Time-invariant system




A time-invariant (TIV) system has a time-dependent system function that is not a direct function of time. Such systems are regarded as a class of systems in the field of system analysis. The time-dependent system function is a function of the time-dependent input function. If this function depends only indirectly on the time-domain (via the input function, for example), then that is a system that would be considered time-invariant. Conversely, any direct dependence on the time-domain of the system function could be considered as a "time-varying system".


Mathematically speaking, "time-invariance" of a system is the following property:[1]:p. 50



Given a system with a time-dependent output function y(t){displaystyle y(t)}y(t), and a time-dependent input function x(t){displaystyle x(t)}x(t); the system will be considered time-invariant if a time-delay on the input x(t+δ){displaystyle x(t+delta )}x(t+delta ) directly equates to a time-delay of the output y(t+δ){displaystyle y(t+delta )}y(t+delta ) function. For example, if time t{displaystyle t}t is "elapsed time", then "time-invariance" implies that the relationship between the input function x(t){displaystyle x(t)}x(t) and the output function y(t){displaystyle y(t)}y(t) is constant with respect to time t{displaystyle t}t:

y(t)=f(x(t),t)=f(x(t)){displaystyle y(t)=f(x(t),t)=f(x(t))}{displaystyle y(t)=f(x(t),t)=f(x(t))}


In the language of signal processing, this property can be satisfied if the transfer function of the system is not a direct function of time except as expressed by the input and output.


In the context of a system schematic, this property can also be stated as follows:


If a system is time-invariant then the system block commutes with an arbitrary delay.

If a time-invariant system is also linear, it is the subject of linear time-invariant theory (linear time-invariant) with direct applications in NMR spectroscopy, seismology, circuits, signal processing, control theory, and other technical areas. Nonlinear time-invariant systems lack a comprehensive, governing theory. Discrete time-invariant systems are known as shift-invariant systems. Systems which lack the time-invariant property are studied as time-variant systems.




Contents






  • 1 Simple example


  • 2 Formal example


  • 3 Abstract example


  • 4 See also


  • 5 References





Simple example


To demonstrate how to determine if a system is time-invariant, consider the two systems:



  • System A: y(t)=tx(t){displaystyle y(t)=t,x(t)}y(t)=t,x(t)

  • System B: y(t)=10x(t){displaystyle y(t)=10x(t)}y(t)=10x(t)


Since system A explicitly depends on t outside of x(t){displaystyle x(t)}x(t) and y(t){displaystyle y(t)}y(t), it is not time-invariant. System B, however, does not depend explicitly on t so it is time-invariant.



Formal example


A more formal proof of why systems A and B above differ is now presented.
To perform this proof, the second definition will be used.


System A:



Start with a delay of the input xd(t)=x(t+δ){displaystyle x_{d}(t)=,!x(t+delta )}x_{d}(t)=,!x(t+delta )

y(t)=tx(t){displaystyle y(t)=t,x(t)}y(t)=t,x(t)

y1(t)=txd(t)=tx(t+δ){displaystyle y_{1}(t)=t,x_{d}(t)=t,x(t+delta )}y_{1}(t)=t,x_{d}(t)=t,x(t+delta )



Now delay the output by δ{displaystyle delta }delta

y(t)=tx(t){displaystyle y(t)=t,x(t)}y(t)=t,x(t)

y2(t)=y(t+δ)=(t+δ)x(t+δ){displaystyle y_{2}(t)=,!y(t+delta )=(t+delta )x(t+delta )}y_{2}(t)=,!y(t+delta )=(t+delta )x(t+delta )



Clearly y1(t)≠y2(t){displaystyle y_{1}(t),!neq y_{2}(t)}y_{1}(t),!neq y_{2}(t), therefore the system is not time-invariant.


System B:



Start with a delay of the input xd(t)=x(t+δ){displaystyle x_{d}(t)=,!x(t+delta )}x_{d}(t)=,!x(t+delta )

y(t)=10x(t){displaystyle y(t)=10,x(t)}y(t)=10,x(t)

y1(t)=10xd(t)=10x(t+δ){displaystyle y_{1}(t)=10,x_{d}(t)=10,x(t+delta )}y_{1}(t)=10,x_{d}(t)=10,x(t+delta )



Now delay the output by δ{displaystyle ,!delta },!delta

y(t)=10x(t){displaystyle y(t)=10,x(t)}y(t)=10,x(t)

y2(t)=y(t+δ)=10x(t+δ){displaystyle y_{2}(t)=y(t+delta )=10,x(t+delta )}y_{2}(t)=y(t+delta )=10,x(t+delta )



Clearly y1(t)=y2(t){displaystyle y_{1}(t)=,!y_{2}(t)}y_{1}(t)=,!y_{2}(t), therefore the system is time-invariant.


More generally, the relationship between the input and output is y(t)=f(x(t),t){displaystyle y(t)=f(x(t),t)}{displaystyle y(t)=f(x(t),t)}, and its variation with time is



dydt=∂f∂t+∂f∂xdxdt{displaystyle {frac {mathrm {d} y}{mathrm {d} t}}={frac {partial f}{partial t}}+{frac {partial f}{partial x}}{frac {mathrm {d} x}{mathrm {d} t}}}{displaystyle {frac {mathrm {d} y}{mathrm {d} t}}={frac {partial f}{partial t}}+{frac {partial f}{partial x}}{frac {mathrm {d} x}{mathrm {d} t}}}.

For time-invariant systems, the system properties remain constant with time, f/∂t=0{displaystyle partial f/partial t=0}{displaystyle partial f/partial t=0}. Applied to Systems A and B above:




fA=tx(t)⟹fA∂t=x(t)≠0{displaystyle f_{A}=tx(t)qquad implies qquad {frac {partial f_{A}}{partial t}}=x(t)neq 0}{displaystyle f_{A}=tx(t)qquad implies qquad {frac {partial f_{A}}{partial t}}=x(t)neq 0} in general, so not time-invariant


fB=10x(t)⟹fB∂t=0{displaystyle f_{B}=10x(t)qquad implies qquad {frac {partial f_{B}}{partial t}}=0}{displaystyle f_{B}=10x(t)qquad implies qquad {frac {partial f_{B}}{partial t}}=0} so time-invariant.



Abstract example


We can denote the shift operator by Tr{displaystyle mathbb {T} _{r}}{mathbb  {T}}_{r} where r{displaystyle r}r is the amount by which a vector's index set should be shifted. For example, the "advance-by-1" system


x(t+1)=δ(t+1)∗x(t){displaystyle x(t+1)=,!delta (t+1)*x(t)}x(t+1)=,!delta (t+1)*x(t)

can be represented in this abstract notation by


x~1=T1x~{displaystyle {tilde {x}}_{1}=mathbb {T} _{1},{tilde {x}}}{tilde  {x}}_{1}={mathbb  {T}}_{1},{tilde  {x}}

where x~{displaystyle {tilde {x}}}{tilde {x}} is a function given by


x~=x(t)∀t∈R{displaystyle {tilde {x}}=x(t),forall ,tin mathbb {R} }{tilde  {x}}=x(t),forall ,tin {mathbb  {R}}

with the system yielding the shifted output


x~1=x(t+1)∀t∈R{displaystyle {tilde {x}}_{1}=x(t+1),forall ,tin mathbb {R} }{tilde  {x}}_{1}=x(t+1),forall ,tin {mathbb  {R}}

So T1{displaystyle mathbb {T} _{1}}{mathbb  {T}}_{1} is an operator that advances the input vector by 1.


Suppose we represent a system by an operator H{displaystyle mathbb {H} }mathbb {H} . This system is time-invariant if it commutes with the shift operator, i.e.,


TrH=HTr∀r{displaystyle mathbb {T} _{r},mathbb {H} =mathbb {H} ,mathbb {T} _{r},,forall ,r}{mathbb  {T}}_{r},{mathbb  {H}}={mathbb  {H}},{mathbb  {T}}_{r},,forall ,r

If our system equation is given by


y~=Hx~{displaystyle {tilde {y}}=mathbb {H} ,{tilde {x}}}{tilde  {y}}={mathbb  {H}},{tilde  {x}}

then it is time-invariant if we can apply the system operator H{displaystyle mathbb {H} }mathbb {H} on x~{displaystyle {tilde {x}}}{tilde {x}} followed by the shift operator Tr{displaystyle mathbb {T} _{r}}{mathbb  {T}}_{r}, or we can apply the shift operator Tr{displaystyle mathbb {T} _{r}}{mathbb  {T}}_{r} followed by the system operator H{displaystyle mathbb {H} }mathbb {H} , with the two computations yielding equivalent results.


Applying the system operator first gives


TrHx~=Try~=y~r{displaystyle mathbb {T} _{r},mathbb {H} ,{tilde {x}}=mathbb {T} _{r},{tilde {y}}={tilde {y}}_{r}}{mathbb  {T}}_{r},{mathbb  {H}},{tilde  {x}}={mathbb  {T}}_{r},{tilde  {y}}={tilde  {y}}_{r}

Applying the shift operator first gives


HTrx~=Hx~r{displaystyle mathbb {H} ,mathbb {T} _{r},{tilde {x}}=mathbb {H} ,{tilde {x}}_{r}}{mathbb  {H}},{mathbb  {T}}_{r},{tilde  {x}}={mathbb  {H}},{tilde  {x}}_{r}

If the system is time-invariant, then


Hx~r=y~r{displaystyle mathbb {H} ,{tilde {x}}_{r}={tilde {y}}_{r}}{mathbb  {H}},{tilde  {x}}_{r}={tilde  {y}}_{r}


See also



  • Finite impulse response

  • Sheffer sequence

  • State space (controls)

  • Signal-flow graph

  • LTI system theory



References





  1. ^ Oppenheim, Alan; Willsky, Alan (1997). Signals and Systems (second edition). Prentice Hall..mw-parser-output cite.citation{font-style:inherit}.mw-parser-output .citation q{quotes:"""""""'""'"}.mw-parser-output .citation .cs1-lock-free a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .citation .cs1-lock-limited a,.mw-parser-output .citation .cs1-lock-registration a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .citation .cs1-lock-subscription a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration{color:#555}.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration span{border-bottom:1px dotted;cursor:help}.mw-parser-output .cs1-ws-icon a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Wikisource-logo.svg/12px-Wikisource-logo.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output code.cs1-code{color:inherit;background:inherit;border:inherit;padding:inherit}.mw-parser-output .cs1-hidden-error{display:none;font-size:100%}.mw-parser-output .cs1-visible-error{font-size:100%}.mw-parser-output .cs1-maint{display:none;color:#33aa33;margin-left:0.3em}.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-format{font-size:95%}.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-left{padding-left:0.2em}.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-right{padding-right:0.2em}








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