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.

 

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?

 

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.

 

I do not agree that the system is causal. Recheck the requirements of the definition of causal.

 

I do not agree that the system is causal. Recheck the requirements of the definition of causal.

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?

 

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]?

 

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]?

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

 

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

 

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?

 

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

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

 

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?

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 )

 

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

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

 

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.

 

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 )

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?

 

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.
Fore 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?

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.

 

Do you know about this subject?

For example for linearity;

I can also show you causal and memoryless. But I didnt understand stable and time invariant for this question.

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.

 

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

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?

 

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.

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?

 

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.

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.

 

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.