# 0.8 9. interferencia intersimbólica (isi)  (Page 3/3)

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Esto implica que al sumar todas las repeticiones de P(f) cada f b estas deben sumar una constante

El pulso resultante tiene simetría vestigial

Este filtro puede representarse como la suma de:

Matemáticamente:

$\begin{array}{c}P\left(f\right)=\prod \left(\frac{f}{{f}_{b}}\right)+{H}_{1}\left(f\right)\\ p\left(t\right)=\frac{1}{{t}_{b}}\text{Sinc}\left(\frac{t}{{t}_{b}}\right)+{h}_{1}\left(t\right)\end{array}$

H 1 (f) es simétrica y par.

$\begin{array}{c}{h}_{1}\left(t\right)=\underset{-\infty }{\overset{\infty }{\int }}{H}_{1}\left(f\right){e}^{\mathrm{j\omega t}}\text{df}=2\underset{0}{\overset{\infty }{\int }}{H}_{1}\left(f\right)\text{Cos}\left(\mathrm{\omega t}\right)\text{df}\\ {h}_{1}\left(t\right)=2\underset{-\beta }{\overset{}{\int }}{H}_{1}\left(f\right)\text{Cos}\left(\mathrm{\omega t}\right)\text{df}+2\underset{}{\overset{+\beta }{\int }}{H}_{1}\left(f\right)\text{Cos}\left(\mathrm{\omega t}\right)\text{df}\\ \begin{array}{cccc}& & & \begin{array}{ccc}& & c\text{.}d\text{.}v:\left\{\begin{array}{c}\begin{array}{ccc}\text{Para}\text{la}\text{primera}\text{Integral}:& f=-x& \end{array}\\ \\ \begin{array}{ccc}\text{Para}\text{la}\text{segunda}\text{Integral}:& f=+x& \end{array}\end{array}\end{array}\end{array}\\ \text{Entonces}:\\ {h}_{1}\left(t\right)=2\underset{0}{\overset{\beta }{\int }}{H}_{1}\left(\frac{\text{fb}}{2}-x\right)\text{Cos}\left(2\pi \left(\frac{\text{fb}}{2}-x\right)\right)\text{tdx}+2\underset{0}{\overset{\beta }{\int }}{H}_{1}\left(\frac{\text{fb}}{2}+x\right)\text{Cos}\left(2\pi \left(\frac{\text{fb}}{2}+x\right)\right)\text{tdx}\\ \\ \begin{array}{cccc}\begin{array}{cc}\begin{array}{cc}& \end{array}& \end{array}& \text{Por}\text{simetría}:& & {H}_{1}\left(\frac{\text{fb}}{2}+x\right)\end{array}=-{H}_{1}\left(\frac{\text{fb}}{2}-x\right)\\ \\ {h}_{1}\left(t\right)=2\underset{0}{\overset{\beta }{\int }}{H}_{1}\left(\frac{\text{fb}}{2}+x\right)\left[\text{Cos}\left(2\pi t\left(\frac{\text{fb}}{2}+x\right)\right)-\text{Cos}\left(2\pi t\left(\frac{\text{fb}}{2}-x\right)\right)\right]\text{dx}\\ \\ {h}_{1}\left(t\right)=-4\underset{0}{\overset{\beta }{\int }}{H}_{1}\left(\frac{\text{fb}}{2}+x\right)\left[\text{Sen}\left({\mathrm{\pi f}}_{b}t\right)\text{.}\text{Sen}\left(2\pi \text{xt}\right)\right]\text{dx}\\ \\ {h}_{1}\left(t\right)=-4\text{Sen}\left({\mathrm{\pi f}}_{b}t\right)\underset{0}{\overset{\beta }{\int }}{H}_{1}\left(\frac{\text{fb}}{2}+x\right)\text{Sen}\left(2\pi \text{xt}\right)\text{dx}\end{array}$

Para cada ntb el término que se encuentra fuera de la integral se anulará. De esta forma, se evita la interferencia.

A partir de este criterio podemos implementar el filtro de simetría vestigial de tipo Coseno Alzado : Este se caracteriza porque puede reducir la ISI. La parte no nula del espectro es un coseno que, en su forma más simple, está alzado (es decir, se encuentra por encima del eje de frecuencia):

## Segundo criterio de nyquist

En este criterio se busca no sólo eliminar la interferencia, también se presenta como objetivo el disminuir el ancho de banda. Esto se hace definiendo, en el transmisor, una interacción conocida entre pulsos vecinos. El sacrificio, en este caso, es un mayor consumo de potencia.

Entonces, en vez de transmitir a k (secuencia original), se enviará y k =a k +a k-1 . De esta forma se pueden enviar dos bits haciendo uso del mismo ancho de banda. Supongamos el siguiente ejemplo:

Secuencia original de bits: 01010011

 Secuencia original 0 1 0 1 0 0 1 1 ak -1 1 -1 1 -1 -1 1 1 yk 0 0 0 0 -2 0 2

El filtro que se coloca en el transmisor pudiera modelarse como:

Teniendo considerada la condición de un sistema con interferencia, ahora se debe tomar en cuenta cuando se introduce ruido AWGN al canal. Supongamos que a la entrada de un sistema de comunicaciones se tiene una secuencia aleatoria, con código de línea NRZ y duración tb. La Densidad Espectral de Potencia sería:

$\begin{array}{c}G\left(f\right)=\frac{{\mid P\left(f\right)\mid }^{2}}{\text{tb}};\\ P\left(f\right)\to \text{Transformada}\text{de}\text{Fourier}\text{de}\text{la}\text{señal}\text{de}\text{entrada}\text{.}\end{array}$

Asumiendo un sistema como sigue:

La salida del sistema sería una sucesión de pulsos y(t), asociada a un pulso de salida p R (t) y a los de entrada:

$\begin{array}{c}{A}_{k}\mid {P}_{R}\left(f\right)\mid =\mid P\left(f\right)\mid \text{.}\mid {H}_{T}\left(f\right)\mid \text{.}\mid {H}_{c}\left(f\right)\mid \text{.}\mid {H}_{R}\left(f\right)\mid \to \text{Ecuación}\left(1\right)\\ \\ \text{Si}\text{la}\text{potencia}\text{de}\text{transmisión}\text{es}:\\ {S}_{T}=\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mid P\left(f\right)\mid }^{2}{\mid {H}_{T}\left(f\right)\mid }^{2}}{\text{tb}}\text{df}\\ \text{Pudiéramos}\text{Expresarla}\text{en}\text{función}\text{de}\text{la}\text{ecuación}\left(1\right):\\ \\ \begin{array}{ccc}& & \end{array}\text{tb}\text{.}{S}_{T}={A}_{{k}^{2}}\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mid {P}_{R}\left(f\right)\mid }^{2}}{\mid {H}_{c}\left(f\right)\mid \text{.}\mid {H}_{R}\left(f\right)\mid }\text{df}\end{array}$

Ahora bien, como nuestro objetivo es maximizar la relación señal a ruido, despejamos el valor de Ak (Amplitud del pulso) y definimos σ² (que debe ser minimizado):

$\begin{array}{c}{A}_{{k}^{2}}=\frac{\text{tb}\text{.}{S}_{T}}{\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mid {P}_{R}\left(f\right)\mid }^{2}}{\mid {H}_{c}\left(f\right)\mid \text{.}\mid {H}_{R}\left(f\right)\mid }\text{df}}\\ y\\ {\sigma }^{2}=\underset{-\infty }{\overset{\infty }{\int }}\text{Gn}\left(f\right){\mid {H}_{R}\left(f\right)\mid }^{2}\text{df}\\ \text{Por}\text{lo}\text{que}:\\ \\ \frac{{A}_{{k}^{2}}}{{\sigma }^{2}}=\frac{\text{tb}\text{.}{S}_{T}}{\underset{-\infty }{\overset{\infty }{\int }}\text{Gn}\left(f\right){\mid {H}_{R}\left(f\right)\mid }^{2}\text{df}\text{.}\underset{-\infty }{\overset{\infty }{\int }}\frac{{\mid {P}_{R}\left(f\right)\mid }^{2}}{\mid {H}_{c}\left(f\right)\mid \text{.}\mid {H}_{R}\left(f\right)\mid }\text{df}}\begin{array}{c}\\ \\ \to \text{Minimizar}\end{array}\end{array}$

A través de la igualdad de Schwartz podemos cumplir el objetivo:

$\begin{array}{c}\mid \underset{-\infty }{\overset{\infty }{\int }}V\left(f\right)\text{.}{W}^{}\left(f\right)\text{df}\mid \le \underset{-\infty }{\overset{\infty }{\int }}{\mid V\left(f\right)\mid }^{2}\text{.}\text{df}\text{.}\underset{-\infty }{\overset{\infty }{\int }}{\mid W\left(f\right)\mid }^{2}\text{.}\text{df}\\ \\ \text{Que}\text{será}\text{igual}\text{cuando}V\left(f\right)=k\text{.}W\left(f\right)\text{.}\text{Si}:\\ \\ W\left(f\right)=\frac{\mid {P}_{R}\left(f\right)\mid }{\mid {H}_{C}\left(f\right)\mid \mid {H}_{R}\left(f\right)\mid }yV\left(f\right)=\mid {H}_{R}\left(f\right)\mid \sqrt{\text{Gn}\left(f\right)}\\ \\ \text{Entonces}:\\ \\ \mid {H}_{R}\left(f\right)\mid \sqrt{\text{Gn}\left(f\right)}=k\left[\frac{\mid {P}_{R}\left(f\right)\mid }{\mid {H}_{C}\left(f\right)\mid \mid {H}_{R}\left(f\right)\mid }\right]\\ \\ \begin{array}{ccc}& & {\mid {H}_{R}\left(f\right)\mid }^{2}=k\left[\frac{\mid {P}_{R}\left(f\right)\mid }{\mid {H}_{C}\left(f\right)\mid \sqrt{\text{Gn}\left(f\right)}}\right]\end{array}\end{array}$

Finalmente, con la ecuación (1) tenemos que:

${\mid {H}_{T}\left(f\right)\mid }^{2}=\frac{{A}_{{k}^{2}}\mid {P}_{R}\left(f\right)\mid \sqrt{\text{Gn}\left(f\right)}}{k\mid {H}_{C}\left(f\right)\mid {\mid {P}_{R}\left(f\right)\mid }^{2}}$

## Simulaciones en labview

El VI correspondiente a la teoría de este módulo puede descargarse a través del siguiente enlace:

#### Questions & Answers

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