# Transfer Function definition

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 $$h(t) = \frac{y(t)}{x(t)}$$

• "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 $h(t)$. Apr 22 '19 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.

• If this is the reason then can we write h(t) = y(t) * (1/x(t)) Apr 22 '19 at 13:21
• 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. Apr 22 '19 at 13:27
• 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. Apr 24 '19 at 2:10
• Multiplication in the Laplace domain is truly a convolution in the time domain. However, the inverse transformation $\mathcal{F}^{-1}\left\{\frac{1}{X(s)}\right\}\neq\frac{1}{x(t)}$. Apr 24 '19 at 7:50