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Time-invariant system
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<h2 id="simple-example">Simple example</h2> <p>To demonstrate how to determine if a system is time-invariant, consider the two systems: </p> <ul><li>System A: y ( t ) = t x ( t ) {\displaystyle y(t)=tx(t)} </li> <li>System B: y ( t ) = 10 x ( t ) {\displaystyle y(t)=10x(t)} </li></ul> <p>Since the System Function y ( t ) {\displaystyle y(t)} for system A explicitly depends on <i>t</i> outside of x ( t ) {\displaystyle x(t)} , it is not <a href="/facts/Time-invariant/NpWUk0YG">time-invariant</a> because the time-dependence is not explicitly a function of the input function. </p><p>In contrast, system B's time-dependence is only a function of the time-varying input x ( t ) {\displaystyle x(t)} . This makes system B <a href="/facts/Time-invariant/NpWUk0YG">time-invariant</a>. </p><p>The Formal Example below shows in more detail that while System B is a Shift-Invariant System as a function of time, <i>t</i>, System A is not. </p> <h2 id="formal-example">Formal example</h2> <p>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. </p> System A: Start with a delay of the input x d ( t ) = x ( t + δ ) {\displaystyle x_{d}(t)=x(t+\delta )} y ( t ) = t x ( t ) {\displaystyle y(t)=tx(t)} y 1 ( t ) = t x d ( t ) = t x ( t + δ ) {\displaystyle y_{1}(t)=tx_{d}(t)=tx(t+\delta )} Now delay the output by δ {\displaystyle \delta } y ( t ) = t x ( t ) {\displaystyle y(t)=tx(t)} y 2 ( t ) = y ( t + δ ) = ( t + δ ) x ( t + δ ) {\displaystyle y_{2}(t)=y(t+\delta )=(t+\delta )x(t+\delta )} Clearly y 1 ( t ) ≠ y 2 ( t ) {\displaystyle y_{1}(t)\neq y_{2}(t)} , therefore the system is not time-invariant. System B: Start with a delay of the input x d ( t ) = x ( t + δ ) {\displaystyle x_{d}(t)=x(t+\delta )} y ( t ) = 10 x ( t ) {\displaystyle y(t)=10x(t)} y 1 ( t ) = 10 x d ( t ) = 10 x ( t + δ ) {\displaystyle y_{1}(t)=10x_{d}(t)=10x(t+\delta )} Now delay the output by δ {\displaystyle \delta } y ( t ) = 10 x ( t ) {\displaystyle y(t)=10x(t)} y 2 ( t ) = y ( t + δ ) = 10 x ( t + δ ) {\displaystyle y_{2}(t)=y(t+\delta )=10x(t+\delta )} Clearly y 1 ( t ) = y 2 ( t ) {\displaystyle y_{1}(t)=y_{2}(t)} , therefore the system is time-invariant. <p>More generally, the relationship between the input and output is </p> y ( t ) = f ( x ( t ) , t ) , {\displaystyle y(t)=f(x(t),t),} <p>and its variation with time is </p> d y d t = ∂ f ∂ t + ∂ f ∂ x d x 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}}.} <p>For time-invariant systems, the system properties remain constant with time, </p> ∂ f ∂ t = 0. {\displaystyle {\frac {\partial f}{\partial t}}=0.} <p>Applied to Systems A and B above: </p> f A = t x ( t ) ⟹ ∂ f A ∂ t = x ( t ) ≠ 0 {\displaystyle f_{A}=tx(t)\qquad \implies \qquad {\frac {\partial f_{A}}{\partial t}}=x(t)\neq 0} in general, so it is not time-invariant, f B = 10 x ( t ) ⟹ ∂ f B ∂ t = 0 {\displaystyle f_{B}=10x(t)\qquad \implies \qquad {\frac {\partial f_{B}}{\partial t}}=0} so it is time-invariant. <h2 id="abstract-example">Abstract example</h2> <p>We can denote the <a href="/facts/Shift_operator/6c6lc0ks">shift operator</a> by T r {\displaystyle \mathbb {T} _{r}} where r {\displaystyle r} is the amount by which a vector's <a href="/facts/Parameter/zFvVFMv9">index set</a> should be shifted. For example, the "advance-by-1" system </p> x ( t + 1 ) = δ ( t + 1 ) ∗ x ( t ) {\displaystyle x(t+1)=\delta (t+1)*x(t)} <p>can be represented in this abstract notation by </p> x ~ 1 = T 1 x ~ {\displaystyle {\tilde {x}}_{1}=\mathbb {T} _{1}{\tilde {x}}} <p>where x ~ {\displaystyle {\tilde {x}}} is a function given by </p> x ~ = x ( t ) ∀ t ∈ R {\displaystyle {\tilde {x}}=x(t)\forall t\in \mathbb {R} } <p>with the system yielding the shifted output </p> x ~ 1 = x ( t + 1 ) ∀ t ∈ R {\displaystyle {\tilde {x}}_{1}=x(t+1)\forall t\in \mathbb {R} } <p>So T 1 {\displaystyle \mathbb {T} _{1}} is an operator that advances the input vector by 1. </p><p>Suppose we represent a system by an <a href="/facts/Operator_(mathematics)/1VWgEdjS">operator</a> H {\displaystyle \mathbb {H} } . This system is time-invariant if it <a href="/facts/Commutative_operation/WaYxJLxd">commutes</a> with the shift operator, i.e., </p> T r H = H T r ∀ r {\displaystyle \mathbb {T} _{r}\mathbb {H} =\mathbb {H} \mathbb {T} _{r}\forall r} <p>If our system equation is given by </p> y ~ = H x ~ {\displaystyle {\tilde {y}}=\mathbb {H} {\tilde {x}}} <p>then it is time-invariant if we can apply the system operator H {\displaystyle \mathbb {H} } on x ~ {\displaystyle {\tilde {x}}} followed by the shift operator T r {\displaystyle \mathbb {T} _{r}} , or we can apply the shift operator T r {\displaystyle \mathbb {T} _{r}} followed by the system operator H {\displaystyle \mathbb {H} } , with the two computations yielding equivalent results. </p><p>Applying the system operator first gives </p> T r H x ~ = T r y ~ = y ~ r {\displaystyle \mathbb {T} _{r}\mathbb {H} {\tilde {x}}=\mathbb {T} _{r}{\tilde {y}}={\tilde {y}}_{r}} <p>Applying the shift operator first gives </p> H T r x ~ = H x ~ r {\displaystyle \mathbb {H} \mathbb {T} _{r}{\tilde {x}}=\mathbb {H} {\tilde {x}}_{r}} <p>If the system is time-invariant, then </p> H x ~ r = y ~ r {\displaystyle \mathbb {H} {\tilde {x}}_{r}={\tilde {y}}_{r}} <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Finite_impulse_response/5u4Gaqcb">Finite impulse response</a></li> <li><a href="/facts/Sheffer_sequence/CpjBBWcM">Sheffer sequence</a></li> <li><a href="/facts/State_space_(controls)/Jf4EanCP">State space (controls)</a></li> <li><a href="/facts/Signal-flow_graph/Tg5fNxgo">Signal-flow graph</a></li> <li><a href="/facts/LTI_system_theory/kEmj0QDC">LTI system theory</a></li> <li><a href="/facts/Autonomous_system_(mathematics)/ahlr3A6o">Autonomous system (mathematics)</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Oppenheim, Alan; Willsky, Alan (1997). Signals and Systems (second ed.). Prentice Hall. <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>