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The reason you're having trouble with this is because the impulse response of a system tells you something considerably different from a system's behavior when it is starting with non-zero initial values -- so no one bothers to treat this in the literature. Problem But I think this method will give wrong answer because due to impulse input there must be ...

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In such cases it can be useful to introduce auxiliary variables at the input of the integrators. For the given diagram you could use a signal $u(t)$ at the input of the second integrator. The equation for its Laplace transform $U(s)$ becomes $$U(s)=\frac{1}{s}\big[X(s)-bY(s)\big]-aY(s)\tag{1}$$ You need another equation relating $U(s)$ to $Y(s)$, but that ...

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Matlab can only deal with transfer functions that are rational ratios of polynomials in $s$ (or $z$, if you're in that domain). So you must approximate. You can either use the responses that @Matt L. has shown you, and convolve by the impulse response at each step, or you can write the partial differential equation (not an ODE) and solve it at each time ...

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The inverse Laplace transform of $G(s)=e^{-\sqrt{s}}$ can be computed analytically: $$g(t)=\mathcal{L}^{-1}\left\{e^{-\sqrt{s}}\right\}=\frac{1}{2t\sqrt{\pi t}}e^{-\frac{1}{4t}},\qquad t>0\tag{1}$$ [I've cross-checked this result by looking it up in one of my old math books.] So now you have an analytical expression for the impulse response $g(t)$. The ...

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