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I'm given the following system \$H\$ $$y(t)=\exp\left(-\intop_{-\infty}^{\infty}x(\tau)\cdot e^{-|t-\tau|}d\tau\right)$$ I proved it's TI (time-invariant) and that it's not linear.

Now we take as input \$y(t)\$ and we pass it through \$H_2\$, we are asked to construct \$H_2\$ such that the combination of \$H\$ and \$H_2\$ will be an LTI system.

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When passing a linear combination of x signals through \$H\$ I get the following:

$$\exp\left(-{\displaystyle \intop_{-\infty}^{\infty}}ax_{1}\left(\tau\right)e^{-\left|t-\tau\right|}\right)\cdot\exp\left(-{\displaystyle \intop_{-\infty}^{\infty}}bx_{2}\left(\tau\right)e^{-\left|t-\tau\right|}d\tau\right)$$

In comparison to passing each part of it through \$H\$ and then multiplying each by the coefficients and adding them together

$$a\cdot\exp\left(-{\displaystyle \intop_{-\infty}^{\infty}}x_{1}\left(\tau\right)e^{-\left|t-\tau\right|}d\tau\right)+b\exp\left(-{\displaystyle \intop_{-\infty}^{\infty}}x_{2}\left(\tau\right)e^{-\left|t-\tau\right|}d\tau\right)$$

And this is where I'm stuck, I don't know how to get from all this information (from the question and my calculations) if a system \$H_2\$ is possible or not, and if it's possible I don't know how to derive it.

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  • \$\begingroup\$ Need to see an attempt at a solution, Use Fourier transform equivalents \$\endgroup\$ Commented Nov 26, 2024 at 17:46
  • \$\begingroup\$ @VoltageSpike we've yet to learn about using the Fourier transform, it's the first practice exercises paper we have since learned about systems just yesterday \$\endgroup\$ Commented Nov 26, 2024 at 17:51
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    \$\begingroup\$ System responses and LTI systems QA's also fit dsp.stackexchange.com section btw. \$\endgroup\$ Commented Nov 26, 2024 at 19:39

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\$H_2=0\$ linearizes any system and makes it also time-invariant. This is not really a joke answer– in the end you're being asked to find a solution to a system of differential equations, and this is one particular solution.

Now, more productively: Let's start with the easy part, the nonlinearity. I'd say pretty intuitively, the \$\exp\$ ruins our day there. What counteracts an \$\exp(…)\$? Right, an application of \$\ln\$: \$\ln(\exp(u))=u\$, and \$u\$ is a constant multiple of \$u\$, so this is linear. If \$u\$ itself is linear in its input, then linear of linear is still linear. \$u(x(t))=-\int\limits_{-\infty}^{\infty}x(\tau)\cdot e^{-|t-\tau|}\mathrm{d}\tau\$, in this case, and we can test:

  • Homogenity: \$u(\alpha\cdot x)=-\int\limits_{-\infty}^{\infty}\alpha x(\tau)\cdot e^{-|t-\tau|}\mathrm{d}\tau=-\alpha\int\limits_{-\infty}^{\infty}x(\tau)\cdot e^{-|t-\tau|}\mathrm{d}\tau=\alpha\cdot u(x)\$, cheeeeck,
  • Additivity: \$u(x_1 + x_2) = u(x_1)+u(x_2)\$ (same calculation. Simply works because the integral is a linear operation.)

So, yeah that thing is linear.

Now, about the time variance: I'll leave this up to you, but I think your proof of time invariance might have room for improvement :) compare this integral with the convolution integral. You're just convolving \$x(t)\$ with \$e^{-\lvert t \rvert}\$

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  • \$\begingroup\$ I did think of ln of course but in my head, it doesn't work, even taking your example of Homogeneity: if the input signal is ax then the output after H is the exp of the integral you wrote, but then with exponent rule,s you can bring it to the form of e^a * e^(the integral) and then passing that through a ln gives us the output a+(the integral) which is different then what you expressed \$\endgroup\$ Commented Nov 26, 2024 at 20:48
  • \$\begingroup\$ correct me if I'm wrong, but \$\ln(\exp(\alpha u)) = \alpha u \ne \alpha + u\$. You applied the ln twice, maybe? \$\endgroup\$ Commented Nov 27, 2024 at 10:25

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