# Señales paso-banda y su representación

Ahora se van a definir distintas maneras de representar las señales paso-banda que nos van a permitir, entre otras aplicaciones, obtener la relación señal a ruido los sistemas de comunicación con ruido, o si no, de forma mas sencilla. Además de ello, nos permitirá obtener la representación paso-bajo de una señal paso-banda, hacer las operaciones pertinentes con ella, y al final, volver a convertirla en una señal paso-banda. ¿Por qué esto? Una razón es que en simulaciones por computador, por ejemplo, no es posible o requeriría muchos recursos el simular señales de alta frecuencia, digamos a 1 MHz, por ello, mediante diferentes representaciones podremos convertirlas en paso-bajo aliviando la carga del procesador y, obviamente, obteniendo el mismo resultado.
Este apartado es eminentemente teórico, siendo los siguientes donde se empiezan a analizar las diferentes modulaciones.

## Transformada de Hilbert

Comentar primeramente que, al ser h(t) real, la convolución de una señal real como x(t) por h(t) dará una señal real, esto es, sin parte imaginaria. Por lo que si x(t) es real, tras pasar el filtro de hilbert, seguirá siendo real.
${\hat {x}}(t)=x(t)*h(t)=\int _{-\infty }^{\infty }{\frac {x(\tau )}{\pi (t-\tau )}}\partial \tau$
Es muy difícil hacer esa convolución, por lo que nos pasamos a frecuencia. Para ello, recordemos la función sign(t):
$\mathbb {F} [\operatorname {sign} (t)]={\frac {1}{j\pi f}}$ Por propiedades: $\mathbb {F} [\operatorname {sign} (f)]={\frac {1}{-j\pi t}}$
${\hat {X}}(f)=X(f)\cdot H(f)=X(f)\left(-j\operatorname {sign} (f)\right)$
${\begin{aligned}&Ej:\\&x(t)=\cos(2\pi f_{0}t)\\&x(t)\to X(f)={\frac {1}{2}}\delta (f-f_{0})+{\frac {1}{2}}\delta (f+f_{0})\\&{\hat {X}}(f)=X(f)\left(-j\operatorname {sign} (f)\right)={\frac {-j}{2}}\delta (f-f_{0})-{\frac {-j}{2}}\delta (f+f_{0})\\&\mathbb {F} [\sin(2\pi f_{0}t)]={\frac {1}{2j}}\delta (f-f_{0})-{\frac {1}{2j}}\delta (f+f_{0})\to j={\frac {1}{-j}}\\&{\hat {x}}(t)=\sin(2\pi f_{0}t)\\\end{aligned}}$
Propiedades:
${\begin{aligned}&H(f)=-j\operatorname {sign} (f)\\&\left|H(f)\right|=1{\text{ }}\forall f\\&\measuredangle H(f)=\left\{{\begin{aligned}&^{\pi }\!\!\diagup \!\!_{2}\;,f>0\\&^{-\pi }\!\!\diagup \!\!_{2}\;,f<0\\\end{aligned}}\right.\\\end{aligned}}$
${\begin{aligned}&\mathbb {F} [{\hat {x}}(t)]={\hat {\hat {x}}}(t)=-x(t)\\&{\hat {\hat {X}}}(f)=X(f)\cdot H^{2}(f)=X(f)\cdot \left[-j\operatorname {sign} (f)\right]^{2}=X(f)\cdot \underbrace {(-\operatorname {sign} ^{2}(f))} _{\operatorname {sign} ^{2}(f)=1}=X(f)\cdot (-1)=-X(f)\\\end{aligned}}$
La DEP de la transformada de Hilbert
$G_{\hat {x}}(f)=G_{x}(f)\cdot \underbrace {\left|H(f)\right|^{2}} _{H(f)=-j\operatorname {sign} (f)}=G_{x}(f)\to R_{\hat {x}}(\tau )=R_{x}(\tau )$
Correlación
${\begin{aligned}&R_{yx}(\tau )=R_{xy}^{*}(\tau )\Rightarrow R_{x}(\tau )=R_{x}^{*}(\tau )\\&R_{yx}(\tau )=R_{x}(\tau )*h(\tau )\to R_{{\hat {x}}x}(\tau )=R_{x}(\tau )*{\frac {1}{\pi t}}\\&R_{x{\hat {x}}}(\tau )=R_{{\hat {x}}x}^{*}(-\tau )=R_{x}^{*}(-\tau )*{\frac {1}{\pi \left(-t\right)}}=R_{x}(\tau )*{\frac {-1}{\pi t}}\\&R_{{\hat {x}}x}(\tau )=-R_{x{\hat {x}}}(\tau )\\\end{aligned}}$

## Señal analitica + y - de x(t)

${\begin{aligned}&x_{+}(t)=x(t)+j{\hat {x}}(t)\\&x_{-}(t)=x(t)-j{\hat {x}}(t)\\&x_{+}(t)+x_{-}(t)=2x(t)\\&X_{+}(f)=X(f)+j\left(-j\operatorname {sign} (f)X(f)\right)=X(f)+X(f)\cdot \operatorname {sign} (f)=X(f)\left(1+\operatorname {sign} (f)\right)\\&X_{+}(f)=X(f)\left(1+\operatorname {sign} (f)\right)\\&X_{-}(f)=X(f)\left(1-\operatorname {sign} (f)\right)\\&X_{+}(f)=\left\{{\begin{aligned}&2X(f),f>0\\&0,f<0\\\end{aligned}}\right.\\&X_{-}(f)=\left\{{\begin{aligned}&0,f>0\\&2X(f),f<0\\\end{aligned}}\right.\\&X_{+}(f)+X_{-}(f)=2X(f)\\&G_{x}(f)={\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {\left|X(f)\right|^{2}}{T}}\\&G_{x_{+}}(f)=\left\{{\begin{aligned}&4G_{x}(f),f>0\\&0,f<0\\\end{aligned}}\right.=2G_{x}(f)\left(1+\operatorname {sign} (f)\right)\\&G_{x_{-}}(f)=\left\{{\begin{aligned}&0,f>0\\&4G_{x}(f),f<0\\\end{aligned}}\right.=2G_{x}(f)\left(1-\operatorname {sign} (f)\right)\\&G_{x_{+}}(f)+G_{x_{-}}(f)=4G_{x}(f)\\&S_{x_{+}}=2S_{x}\\&S_{x_{-}}=2S_{x}\\&R_{x_{+}x_{-}}(\tau )=0\to G_{x_{+}x_{-}}(f)=0\\\end{aligned}}$

## Envolvente compleja

La envolvente compleja se define como (se toma x(t) como señal paso-banda):
${\begin{aligned}&{\tilde {x}}(t)=x_{+}(t)e^{-j\omega _{c}t}\\&{\tilde {X}}(f)=X_{+}(f+f_{c})\\&R_{\tilde {x}}(\tau )={\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{{\tilde {x}}(t+\tau )\cdot {\tilde {x}}^{*}(t)\partial t=}{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{\left[x_{+}(t+\tau )e^{-j\omega _{c}\left(t+\tau \right)}\right]\cdot \left[x_{+}(t)e^{-j\omega _{c}t}\right]^{*}\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{\left[x_{+}(t+\tau )e^{-j\omega _{c}t}e^{-j\omega _{c}\tau }\right]\cdot x_{+}^{*}(t)e^{+j\omega _{c}t}\partial t=}{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{x_{+}(t+\tau )x_{+}^{*}(t)e^{-j\omega _{c}\tau }\partial t=}\\&R_{\tilde {x}}(\tau )=R_{x_{+}}(\tau )e^{-j\omega _{c}\tau }\to G_{\tilde {x}}(f)=G_{x_{+}}(f+f_{c})\\&G_{\tilde {x}}(f)=G_{x_{-}}(f-f_{c})\\&S_{\tilde {x}}=\int _{-\infty }^{\infty }{G_{x_{+}}(f+f_{c})\partial f}=2S_{x}\\\end{aligned}}$
${\begin{aligned}&{\tilde {x}}(t)=x_{+}(t)e^{j\omega _{c}t}\\&{\tilde {x}}(t)=x_{I}(t)+jx_{Q}(t)=e(t)e^{j\varphi (t)}\\&x_{I}(t)={\text{Componente en fase de }}x(t){\text{ }}(I:{\text{ In phase)}}\\&x_{Q}(t)={\text{Componente en cuadratura de }}x(t){\text{ }}(Q:{\text{ Quadrature)}}\\&e(t)={\text{ Envolvente (el modulo) de }}x(t)\\&\varphi (t)={\text{ Fase de }}x(t){\text{ }}\\&x_{+}(t)=x(t)+j{\hat {x}}(t)\\&x(t)=\Re \left\{x_{+}(t)\right\}=\Re \left\{{\tilde {x}}(t)e^{+j\omega _{c}t}\right\}=\Re \left\{\left[x_{I}(t)+jx_{Q}(t)\right]e^{+j\omega _{c}t}\right\}=\\&\Re \left\{\left[x_{I}(t)+jx_{Q}(t)\right]\left[\cos(\omega _{c}t)+j\sin(\omega _{c}t)\right]\right\}=x_{I}(t)\cos(\omega _{c}t)-x_{Q}(t)\sin(\omega _{c}t)\\&x(t)=\Re \left\{{\tilde {x}}(t)e^{+j\omega _{c}t}\right\}=\Re \left\{\underbrace {e(t)} _{\begin{smallmatrix}{\text{valor}}\\{\text{absoluto}}\end{smallmatrix}}e^{j\varphi (t)}e^{+j\omega _{c}t}\right\}=e(t)\cos(\omega _{c}t+\varphi (t))\\&x(t)=x_{I}(t)\cos(\omega _{c}t)-x_{Q}(t)\sin(\omega _{c}t)=e(t)\cos(\omega _{c}t+\varphi (t))\\\end{aligned}}$
Mas propiedades
${\begin{aligned}&{\tilde {x}}(t)=x_{+}(t)e^{-j\omega _{c}t};{\text{ }}x_{+}(t)=x(t)+j{\hat {x}}(t)\\&{\tilde {x}}(t)=x_{I}(t)+jx_{Q}(t)=e(t)e^{j\varphi (t)}\\&e(t)={\sqrt {x_{I}^{2}(t)+x_{Q}^{2}(t)}}=\left|{\tilde {x}}(t)\right|=\left|x_{+}(t)e^{-j\omega _{c}t}\right|=\left|x_{+}(t)\right|={\sqrt {x^{2}(t)+{\hat {x}}^{2}(t)}}\\&\varphi (t)=\arctan \left({\frac {x_{Q}(t)}{x_{I}(t)}}\right)\\&\Re \left\{{\tilde {x}}(t)\right\}=x_{I}(t)=\Re \left\{e(t)e^{j\varphi (t)}\right\}=e(t)\cos \left(\varphi (t)\right)\to x_{Q}(t)=e(t)\sin \left(\varphi (t)\right)\\\end{aligned}}$
Características importantes:
Si tomamos x(t) como señal paso-banda (una señal modulada), entonces tenemos que:
${\begin{aligned}&x(t):{\text{ Se }}\!\!{\tilde {\mathrm {n} }}\!\!{\text{ al paso-banda}}\\&{\tilde {x}}(t),x_{I}(t),x_{Q}(t),e(t){\text{ : Se }}\!\!{\tilde {\mathrm {n} }}\!\!{\text{ ales paso-bajo}}\\\end{aligned}}$
Ahora:
${\begin{aligned}&x_{+}(t)=x_{-}^{*}(t)\\&x_{I}(t)=\Re \left\{{\tilde {x}}(t)\right\}={\frac {{\tilde {x}}(t)+{\tilde {x}}^{*}(t)}{2}}={\frac {x_{+}(t)e^{-j\omega _{c}t}+x_{+}^{*}(t)e^{+j\omega _{c}t}}{2}}\\&R_{x_{I}}(\tau )={\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{x_{I}(t+\tau )\cdot x_{I}^{*}(t)\partial t}=\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{\left[{\frac {x_{+}(t+\tau )e^{-j\omega _{c}\left(t+\tau \right)}+x_{+}^{*}(t+\tau )e^{+j\omega _{c}\left(t+\tau \right)}}{2}}\right]\cdot \left[{\frac {x_{+}(t)e^{-j\omega _{c}t}+x_{+}^{*}(t)e^{+j\omega _{c}t}}{2}}\right]^{*}\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot {\frac {1}{4}}\int _{-\infty }^{\infty }{\left[x_{+}(t+\tau )e^{-j\omega _{c}\left(t+\tau \right)}+x_{+}^{*}(t+\tau )e^{+j\omega _{c}\left(t+\tau \right)}\right]\cdot \left[x_{+}^{*}(t)e^{+j\omega _{c}t}+x_{+}(t)e^{-j\omega _{c}t}\right]\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot {\frac {1}{4}}\int _{-\infty }^{\infty }{x_{+}(t+\tau )x_{+}^{*}(t)e^{-j\omega _{c}\tau }+\overbrace {x_{+}(t+\tau )x_{+}(t)e^{-j\omega _{c}\left(2t+\tau \right)}} ^{\text{no entra en area de integracion}}+...}\\&\overbrace {x_{+}^{*}(t+\tau )x_{+}^{*}(t)e^{+j\omega _{c}\left(2t+\tau \right)}} ^{{\text{ }}\!\!{\hat {\mathrm {i} }}\!\!{\text{ dem}}}+x_{+}^{*}(t+\tau )x_{+}(t)e^{+j\omega _{c}\tau }\partial t=\\&R_{x_{I}}(\tau )=^{1}\!\!\diagup \!\!_{4}\;\left(R_{x_{+}}(\tau )e^{-j\omega _{c}\tau }+R_{x_{-}}(\tau )e^{+j\omega _{c}\tau }\right)\\&G_{x_{I}}(f)=^{1}\!\!\diagup \!\!_{4}\;\left(G_{x_{+}}(f+f_{c})+G_{x_{-}}(f-f_{c})\right)\\&G_{x_{+}}(f)+G_{x_{-}}(f)=4G_{x}(f)\\&S_{x_{I}}=\int _{-\infty }^{\infty }{G_{x_{I}}(f)\partial f=S_{x}}\\&x_{Q}(t)=\Im \left\{{\tilde {x}}(t)\right\}={\frac {{\tilde {x}}(t)-{\tilde {x}}^{*}(t)}{2j}}\to G_{x_{Q}}(f)=^{1}\!\!\diagup \!\!_{4}\;\left(G_{x_{+}}(f+f_{c})+G_{x_{-}}(f-f_{c})\right)\\\end{aligned}}$
Ahora, sacaremos la correlación cruzada entre la señal en fase y, la señal en cuadratura:
${\begin{aligned}&R_{x_{Q}x_{I}}(\tau )={\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{x_{Q}(t+\tau )\cdot x_{I}^{*}(t)\partial t}=\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{\left[{\frac {x_{+}(t+\tau )e^{-j\omega _{c}\left(t+\tau \right)}-x_{+}^{*}(t+\tau )e^{+j\omega _{c}\left(t+\tau \right)}}{2j}}\right]\cdot \left[{\frac {x_{+}(t)e^{-j\omega _{c}t}+x_{+}^{*}(t)e^{+j\omega _{c}t}}{2}}\right]^{*}\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot {\frac {1}{4j}}\int _{-\infty }^{\infty }{\left[x_{+}(t+\tau )e^{-j\omega _{c}\left(t+\tau \right)}-x_{+}^{*}(t+\tau )e^{+j\omega _{c}\left(t+\tau \right)}\right]\cdot \left[x_{+}^{*}(t)e^{+j\omega _{c}t}+x_{+}(t)e^{-j\omega _{c}t}\right]\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot {\frac {1}{4j}}\int _{-\infty }^{\infty }{x_{+}(t+\tau )x_{+}^{*}(t)e^{-j\omega _{c}\tau }+\overbrace {x_{+}(t+\tau )x_{+}(t)e^{-j\omega _{c}\left(2t+\tau \right)}} ^{\text{no entra en area de integracion}}+...}\\&\overbrace {-x_{+}^{*}(t+\tau )x_{+}^{*}(t)e^{+j\omega _{c}\left(2t+\tau \right)}} ^{{\text{ }}\!\!{\hat {\mathrm {i} }}\!\!{\text{ dem}}}-\overbrace {x_{+}^{*}(t+\tau )x_{+}(t)} ^{=x_{-}(t+\tau )x_{-}^{*}(t)}e^{+j\omega _{c}\tau }\partial t=\\&R_{x_{Q}x_{I}}(\tau )={\frac {1}{4j}}\left(R_{x_{+}}(\tau )e^{-j\omega _{c}\tau }-R_{x_{-}}(\tau )e^{+j\omega _{c}\tau }\right)\\&R_{x_{I}x_{Q}}(\tau )=R_{x_{Q}x_{I}}^{*}(-\tau )={\frac {1}{4\left(-j\right)^{*}}}\left(R_{x_{+}}(-\tau )e^{+j\omega _{c}\tau }-R_{x_{-}}(-\tau )e^{-j\omega _{c}\tau }\right)^{*}=\\&{\frac {1}{4j}}\left(\overbrace {R_{x_{+}}^{*}(-\tau )} ^{R_{x_{+}}(\tau )}e^{-j\omega _{c}\tau }-\overbrace {R_{x_{-}}^{*}(-\tau )} ^{R_{x_{-}}(\tau )}e^{+j\omega _{c}\tau }\right)\Rightarrow \\&R_{x_{I}x_{Q}}(\tau )=R_{x_{Q}x_{I}}(\tau )\\\end{aligned}}$

## Correlación de una señal paso-banda

Con la representación de una señal paso-banda en sus componentes de fase y cuadratura, tenemos que, si obtenemos su correlación:
${\begin{aligned}&R_{x_{m}}(\tau )={\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{x_{m}^{*}(t)\cdot x_{m}(t+\tau )\partial t=}\\&x_{m}(t)=x_{I}(t)\cos(\omega _{c}t)-x_{Q}(t)\sin(\omega _{c}t)\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\int _{-\infty }^{\infty }{\left[x_{I}(t)\cos(\omega _{c}t)-x_{Q}(t)\sin(\omega _{c}t)\right]^{*}\cdot \left[x_{I}(t+\tau )\cos \left(\omega _{c}\left(t+\tau \right)\right)-x_{Q}(t)\sin \left(\omega _{c}\left(t+\tau \right)\right)\right]\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\int _{-\infty }^{\infty }{x_{I}^{*}(t)\cos(\omega _{c}t)x_{I}(t+\tau )\cos \left(\omega _{c}\left(t+\tau \right)\right)}+\\&-x_{Q}^{*}(t)\sin(\omega _{c}t)x_{I}(t+\tau )\cos \left(\omega _{c}\left(t+\tau \right)\right)+\\&x_{I}^{*}(t)\cos(\omega _{c}t)\left(-x_{Q}(t+\tau )\sin \left(\omega _{c}\left(t+\tau \right)\right)\right)+\\&-x_{Q}^{*}(t)\sin(\omega _{c}t)\left(-x_{Q}(t+\tau )\sin \left(\omega _{c}\left(t+\tau \right)\right)\right)\partial t=\left\{{\text{Relaciones Trigonometricas}}\right\}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\int _{-\infty }^{\infty }\\&x_{I}^{*}(t)x_{I}(t+\tau )\left({\frac {\cos \left(\omega _{c}\tau \right)+\overbrace {\cos \left(\omega _{c}\left(2t+\tau \right)\right)} ^{{\text{fuera del area de }}\int}}{2}}\right)+\leftarrow \left\{\cos a\cos b={\frac {\cos(a+b)+\cos(a-b)}{2}}\right\}\\&-x_{Q}^{*}(t)x_{I}(t+\tau )\left({\frac {\sin(-\omega _{c}\tau )+\overbrace {\sin \left(\omega _{c}\left(2t+\tau \right)\right)} ^{{\text{fuera del area de }}\int}}{2}}\right)\leftarrow \left\{\sin a\cos b={\frac {\sin(a+b)+\sin(a-b)}{2}}\right\}\\&-x_{I}^{*}(t)x_{Q}(t+\tau )\left({\frac {\sin \left(\omega _{c}\tau \right)+\overbrace {\sin \left(\omega _{c}\left(2t+\tau \right)\right)} ^{{\text{fuera del area de }}\int}}{2}}\right)+\leftarrow \left\{\sin a\cos b={\frac {\sin(a+b)+\sin(a-b)}{2}}\right\}\\&x_{Q}^{*}(t)x_{Q}(t+\tau )\left({\frac {\cos \left(\omega _{c}\tau \right)-\overbrace {\cos \left(\omega _{c}\left(2t+\tau \right)\right)} ^{{\text{fuera del area de }}\int}}{2}}\right)\partial t\leftarrow \left\{\sin a\sin b={\frac {\cos(a-b)-\cos(a+b)}{2}}\right\}=\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\int _{-\infty }^{\infty }{x_{I}^{*}(t)x_{I}(t+\tau ){\frac {\cos \left(\omega _{c}\tau \right)}{2}}}+x_{Q}^{*}(t)x_{I}(t+\tau ){\frac {\sin \left(\omega _{c}\tau \right)}{2}}\\&-x_{I}^{*}(t)x_{Q}(t+\tau ){\frac {\sin \left(\omega _{c}\tau \right)}{2}}+x_{Q}^{*}(t)x_{Q}(t+\tau ){\frac {\cos \left(\omega _{c}\tau \right)}{2}}=\\&R_{x_{I}}(\tau ){\frac {\cos \left(\omega _{c}\tau \right)}{2}}+R_{x_{I}x_{Q}}(\tau ){\frac {\sin \left(\omega _{c}\tau \right)}{2}}-R_{x_{Q}x_{I}}(\tau ){\frac {\sin \left(\omega _{c}\tau \right)}{2}}+R_{x_{Q}}(\tau ){\frac {\cos \left(\omega _{c}\tau \right)}{2}}\\\end{aligned}}$
${\begin{aligned}&R_{x_{I}x_{Q}}(\tau )=R_{x_{Q}x_{I}}(\tau )\to \\&R_{x_{m}}(\tau )=R_{x_{I}}(\tau ){\frac {\cos \left(\omega _{c}\tau \right)}{2}}+R_{x_{Q}}(\tau ){\frac {\cos \left(\omega _{c}\tau \right)}{2}}\\\end{aligned}}$

## Relación entre salida paso-banda y salida envolvente compleja

${\begin{aligned}&Y(f)=X(f)\cdot H(f)\\&{\tilde {Y}}(f)\propto X(f)\cdot H(f)?\\&X_{+}(f)=\left\{{\begin{aligned}&2X(f),f>0\\&0,f<0\\\end{aligned}}\right.\\&Y(f)=X(f)\cdot H(f)\\&x(t)=\Re \left\{x_{+}(t)\right\}=\Re \left\{{\tilde {x}}(t)e^{+j\omega _{c}t}\right\}={\frac {x_{+}(t)+x_{+}^{*}(t)}{2}}={\frac {{\tilde {x}}(t)e^{+j\omega _{c}t}+{\tilde {x}}^{*}(t)e^{-j\omega _{c}t}}{2}}\\&X_{+}(f)={\tilde {X}}(f-f_{c})\\&X(f)={\frac {X_{+}(f)+X_{+}^{*}(-f)}{2}}={\frac {{\tilde {X}}(f-f_{c})+{\tilde {X}}^{*}(-f-f_{c})}{2}}={\frac {{\tilde {X}}\left(f-f_{c}\right)+{\tilde {X}}^{*}\left(-\left(f+f_{c}\right)\right)}{2}}\\&X(f)\cdot H(f)={\frac {{\tilde {X}}\left(f-f_{c}\right)+{\tilde {X}}^{*}\left(-\left(f+f_{c}\right)\right)}{2}}\cdot {\frac {{\tilde {H}}\left(f-f_{c}\right)+{\tilde {H}}^{*}\left(-\left(f+f_{c}\right)\right)}{2}}\\&{\text{Suponiendo se }}\!\!{\tilde {\mathrm {n} }}\!\!{\text{ al paso-banda}}\\&Y(f)=X(f)\cdot H(f)\\&{\frac {{\tilde {Y}}\left(f-f_{c}\right)+{\tilde {Y}}^{*}\left(-\left(f+f_{c}\right)\right)}{2}}={\frac {{\tilde {X}}\left(f-f_{c}\right)+{\tilde {X}}^{*}\left(-\left(f+f_{c}\right)\right)}{2}}\cdot {\frac {{\tilde {H}}\left(f-f_{c}\right)+{\tilde {H}}^{*}\left(-\left(f+f_{c}\right)\right)}{2}}\\&{\frac {{\tilde {Y}}\left(f-f_{c}\right)+{\tilde {Y}}^{*}\left(-\left(f+f_{c}\right)\right)}{2}}={\frac {{\tilde {X}}\left(f-f_{c}\right)\cdot {\tilde {H}}\left(f-f_{c}\right)+{\tilde {X}}^{*}\left(-\left(f+f_{c}\right)\right)\cdot {\tilde {H}}^{*}\left(-\left(f+f_{c}\right)\right)}{4}}\to \\&{\frac {{\tilde {Y}}\left(f-f_{c}\right)}{2}}={\frac {{\tilde {X}}\left(f-f_{c}\right)\cdot {\tilde {H}}\left(f-f_{c}\right)}{4}}\to {\frac {{\tilde {Y}}\left(f\right)}{2}}={\frac {{\tilde {X}}\left(f\right)\cdot {\tilde {H}}\left(f\right)}{4}}\\&{\tilde {Y}}\left(f\right)={\frac {{\tilde {X}}\left(f\right)\cdot {\tilde {H}}\left(f\right)}{2}}\\\end{aligned}}$
* Proyecto: Departamento de Teoría de la Señal y Comunicaciones
* Anterior: Esquema de un sistema de comunicacion con ruido y Conceptos — Señales paso-banda y su representación — Siguiente: Modulaciones lineales