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In the slide 6 of the page about propagation delay of a comparator, the time relationship between input and output is as below. However, I am not sure how the function between input and output voltage is derived. Thank you.

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  • \$\begingroup\$ Perhaps is it measured? \$\endgroup\$ – Marko Buršič May 18 '16 at 9:06
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The frequency domain transfer function is just a fairly simplified model. It makes a critical assumption that there is a single dominant pole (any other poles are much higher in frequency) and no zeroes, which makes this a first-order model—it can usefully predict real-life behaviour as long as that assumption is true, which it might be for certain circuits. It also specifies \$A_v(0)\$ is the gain of the comparator at DC, which is another key parameter of the comparator.

The time domain equation corresponds to a unit step function \$V_{in} u(t)\$ (Laplace: \$\frac{V_{in}}s\$) applied differentially to the inputs of the comparator. You can obtain the time domain equation by one of: a) multiplying the transfer function by the Laplace transform of the unit step, then taking the inverse Laplace transform; b) taking the inverse Laplace transform of the transfer function (to get the impulse response), then convolving that with the time-domain unit step input. (a is easier.)

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  • \$\begingroup\$ Thank you. I tried that but the inverse Laplace transform of the frequency domain transfer funtion is different from the one above. It is Av(0)/tau * e^(-t/tau). The result is different and where the number 1 come from the expression? \$\endgroup\$ – anhnha May 18 '16 at 9:20
  • \$\begingroup\$ Sorry, my mistake—the inverse Laplace of the transfer function only gives the impulse response. Here it looks like the time domain function v_o(t) is assuming you have a step function input of magnitude V_In. See my edit. \$\endgroup\$ – Laogeodritt May 18 '16 at 16:49

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