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What I don't understand is what exp(.) and ln(.) mean. For example y(t) and x(t) have exp(.) between them. What exactly does that mean? y(t)=?

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    $\begingroup$ This question appears to be homework. Complete answers to homework are off-topic, but specific questions about homework are acceptable if they include enough detail. Please edit the question to include more background about what you don't understand. $\endgroup$ – Marcus Müller Aug 11 '20 at 20:20
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    $\begingroup$ For extra credit, explain to your professor why it's invalid to apply the concept of "impulse response" to that system. For extra-extra credit, explain why doing the $\ln(\cdot)$ operation first is slightly less invalid. For super-extra credit, write out a system description that properly uses the rules of real analysis to show that the impulse response that the professor thinks will happen will, in fact, happen. $\endgroup$ – TimWescott Aug 11 '20 at 21:49
  • $\begingroup$ @TimWescott the z(x) system looks pretty LTI to me, what's wrong with describing it with an impulse response? (the y(x) and the z(y) systems certainly don't). $\endgroup$ – Marcus Müller Aug 12 '20 at 8:34
  • $\begingroup$ It's the interior nonlinearities that are giving me gas. If you were to flop that block diagram in front of my eyes in a work environment and say that's a description of the system, I'd want to make very sure that any of the inevitable undocumented low- (and possibly band-) pass processes acting on $y(t)$ are fast enough that they don't matter for the task at hand -- and I might ask you to document the upper bound of frequency and amplitude where that assumption would still be valid. $\endgroup$ – TimWescott Aug 12 '20 at 15:25
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    $\begingroup$ Given the edit, this should be re-opened. In a block diagram, $f(\cdot)$ means "act on the input with the function $f$", (usually in the time domain). So in the block diagram above, $y(t) = \exp(x(t))$. $\endgroup$ – TimWescott Aug 12 '20 at 15:27
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Hint:

Write the relationship between $z(t)$ and $x(t)$ and then simplify it a bit.

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As suggested by the notation $\exp(\cdot)$, the argument is point-wise, and sometimes called a scalar operation. So for each $x(t)$, you will replace the $\cdot$ by $x(t)$, hence for instance $\exp(x(t))$. The rest could flow.

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What I don't understand is what exp(.) and ln(.) mean. For example y(t) and x(t) have exp(.) between them. What exactly does that mean?

$f(\cdot)$ inside a block means that you should apply $f$ in the time domain to the input.

y(t)=?

So $y(t) = e^{x(t)}$, $z(t) = \ln y(t)$. In theory, in turn that means that $z(t) = x(t)$. In practice there's all sorts of difficulties, because of the nonlinearities "under the hood". But to the extent that practice follows theory, it'll fly.

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