there.
    not memoryless is so easy for me. Because this problem using past values, future values. So it is not memoryless.
     
  20. WBahn

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    What is the definition that you are using for a "memoryless" system? Conventional usage requires that the current output not be dependent on any past of future values of the input.
     
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