Question

Question - 9 : Signals :

How this system is Time-Invariant??

\(\color{green}{\textsf{y(t) = x(t - 5) - x(3 - t)}}\)

Answer 1

so you're telling me that you know it's time invariant
Yes?

Answer 2

Yes, okay, so to test time invariance, we need to shift the time
and see what happens
so we'll let t-->t+delta
or something like this
Are you familiar with that?

Answer 3

haha, see - I thought so!

Answer 4

yes, continue...

Answer 5

I like your attitude. I want you to do it yourself as well... :)

Answer 6

so good so far
although...

t--->t+delta

so -t --> -t - delta

Answer 7

sure... I'll shift my attention in time xD

Answer 8

Hm, okay, so when you substitute t+delta for t
and you have the x(3-t) both the t and the delta pick up the minus sign
Is that clear?

Answer 9

Otherwise, it's perfectly correct. You time shifted (delayed, in fact) the input and now you've written what the output y(t) looks like in this case--- y_1(t) in particular.
That's right...

Answer 10

Not if you time shift the output y(t) similarly. That is: y(t+delta)... the question is what does that look like?
And do the forms agree?

Answer 11

If they agree, then it's time invariant

Answer 12

catch me if you can... ;)

What this means... is that it doesn't really matter where you begin in time, the relationship between the input and output remains the same.

Answer 13

I have to leave soon as its really late here... so my apologies if I'm going ''father than the speed of light'' lol

Answer 14

I never intended to be a ball or become a ball, I'm who I'm.

Answer 15

I'm glad you've finally got this, you need to catch up a bit more.

Answer 16

So... we want to consider what it looks like in terms of x not x1
You'll see when we time shift the output
Can you do that?

Answer 17

lol... up there ^ i meant faster* not father... sorry typo >.<
I'm so tired.

Answer 18

okie.

Answer 19

Its fine, I like to help. :)

Answer 20

but if you insist, I will... not now but later. xD

Answer 21

Also uh Hm, you don't have to do it the way I want. You can do it anyway that you can solve the problem. I'm just providing a suggested path

Answer 22

o.O ?

Answer 23

I guess if it's a delay or advance depends on what you think the sign of delta is
But yes, let's assume delta > 0, so it's an advance

Answer 24

Yes, you're right. So
y(t+delta) = y1(t)
and therefore, it's time invariant

Answer 25

it's not there's just a sign error

Answer 26

Nono...it's time invariant

Answer 27

there's no disagreement between the answers
It's just there were some sign errors

Answer 28

I will write it here:

Answer 29

So...

(1) y1(t) --> x1(t+d)

(2) y1(t) --> x( (t+d) - 5) - x( 3 - (t+d) )

(3) y1(t) --> x( t+ d - 5) - x ( 3 - t - d)

(4) y2(t) --> y(t+d) --> x( (t+d) - 5) - x ( 3 - (t+d)

(5) y2(t) --> x( t + d - 5) - x ( 3 - t - d )

(6) y1(t) = y2(t)

Answer 30

I'm missing an end parenthesis in line (4)

Answer 31

(4) y2(t) --> y(t+d) --> x( (t+d) - 5) - x ( 3 - (t+d) )

Answer 32

so that's it... we advanced the input and the output looks as in line (3). Then we advanced the output similarly and again it looks the same. So we conclude time invariance.

Answer 33

"it" being the expression for the output y1(t) and y2(t)

Answer 34

I suppose it is subtle and tricky. Conceptually: we advance the input in time, and we ask what does the output look like? Okay, it looks like this:

Answer 35

Then we simply take the output and advance it in time, what does that look like?

Answer 36

Oh, they look the same.
Therefore, in this precise sense, we say the system is time invariant

Answer 37

It means that the particular zero of time is unimportant. The form of the relationship between the input and the output signal does not vary with when we start counting time.

Answer 38

sure, take all the time you need & good luck!

Answer 39

I have time, one sec.

Answer 40

yes, time invariant

Answer 41

Alright then, good luck with the rest!

Answer 42

Good.

Answer 43

My pleasure. G'bye!

Answer 44

@waterineyes is your previous question about SIgnals ...I haven't particpated yet of what diagrams sourcing you are talking about in that post?

Answer 45

yes?

Answer 46

@Miracrown what are you doing above.. :P

Answer 47

@SithsAndGiggles

Answer 48

@waterineyes why did you deleted are your comments?
above ^

Answer 49

The discussion was going long giving no fruitful result..

Answer 50

@SithsAndGiggles I think this system is a Time-Variant, what do you think??

Answer 51

I'd have to say that I don't know enough about time (in)variance to be considered an authority on it. Sorry!

Answer 52

For Input Shift I got:

\[y_1(t) = x(t + \delta - 5) - x(-t + \delta +3)\]

For Output Shift:

\[y_2(t) = x(t + \delta - 5) - x(-t - \delta +3)\]

Answer 53

And I am alone again.. :(

Answer 54

Closing this question.. :( Starting a new question.. :)