# Funciones análogicas de transferencia

This page is a translated version of the page Analog transfer functions and the translation is 57% complete.

Other languages:
English • ‎español
Series Geophysical References Series Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing Enders A. Robinson and Sven Treitel 15 http://dx.doi.org/10.1190/1.9781560801610 9781560801481 SEG Online Store

The transfer function (or system function) of an analog system is defined as the Laplace transform of the impulse response. If h(t) is the impulse response, then the transfer function is the Laplace transform

 {\displaystyle {\begin{aligned}H\left(s\right)=\int \limits _{-\infty }^{\infty }{h}\left(t\right)e^{-st}dt\;\;\;\ \mathrm {where} \;\;\;\;s=\sigma +i\omega .\end{aligned}}} (38)

The region of convergence is a vertical strip ${\displaystyle {\sigma }_{1}<\sigma <{\sigma }_{2}}$. Within this strip, H(s) has no poles or other singularities (Figure 5). For causal functions, we use the one-sided Laplace transform

 {\displaystyle {\begin{aligned}F\left(s\right)=\int \limits _{0}^{\infty }{f}\left(t\right)e^{-st}dt=1\left\{f\left(t\right)\right\}\mathrm {\;} \;\;\;{where}\;\;\;\;\;s=\sigma +i\omega .\end{aligned}}} (39)

The region of convergence of F(s) is the half-plane ${\displaystyle \sigma >{\sigma }_{1}}$. The symbol L is used to denote the one-sided Laplace transform.

The analog ARMA system is given by the differential equation

 {\displaystyle {\begin{aligned}y^{\left(p\right)}+{\alpha }_{1}y^{\left(p-1\right)}+\ldots +{\alpha }_{p}y={\beta }_{\rm {o}}u^{\left(q\right)}+{\beta }_{1}u^{\left(q-1\right)}+\ldots +{\beta }_{q}u,\end{aligned}}} (40)

where ${\displaystyle y^{\left(n\right)}}$ denotes ${\displaystyle d^{n}y/dt^{n}}$ and ${\displaystyle u^{\left(n\right)}}$ denotes ${\displaystyle d^{n}y/dt^{n}}$. Let us now assume that the input is causal — that is, ${\displaystyle u\left(t\right)=0}$ for t < 0 — so that the output also is causal. We also assume that the input and output have zero initial conditions; that is,

 {\displaystyle {\begin{aligned}u\left(0\right){\rm {=0,\ }}u^{\left(1\right)}\left(0\right){\rm {=0,\ }}u^{\left(2\right)}\left(0\right){\rm {=0,\ldots ,\ }}u^{\left(q-1\right)}\left(0\right)=0\\y\left(0\right){\rm {=0,\ }}y^{\left(1\right)}\left(0\right){\rm {=0,\ }}y^{\left(2\right)}\left(0\right){\rm {=0,\ldots ,\ }}y^{\left(p-1\right)}\left(0\right)=0.\end{aligned}}} (41)

If we take the (one-sided) Laplace transform of the above differential equation, we obtain

 {\displaystyle {\begin{aligned}\left(s^{p}+{\alpha }_{1}s^{p-1}+\ldots +{\alpha }_{p}\right){Y}\left(s\right)=\left({\beta }_{0}s^{q}+{\beta }_{1}s^{q-1}+\ldots +{\beta }_{q}\right)U\left(s\right),\end{aligned}}} (42)

where U(s) is the Laplace transform of the input and Y(s) is the Laplace transform of the output. In the case in which the input is the Dirac delta function ${\displaystyle \delta \left(t\right)}$, which has a Laplace transform equal to 1, then the output is the impulse response function h(t), with the Laplace transform H(s). In such a case, the above equation becomes

 {\displaystyle {\begin{aligned}\left(s^{p}+{\alpha }_{1}s^{p-1}+\ldots +{\alpha }_{p}\right)H\left(s\right)\;\;\;\\=\left({\beta }_{0}s^{q}+{\beta }_{1}s^{q-1}+\ldots +{\beta }_{q}\right).\end{aligned}}} (43)

If we define the polynomials ${\displaystyle \alpha \left(s\right)}$ and ${\displaystyle \beta \left(s\right)}$ as

 {\displaystyle {\begin{aligned}\alpha \left(s\right)=s^{p}+{\alpha }_{1}s^{p-1}+\ldots +{\alpha }_{p}\;\;\;\;\\\beta \left(s\right)={\beta }_{0}s^{q}+{\beta }_{1}s^{q-1}+\ldots +{\beta }_{q},\end{aligned}}} (44)
Figure 5.  Vertical strip of convergence of a Laplace transform.

then we see that H(s) is

 {\displaystyle {\begin{aligned}H\left(s\right)={\frac {\beta \left(s\right)}{\alpha \left(s\right)}}.\end{aligned}}} (45)

The function H(s), which is the Laplace transform of the impulse response h(t), is the transfer function. If we factor the polynomials ${\displaystyle \alpha \left(s\right)}$ and ${\displaystyle \beta \left(s\right)}$, we can write H(s) in its factored form as

 {\displaystyle {\begin{aligned}H\left(s\right)={\frac {{\beta }_{0}\left(s-b_{1}\right)\left(s-b_{2}\right)...\left(s-b_{q}\right)}{\left(s-a_{1}\right)\left(s-a_{2}\right)...\left(s-a_{p}\right)}}.\end{aligned}}} (46)

The constants ${\displaystyle a_{1}}$, ${\displaystyle a_{2}}$, ..., ${\displaystyle a_{p}}$ are the poles of H(s), and the constants ${\displaystyle b_{1}}$, ${\displaystyle b_{2}}$, ..., ${\displaystyle b_{q}}$ are the zeros. In the case in which the ${\displaystyle {\alpha }_{k}}$ and ${\displaystyle \beta _{k}}$ coefficients are real, it follows that all complex ${\displaystyle a_{i}}$ and ${\displaystyle b_{i}}$ must occur in complex-conjugate pairs. Because s corresponds to differentiation, it follows that ${\displaystyle s^{-1}}$ corresponds to integration. Thus, we can write H(s) as

 {\displaystyle {\begin{aligned}H\left(s\right)={\frac {{\beta }_{0}+{\beta }_{1}s^{-1}+{\beta }_{2}s^{-2}+\ldots +{\beta }_{q}s^{-q}}{1+{\alpha }_{1}s^{-1}+{\alpha }_{2}s^{-2}+\ldots +{\alpha }_{p}s^{-p}}}s^{-\left(p-q\right)}\end{aligned}}} (47)

when we want to implement the system by means of integrating circuits.

## Sigue leyendo

Sección previa Siguiente sección