proving time invarianceof a system.. - Electronic Circuits Projects Diagrams Free

transgalactic · 15th July 2009, 01:47 PM

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

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

so what now??
how does it prove that

$ M_{f(t-x)}=M_{f(t)} $
??
??

dknguyen · 15th July 2009, 09:30 PM

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.

MrAl · 16th July 2009, 02:04 AM

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.

Tesla23 · 16th July 2009, 02:47 AM

Quote:

Originally Posted by transgalactic

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

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

so what now??
how does it prove that

$ M_{f(t-x)}=M_{f(t)} $
??
??

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).

transgalactic · 16th July 2009, 06:00 AM

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 $y(t-\tau)$ step we subtract $\tau$ from every t including the intervals of the integral.
in the $L(x)(t-\tau)$ step we subtract $\tau$ only from the variable inside x function
then we if we get $y(t-\tau) =L(x)(t-\tau)$ then its time invariant

but examples 2 and 3 are not following this pattern

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

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

Tesla23 · 16th July 2009, 06:41 AM

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

$ M_f(t)=0.5\int_{t-1}^{t+1}f(u)du $

is independent of t?

MrAl · 16th July 2009, 07:25 AM

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.

transgalactic · 16th July 2009, 07:53 AM

Quote:

Originally Posted by Tesla23

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

$ M_f(t)=0.5\int_{t-1}^{t+1}f(u)du $

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

??

dknguyen · 16th July 2009, 07:56 AM

The proof here might help you figure out what you need to do for your equation:
Time-invariant system - Wikipedia, the free encyclopedia

Tesla23 · 16th July 2009, 08:03 AM

Quote:

Originally Posted by MrAl

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:

$ 0.5\int_{t-1}^{t+1}f(u-x)du=0.5\int_{t-1-x}^{t-x+1}f(s)ds $

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

dknguyen · 16th July 2009, 08:17 AM

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.

transgalactic · 16th July 2009, 10:07 AM

Quote:

Originally Posted by dknguyen

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

dknguyen · 16th July 2009, 09:11 PM

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.

MrAl · 17th July 2009, 02:44 AM

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:

transgalactic · 17th July 2009, 07:27 AM

thanks

i understand it now

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15th July 2009, 01:47 PM	#1
transgalactic	proving time invarianceof a system.. i need to prove that $<br /> M_{f(t)}=0.5\int_{t-1}^{t+1}f(u)du<br />$ is time invarient ? i know that i need to get the same result for a shift in time $<br /> M_{f(t-x)}=0.5\int_{t-1-x}^{t-x+1}f(u-x)du<br />$ u-x=s -> du=ds $<br /> M_{f(t-x)}=0.5\int_{t-1-x}^{t-x+1}f(s)ds<br />$ so what now?? how does it prove that $<br /> M_{f(t-x)}=M_{f(t)} <br />$ ?? ?? Last edited by transgalactic; 15th July 2009 at 02:18 PM.

15th July 2009, 09:30 PM	#2
dknguyen	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. __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^ Last edited by dknguyen; 15th July 2009 at 09:34 PM.

16th July 2009, 02:04 AM	#3
MrAl	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 by MrAl; 16th July 2009 at 02:12 AM.

16th July 2009, 06:00 AM	#5
transgalactic	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 $y(t-\tau)$ step we subtract $\tau$ from every t including the intervals of the integral. in the $L(x)(t-\tau)$ step we subtract $\tau$ only from the variable inside x function then we if we get $y(t-\tau) =L(x)(t-\tau)$ then its time invariant but examples 2 and 3 are not following this pattern in the second example in the $L(x)(t-\tau)$ step they dont substitute the t with t-tau i expect it to be $t-2\tau$ inside the x function ?? and in the 4th they add T instead of subtracting it why?? Last edited by transgalactic; 16th July 2009 at 07:44 AM.

16th July 2009, 06:41 AM	#6
Tesla23	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 $<br /> M_f(t)=0.5\int_{t-1}^{t+1}f(u)du<br />$ is independent of t?

16th July 2009, 07:25 AM	#7
MrAl	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.

16th July 2009, 07:56 AM	#9
dknguyen	The proof here might help you figure out what you need to do for your equation: Time-invariant system - Wikipedia, the free encyclopedia __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^

16th July 2009, 08:17 AM	#11
dknguyen	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. __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^ Last edited by dknguyen; 16th July 2009 at 08:26 AM.

16th July 2009, 09:11 PM	#13
dknguyen	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. __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^ Last edited by dknguyen; 16th July 2009 at 09:22 PM.

17th July 2009, 02:44 AM	#14
MrAl	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: Attached Thumbnails Last edited by MrAl; 17th July 2009 at 02:45 AM.

17th July 2009, 07:27 AM	#15
transgalactic	thanks i understand it now