Editing Time-invariant system (section)

== 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.

:<u>System A:</u> Start with a delay of the input <math>x_d(t) = x(t + \delta)</math>
::<math>y(t) = t x(t)</math>
::<math>y_1(t) = t x_d(t) = t x(t + \delta)</math>
:Now delay the output by <math>\delta</math>
::<math>y(t) = t x(t)</math>
::<math>y_2(t) = y(t + \delta) = (t + \delta) x(t + \delta)</math>
:Clearly <math>y_1(t) \ne y_2(t)</math>, therefore the system is not time-invariant.

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

More generally, the relationship between the input and output is 

:<math> y(t) = f(x(t), t),</math>

and its variation with time is

:<math>\frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \frac{\mathrm{d} x}{\mathrm{d} t}.</math>

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

:<math> \frac{\partial f}{\partial t} =0.</math> 

Applied to Systems A and B above:

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