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\$x(t)=e^{st}\$ is the input to the system. \$x(t)\$ is not going to present in this question anything else than the exponential \$e^{st}\$ where t is time and s is a complex valued parameter.

\$H(s)\$ is the transfer function of the system. \$y(t)\$ is the output of the system

$$ y(t)=x(t)*h(t) = \int_{-\infty}^\infty h(\tau) x(t-\tau) d\tau = \int_{-\infty}^\infty h(\tau) e^{st}e^{-s\tau}d\tau $$ $$ y(t) = e^{st}\int_{-\infty}^\infty h(\tau) e^{-s\tau}d\tau = e^{st}\mathcal{L}\{h(\textbf{t})\} = x(\textbf{t})H(s). $$

This is my derivation of the equation \$y(t)=H(s)x(t)\$. Is it correct? if this derivation is correct, That means the overall response is \$H(s)x(t)\$. But according to my understanding, \$H(s)x(t)\$ only gives the forced response. The overall response should be calculated by inverse Laplace transform of \$H(s)X(s)\$. Where does this discrepancy come from? Thanks!

BTW, this is not derived by me,this derivation was done by Professor Ali Hajimiri at time 4:09

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  • \$\begingroup\$ Use latex i.e. \$x(t) = e^{st}\$ becomes \$x(t) = e^{st}\$ - reason - your handwriting is a scrawl. \$\endgroup\$
    – Andy aka
    Commented Jul 3, 2020 at 11:21
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    \$\begingroup\$ Your input signal uses the symbol \$s\$, which also happens to be the name of the frequency domain variable used in Laplace transform. Now you have two variables with same name, but possibly different meanings. What would happen to the derivation if the input used a different symbol, say \$x(t) = e^{rt}\$ ? \$\endgroup\$
    – AJN
    Commented Jul 3, 2020 at 11:36
  • \$\begingroup\$ To my knowledge, simplified expressions contain either the time variable (t) or the frequency variable (s). Both won't appear together. Either you are representing your signals & systems in time domain or in frequency domain, but not both. Expressions in frequency domain may contain specific time instances like \$t_0\$, but not t. Vice versa for time domain representations. \$\endgroup\$
    – AJN
    Commented Jul 3, 2020 at 11:39
  • \$\begingroup\$ Alternately, why does the time domain representation of your input signal \$x(t)\$ depend on the frequency variable \$s\$ ? Can you plot \$x(t)\$ while allowing \$s\$ to hold any value value from \$-\infty\$ to \$+\infty\$ ? (i.e. you are not allowed to freeze the value of \$s\$ to any particular number). \$\endgroup\$
    – AJN
    Commented Jul 3, 2020 at 11:40
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    \$\begingroup\$ Repeat your derivation with \$x(t) = e^{j\omega_0 t}\$ where \$\omega_0\$ is a specific frequency (say 5 rad/s) instead of being able to hold any value from \$-\infty\$ to \$+\infty\$. You may be confusing a signal having a specific complex frequency \$s_0 = \sigma_0 + j\omega_0\$ with the frequency variable \$s\$. Think about how you would plot the signal. \$\endgroup\$
    – AJN
    Commented Jul 3, 2020 at 11:44

2 Answers 2

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Initially, you're pretending that '\$s\$' is a run of the mill, real valued parameter, and then later in the analysis you say, oh no! it's really the Laplace operator. This means that your very first equation is mixing the time domain and the Laplace domain in the same expression, \$e^{st}\$, which is not valid.

Your analysis is not an accurate version of the youtube video, which actually shows that convolution in the time domain is equivalent to multiplication in the Laplace domain.

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  • \$\begingroup\$ No, it is not a real valued parameter, it is intended to be a complex input \$\endgroup\$
    – Xiutao
    Commented Jul 3, 2020 at 16:08
  • \$\begingroup\$ For example, v(t)=e^(jwt) \$\endgroup\$
    – Xiutao
    Commented Jul 3, 2020 at 16:13
  • \$\begingroup\$ exactly, i don't see any problem \$\endgroup\$
    – Xiutao
    Commented Jul 3, 2020 at 16:16
  • \$\begingroup\$ So \$x(t)= cos\:\omega t+jsin\:\omega t\$? A real signal cannot be a complex function of time. \$\endgroup\$
    – Chu
    Commented Jul 3, 2020 at 16:17
  • \$\begingroup\$ It can. If not, why transfer function is H(jw)? \$\endgroup\$
    – Xiutao
    Commented Jul 3, 2020 at 16:20
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Okay, I have finally understood. This derivation is correct except that the laplace transform of the impulse response is not the transfer function. And when doing the laplace transform the upper limit should be negative infinity and the lower bound should be t.

So basically, this derivation shows that the response of a system can be obtained by multiplying input x(t) and the laplace transform of the impulse response of the system with the upper bound going into negative infinity and the lower bound being t.

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