To find the transfer function of a channel we say that it is $$ H(s) = \frac{y(s)}{x(s)}|x(s)=0 for <0 $$ Why we do not define it like $$ h(t) = \frac{y(t)}{x(t)} $$

  • $\begingroup$ "Why do we not define it like $$h(t) = \frac{y(t)}{x(t)}?$$ Because $x(t)$ might have value $0$ for some choices of $t$ (think of sinusoidal signals that take on value $0$ frequently) and besides, the $h(t)$ as you would like to define it is pretty meaningless: the output $y(t)$ is not the input $x(t)$ multiplied by some $h(t)$. $\endgroup$ – Dilip Sarwate Apr 22 at 14:21

We do not define it in that manner. Note the difference between the time and Laplace domains.

Generally, we characterize a system by the response produced when the system is excited by an impulse response $\delta(t)$. Given the output $y(t)$ for such signal we ask which function $h(t)$ did we convolve with the $\delta(t)$ to produce it. Formulation: $$\delta(t)*h(t)=y(t)$$

Here $*$ represent convolution and not a product. When we find $h(t)$ we can produce the output for any other input $x(t)$ by means of convolution: $$x(t)*h(t)=y(t)$$

If we transform into the Laplace domain, a convolution is translated into multiplication:

$$\mathcal{F}\{x(t)*h(t)\}=\mathcal{F}\{y(t)\}$$ $$\mathcal{F}\{x(t)\}\cdot \mathcal{F}\{h(t)\}=\mathcal{F}\{y(t)\}$$ $$X(s)\cdot H(s)=Y(s)$$ $$H(s)=\frac{Y(s)}{X(s)}$$

It is a convention that the capital letters represent the transformed signals. Also, $t$ has changed into $s$ because we are now in the Laplace domain. Since the convolution transformed into a product we may now divide both sides by $X(s)$. We could not do this on the time domain because there we had a convolution operator and not a product.

  • $\begingroup$ If this is the reason then can we write h(t) = y(t) * (1/x(t)) $\endgroup$ – Avinash Baldi Apr 22 at 13:21
  • $\begingroup$ No, we cannot! This is not a legal operation with convolution. You should take a pick in the Wikipedia entry for convolution to get a sense of it. $\endgroup$ – havakok Apr 22 at 13:27
  • $\begingroup$ I'm saying that if H(s) = Y(s)/X(s) then this can be written in the form H(s)=Y(s).1/(X(s))..i.e. in multiplication in s-domain and multiplication in s-domain is a convolution in time domain. $\endgroup$ – Avinash Baldi Apr 24 at 2:10
  • $\begingroup$ Multiplication in the Laplace domain is truly a convolution in the time domain. However, the inverse transformation $\mathcal{F}^{-1}\left\{\frac{1}{X(s)}\right\}\neq\frac{1}{x(t)}$. $\endgroup$ – havakok Apr 24 at 7:50

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