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enter image description here

Another basic question to understand. It is well known that the transfer function of a low pass RC filter in Laplace domain is \$V_o(s)/V_i(s)=1/(1+sRC)\$. Now sticking to this format and taking inverse Laplace transform \$V_o(t)/V_i(t)=e^{-(t/RC)}\$. All text books have the time domain equation as \$V_o(t)/V_i(t)=1-e^{-(t/RC)}\$ considering a step response as shown in the picture attached. But if i was to strictly follow \$V_o(s)/V_i(s)\$ and its time domain counterpart, without specifically mentioning that \$V_i(s)\$ or \$V_i(t)\$ is a step response then how are the first two equations similar? What am I missing here?

Also if i am correct if \$V_o(t)/V_i(t)=1-e^{-(t/RC)}\$ then \$V_o(s)/V_i(s)=1/s(1+sRC)\$ and not what is shown in \$V_o(s)/V_i(s)=1/(1+sRC)\$. This is really driving me up the wall.

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  • \$\begingroup\$ one input is 'impulse'; another is 'step' \$\endgroup\$
    – user361690
    Jan 12 at 17:20

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Vo(s)/Vi(s) = 1/(1+sRC) doesn't state that Vo(t)/Vi(t) is exp(-t/RC). It states that the impulse response of the filter is exp(-t/RC) i.e. the output in case the input is Dirac's delta would be exp(-t/RC). There's no simple rule for the s-domain presentation of the ratio of time domain functions Vo and Vi.

The output time domain function can be calculated from an arbitary input time domain function as convolution integral of the impulse response and input time domain function, so the impulse response well is a kind of equivalent for the transfer function. But in time domain the simple multiplication must be replaced by convolution integral.

Convolution of time domain functions is often written with an operator symbol which resembles multiplication symbol. That's because it has the same role in time domain as multiplication has in s-domain.

You are right that your textbook snippet looks obscure. But we cannot see more than you have uploaded, so the theta in the formulas and what the text says is unknown. Thus there's not enough evidence to say the book is nonsense.

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  • \$\begingroup\$ Although your explanation was simple and appropriate to this question, i am confused more than ever. Hence I raised another question, to which it would be nice if you could answer or provide an explanation. The link: electronics.stackexchange.com/questions/490607/… \$\endgroup\$
    – RAN
    Apr 2, 2020 at 21:43

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