Discrete-Time Signals Properties

Thread Starter

jessica99

Joined Mar 16, 2015
13
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 just said my idea. But it isnt have basis. I said that I didnt understand abour stability.
it is not about BIBO. Its asked just this system is stable?

But Papabravo said that:
I think this system is stable.
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 gin 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.
 

Thread Starter

jessica99

Joined Mar 16, 2015
13
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?
Time invariant is if x[n] output is-->y[n]
x[n-n0] output is -->y[n-n0]
But I didnt proof that.
 

WBahn

Joined Mar 31, 2012
29,978
If the criteria is that y[n] be bounded for all n (which is known as BO stable, or Bounded Output stable), then the system is not stable. What is the output if x[15] = ∞?
The usual criteria is that the output remain bounded for all n provided that the input in bounded for all n; that is known as BIBO stable, or Bounded Input, Bounded Output stable).
 

WBahn

Joined Mar 31, 2012
29,978
Time invariant is if x[n] output is-->y[n]
x[n-n0] output is -->y[n-n0]
But I didnt proof that.
One thing to keep in mind is that to prove that it is NOT time invariant you only have to find a single example for which it is not time-invariant. But in order to prove that it IS time invariant, you have to prove that it is time-invariant for ALL values of n. The first you can do by counter-example, the second you must do by proof.
 

Thread Starter

jessica99

Joined Mar 16, 2015
13
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.
Memoryless systems are not depend on future or past values. This systems has future and past values. So it is not memoryless.
 
Top