# 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 Moderator

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 Thread Starter New Member

Mar 16, 2015
13
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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 Expert

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 Thread Starter New Member

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

Feb 24, 2006
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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 Thread Starter New Member

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

8. ### jessica99 Thread Starter New Member

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 Expert

Feb 24, 2006
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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 Expert

Feb 24, 2006
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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 Thread Starter New Member

Mar 16, 2015
13
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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 Thread Starter New Member

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

13. ### Papabravo Expert

Feb 24, 2006
10,340
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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 Expert

Feb 24, 2006
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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 Thread Starter New Member

Mar 16, 2015
13
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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 Moderator

Mar 31, 2012
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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 Moderator

Mar 31, 2012
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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 Moderator

Mar 31, 2012
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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 Thread Starter New Member

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 Moderator

Mar 31, 2012
18,087
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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.