Question

Question : 4 - Signals:

How to check whether the given system is \(\color{green}{\text{Time Invariant}}\) or \(\color{red}{\text{Time Variant}}\)??

Answer 1

Perhaps a working example would help?

Answer 2

Have you actually SEEN "time-invariance"? In the strictest sense, man made time-invariance doesn't exist. Systems degrade.

Answer 3

I am just learning the case of LTI Systems, so I want to know the method of checking Time Invariance or Variance of a system..

Answer 4

Let it be:

y(t) = x(-t)...

Answer 5

Okay, lag the system and see if you get a different answer.

y(t-T) = x(-(t-T))

Is that the same response?

Answer 6

I did not get what you are trying to say..

Answer 7

y(t) = x(-t)

y(t-T) = x(-(t-T)) = x(-t + T)

Now, with what I have to compare it??

Answer 8

What does x(-t) mean? Is that 'x' a function or is that 'x' multiplied by "-t"?

Answer 9

y(t) = output of a system.

x(t) = Input given.. x(t) is a signal or x is function on independent variable 't'..

Answer 10

So, my when I am giving x(t) as an Input, my system is returning x(-t) as Output. :)

Answer 11

So, my System is designed to do Time-Reversal function..

Answer 12

Well, how can Time Reversal be Time-Invariant?

The whole question is on whether you get a different response at a different time. If all youg3et is reversal, and that's all it does, I'm not seeing how that is time-variant.

Answer 13

This you are just saying by manual analyzing..

That's why I said, I know the method to check Time Variance or Invariance of a system..

Any I have never said the system is Time Variant..

Answer 14

May be you know or not: There is certain procedure to test whether a system is time invariant or not, I want to know that only.. :)

Answer 15

We poked at that, earlier.

I found your specific problem. http://vimeo.com/6808745

Answer 16

How to check it for:

\[y(t) = \int\limits_{-10}^{10}x(t).dt\]

Answer 17

Let me try :

\[x_1(t) = x_1(t - t_0)\]
\[y_1(t) = \int\limits_{-10}^{10}x_1(t) \cdot dt \implies y_1(t) = \int\limits_{-10}^{10}x(t - t_0) \cdot dt\]

Answer 18

Now, substitute : \(t - t_0 = \lambda \implies dt = d\lambda\)

\[y_1(t) = \int\limits_{-10 - t_0}^{10 - t_0} x(\lambda) \cdot d \lambda\]

Answer 19

While :

\[y (t - t_0) = \int\limits_{-10}^{10}x(t - t_0) \cdot dt \color{green}{\implies y_1(t) \ne y(t - t_0)}\]

So, given system is \(\color{blue}{\text{Time-Variant..}}\)

Am I true in doing that??

Answer 20

@tkhunny

Answer 21

Sorry, spent the day in the hospital...

That looks like it!

I am a little curious where you got [-10,10]. Is that in the problem statement?

Answer 22

Sorry, are you alright??

The limits are from -10 to 10 in the question itself..

Answer 23

Fair enough. That would have been helpful information up front. :-)

Getting better. Thanks.

Answer 24

I have one doubt left..

In \(y_1(t)\), my integrand is \(x(t)\)

And in \(y(t - t_0)\), it is \(x(t - t_0)\), is it right there?