Regarding the inverse of the system $dx(t)/dt$

Question

The LTI system $\int_{-\infty}^t x(\tau) d\tau$ is invertible with inverse system being $\frac{dx(t)}{dt}$ which is proved with its $$h(t)=u(t)$$ and $$H(s)=1/s,{so}\space H^{-1}(s)=s$$ where $H^{-1}(s)$ is the inverse system transfer function, so $$h^{-1}(t)=\delta'(t), which~is~the~ doublet{\space}function~\frac{d\delta(t)}{dt}$$ where $\delta(t)$ is the delta function

i.e., $y(t)=x'(t)=\frac{dx(t)}{dt}$ is the inverse system of the integrator

All the while the system $dx(t)/dt$ has shown to not have any inverse, on the basis of the argument that any constant function $f(t)=c$ applied to the differentiator give the same "0" output, and also that the associative property in the convolution does not work for integrator as given in this answer: Why aren't the integrator and the differentiator inverse systems?.

My question is, why can't be the integrator the inverse system of the differentiator as one can prove this to be the case by going through the same line of procedure as above in reverse. Also, how to know for what system does the associative property of the convolution fail to work.

Please do point out any mistakes in the above argument, thank you.

Answer 1

The multiplication of transfer functions and the cancellation of common factors in the numerator and denominator implicitly assumes that the corresponding convolution integrals converge and that their order can be interchanged. This is generally not the case when we're dealing with unstable systems. When dealing with unstable (or marginally stable) systems, we have to take a closer look in the time domain.

Note that both the ideal integrator and the ideal differentiator are unstable systems (in the bounded input - bounded output sense). Assuming that for a certain input signal $x(t)$ the integral

$$w(t)=\int_{-\infty}^{t}x(\tau)d\tau\tag{1}$$

converges, we have

$$y(t)=\frac{dw(t)}{dt}=\frac{d}{dt}\int_{-\infty}^{t}x(\tau)d\tau=x(t)\tag{2}$$

Consequently, the differentiator is the inverse of the integrator. However, interchanging the order of the two systems doesn't give the same result:

$$y(t)=\int_{-\infty}^{t}\frac{dx(\tau)}{d\tau}d\tau=x(t)-x(-\infty)\tag{3}$$

Intuitively, if the first system in a chain has zeros on the imaginary axis (like the differentiator at $s=0$), those frequencies are removed from the input signal and cannot be recovered. A following system cancelling these zeros on the imaginary axis is necessarily unstable (marginally stable) because of its poles on the imaginary axis.

Answer 2

The LTI system $\int_{-\infty}^t x(\tau) d\tau$ is invertible with inverse system being $\frac{dx(t)}{dt}$

It's invertible ONLY if the integral converges in the first place, which it doesn't for any signal with a non-zero mean. You cannot invert something that doesn't exist.

My question is, why can't be the integrator the inverse system of the differentiator

For the same reason that $\frac{1}{x}$ is or isn't the inverse of $x$. $\frac{1}{x}$ IS the inverse of $x$ but only for $x \neq 0$. For $x = 0$ $\frac{1}{x}$ is undefined and hence the inverse doesn't exist.

Also, how to know for what system does the associative property of the convolution fail to work.

For the convolution integrals to be associative all integrals involved must be convergent.