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Convolution is is defined as the integral of the product of the two functions after one is reversed and shifted. And in a system where its transfer function is g and the input is f, the output is the convolution of these two functions.

And if a signal is buffered, an ideal buffer should output the input exactly the same.

Lets say the input is f(t) and the transfer function of the buffer is g(t), so convolution of f and g should yield f as the output.

Can we then say the ideal buffer is a Dirac impulse at origin as a function of time?

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    \$\begingroup\$ the transfer function of an ideal buffer in laplace domain is : G(s)= 1. \$\endgroup\$ Oct 19, 2018 at 10:07

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You can say that the impulse response of an ideal buffer is a Dirac pulse. Which is actually quite obvious, because the impulse response has to return the impulse response again in its entirety.

Other variations on the matter:

$$\begin{align} y(t) &= u(t)\\ &\Updownarrow \mathcal{L}\{\}\\ Y(s) &= U(s)\\ &\Downarrow\\ G(s) &= \frac{Y(s)}{U(s)}=1 \end{align}$$

If you want to express this linear function in the time domain, then you would indeed get

$$\begin{align} G(s) &= 1\\ &\Updownarrow \mathcal{L}^{-1}\{\}\\ g(t) &= \delta(t) \end{align}$$

Which has the property:

$$f(t) * g(t) = f(t)$$

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