I read this in a book: enter image description here

here ${\cal L}[x(t-a)u(t-a)]$ is the laplace transform of a time shifted function $x(t)$ shifted by $a$ seconds and we know that transfer functions have the formula $H(s)=Y(s)/X(s)$. Also, $y(t)$ = Laplace inverse of $Y(s)$, so shouldn't $y(t) = {\cal L}^{-1}[H(s)X(s)e^{-as}]$ instead of the given result in equation 14.62

  • $\begingroup$ Which book is that? $\endgroup$
    – Matt L.
    Jun 13, 2022 at 15:16
  • $\begingroup$ Circuits and networks (analysis and synthesis) by some indian authors - Shyammohan and Sudhakar $\endgroup$
    – Sam1470
    Jun 14, 2022 at 9:33

1 Answer 1


The notation in the book is confusing. They use $y(t)$ to denote the output signal corresponding to the original input signal $x(t)$. However, they use $Y(s)$ as the Laplace transform of the response to the delayed input, i.e., $Y(s)$ is not the Laplace transform of $y(t)$ (but of $y(t-a)$).


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