Representación en tiempo y frecuencia
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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:
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{\displaystyle X_{SSB}(f)=X_{DSB}(f)\cdot H^{u}(f)}
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Intentemos conseguir primero la función envolvente compleja de la señal SSB. Una vez lograda, la transformaremos en paso-banda.
Recordemos la propiedad de:
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{\displaystyle {\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}}}
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{\displaystyle {\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...
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{\displaystyle {\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:
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{\displaystyle {\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:
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{\displaystyle {\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:
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{\displaystyle {\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:
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{\displaystyle {\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
editar
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{\displaystyle {\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:
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{
Relaciones Trigonometricas
}
lim
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⏞
fuera del area de
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+
←
{
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⏞
fuera del area de
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←
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⏞
fuera del area de
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+
←
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sin
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}
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⏞
fuera del area de
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∂
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{
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cos
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cos
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}
=
{\displaystyle {\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}}}
lim
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1
2
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{
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→
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2
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x
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x
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)
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x
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{\displaystyle {\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:
X
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]
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x
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G
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{\displaystyle {\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}}}
P
S
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B
=
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R
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=
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{\displaystyle {\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
editar
ACTUALIZAR DIBUJO PARA BANDA LATERAL
x
(
t
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I
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c
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c
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{
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^
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sin
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c
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y
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s
R
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n
R
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t
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S
╱
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R
=
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G
n
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⋅
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R
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2
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n
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η
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∫
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∞
∞
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n
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f
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2
η
2
β
T
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η
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→
β
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W
→
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η
W
S
S
S
B
=
A
c
2
4
S
x
(
S
╱
N
)
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N
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β
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W
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A
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2
4
S
x
η
W
{\displaystyle {\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
(
S
╱
N
)
D
{\displaystyle \left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{D}}
s
R
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=
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2
(
x
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)
cos
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(
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sin
(
ω
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)
)
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)
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(
ω
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)
y filtrar
→
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s
R
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)
cos
(
ω
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t
)
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2
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cos
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ω
c
t
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−
x
^
(
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)
sin
(
ω
c
t
)
)
cos
(
ω
c
t
)
=
A
R
2
(
x
(
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)
cos
(
ω
c
t
)
cos
(
ω
c
t
)
−
x
^
(
t
)
sin
(
ω
c
t
)
cos
(
ω
c
t
)
)
=→
{
cos
a
cos
b
=
cos
(
a
+
b
)
+
cos
(
a
−
b
)
2
sin
a
cos
b
=
sin
(
a
+
b
)
+
sin
(
a
−
b
)
2
}
s
D
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t
)
=
A
R
2
x
(
t
)
(
cos
(
2
ω
c
t
)
⏞
eliminado por el filtro
+
1
2
)
−
x
^
(
t
)
(
sin
(
2
ω
c
t
)
⏞
eliminado por el filtro
+
0
2
)
=
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R
4
x
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s
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4
x
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x
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c
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Q
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n
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−
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sin
(
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)
→
y
R
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s
R
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+
n
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s
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sin
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n
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=
n
R
(
t
)
⋅
cos
(
ω
c
t
)
y filtrar
n
D
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=
(
n
R
I
(
t
)
cos
(
ω
c
t
)
−
n
R
Q
(
t
)
sin
(
ω
c
t
)
)
⋅
cos
(
ω
c
t
)
→
{
cos
a
cos
b
=
cos
(
a
+
b
)
+
cos
(
a
−
b
)
2
sin
a
cos
b
=
sin
(
a
+
b
)
+
sin
(
a
−
b
)
2
}
n
D
(
t
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=
n
R
I
(
t
)
(
cos
(
2
ω
c
t
)
⏞
eliminado por el filtro
+
1
2
)
−
n
R
Q
(
t
)
(
sin
(
2
ω
c
t
)
⏞
eliminado por el filtro
+
0
2
)
n
D
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)
=
n
R
I
(
t
)
2
{\displaystyle {\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}}}
y
D
(
t
)
=
s
D
(
t
)
+
n
D
(
t
)
s
D
(
t
)
=
A
R
4
x
(
t
)
n
D
(
t
)
=
n
R
I
(
t
)
2
(
S
╱
N
)
D
=
S
D
N
D
G
Y
D
(
f
)
=
lim
T
→
∞
|
Y
D
(
f
)
|
2
T
→
S
D
=
A
R
2
S
x
16
N
D
=
N
R
I
4
→
{
Propiedades:
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{\displaystyle {\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
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Para comparar lo eficaz de nuestra modulación, podremos la relación señal a ruido de detección en función del factor de calidad.
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{\displaystyle {\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.
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{\displaystyle \left(\beta _{T}=W\right)}
Detección por envolvente
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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:
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{\displaystyle {\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}}}
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6be812d4a3d755e0cdbe44257721592632f56f4)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c7b3e8975bf1f93a44e2525967e86219369bde5)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a781f4b27ca3f073317cd46e4a44fba0a961529c)
