Is the system $y\left(t\right)=\int _{t^3-1}^{t^3}x\left(s\right)ds\:$ invertible?

Question

I have the following system: $$y\left(t\right)=\int _{t^3-1}^{t^3}x\left(s\right)ds\:$$ I was told to determine if it's invertible system, casual system, memoryless system and linear system. I was able to tell that the system is linear, not memoryless and not casual. However I have no clue how to determine if it's invertible or not so I would like to get the answer with explanation and if there is a general way to find out about invertibility of systems that equals to integral with limits.

Answer 1

About Invertible Systems

System: $$ y(t)=H\{x(t),t\}=\int_{t^3-1}^{t^3} x(s)ds $$

Single Input, Single Output, Non Zero

This is evident. The system has one input, one output, and is different from the zero trivial system ($H\{x(t)\}=0 \forall x$).

Linear

The system is Linear, since the integral operator is linear in $x$: $$ H\{ax(t),t\}=\int_{t^3-1}^{t^3} ax(s)ds=a\int_{t^3-1}^{t^3} x(s)ds=aH\{x(t),t\} \forall a\\ H\{x_1(t)+x_2(t),t\} =\int_{t^3-1}^{t^3} x_1(s)+x_2(s)ds\\ =\int_{t^3-1}^{t^3} x_1(s)ds+\int_{t^3-1}^{t^3} x_2(s)ds \\ =H\{x_1(t),t\}+H\{x_2(t),t\} \forall x_1 \ x_2 $$

Time Variant

The system is Time Variant since the response on time delays in inputs differs from time delays in outputs $$ H\{x(t-dt),t-dt\}=\int_{(t-dt)^3-1}^{(t-dt)^3} x(s-dt)ds \\ =\int_{(t-dt)^3-1+dt}^{(t-dt)^3+dt} x(s)ds \\ \ne H\{x(t-dt),t\}=\int_{t^3-1}^{t^3} x(s-dt)ds \\ =\int_{t^3-1+dt}^{t^3+dt} x(s)ds $$

Dynamic

The system is Not Static, Dynamic or With Memory, meaning all the same, since it NOT depends only on the present input. On our case, it depends on past and future inputs, according the interval range $[t^3-1 \ t^3]$ including values less or greater than $t$. $$ H\{x(t),t\}=\int_{t^3-1}^{t^3}x(s)ds=H\{x(t^3-1),...,x(t^3),t\} $$

Not Causal

The system is Not Causal, since the output does NOT depends exclusively on past inputs. Only for the time ranges with $t\le 1$ we have a Causal behaviour $[t^3-1 \ t^3] \le t$: $$ H\{x(t),t\}=\int_{t^3-1}^{t^3}x(s)ds=H\{x(\tau)|_{\tau<t},t\} $$

Invertible

The system is Invertible, since there exists a system $G=H^{-1}\{y(t),t\}$, for all $y(t)=H\{x(t),t\}$ which allows to recover the original $x(t)$. $$ x(t)=H^{-1}\{H\{x(t),t\},t\} \forall x $$

In our case, since the system is based on an integral, we can simply obtain an inverse by derivating and changing variables. As you could expect, this is not the general case, moreover for a Time Varying system.

In fact, this expression gives us straightfordwardly the value of $x(t)$ with respect the delayed version $x(t-1)$: $$ \frac{d}{dt}y(t^{1/3})=\frac{d}{dt}\int_{t-1}^{t}x(s)ds=x(t)-x(t-1) $$

Hence, this sum returns our requested input, solving the inversion, only requiring to know the initial input $x_0$. $$ H^{-1}(y(t))=\sum_{k=1}^N\frac{d}{dt}y((t-k)^{1/3})\\ =\sum_{k=1}^N x(t)-x(t-k)=x(t)-x_0 $$

Systems such as $H\{x(t)\}=tx(t), t\ne 0$ are genuinely Invertible since its inverse $H^{-1}\{y(t)\}=\frac 1t y(t)$ does not depend on any initial condition, but for all dynamical systems based on integration, we can name them as invertible accepting to constraint the initial input to a known value $x_0$.

Definition

Equivalently, a System is Invertible if $H$ is injective, i.e. maps different outputs with different inputs (Ref.).

Invertible, Matrices and Rank

Remember all Linear Systems are Homeomorphic to a Matrix representation, which have a Rank. If that Matrix has non zero Rank, the system will be invertible.

Integrals, Derivatives, time scalig, IIR are (non trivially) linked to a Matrix representation.

Hence, Non Zero Gain SISO Linear systems will be Invertible. A System with Matrix representation with Non Zero Rank, will be Invertible too (Ref.).

Invertible does NOT imply Causality ($H\{x(t)\}=x(t+1)$), Stability ($H\{x(t)\}=\frac{d}{dt}x(t)$) nor evidently Linearity.

Final Remark

This is a Linear System, Non Zero Gain and Single Input Single Output. Can you finally conclude from there, that the system is Invertible?