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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.

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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.

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  • $\begingroup$ you can upvote as well if you found the answer correct. $\endgroup$ – Fat32 Nov 25 '19 at 14:26

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