Question

Question : 5 Signals :

\[\large \color{green}{y(t) = \frac{1}{T} \int\limits_{t - \frac{T}{2}}^{t + \frac{T}{2}}x(t) \cdot dt}\]

\(\text{How this system is Time-Invariant ??}\)

Answer 1

Am I doing it right??

Answer 2

\[x_1(t) \longrightarrow x(t - t_0)\]

\[y_1(t) \longrightarrow \frac{1}{T} \int\limits_{t - \frac{T}{2}}^{t+ \frac{T}{2}} x_1(t) dt \implies \frac{1}{T} \int\limits_{t - \frac{T}{2}}^{t+ \frac{T}{2}} x(t - t_0) dt\]

Answer 3

Put : \(t - t_0 = \lambda\) such that \(dt = d \lambda\)

Answer 4

\[y_1(t) \longrightarrow \frac{1}{T} \int\limits_{t - t_0- \frac{T}{2}}^{t - t_0+ \frac{T}{2}} x(\lambda) d \lambda\]

Answer 5

While :

\[y(t - t_0) = \frac{1}{T} \int\limits_{t - t_0 -\frac{T}{2}}^{t - t_0+ \frac{T}{2}} x(t - t_0) dt\]

Answer 6

So:

\[y_1(t) \longrightarrow y(t-t_0)\]

So, the system is Time-Invariant..

Answer 7

To what extent I am right in doing these steps???

Answer 8

well you use the sub \[t-t_0=\lambda\]
So That would mean\[t=\lambda+t_0\]

Answer 9

Yes..

Answer 10

Where you are saying??

Answer 11

I mean where you are saying to replace it with \(\lambda + t_0\) ??

Answer 12

Oh I guess you are saying that x(t) is equivalent to x(t-t_0)

Answer 13

I'm just looking at your integral transformation from something in terms of t to something in terms of lambda

Answer 14

why is t replaced with t-t_0?

Answer 15

and not lambda+t_0

Answer 16

Yes, \(x_1(t)\) = \(x(t - t_0)\)

Answer 17

But the limits...why is t replaced with t-t_0

Answer 18

It is for substitution I did..

Answer 19

So t=t-t_0 is your substitution?

Answer 20

\[t - t_0 = \lambda\]

Now change limits..

Answer 21

I'm just trying to figure out why the t was replaced with t-t_0

Answer 22

No.. \(t - t_0 = \lambda\) is my substitution..

Answer 23

Yes, it is for checking time invariance or variance of a system..

Answer 24

1. First you check if your input was \(x(t - t_0)\), then your output is changing correspondingly with same amount or not..

Answer 25

\[If \quad x(t ) \longrightarrow x(t -t_0) \quad then \quad y(t) \longrightarrow y(t-t_0)\]

Answer 26

If output is changing by same amount as the input is, then the system is time invariant, and input and output relationship is maintained.. :)

Answer 27

Are you getting Ma'm what I am trying to do here??

You can very well tell me if I am wrong anywhere according to you..
I am seriously not sure, how to check Time-Invariance of any system.. :(

Answer 28

I'm looking to see if I can find a section on this in my calculus book.

Answer 29

Oh wait..

It is from Signals And Systems..

Answer 30

I am not sure if you will find it in your Calculus book or not.. :)

Answer 31

Ok then I will google signals and systems integrals

Answer 32

That will be nice.. :)

Answer 33

Hey so that capital T outside is just a constant...It is not t right?

Answer 34

Getting something to work on Ma'm??

I need to go now, time is not permitting me to stay here more.. :(

Answer 35

Oh yes, it is just constant.. Capital T is everywhere constant..

Answer 36

A signal is represented as : \(x(t)\), where t is time parameter..

As it is One Dimensional Signal, only t is variable here and others are constants..

Answer 37

http://en.wikibooks.org/wiki/Signals_and_Systems/Time_Domain_Analysis


We want to check to see if these two integrals are equal:

\[\frac{1}{T}\int\limits_{t-\pi/2}^{t+\pi/2}x(t+d) dt \]
\[\frac{1}{T}\int\limits_{t+d-\pi/2}^{t+d+\pi/2}x(t+d) dt \]
and yes I used d instead to t_0

That is what I have come to so far...
And we would need to use transformations to see if these are indeed equal.

Answer 38

I don't think they are equal.. :(

Answer 39

Well I would have to do transformations to see...
But I think you already have gotten these integrals above.

Answer 40

No, I have got something different..

Answer 41

I don't see any difference.

Answer 42

I have d instead of -t_0
this is just a constant

Answer 43

anyways I do not think the first integral equals the second integral

Answer 44

I mean:

Look at my \(y_1(t)\) and \(y(t-t_0)\) expressions..

Answer 45

Yes, that is what I think, Oh so, we think the same.. :P

Answer 46

There \(y_1(t)\) and \(y(t-t_0)\) are equal I think..

Answer 47

yes

Answer 48

The only difference is instead of them using -t_0 they put +d

this is not really a difference though

Answer 49

wait wait
I think I'm wrong
y_1 =y_2 i think

Answer 50

I will show you

Answer 51

Difference?

Which two expressions you are looking difference in??

Answer 52

So we are comparing \[y_2=\frac{1}{T}\int\limits_{t-t_0-\pi/2}^{t-t_0+\pi/2} x(t-t_0) dt \\ \text{ and } y_1=\frac{1}{T}\int\limits_{t-\pi/2}^{t+\pi/2} x(t-t_0) dt \]
I will go back and use -t_0 instead +d
So you see which of your two integrals I'm comparing
I will use the sub on the first integral \[\lambda=t-t_0 => d \lambda= d t \]
so we have
\[y_2(\lambda)=\frac{1}{T}\int\limits_{\lambda-\pi/2}^{\lambda+\pi/2}x(\lambda) d \lambda \]

So if our input was t we would have
\[y_2(t)=\frac{1}{T}\int\limits_{t-\pi/2}^{t+\pi/2}x(t) dt \]
But you told me x(t)=x(t-t0)
so y_1=y_2

Answer 53

so they are time in-variant

Answer 54

Wait, let me take me some time for viewing it properly.. :)

Answer 55

That's the thing I was looking for.. :)

Real help.. :)

Thank you Ma'm.. :)

Answer 56

I'm glad I could help. I guess it makes perfect sense?

Answer 57

Just perfect, nothing greater than that.. :)

Answer 58

nice

Answer 59

I have many more with having doubts in to work on.. :)

May be next time I will pick them.. :)