The time-invariant system is defined as following:

If y(t) = (H(x))(t) and y(t − τ ) = (H(x))(t− τ ) Then H is a time-invariant system.

Has this definition taken system initial stats account? At t=0 and t=τ, the same system could have different states, due to a discharging capacitor and etc. Then y(t) and y(t-τ) could be different for a time-invariant system.

So when taking initial states into account, a time-invariant system could be no longer time-invariant. Is my understanding correct?

  • \$\begingroup\$ When you say "time-invariant system", do you mean a linear time invariant system (LTI), or a generic time-invariant system (i.e. a system that may be also non-linear). \$\endgroup\$ – Lorenzo Donati supports Monica Aug 26 '15 at 13:45
  • \$\begingroup\$ @LorenzoDonati A general time-invariant system actually. \$\endgroup\$ – richieqianle Aug 26 '15 at 13:55

Has this definition taken system initial stats account?

If it doesn't hold for any τ, it isn't time invariant! That's the definition, and it's enough to answer your question:

At t=0 and t=τ, the same system could have different states, due to a discharging capacitor and etc. Then y(t) and y(t-τ) could be different for a time-invariant system.

They are different, so the system MUST be time variant!

Really, that's it.

The story here is you are mixing a property of linear systems with time invariant systems:

A linear system with no initial state is time invariant. Because it is linear. A linear system with a initial state is time variant. Because the initial state doesn't shift with the input.

Yet, the system is linear. That means you can decompose the system in 2 parts:

  1. The Homogeneous State Response (u=0,x(0)!=0)

  2. The Forced State Response (u!=0,x(0)=0)

The output of the system is the sum of those two part (remember, it's linear!). The system is only fully time invariant if the state response is 0, otherwise it doesn't fit the definition!

Still, the second system is time-invariant, and you can always use time invariant techniques over it. You just have to remember to sum the response coming from 1

  • \$\begingroup\$ More reading over here web.mit.edu/2.14/www/Handouts/StateSpaceResponse.pdf I had to look for the names \$\endgroup\$ – MdxBhmt Aug 26 '15 at 18:02
  • \$\begingroup\$ IIRC: A LTI system is a system that has it's coefficients time invariant - a subtle difference. \$\endgroup\$ – MdxBhmt Aug 26 '15 at 18:04
  • \$\begingroup\$ Do you mean that when we say a system is timeinvariant, we are also assuming the initial states are zero? \$\endgroup\$ – richieqianle Aug 27 '15 at 2:27
  • \$\begingroup\$ Exactly. I'm trying to find a good resource with all definition so I can justify this better - requiring 'state=0' before determining if its Time invariant feels iffy and not general - maybe it doesn't matter because a time invariant system may be equivalent to linear system ( so you can always separate the initial state as a separate concern) \$\endgroup\$ – MdxBhmt Aug 27 '15 at 10:25

Okay, I am a bit new here, so pardon my mistakes of the answer aesthetics.


A system is time invariant if e(t)→r(t) implies that e(t±T)→r(t±T). where e(t), r(t) and t bears the conventional meaning of excitation, response, and time.Here T is the finite delayed time.Plus-minus sign is non-causal signal.

and that's all.

To understand it we need some explanation right?

The definition written above could be explained for a Linear System (superposition and proportionality), we excite the system initially at t = 0 with e(t) and the system responded with r(t).Again when the excitation is introduced at t=T and if the shape of the response is same compared to the first case except a time delay of T, then we could say the system is time invariant.

To address your question I have taken the liberty to modify the frame of the question.your question is basically

Can a system be time invariant which contains time varying elements?

You got it correct, the obvious answer is NO.

Another way of looking at this concept is through the fact that the time invariant system contains only elements that doesn't varies with time.

Linear systems need not be time invariant.

  • \$\begingroup\$ Welcome to this great forum and thanks for your answer. The aesthetics is quite good. If you check this link web.mit.edu/2.14/www/Handouts/StateSpaceResponse.pdf i think they usually assume the time variant elements, which are states to be zero when determine a system is time variant or not. \$\endgroup\$ – richieqianle Aug 28 '15 at 16:50
  • \$\begingroup\$ Thank you.Yes, when the system is relaxed that is there are no magnetic energy in inductors and no charge in capacitors at time t=0-. The output of e(t)=0 and T will have same shape of r(t) at 0 with the only difference of delay of time T between them.(provided the system is causal right?) \$\endgroup\$ – kaustav Aug 28 '15 at 17:05
  • \$\begingroup\$ Thanks for your reply. May I ask how a system could be non-causal? \$\endgroup\$ – richieqianle Aug 29 '15 at 1:59
  • \$\begingroup\$ I would like to address the system in informal way.A system which has no memory is called a causal signal.We get output only when, the system is under some excitation. \$\endgroup\$ – kaustav Aug 29 '15 at 14:09
  • \$\begingroup\$ I thought causal systems depend on past and current inputs but not future ones. If so, it can has memory i think. \$\endgroup\$ – richieqianle Aug 29 '15 at 16:25

First, I think you meant to write \$y(t)=H(x(t))\$. Such a system has no memory, i.e it has no states. \$y(t)\$ is computed using instantaneous values of \$x(t)\$, so it does not depend on previous values and the initial conditions do not come into play. If \$x(t)\$ is delayed by any amount then \$y(t)\$ is also delayed by the same amount. Thus you have a memoryless, time-invariant system.

  • 1
    \$\begingroup\$ I believe his using notation correctly, H being an operator over input x \$\endgroup\$ – MdxBhmt Aug 26 '15 at 15:57
  • \$\begingroup\$ Even if H is an operator, it acts on the input x(t) which is a function of time, i.e. H(x(t))? \$\endgroup\$ – Suba Thomas Aug 26 '15 at 16:12
  • \$\begingroup\$ I would say that's correct, but often omitted. \$\endgroup\$ – MdxBhmt Aug 26 '15 at 17:51
  • \$\begingroup\$ No, I think the notation is perfect: H is an operator that maps an entire input function x to an entire output function H(y), whose value at time t is (H(x))(t). Really mathematical stuff, though (usually found in advanced signal analysis/theory books, seldom in material for undergraduates). \$\endgroup\$ – Lorenzo Donati supports Monica Aug 26 '15 at 19:00

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