# SSB
## Representación en tiempo y frecuencia

La modulación SSB (Single Side Band) es una modulación que transmite la información usando una solo banda de frecuencia (de ahí su nombre), esto es, el ancho de banda necesario para una señal SSB será la mitad que la de una DSB (o AM). Veamos primero el espectro de una señal DSB.
La transformada de Fourier de una señal modulada en SSB sería:
Ahora, debemos conseguir la ecuación que represente en frecuencia a una señal SSB.
Tenemos 2 tipos de señal SSB: alta y baja, que dependen de la parte del espectro que elijamos.
Podemos ver que:
$X_{SSB}(f)=X_{DSB}(f)\cdot H^{u}(f)$
Intentemos conseguir primero la función envolvente compleja de la señal SSB. Una vez lograda, la transformaremos en paso-banda.
Recordemos la propiedad de:
${\begin{aligned}&Y(f)=X(f)\cdot H(f)\to \\&{\tilde {Y}}(f)={\frac {1}{2}}{\tilde {X}}(f)\cdot {\tilde {H}}(f)\\\end{aligned}}$
${\begin{aligned}&{\tilde {X}}_{SSB}(f)={\frac {1}{2}}{\tilde {X}}_{DSB}(f)\cdot {\tilde {H}}^{u}(f)\\&{\tilde {X}}_{DSB}(f)=?\\&x_{DSB}(t)=A_{c}x(t)\cos(\omega _{c}t)\\&{\hat {x}}_{DSB}(t)=A_{c}x(t)\sin(\omega _{c}t)\\&x_{DSB}^{+}(t)=x_{DSB}(t)+j{\hat {x}}_{DSB}(t)=A_{c}x(t)\cos(\omega _{c}t)+jA_{c}x(t)\sin(\omega _{c}t)=\\&A_{c}x(t)\left(\cos(\omega _{c}t)+j\sin(\omega _{c}t)\right)=A_{c}x(t)e^{+j\omega _{c}t}\\&{\tilde {x}}_{DSB}(t)=x_{DSB}^{+}(t)\cdot e^{-j\omega _{c}t}=A_{c}x(t)\to {\tilde {X}}_{DSB}(f)=A_{c}X(f)\\\end{aligned}}$
Ahora...
${\begin{aligned}&H^{u}(f)=\prod {\left({\frac {f-fc-^{W}\!\!\diagup \!\!_{2}\;}{W}}\right)}\\&H^{l}(f)=\prod {\left({\frac {f-fc+^{W}\!\!\diagup \!\!_{2}\;}{W}}\right)}\\&{\tilde {x}}(t)=x^{+}(t)e^{-j\omega _{c}t};x^{+}(t)=x(t)+j{\hat {x}}(t)\\&{\tilde {H}}^{u}(f)=2\prod {\left({\frac {f-^{W}\!\!\diagup \!\!_{2}\;}{W}}\right)}=\prod {\left({\frac {f}{2W}}\right)\left(1+\operatorname {sign} (f)\right)}\\\end{aligned}}$
Por lo que:
${\begin{aligned}&{\tilde {H}}^{u}(f)=\prod {\left({\frac {f}{2W}}\right)\left(1+\operatorname {sign} (f)\right)}\\&{\tilde {X}}_{DSB}(f)=A_{c}X(f){\text{ ; }}\beta _{T}=2W\\&{\tilde {X}}_{SSB}(f)={\frac {1}{2}}{\tilde {X}}_{DSB}(f)\cdot {\tilde {H}}^{u}(f)\\&{\tilde {X}}_{SSB}(f)={\frac {1}{2}}A_{c}\underbrace {X(f)\prod {\left({\frac {f}{2W}}\right)}} _{\beta _{T}=2W}\left(1+\operatorname {sign} (f)\right)=\\&{\tilde {X}}_{SSB}(f)={\frac {1}{2}}A_{c}\underbrace {X(f)\left(1+\operatorname {sign} (f)\right)} _{X^{+}(f)}={\frac {1}{2}}A_{c}X^{+}(f)\\&{\tilde {X}}_{SSB}(f)={\frac {A_{c}}{2}}X^{+}(f)\to {\tilde {x}}_{SSB}(t)={\frac {A_{c}}{2}}x^{+}(t)\\\end{aligned}}$
Por propiedades, nosotros sabemos que:
${\begin{aligned}&{\tilde {x}}(t)=x^{+}(t)e^{-j\omega _{c}t}\to x^{+}(t)={\tilde {x}}(t)\cdot e^{+j\omega _{c}t}\\&x(t)=\Re \left\{x^{+}(t)\right\}\\\end{aligned}}$
Por lo que:
${\begin{aligned}&{\tilde {x}}_{SSB}(t)={\frac {A_{c}}{2}}x^{+}(t)\to \\&x_{SSB}^{+}(t)=\left({\frac {A_{c}}{2}}x^{+}(t)\right)e^{+j\omega _{c}t}\\&x_{SSB}(t)=\Re \left\{x_{SSB}^{+}(t)\right\}={\frac {A_{c}}{2}}\Re \left\{x^{+}(t)e^{+j\omega _{c}t}\right\}={\frac {A_{c}}{2}}\Re \left\{\left(x(t)+j{\hat {x}}(t)\right)\left(\cos \left(\omega _{c}t\right)+j\sin \left(\omega _{c}t\right)\right)\right\}=\\&x_{SSB}^{u}(t)={\frac {A_{c}}{2}}\left(x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right)\\\end{aligned}}$
La función matemática en frecuencia de la señal SSB será:
${\begin{aligned}&x_{SSB}^{u}(t)={\frac {A_{c}}{2}}\left(x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right)\\&x_{SSB}^{l}(t)={\frac {A_{c}}{2}}\left(x(t)\cos(\omega _{c}t)+{\hat {x}}(t)\sin(\omega _{c}t)\right)\\&\mathbb {F} \left[x_{SSB}(t)\right]=X_{SSB}(f)\\&X_{SSB}^{u}(f)={\frac {A_{c}}{2}}\left[X(f-f_{c}){\frac {\left(1+\operatorname {sign} (f-f_{c})\right)}{2}}\right]+\left[X(f+f_{c}){\frac {\left(1-\operatorname {sign} (f+f_{c})\right)}{2}}\right]=\\&X_{SSB}^{u}(f)={\frac {A_{c}}{4}}\left[X(f-f_{c})\left(1+\operatorname {sign} (f-f_{c})\right)\right]+\left[X(f+f_{c})\left(1-\operatorname {sign} (f+f_{c})\right)\right]\\&X_{SSB}^{l}(f)={\frac {A_{c}}{4}}\left[X(f-f_{c})\left(1-\operatorname {sign} (f-f_{c})\right)\right]+\left[X(f+f_{c})\left(1+\operatorname {sign} (f+f_{c})\right)\right]\\\end{aligned}}$
Como se ve, la representación en frecuencia podíamos obtenerla a partir del dibujo de frecuencia y usando las representaciones vistas (señal analítica, hilbert...)

## Propiedades y DEP

${\begin{aligned}&x(t)=x_{I}(t)\cos(\omega _{c}t)-x_{Q}(t)\sin(\omega _{c}t)\to \\&s(t)=x_{SSB}^{u}(t)={\frac {A_{c}}{2}}\left(x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right)\\&s_{I}(t)={\frac {A_{c}}{2}}x(t)\\&s_{Q}(t)=-{\frac {A_{c}}{2}}{\hat {x}}(t)\\&e(t)={\sqrt {s_{I}^{2}(t)+s_{Q}^{2}(t)}}={\frac {A_{c}}{2}}{\sqrt {x^{2}(t)+{\hat {x}}^{2}(t)}}\\&{\tilde {s}}(t)={\frac {A_{c}}{2}}x^{+}(t)\\\end{aligned}}$
Para sacar la Densidad Espectral de Potencia, sacaremos primeramente la autocorrelación:
${\begin{aligned}&R_{SSB}(\tau )={\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{x_{SSB}^{*}(t)\cdot x_{SSB}(t+\tau )\partial t=}\\&x_{SSB}(t)={\frac {A_{c}}{2}}\left(x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right)\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \left({\frac {A_{c}}{2}}\right)^{2}\int _{-\infty }^{\infty }{\left[x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right]^{*}\cdot \left[x(t+\tau )\cos \left(\omega _{c}\left(t+\tau \right)\right)-{\hat {x}}(t+\tau )\sin \left(\omega _{c}\left(t+\tau \right)\right)\right]\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \left({\frac {A_{c}}{2}}\right)^{2}\int _{-\infty }^{\infty }\\&x^{*}(t)\cos(\omega _{c}t)x(t+\tau )\cos \left(\omega _{c}\left(t+\tau \right)\right)+\\&-{\hat {x}}^{*}(t)\sin(\omega _{c}t)x(t+\tau )\cos \left(\omega _{c}\left(t+\tau \right)\right)+\\&x^{*}(t)\cos(\omega _{c}t)\left(-{\hat {x}}(t+\tau )\sin \left(\omega _{c}\left(t+\tau \right)\right)\right)+\\&-{\hat {x}}^{*}(t)\sin(\omega _{c}t)\left(-{\hat {x}}(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}}\cdot \left({\frac {A_{c}}{2}}\right)^{2}\int _{-\infty }^{\infty }\\&x^{*}(t)x(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\}\\&-{\hat {x}}^{*}(t)x(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^{*}(t)-{\hat {x}}(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\}\\&-{\hat {x}}^{*}(t)-{\hat {x}}(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\}=\\&\\\end{aligned}}$
${\begin{aligned}&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \left({\frac {A_{c}}{2}}\right)^{2}\int _{-\infty }^{\infty }{\frac {1}{2}}\left({\begin{aligned}&x^{*}(t)x(t+\tau )\cos \left(\omega _{c}\tau \right)-{\hat {x}}^{*}(t)x(t+\tau )\overbrace {\sin(-\omega _{c}\tau )} ^{-\sin \left(\omega _{c}\tau \right)}\\&-x^{*}(t){\hat {x}}(t+\tau )\sin \left(\omega _{c}\tau \right)+{\hat {x}}^{*}(t){\hat {x}}(t+\tau )\cos \left(\omega _{c}\tau \right)\\\end{aligned}}\right)=\\&\left({\frac {A_{c}}{2}}\right)^{2}{\frac {1}{2}}\left(R_{x}\left(\tau \right)\cos \left(\omega _{c}\tau \right)+R_{x{\hat {x}}}\left(\tau \right)\sin \left(\omega _{c}\tau \right)-R_{{\hat {x}}x}\left(\tau \right)\sin \left(\omega _{c}\tau \right)+R_{\hat {x}}\left(\tau \right)\cos \left(\omega _{c}\tau \right)\right)=\\&\left\{{\begin{aligned}&R_{{\hat {x}}x}\left(\tau \right)=-R_{x{\hat {x}}}\left(\tau \right)\\&R_{x}\left(\tau \right)=R_{\hat {x}}\left(\tau \right)\\\end{aligned}}\right\}\to \\&R_{SSB}(\tau )=\left({\frac {A_{c}^{2}}{4}}\right){\frac {1}{2}}\left(2R_{x}\left(\tau \right)\cos \left(\omega _{c}\tau \right)-2R_{{\hat {x}}x}\left(\tau \right)\sin \left(\omega _{c}\tau \right)\right)=\\&R_{SSB}(\tau )=\left({\frac {A_{c}^{2}}{4}}\right)\left(R_{x}\left(\tau \right)\cos \left(\omega _{c}\tau \right)-R_{{\hat {x}}x}\left(\tau \right)\sin \left(\omega _{c}\tau \right)\right)\\\end{aligned}}$
Una vez obtenida la autocorrelación, la DEP será:
${\begin{aligned}&X_{SSB}^{u}(f)={\frac {A_{c}}{4}}\left[X(f-f_{c})\left(1+\operatorname {sign} (f-f_{c})\right)\right]+\left[X(f+f_{c})\left(1-\operatorname {sign} (f+f_{c})\right)\right]\\&R_{SSB}(\tau )=\left({\frac {A_{c}^{2}}{4}}\right)\left(R_{x}\left(\tau \right)\cos \left(\omega _{c}\tau \right)-R_{{\hat {x}}x}\left(\tau \right)\sin \left(\omega _{c}\tau \right)\right)\\&G_{SSB}(f)=\mathbb {F} \left[R_{SSB}(\tau )\right]=\left({\frac {A_{c}}{4}}\right)^{2}\cdot \left[G_{x}(f-f_{c})\left(1+\operatorname {sign} (f-f_{c})\right)^{2}\right]+\left[G_{x}(f+f_{c})\left(1-\operatorname {sign} (f+f_{c})\right)^{2}\right]=\\&G_{SSB}(f)={\frac {A_{c}^{2}}{8}}\left[G_{x}(f-f_{c})\left(1+\operatorname {sign} (f-f_{c})\right)\right]+\left[G_{x}(f+f_{c})\left(1-\operatorname {sign} (f+f_{c})\right)\right]\\\end{aligned}}$
Y la potencia será:
${\begin{aligned}&P_{SSB}=S_{SSB}=R_{SSB}(0)=\left({\frac {A_{c}^{2}}{4}}\right)\left(R_{x}\left(0\right)\cos \left(0\right)-R_{{\hat {x}}x}\left(0\right)\sin \left(0\right)\right)=\\&S_{SSB}={\frac {A_{c}^{2}}{4}}S_{x}\\&S_{DSB}={\frac {A_{c}^{2}}{2}}S_{x}\to S_{SSB}={\frac {S_{DSB}}{2}}\\\end{aligned}}$
Tiene sentido, pues si recordamos, la potencia de una señal DSB era el doble, pues necesitaba el doble de ancho de banda.

## Relación señal a ruido de una señal SSB, detección coherente

ACTUALIZAR DIBUJO PARA BANDA LATERAL
${\begin{aligned}&x(t)=x_{I}(t)\cos(\omega _{c}t)-x_{Q}(t)\sin(\omega _{c}t)\to \\&s(t)=x_{SSB}^{u}(t)={\frac {A_{c}}{2}}\left(x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right)\\&s_{R}(t)=x_{SSB}(t)\cdot^{g_{T}}\!\!\diagup \!\!_{L}\;\to \left\{A_{R}=A_{c}{\frac {g_{T}}{L}}\right\}\to \\&s_{R}(t)={\frac {A_{R}}{2}}\left(x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right)\\&y_{R}(t)=s_{R}(t)+n_{R}(t)\\&\left(^{S}\!\!\diagup \!\!_{N}\;\right)_{R}=?\\&G_{n_{R}}(f)=G_{n}(f)\cdot \left|H_{R}(f)\right|^{2}\to G_{n}(f)={\frac {\eta }{2}}\to N_{R}=\int _{-\infty }^{\infty }{G_{n_{R}}(f)\partial f}\\&N_{R}=2{\frac {\eta }{2}}\beta _{T}=\eta \beta _{T}\to \beta _{T}=W\to N_{R}=\eta W\\&S_{SSB}={\frac {A_{c}^{2}}{4}}S_{x}\\&\left(^{S}\!\!\diagup \!\!_{N}\;\right)_{R}={\frac {S_{R}}{N_{R}}}={\frac {S_{R}}{\eta \beta _{T}}}={\frac {S_{R}}{\eta W}}={\frac {{\frac {A_{R}^{2}}{4}}S_{x}}{\eta W}}\\\end{aligned}}$
Calculemos ahora, la relación señal a ruido en detección $\left(^{S}\!\!\diagup \!\!_{N}\;\right)_{D}$
${\begin{aligned}&s_{R}(t)={\frac {A_{R}}{2}}\left(x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right)\\&y_{R}(t)=s_{R}(t)+n_{R}(t)\\&y_{D}(t)=s_{D}(t)+n_{D}(t)=y_{R}(t)\cos(\omega _{c}t){\text{ y filtrar}}\to \\&s_{D}(t)=s_{R}(t)\cos(\omega _{c}t)={\frac {A_{R}}{2}}\left(x(t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\right)\cos(\omega _{c}t)=\\&{\frac {A_{R}}{2}}\left(x(t)\cos(\omega _{c}t)\cos(\omega _{c}t)-{\hat {x}}(t)\sin(\omega _{c}t)\cos(\omega _{c}t)\right)=\to \left\{{\begin{aligned}&\cos a\cos b={\frac {\cos(a+b)+\cos(a-b)}{2}}\\&\sin a\cos b={\frac {\sin(a+b)+\sin(a-b)}{2}}\\\end{aligned}}\right\}\\&s_{D}(t)={\frac {A_{R}}{2}}x(t)\left({\frac {\overbrace {\cos(2\omega _{c}t)} ^{\text{eliminado por el filtro}}+1}{2}}\right)-{\hat {x}}(t)\left({\frac {\overbrace {\sin(2\omega _{c}t)} ^{\text{eliminado por el filtro}}+0}{2}}\right)={\frac {A_{R}}{4}}x(t)\\&s_{D}(t)={\frac {A_{R}}{4}}x(t)\\&n_{D}(t)=???\\&x(t)=x_{I}(t)\cos(\omega _{c}t)-x_{Q}(t)\sin(\omega _{c}t)\to \\&n(t)=n_{I}(t)\cos(\omega _{c}t)-n_{Q}(t)\sin(\omega _{c}t)\to \\&n_{R}(t)=n_{R_{I}}(t)\cos(\omega _{c}t)-n_{R_{Q}}(t)\sin(\omega _{c}t)\to \\&y_{R}(t)=s_{R}(t)+n_{R}(t)=s_{R}(t)+n_{R_{I}}(t)\cos(\omega _{c}t)-n_{R_{Q}}(t)\sin(\omega _{c}t)\\&n_{D}(t)=n_{R}(t)\cdot \cos(\omega _{c}t){\text{ y filtrar}}\\&n_{D}(t)=\left(n_{R_{I}}(t)\cos(\omega _{c}t)-n_{R_{Q}}(t)\sin(\omega _{c}t)\right)\cdot \cos(\omega _{c}t)\to \left\{{\begin{aligned}&\cos a\cos b={\frac {\cos(a+b)+\cos(a-b)}{2}}\\&\sin a\cos b={\frac {\sin(a+b)+\sin(a-b)}{2}}\\\end{aligned}}\right\}\\&n_{D}(t)=n_{R_{I}}(t)\left({\frac {\overbrace {\cos(2\omega _{c}t)} ^{\text{eliminado por el filtro}}+1}{2}}\right)-n_{R_{Q}}(t)\left({\frac {\overbrace {\sin(2\omega _{c}t)} ^{\text{eliminado por el filtro}}+0}{2}}\right)\\&n_{D}(t)={\frac {n_{R_{I}}(t)}{2}}\\\end{aligned}}$
${\begin{aligned}&y_{D}(t)=s_{D}(t)+n_{D}(t)\\&s_{D}(t)={\frac {A_{R}}{4}}x(t)\\&n_{D}(t)={\frac {n_{R_{I}}(t)}{2}}\\&\left(^{S}\!\!\diagup \!\!_{N}\;\right)_{D}={\frac {S_{D}}{N_{D}}}\\&G_{Y_{D}}(f)={\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {\left|Y_{D}(f)\right|^{2}}{T}}\to S_{D}={\frac {A_{R}^{2}S_{x}}{16}}\\&N_{D}={\frac {N_{R_{I}}}{4}}\to \left\{{\text{Propiedades: }}S_{x}=S_{x_{I}}=S_{x_{Q}}\right\}\\&N_{R}=\eta \beta _{T}=\eta W\to N_{D}={\frac {\eta W}{4}}\\&\left(^{S}\!\!\diagup \!\!_{N}\;\right)_{D}={\frac {S_{D}}{N_{D}}}={\frac {\frac {A_{R}^{2}S_{x}}{16}}{\frac {\eta W}{4}}}={\frac {A_{R}^{2}S_{x}}{4\eta W}}\\\end{aligned}}$

## Comparación mediante factor de calidad

Para comparar lo eficaz de nuestra modulación, podremos la relación señal a ruido de detección en función del factor de calidad.
${\begin{aligned}&\gamma ={\frac {S_{R}}{\eta W}}\\&S_{R}={\frac {A_{R}^{2}}{4}}S_{x}\\&\left(^{S}\!\!\diagup \!\!_{N}\;\right)_{D}={\frac {A_{R}^{2}S_{x}}{4\eta W}}={\frac {S_{R}}{\eta W}}=\gamma \\\end{aligned}}$
Con lo que la calidad de la modulación es igual al factor de calidad, igual que ocurría en DSB, salvo que ahora el ancho de banda necesitado es la mitad.
$\left(\beta _{T}=W\right)$

## Detección por envolvente

Seria posible también usar detección por envolvente, con todas las ventajas que ello conlleva (simplicidad, eficacia...) siendo su análisis análogo al hecho en AM. Para ello, lo único que tenemos que hacer:
${\begin{aligned}&x^{'}(t)=A_{c}\left(1+mx(t)\right)\\&x_{SSB}^{u}(t)={\frac {A_{c}}{2}}\left(x^{'}(t)\cos(\omega _{c}t)-{\hat {x}}^{'}(t)\sin(\omega _{c}t)\right)\\\end{aligned}}$
* Proyecto: Departamento de Teoría de la Señal y Comunicaciones
* Anterior: AM — SSB — Siguiente: VSB