Why is the function y=x(2t) not time-invariant?

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Discussion Overview

The discussion revolves around the function y=x(2t) and its property of time-invariance. Participants explore the implications of shifting the input signal x1(t) and the resulting output signal, questioning why the expected output does not align with the shifted input. The conversation includes mathematical reasoning and conceptual clarifications related to time-scaling in signal processing.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant expresses confusion about the time-invariance of the function, noting that shifting the input x1(t) does not yield the expected output y1(t-2).
  • Another participant suggests that the system represents time-scaling, indicating that any time shift in the input will also be compressed, leading to a lack of time invariance.
  • A participant questions the nature of the images referenced in the discussion, suggesting that the output y may require knowledge of future values of x, which raises concerns about causality.
  • One contributor speculates about the relationship between the operators y and x, pondering if changing the scaling factor would affect the time-invariance property.
  • Another participant corrects an earlier claim about the output function, clarifying the process of replacing t with 2t in the input signal to derive the output.
  • Several participants express their struggles with the concept, indicating that they find the topic challenging and are seeking further understanding.

Areas of Agreement / Disagreement

Participants generally express confusion and uncertainty about the time-invariance property of the function. Multiple competing views and interpretations of the mathematical relationships remain unresolved.

Contextual Notes

Some participants note that the discussion involves complex mathematical reasoning and may depend on specific definitions of time-invariance and scaling. There are indications of unresolved assumptions regarding the nature of the signals and the implications of the transformations applied.

Who May Find This Useful

This discussion may be of interest to students and practitioners in signal processing, particularly those grappling with concepts of time-invariance and time-scaling in systems theory.

Bassalisk
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I am a bit confused.

http://pokit.org/get/bedc0ac7e1d17e01d7d58b021b81663c.jpg

The function is y=x(2t)

and the point of it is to show the property of time-invariance.(which we should fail in this example, because it isn't time invariant.)

Input signal is x1(t)

Output signal is shown for that signal y1(t)

But when you shift it, x1(t-2) it doesn't give y1(t-2).

I am both ok and not ok with that. First let's go over the math.

If I shift my signal x1(t) by -2

I get y_2(t)=x(2(t-2))

Is that correct?

But that is y_2(t)=x(2(t-2))=x(2t-4) and that is original signal, shifted by 4, and then scaled by 2, which is not what it should be.

y2(0)=x(-4) which is 0, not 1.

I am probably stuck on something simple, but nevertheless I am stuck.
 
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bumpity
 
i don't know anything about that notation.

is it analog or digital?
 
jim hardy said:
i don't know anything about that notation.

is it analog or digital?

I believe its analog.
 
This might help.

Its a quote from Oppenheim, signals and systems.

This system represents a time scaling. That is, y(t) is a time-compressed (by a factor of 2) version of x(t).
Intuitively, then, any time shift in the input will also be compressed by a factor of 2, and it is for this reason
that the system is not time invariant.

To demonstrate this by counterexample, consider the input x1(t) shown in the Figure 1 47(a) and the resulting output y1(t) depicted in figure (b).

If we then shift the input by 2 i.e. consider x2(t)=x1(t-2) as shown in figure (c) - we obtain the resulting output

y2(t)=x2(2t) shown in figure (d)

Comparing figures (d) and (e) we see that the y2(t) <>y1(t-2)
 
it looks like the time axis is compressed towards the origin between a) and b), and between c) and d). Doing that means that every point on the time axis is mapped to a new point, except the origin, which remains unchanged. This makes t=0 special. I think the idea of a time invariant system is that no time t is special, and you are free to choose your origin wherever you like.

What are these images supposed to represent? Input and output signals of some system? I think the output y requires knowledge of the future of x which violates causality
 
i'm a plodder and haven't ruled out that the paradox might stem from 2 X 2 = 2 + 2

is operator y half of operator x?
and offset also 2?
That's too many twos for me.

what if operator y were one-third of operator x ?
or 1/e ?

please excuse if that's a dumb question but my simple mind has to rule out the dumb things first. And i do just plod.

thanks for your tolerance.

old jim
 
I am stuck just like you guys. I think I will have just skip this one, and accept it.

This represents a system with function y=x(2t).
 
This is wrong:

If I shift my signal x1(t) by -2

I get y2(t)=x(2(t−2))
----------------

The function is H{x(t)}=x(2t)
What it means is take any occurrence of t in your input, and replace it with 2t.

So we look at x2(t)=x1(t-2)
And we want to get its output: we replace any t with 2t:
so we get y2(t)=x2(2t)=x1(2t-2)=x1(2(t-1))

Which agrees with the plots in your post

Think hard about it, and do more examples, I remember these kind of things were really a pain in the *** in my signals & systems course
 
  • #10
elibj123 said:
This is wrong:

If I shift my signal x1(t) by -2

I get y2(t)=x(2(t−2))
----------------

The function is H{x(t)}=x(2t)
What it means is take any occurrence of t in your input, and replace it with 2t.

So we look at x2(t)=x1(t-2)
And we want to get its output: we replace any t with 2t:
so we get y2(t)=x2(2t)=x1(2t-2)=x1(2(t-1))

Which agrees with the plots in your post

Think hard about it, and do more examples, I remember these kind of things were really a pain in the *** in my signals & systems course

I will. This was just a brainer. I didn't have a lot of time to tackle this. I did in the end, get a very good mark on this course 9/10.

Thank you, I will give it some deep thought tomorrow. Gotta study Theory of information now :D
 

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