Time-invariant system



In control theory, a time-invariant (TI) 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:


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

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, as shown in the figure to the right:


 * 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.

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


 * System A: $$y(t) = t x(t)$$
 * System B: $$y(t) = 10 x(t)$$

Since the System Function $$y(t)$$ for system A explicitly depends on t outside of $$x(t)$$, it is not time-invariant because the time-dependence is not explicitly a function of the input function.

In contrast, system B's time-dependence is only a function of the time-varying input $$x(t)$$. This makes system B time-invariant.

The Formal Example below shows in more detail that while System B is a Shift-Invariant System as a function of time, t, System A is not.

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 $$x_d(t) = x(t + \delta)$$
 * $$y(t) = t x(t)$$
 * $$y_1(t) = t x_d(t) = t x(t + \delta)$$
 * Now delay the output by $$\delta$$
 * $$y(t) = t x(t)$$
 * $$y_2(t) = y(t + \delta) = (t + \delta) x(t + \delta)$$
 * Clearly $$y_1(t) \ne y_2(t)$$, therefore the system is not time-invariant.


 * System B: Start with a delay of the input $$x_d(t) = x(t + \delta)$$
 * $$y(t) = 10 x(t)$$
 * $$y_1(t) = 10 x_d(t) = 10 x(t + \delta)$$
 * Now delay the output by $$\delta$$
 * $$y(t) = 10 x(t)$$
 * $$y_2(t) = y(t + \delta) = 10 x(t + \delta)$$
 * Clearly $$y_1(t) = y_2(t)$$, therefore the system is time-invariant.

More generally, the relationship between the input and output is


 * $$ y(t) = f(x(t), t),$$

and its variation with time is


 * $$\frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \frac{\mathrm{d} x}{\mathrm{d} t}.$$

For time-invariant systems, the system properties remain constant with time,


 * $$ \frac{\partial f}{\partial t} =0.$$

Applied to Systems A and B above:


 * $$ f_A = t x(t) \qquad \implies \qquad \frac{\partial f_A}{\partial t} = x(t) \neq 0 $$ in general, so it is not time-invariant,
 * $$ f_B = 10 x(t) \qquad \implies \qquad \frac{\partial f_B}{\partial t} = 0 $$ so it is time-invariant.

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


 * $$x(t+1) = \delta(t+1) * x(t)$$

can be represented in this abstract notation by


 * $$\tilde{x}_1 = \mathbb{T}_1 \tilde{x}$$

where $$\tilde{x}$$ is a function given by


 * $$\tilde{x} = x(t) \forall t \in \R$$

with the system yielding the shifted output


 * $$\tilde{x}_1 = x(t + 1) \forall t \in \R$$

So $$\mathbb{T}_1$$ is an operator that advances the input vector by 1.

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


 * $$\mathbb{T}_r \mathbb{H} = \mathbb{H} \mathbb{T}_r \forall r$$

If our system equation is given by


 * $$\tilde{y} = \mathbb{H} \tilde{x}$$

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

Applying the system operator first gives


 * $$\mathbb{T}_r \mathbb{H} \tilde{x} = \mathbb{T}_r \tilde{y} = \tilde{y}_r$$

Applying the shift operator first gives


 * $$\mathbb{H} \mathbb{T}_r \tilde{x} = \mathbb{H} \tilde{x}_r$$

If the system is time-invariant, then


 * $$\mathbb{H} \tilde{x}_r = \tilde{y}_r$$