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Is the following system considered time invariant??

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johnharris40

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y[n] = x[n] + x[n-1] where n > 0
y[n] = 0 otherwise

I'm thinking it's not, because if you were to shift n by n - n', there's a possibility that certain shifted values will then take on 0 and be erased. Is this true?
 

m = n - n'
results in:
y[m] = x[m] + x[m-1] where m > 0
y[m] = 0 otherwise
same response but delayed so it is time invariant.
Please somebody correct me if I am wrong
 

A counterexample shows that it's not TI:

Let x[n]=1 for n=1; x[n]=0 otherwise. Then y[n] has values different from 0.
If you apply the same x but at an earlier time (at least 2 samples), y[n] is 0 for all n.

Regards

Z
 

zorro said:
A counterexample shows that it's not TI:

Let x[n]=1 for n=1; x[n]=0 otherwise. Then y[n] has values different from 0.
If you apply the same x but at an earlier time (at least 2 samples), y[n] is 0 for all n.
Click to expand...

I don't agree with you. I think albert22 is right.
Also in you example, moving the stimulus from n to n+n' the output y will be affected by it at time n+n'. y(n+n') = x(n+n') + x(n+n'-1)

In order not to have a TI system you need somthing related to the absolute time (or sample). For instance a system like:

y(n) = 3*n+x(n)

in this case at output level --> y(n+n') = 3*(n+n') + x(n+n')
but at the stimulus side (let me call the output related to it z, instead of y) --> z(n+n') = 3*n + x(n+n')
since y(n+n') <> z(n+n') the system is not TI.

This because the first term "3*n" doesn't follows the stimulus but it is fixed in the time sequence.
 
Time invariance means that the input-output relationship doesn't change under time shifts.
If time-shifting an input by a certain amount you get something different to the time-shifted version of the output, it is not TI.
Of course, in any real system time invariance holds in restricted time domain (e.g. since switch-on to switch-off).
Regards

Z
 

I really don't understand.
If you shift x(n) by whatever quantity the output will just follows x. From y point of view, y(n) and y(n+n') will have exactly the same shape just shifted by n'.
Zorrom could you, please, post a numerical example ?
 

The system is time invariant. Albert posted a nearly complete proof.

First, delay the input:
Suppose x'[n] is a delayed version of x[n]: x'[n] = x[n-n']
Then y1[n] = x'[n] + x'[n-1] = x[n-n'] + x[n-n'-1]

Second, delay the output by the same amount:
y2[n] = y[n-n'] = x[n-n']+x[n-n'-1]

Thus:
y1[n] = y2[n]
 
albbg said:
Zorrom could you, please, post a numerical example ?
Click to expand...
I try to show the example of my post #3:

Let's take the system described in post #1.
Suppose the input sequence x[n] ...,0,0,0,1,0,0,0... where the "1" occupies the position n=1 . The x and its corresponding output y[n] are shown here:

seq0.png

If the system were time invariant, shifting x in time would produce the same shift in y, as follows:

seq1.png

Instead, the system described in post #1 gives y[n] like this:

seq2.png

So, in is not time invariant.
Regards

Z


ERRATUM: "-1" should be "1" in the first two cases. See posts #10 and #12 below.
 
Last edited:

Why ?
In both cases the y-sequence ...00011000... will be generated, more or less shifted. How did you find y(2)=-1 (first table) ? There is an addition and everything is positive.
 

I agree with zorro, that y[n] = 0 seems to define time variant behaviour. But I wonder if the definition is intended this way? May be it's just a badly considered attempt to define a causal system.
 

albbg said:
Why ?
... the y-sequence ...00011000... will be generated, ...
Click to expand...
I'm sorry. I made a mistake: I took y[n] = x[n] - x[n-1] instead of y[n] = x[n] + x[n-1] .
"-1" must be changed by "1" in the two first cases.

Anyway, in the the third case he output is "all zeros" because, according to specification, y[n] = 0 for all n<=0.
The output is 0 for n<=0 (despite of the input), and x[n]+x[n-1] for n>0 .
The system varies its behaviour with time. So, it is time variant.

Regards

Z
 
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    albbg

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OK, I've understood I'm wrong. Sorry.

I focused on y =x[n] + x[n-1] (that alone is TI) forgetting the additional y[n] = 0 n<0 then for negative shifts the output will be zero regardless the input.

Thank you zorro you was right.
 

I am learning from this discussion. Let me ask some questions:
When we shift the response. Should we shift y[n] = 0 ?, In this case I think the system is time invariant.
If we not shift y[n] = 0 is the system causal ?
Could a non causal system be time invariant ?
 

Hi,

albert22 said:
When we shift the response. Should we shift y[n] = 0 ?...
If we not shift y[n] = 0 is the system causal ?
Click to expand...

The questions are not clear to me. But maybe the answer is here:
When we shift x(t), it is the signal that is shifted, not the time reference.
Take a simple example in continuous time. Suppose se have a switch that is turned on at midnight, i.e. at t=0 according to our time reference. We describe the system as:

y(t) = x(t) for t>0
y(t) = 0 otherwise

If x(t) is a pulse starting at t=1 and ending at t=2, then y(t)=x(t). But if we shift x(t) by 3 time units in advance (i.e. the pulse lasts while the switch is open), then the output is not the shifted version of the original one.
The system is time variant.
Nevertheless, we could say that it is time invariant restricted to t>0, as I pointed out in post #5.

Could a non causal system be time invariant ?
Click to expand...

Yes. For example: y(n)=x(n+1) is not causal (the output anticipates the input) and it is time invariant.

Regards

Z
 
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