# System memory, causality, stability

im new into systems and im supposed to solve if the system has memory, us causal, linear, stationery, BIBO stable...The problem is i have never had experience with this type of system where the actual m is squared. I tried to put in some reasoning and my guess is that system does have memory (it is integral after all), but i cant decide if it is causal. I think i proved successfully that this system should be linear. I tried to prove that system is stationary but in the very first step i dont even know how to substitute , and when it comes to BIBO stability i am lost because i cant decide if i should suppose that is boundary or is boundary.

It's easy to show that the system defined by

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

is

Not Causal : it's because for for any value of $$t$$ the output $$y(t)$$ will always depend on future values of the input $$x(t)$$. Because while the dummy variable $$\tau$$ used in the integration ranges from $$-\infty$$ to $$t$$. The argument of the input function $$x(t)$$ will range from $$0$$ to $$\infty$$ due squaring. Hence for any finite value of $$t$$, th eoutput $$y(t)$$ will depend on value for $$x(t)$$ u to time infinity. Hence non-cusal.

Has Memory: as you guessed to due integration, current value of the output $$y(t)$$ depends on values other than current time $$t$$.

Linear: follows from the linearity of the integral operator.

Time-Varying : Lets call shifted input as $$x_2(t) = x(t-d)$$ and compute the corresponding output $$y_2(t)$$ for the shifted input:

$$y_2(t) = \int_{-\infty}^{t} x_2(\tau^2) d\tau = \int_{-\infty}^{t} x(\tau^2 - d) d\tau$$

then also compute the shifted version of $$y(t)$$ by $$d$$ as $$y_d(t) = y(t-d) = \int_{-\infty}^{\tau = t-d} x(\tau^2) d\tau$$

Let $$\beta = \tau + d$$ and $$\tau = \beta - d$$; substitude into the integral: $$y(t-d) = \int_{-\infty}^{ t} x((\beta - d)^2) d\beta = \int_{-\infty}^{ t} x(\tau^2 - 2\tau d +d^2) d\tau$$

clearly $$y_2(t) \neq y(t-d)$$ and system is shift-varying.

Unstable: by counter-example let $$x(t) = u(t)$$ and see that output goes unbounded to infinity while the input is bounded for all $$t$$. Hence unstable in the BIBO sense.

• you can upvote as well if you found the answer correct. Nov 25, 2019 at 14:26