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proving time invarianceof a system..

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i need to prove that
[latex]
M_{f(t)}=0.5\int_{t-1}^{t+1}f(u)du
[/latex]
is time invarient
?

i know that
i need to get the same result for a shift in time
[latex]
M_{f(t-x)}=0.5\int_{t-1-x}^{t-x+1}f(u-x)du
[/latex]
u-x=s -> du=ds
[latex]
M_{f(t-x)}=0.5\int_{t-1-x}^{t-x+1}f(s)ds
[/latex]

so what now??
how does it prove that

[latex]
M_{f(t-x)}=M_{f(t)}
[/latex]
??
??
 
Last edited:

dknguyen

Well-Known Member
Most Helpful Member
Prove by induction is how I would try going about it. I suck at math proofs though so take my advice critically.

Or maybe take what you have right now and break up boundaries of itnegration and turn it into a sum of integrals and see what you can get from that.
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
Hi,


I dont know if this helps or not yet, but you do not need to show that
the output is the *same* with a shifted input (as you seem to be saying),
you need to show that the output is *shifted* by the same amount as the
shifted input. In other words, for no shifted input then we get no shifted output,
but for a shifted input then we get a shifted output too. If we get a
different output (other than shifted by the same amount) then the system
is not time invariant.


For:
y(t)=F(x(t))

we also get:
y(t-d)=F(x(t-d))

Note we do *not* get y(t)=F(x(t-d)).

Hope that helps.
 
Last edited:

Tesla23

Member
i need to prove that
[latex]
M_{f(t)}=0.5\int_{t-1}^{t+1}f(u)du
[/latex]
is time invarient
?

i know that
i need to get the same result for a shift in time
[latex]
M_{f(t-x)}=0.5\int_{t-1-x}^{t-x+1}f(u-x)du
[/latex]
u-x=s -> du=ds
[latex]
M_{f(t-x)}=0.5\int_{t-1-x}^{t-x+1}f(s)ds
[/latex]

so what now??
how does it prove that

[latex]
M_{f(t-x)}=M_{f(t)}
[/latex]
??
??
If I understand the question correctly, you are asking when the mean of f(u) over [t-1, t+1] is independent of t. You should be able to show that this is only true if f(u) is periodic with period 2, i.e. f(u+2)=f(u).
 
the correct to my question if the first post in this thread is that it time invariant

here are similar solved questions
http://i27.tinypic.com/2w1z1uo.gif
from the first and 4th examples there i got the idea that
in the [latex]y(t-\tau)[/latex] step we subtract [latex]\tau[/latex] from every t including the intervals of the integral.
in the [latex]L(x)(t-\tau)[/latex] step we subtract [latex]\tau[/latex] only from the variable inside x function
then we if we get[latex]y(t-\tau) =L(x)(t-\tau)[/latex] then its time invariant


but examples 2 and 3 are not following this pattern

in the second example
in the [latex]L(x)(t-\tau)[/latex] step they dont substitute the t with t-tau
i expect it to be [latex]t-2\tau[/latex] inside the x function
??

and in the 4th they add T instead of subtracting it
why??
 
Last edited:

Tesla23

Member
I'm sorry, but I find what you are saying very confusing.

Can you state what you are actually trying to do, before you start copying in bits of other problems. Are you trying to find the conditions under which

[latex]
M_f(t)=0.5\int_{t-1}^{t+1}f(u)du
[/latex]

is independent of t?
 

MrAl

Well-Known Member
Most Helpful Member
Hi,


He is trying to prove if a system is time invariant or not.

Time invariance is not that the function does not depend on time t,
but that it is independent of the *starting* time of the input function.
In other words, if we start the input at t=0 and we get response y(t),
then if we instead start the input at t=tau we get the response y(t-tau)
then the system is time invariant. If we get anything other than y(t-tau)
as output, the system is not time invariant.
 
I'm sorry, but I find what you are saying very confusing.

Can you state what you are actually trying to do, before you start copying in bits of other problems. Are you trying to find the conditions under which

[latex]
M_f(t)=0.5\int_{t-1}^{t+1}f(u)du
[/latex]

is independent of t?
yes it is
its like in my example but here x() if called f()

regarding to MrAl's post:
"if we start the input at t=0 and we get response y(t),"

"then if we instead start the input at t=tau we get the response y(t-tau)
then the system is time invariant"

ok so lets look the the 4th example
you just say to substitute t for tau

but we dont get y(t-tau) that way

??
 
Last edited:

Tesla23

Member
He is trying to prove if a system is time invariant or not.
Ah, now I re-read it you're right. This is simply the change of variable:

[latex]
0.5\int_{t-1}^{t+1}f(u-x)du=0.5\int_{t-1-x}^{t-x+1}f(s)ds
[/latex]

and the proof that the mean being independent of time required periodicity was so neat!
 
Last edited:

dknguyen

Well-Known Member
Most Helpful Member
I think I got the answer...it's so ghetto the way I did it though. I made two versions of the equation M(t):

Equation 1. Shifted output by making M(t) -> M(t+σ). This basically just changes the boundaries of integration. I did no more work on the formula after that.

Equation 2. Shifted input by making u -> u+σ and plugging that back into M(t). Now I worked to make Equation 2 look like Equation 1. THe key thing I had to do was make d(u) -> d(u+σ) = (du +dσ) = du (since σ is a constant.) I assumed I integrated the function, expanded it out to evalate between the integration boundaries, then repackaged it back into an bounded integral to get Equation 1.

If Equation (shifted output) equals Equation 2 (shifted input), then it means shifting the output, t in M(t) has the same effect as shifting the input. u in f(u) which means time invariant.
 
Last edited:
I think I got the answer...it's so ghetto the way I did it though. I made two versions of the equation M(t):

Equation 1. Shifted output by making M(t) -> M(t+σ). This basically just changes the boundaries of integration. I did no more work on the formula after that.

Equation 2. Shifted input by making u -> u+σ and plugging that back into M(t). Now I worked to make Equation 2 look like Equation 1. THe key thing I had to do was make d(u) -> d(u+σ) = (du +dσ) = du (since σ is a constant.) I assumed I integrated the function, expanded it out to evalate between the integration boundaries, then repackaged it back into an bounded integral to get Equation 1.

If Equation (shifted output) equals Equation 2 (shifted input), then it means shifting the output, t in M(t) has the same effect as shifting the input. u in f(u) which means time invariant.
"M(t) -> M(t+σ). This basically just changes the boundaries of integration"(equation1)
"u -> u+σ and plugging that back into M(t)"(equation2)
i cant see any difference betwen the two
in both you substitute u with u+σ
i cant follow your method on my examples
can you show how you apply it on example 4
and on example 3
 
Last edited:

dknguyen

Well-Known Member
Most Helpful Member
No, in 1 you substitute t -> t+σ (this only changes the boundaries of integration).
In 2 you substitute u -> u+σ (this only changes the the infintesmal element you are integrating over...u and du.

Then if you manipulate 2, you can end up making it look exactly like 1 again. THe thing to not get mixed up about this is that the boundaries of integration is TIME (I'm sure can see why since there is a 't' in there). But the thing not to miss is that u is also time but a different variable of time than 't'. You can tell because after you integrate you end up substituting u with the boundaries of integration (which is time because of the 't'. It's just that in math convention you can't use the same variable for two different things.

So shifting t is time shifting the equation's output M(t). But shifting u is time shifting the input f(u). So you are shifting time in both cases, but not the same time variable. If it is time invariant, you should get the same result either way. It's almost like working backwards from solution 1 ("I have an output, let's find out what input is required to get this output) and working forward from solution 2 ("I have an input, let's see how this input changes the output" to see if they meet in the middle. If they meet in the middle, then solution 1 and 2 are the same thing.

...I think.
 
Last edited:

MrAl

Well-Known Member
Most Helpful Member
Hi again,


Here is a graphical illustration of time invariance (and time variance) of a
network for those interested in this view which makes things a little clearer.

The math behind this is as follows:

The first pulse is made from two unit steps, one plus and one minus,
spaced 1 second apart. This creates a pulse from t=0 seconds to
t=1 seconds. This input is then convolved with the impulse
response of the network, and the output becomes y1(t).

The second pulse is also made from two unit steps, spaced 1 second apart,
but also delayed by 1 second (tau). This creates a pulse from t=1 second to
t=2 seconds. This new input is then convolved with the impulse response
of the network, and the output becomes y2(t).

We next substitute t-tau into y1 to get y1(t-tau) instead of y1(t).
Now if y1(t-tau)=y2(t) the network is time invariant.

The only caution when doing the math is that there are two parts to each
pulse response, the charging time and the discharging time, which have to be
handled separately.


As shown graphically, when we apply non zero initial conditions we end
up with a time variant network.

Here's the graphic:
 

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Last edited:

nguyen12

New Member
Basically time invariant system

Anybody plz explain!
i know that the time invariant system which if you time shift the input, get the same output, but with the same time shift.
ex: y[n]=nx[n]
solution
x[n] delay k--->x[n-k] and then transform---->nx[n-k]
x[n] transform ----->y[n]=nx[n] and then delay k-------->(n-k)x[n-k]( why not nx[n-k]).
i really do not understand.
another ex: y[n]=x[n*n]
if i do the same above,it will
x[n]----->x[n-k]------->x[(n-k)*(n-k)]
x[n]----->x[n*n]------>???
thank so much
 
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