La modulación AM (Amplitude Modulation ) es un tipo de modulación lineal, siendo una de las modulación analógicas mas usadas, por no decir la que mas abunda en los sistemas de comunicaciones.
La diferencia entre esta modulación y la DSB es que en esta última no tenemos energía en la portadora.
Representación en tiempo y frecuencia
editar
Su representación temporal es:
x
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cos
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x
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o
x
m
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:
modulador
m: modulation index
, indice de modulacion
{\displaystyle {\begin{aligned}&x_{AM}(t)=A_{c}\left(1+mx(t)\right)\cos(\omega _{c}t)\\&x(t){\text{ o }}\,x_{m}(t):{\text{ modulador}}\\&{\text{m: modulation index}}{\text{, indice de modulacion}}\\\end{aligned}}}
El índice de modulación nos dice indirectamente la amplitud de la portadora, y varia de 0 a 1, dandose su valor normalmente en porcentajes de 0% a 100%.
En la función se considerá que
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{\displaystyle \left|x(t)\right|\leq 1\to S_{x}\leq 1}
.
También, debido a que los canales bloquean el paso de continua, nuestra señal no tendrá continua:
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∞
∞
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{\displaystyle m_{x(t)}=0\to {\frac {1}{T}}\int _{-\infty }^{\infty }{x(t)\partial t}=0}
Si vemos su representación en frecuencia:
x
A
M
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=
A
c
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1
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m
x
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)
cos
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⋅
cos
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m
⋅
x
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cos
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cos
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e
j
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x
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m
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X
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{\displaystyle {\begin{aligned}&x_{AM}(t)=A_{c}\left(1+mx(t)\right)\cos(\omega _{c}t)=A_{c}\cdot \cos(\omega _{c}t)+m\cdot x(t)\cos(\omega _{c}t)\\&\cos \left(\omega _{c}t\right)={\frac {e^{j\omega _{c}t}+e^{-j\omega _{c}t}}{2}}\\&\mathbb {F} \left[x_{AM}(t)\right]=X_{AM}(f)={\frac {A_{c}}{2}}\left(\delta \left(f-f_{c}\right)+\delta (f+f_{c})\right)+{\frac {A_{c}}{2}}m\left(X\left(f-f_{c}\right)+X(f+f_{c})\right)\\\end{aligned}}}
Como se ha dicho, en AM tenemos potencia en la portadora lo cual nos permitirá demodular usando detección por envolvente.
x
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cos
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sin
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s
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cos
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e
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si
m
≤
1
{\displaystyle {\begin{aligned}&x(t)=x_{I}(t)\cos(\omega _{c}t)-x_{Q}(t)\sin(\omega _{c}t)\to \\&s(t)=x_{AM}(t)=A_{c}\left(1+mx(t)\right)\cos(\omega _{c}t)\\&s_{I}(t)=A_{c}\left(1+mx(t)\right)\\&s_{Q}(t)=0\\&e(t)=A_{c}\left|\left(1+mx(t)\right)\right|\\&A_{c}\left|\left(1+mx(t)\right)\right|=A_{c}\left(1+mx(t)\right){\text{ si }}m\leq 1\\\end{aligned}}}
Para sacar la Densidad Espectral de Potencia, sacaremos primeramente la autocorrelación:
R
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lim
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→
∞
1
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⋅
∫
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∞
∞
x
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∗
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⋅
x
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M
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∂
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lim
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→
∞
1
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⋅
∫
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∞
∞
[
A
c
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m
x
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cos
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c
t
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∗
⋅
[
A
c
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x
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t
+
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cos
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c
(
t
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]
∂
t
=
lim
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→
∞
1
T
⋅
A
c
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∫
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∞
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∗
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cos
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cos
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lim
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→
∞
1
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⋅
A
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∞
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cos
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+
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∂
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=
{
cos
a
cos
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=
cos
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+
cos
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)
2
}
lim
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→
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∞
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cos
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⏞
fuera del area de
∫
+
cos
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⏞
no depende de t
2
)
∂
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=
lim
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→
∞
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cos
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∫
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∞
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lim
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cos
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∫
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∞
∞
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∗
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∂
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cos
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lim
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∞
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∗
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⏞
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2
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R
x
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cos
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{\displaystyle {\begin{aligned}&R_{AM}(\tau )={\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{x_{AM}^{*}(t)\cdot x_{AM}(t+\tau )\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{\left[A_{c}\left(1+mx(t)\right)\cos(\omega _{c}t)\right]^{*}\cdot \left[A_{c}\left(1+mx(t+\tau )\right)\cos \left(\omega _{c}\left(t+\tau \right)\right)\right]\partial t=}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot A_{c}^{2}\int _{-\infty }^{\infty }{\left(1+mx^{*}(t)\right)\left(1+mx(t+\tau )\right)\cos(\omega _{c}t)\cdot \cos \left(\omega _{c}t+\omega _{c}\tau \right)\partial t}=\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot A_{c}^{2}\int _{-\infty }^{\infty }{\left(1+mx(t+\tau )+mx^{*}(t)+m^{2}x^{*}(t)x(t+\tau )\right)\cos(\omega _{c}t)\cdot \cos \left(\omega _{c}t+\omega _{c}\tau \right)\partial t}=\\&\left\{\cos a\cos b={\frac {\cos(a+b)+\cos(a-b)}{2}}\right\}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot A_{c}^{2}\int _{-\infty }^{\infty }{\left(1+mx(t+\tau )+mx^{*}(t)+m^{2}x^{*}(t)x(t+\tau )\right)\left({\frac {\overbrace {\cos(2\omega _{c}t+\tau )} ^{{\text{fuera del area de }}\int {}}+\overbrace {\cos(\omega _{c}\tau )} ^{\text{no depende de t}}}{2}}\right)\partial t}=\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot {\frac {A_{c}^{2}}{2}}\cos(\omega _{c}\tau )\int _{-\infty }^{\infty }{\left(1+mx(t+\tau )+mx^{*}(t)+m^{2}x^{*}(t)x(t+\tau )\right)\partial t}=\\&\left\{m_{x}=0,{\frac {1}{T}}\int _{-\infty }^{\infty }{x(t)\partial t}\right\}\\&{\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot {\frac {A_{c}^{2}}{2}}\cos(\omega _{c}\tau )\int _{-\infty }^{\infty }{\left(1+m^{2}x^{*}(t)x(t+\tau )\right)\partial t}=\\&{\frac {A_{c}^{2}}{2}}\cos(\omega _{c}\tau )\left(1+m^{2}\overbrace {\cdot {\underset {T\to \infty }{\mathop {\lim } }}\,{\frac {1}{T}}\cdot \int _{-\infty }^{\infty }{x^{*}(t)x(t+\tau )\partial t}} ^{R_{x}(\tau )}\right)={\frac {A_{c}^{2}}{2}}\left(1+m^{2}R_{x}(\tau )\right)\cos(\omega _{c}\tau )\\\end{aligned}}}
Por lo que la DEP:
R
A
M
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τ
)
=
A
c
2
2
(
1
+
m
2
R
x
(
τ
)
)
cos
(
ω
c
τ
)
F
[
R
A
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)
]
=
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G
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4
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m
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)
{\displaystyle {\begin{aligned}&R_{AM}(\tau )={\frac {A_{c}^{2}}{2}}\left(1+m^{2}R_{x}(\tau )\right)\cos(\omega _{c}\tau )\\&\mathbb {F} \left[R_{AM}(\tau )\right]=G_{AM}(f)\\&G_{AM}(f)={\frac {A_{c}^{2}}{4}}\left(\delta (f-f_{c})+\delta (f+f_{c})\right)+{\frac {A_{c}^{2}}{4}}\cdot m^{2}\left(G_{x}(f-f_{c})+G_{x}(f+f_{c})\right)\\\end{aligned}}}
Terminando, la potencia de la señal será:
R
A
M
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τ
)
=
A
c
2
2
(
1
+
m
2
R
x
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τ
)
)
cos
(
ω
c
τ
)
P
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S
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A
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2
(
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+
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)
{\displaystyle {\begin{aligned}&R_{AM}(\tau )={\frac {A_{c}^{2}}{2}}\left(1+m^{2}R_{x}(\tau )\right)\cos(\omega _{c}\tau )\\&P_{AM}=S_{AM}=R_{AM}\left(\tau =0\right)\\&S_{AM}={\frac {A_{c}^{2}}{2}}\left(1+m^{2}S_{x}\right)\\\end{aligned}}}
Relación señal a ruido de una señal AM, detección coherente
editar
s
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cos
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⋅
g
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→
{
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}
→
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cos
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+
n
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n
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η
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→
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=
∫
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∞
∞
G
n
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∂
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R
=
2
∫
0
∞
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⋅
∏
(
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)
∂
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S
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=
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c
2
2
(
1
+
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2
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x
)
(
S
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R
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N
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=
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η
β
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=
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=
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2
(
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+
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2
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x
)
η
2
W
{\displaystyle {\begin{aligned}&s(t)=x_{AM}(t)=A_{c}\left(1+mx(t)\right)\cos(\omega _{c}t)\\&s_{R}(t)=x_{AM}(t)\cdot {}^{g_{T}}\!\!\diagup \!\!{}_{L}\;\to \left\{A_{R}=A_{c}{\frac {g_{T}}{L}}\right\}\to \\&s_{R}(t)=A_{R}\left(1+mx(t)\right)\cos(\omega _{c}t)\\&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\int _{0}^{\infty }{{\frac {\eta }{2}}\cdot \prod {\left({\frac {f-f_{c}}{2W}}\right)}\partial f}=2{\frac {\eta }{2}}2W=\eta 2W\\&S_{AM}={\frac {A_{c}^{2}}{2}}\left(1+m^{2}S_{x}\right)\\&\left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{R}={\frac {S_{R}}{N_{R}}}={\frac {S_{R}}{\eta \beta _{T}}}={\frac {S_{R}}{\eta 2W}}={\frac {{\frac {A_{R}^{2}}{2}}\left(1+m^{2}S_{x}\right)}{\eta 2W}}\\\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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+
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x
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cos
(
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c
t
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n
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|
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|
→
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≤
1
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+
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(
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s
D
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→
e
s
R
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⋅
DC-Block
→
A
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m
x
(
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s
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=
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m
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(
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2
n
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x
(
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=
x
I
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cos
(
ω
c
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)
−
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(
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)
sin
(
ω
c
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)
n
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=
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cos
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−
n
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(
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sin
(
ω
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)
→
n
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=
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I
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cos
(
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t
)
−
n
R
Q
(
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)
sin
(
ω
c
t
)
n
D
(
t
)
=
n
R
(
t
)
⋅
cos
(
ω
c
t
)
y filtro
|
H
L
P
F
(
f
)
|
2
n
D
(
t
)
=
(
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
)
=
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
(
t
)
=
n
R
I
(
t
)
2
{\displaystyle {\begin{aligned}&s_{R}(t)=A_{R}\left(1+mx(t)\right)\cos(\omega _{c}t)\\&y_{R}(t)=s_{R}(t)+n_{R}(t)\\&y_{D}(t)=s_{D}(t)+n_{D}(t)\\&e_{s_{R}}(t)=\left|A_{R}\left(1+mx(t)\right)\right|{\xrightarrow[{m\leq 1}]{}}A_{R}\left(1+mx(t)\right)\\&s_{D}(t)\to e_{s_{R}}(t){\text{ }}\cdot {\text{ DC-Block}}\to A_{R}mx(t)\\&s_{D}(t)={\frac {A_{R}mx(t)}{2}}\\&n_{D}(t)=?\\&x(t)=x_{I}(t)\cos \left(\omega _{c}t\right)-x_{Q}(t)\sin(\omega _{c}t)\\&n(t)=n_{I}(t)\cos \left(\omega _{c}t\right)-n_{Q}(t)\sin(\omega _{c}t)\to n_{R}(t)=n_{R_{I}}(t)\cos \left(\omega _{c}t\right)-n_{R_{Q}}(t)\sin(\omega _{c}t)\\&n_{D}(t)=n_{R}(t)\cdot \cos(\omega _{c}t){\text{ y filtro }}\left|H_{LPF}(f)\right|^{2}\\&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
m
x
(
t
)
2
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
m
2
S
x
4
N
D
=
N
R
I
4
→
{
Propiedades:
S
x
=
S
x
I
=
S
x
Q
}
N
R
=
η
β
T
=
η
2
W
→
N
D
=
(
η
2
W
)
4
(
S
╱
N
)
D
=
S
D
N
D
=
A
R
2
m
2
S
x
4
(
η
2
W
╱
4
)
=
A
R
2
m
2
S
x
η
2
W
{\displaystyle {\begin{aligned}&y_{D}(t)=s_{D}(t)+n_{D}(t)\\&s_{D}(t)={\frac {A_{R}mx(t)}{2}}\\&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}m^{2}S_{x}}{4}}\\&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 2W\to N_{D}={\frac {\left(\eta 2W\right)}{4}}\\&\left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{D}={\frac {S_{D}}{N_{D}}}={\frac {\frac {A_{R}^{2}m^{2}S_{x}}{4}}{\left({}^{\eta 2W}\!\!\diagup \!\!{}_{4}\;\right)}}={\frac {A_{R}^{2}m^{2}S_{x}}{\eta 2W}}\\\end{aligned}}}
Para comparar lo eficaz de nuestra modulación, podremos la relación señal a ruido de detección en función del factor de calidad.
γ
=
S
R
η
W
S
R
=
A
R
2
2
(
1
+
m
2
S
x
)
(
S
╱
N
)
D
=
A
R
2
m
2
S
x
η
2
W
→
A
R
2
m
2
S
x
2
1
η
W
⋅
S
R
S
R
=
A
R
2
m
2
S
x
2
1
η
W
⋅
S
R
=
A
R
2
2
(
1
+
m
2
S
x
)
S
R
=
A
R
2
2
(
1
+
m
2
S
x
)
→
(
S
╱
N
)
D
=
A
R
2
m
2
S
x
2
1
A
R
2
2
(
1
+
m
2
S
x
)
S
R
η
W
=
m
2
S
x
1
+
m
2
S
x
S
R
η
W
=
m
2
S
x
1
+
m
2
S
x
γ
{\displaystyle {\begin{aligned}&\gamma ={\frac {S_{R}}{\eta W}}\\&S_{R}={\frac {A_{R}^{2}}{2}}\left(1+m^{2}S_{x}\right)\\&\left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{D}={\frac {A_{R}^{2}m^{2}S_{x}}{\eta 2W}}\to {\frac {A_{R}^{2}m^{2}S_{x}}{2}}{\frac {1}{\eta W}}\cdot {\frac {S_{R}}{S_{R}}}={\frac {A_{R}^{2}m^{2}S_{x}}{2}}{\frac {1}{\eta W}}\cdot {\frac {S_{R}={\frac {A_{R}^{2}}{2}}\left(1+m^{2}S_{x}\right)}{S_{R}={\frac {A_{R}^{2}}{2}}\left(1+m^{2}S_{x}\right)}}\to \\&\left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{D}={\frac {A_{R}^{2}m^{2}S_{x}}{2}}{\frac {1}{{\frac {A_{R}^{2}}{2}}\left(1+m^{2}S_{x}\right)}}{\frac {S_{R}}{\eta W}}={\frac {m^{2}S_{x}}{1+m^{2}S_{x}}}{\frac {S_{R}}{\eta W}}={\frac {m^{2}S_{x}}{1+m^{2}S_{x}}}\gamma \\\end{aligned}}}
En el mejor de los casos
m
=
1
→
m
2
S
x
1
+
m
2
S
x
γ
=
S
x
1
+
S
x
γ
→
S
x
≤
1
γ
2
{\displaystyle m=1\to {\frac {m^{2}S_{x}}{1+m^{2}S_{x}}}\gamma ={\frac {S_{x}}{1+S_{x}}}\gamma {\xrightarrow[{S_{x}\leq 1}]{}}{\frac {\gamma }{2}}}
Con lo que, el mejor de los casos, desperdicíamos la mitad de la potencia, debido a que tenemos energía en la portadora.
Relación señal a ruido de una señal AM, detección por envolvente
editar
(
S
╱
N
)
D
=
?
y
R
(
t
)
=
s
R
(
t
)
+
n
R
(
t
)
n
R
(
t
)
=
n
R
I
(
t
)
cos
(
ω
c
t
)
−
n
R
Q
(
t
)
sin
(
ω
c
t
)
s
R
(
t
)
=
A
R
(
1
+
m
x
(
t
)
)
cos
(
ω
c
t
)
e
s
R
(
t
)
=
|
A
R
(
1
+
m
x
(
t
)
)
|
→
m
≤
1
A
R
(
1
+
m
x
(
t
)
)
s
D
(
t
)
=
e
s
R
(
f
)
⋅
DC-Block
→
A
R
m
x
(
t
)
y
R
(
t
)
=
y
R
I
(
t
)
cos
(
ω
c
t
)
−
y
R
Q
(
t
)
sin
(
ω
c
t
)
y
R
(
t
)
=
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
)
⏟
y
R
I
(
t
)
cos
(
ω
c
t
)
−
n
R
Q
(
t
)
⏟
y
R
Q
(
t
)
sin
(
ω
c
t
)
y
D
(
t
)
=
e
y
R
(
t
)
=
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
)
2
+
n
R
Q
2
(
t
)
{\displaystyle {\begin{aligned}&\left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{D}=?\\&y_{R}(t)=s_{R}(t)+n_{R}(t)\\&n_{R}(t)=n_{R_{I}}(t)\cos(\omega _{c}t)-n_{R_{Q}}(t)\sin(\omega _{c}t)\\&s_{R}(t)=A_{R}\left(1+mx(t)\right)\cos(\omega _{c}t)\\&e_{s_{R}}(t)=\left|A_{R}\left(1+mx(t)\right)\right|{\xrightarrow[{m\leq 1}]{}}A_{R}\left(1+mx(t)\right)\\&s_{D}(t)=e_{s_{R}}(f){\text{ }}\cdot {\text{ DC-Block }}\to A_{R}mx(t)\\&y_{R}(t)=y_{R_{I}}(t)\cos(\omega _{c}t)-y_{R_{Q}}(t)\sin(\omega _{c}t)\\&y_{R}(t)=\underbrace {\left(A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)\right)} _{y_{R_{I}}(t)}\cos(\omega _{c}t)-\underbrace {n_{R_{Q}}(t)} _{y_{R_{Q}}(t)}\sin(\omega _{c}t)\\&y_{D}(t)=e_{y_{R}}(t)={\sqrt {\left(A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)\right)^{2}+n_{R_{Q}}^{2}(t)}}\\\end{aligned}}}
Ahora, hay dos escenarios posibles: señal con poco ruido, y señal con mucho ruido, que dependen directamente de lo buena que sea la relación señal a ruido en recepcion
(
S
╱
N
)
R
{\displaystyle \left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{R}}
(
S
╱
N
)
R
↑↑
y
D
(
t
)
=
e
y
R
(
t
)
=
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
)
2
+
n
R
Q
2
(
t
)
{\displaystyle {\begin{aligned}&\left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{R}\uparrow \uparrow \\&y_{D}(t)=e_{y_{R}}(t)={\sqrt {\left(A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)\right)^{2}+n_{R_{Q}}^{2}(t)}}\\\end{aligned}}}
n
R
Q
2
(
t
)
≪
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
)
2
y
D
(
t
)
=
e
y
R
(
t
)
≈
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
⏟
y
R
I
(
t
)
)
2
y
D
(
t
)
=
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
y filtro DC
y
D
(
t
)
=
A
R
m
x
(
t
)
+
n
R
I
(
t
)
S
D
=
A
R
2
m
2
S
x
N
D
=
N
R
I
=
N
R
=
η
β
T
=
η
2
W
(
S
╱
N
)
D
=
A
R
2
m
2
S
x
η
2
W
=
m
2
S
x
1
+
m
2
S
x
γ
{\displaystyle {\begin{aligned}&n_{R_{Q}}^{2}(t)\ll \left(A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)\right)^{2}\\&y_{D}(t)=e_{y_{R}}(t)\approx {\sqrt {\left(\underbrace {A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)} _{y_{R_{I}}(t)}\right)^{2}}}\\&y_{D}(t)=A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t){\text{ y filtro DC}}\\&y_{D}(t)=A_{R}mx(t)+n_{R_{I}}(t)\\&S_{D}=A_{R}^{2}m^{2}S_{x}\\&N_{D}=N_{R_{I}}=N_{R}=\eta \beta _{T}=\eta 2W\\&\left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{D}={\frac {A_{R}^{2}m^{2}S_{x}}{\eta 2W}}={\frac {m^{2}S_{x}}{1+m^{2}S_{x}}}\gamma \\\end{aligned}}}
Sale lo misma relación señal a ruido que la detección coherente, con la ventaja añadida de usar un detector por envolvente, que es mucho mas barato y sencillo.
(
S
╱
N
)
R
↓↓
y
R
(
t
)
=
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
)
⏟
y
R
I
(
t
)
cos
(
ω
c
t
)
−
n
R
Q
(
t
)
⏟
y
R
Q
(
t
)
sin
(
ω
c
t
)
y
D
(
t
)
=
e
y
R
(
t
)
=
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
)
2
+
n
R
Q
2
(
t
)
{\displaystyle {\begin{aligned}&\left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{R}\downarrow \downarrow \\&y_{R}(t)=\underbrace {\left(A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)\right)} _{y_{R_{I}}(t)}\cos(\omega _{c}t)-\underbrace {n_{R_{Q}}(t)} _{y_{R_{Q}}(t)}\sin(\omega _{c}t)\\&y_{D}(t)=e_{y_{R}}(t)={\sqrt {\left(A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)\right)^{2}+n_{R_{Q}}^{2}(t)}}\\\end{aligned}}}
Para dibujar la representación mediante vectores, lo representaremos mediante la envolvente compleja:
x
~
(
t
)
=
x
I
(
t
)
+
j
x
Q
(
t
)
=
e
(
t
)
⋅
e
j
ϕ
(
t
)
y
~
R
(
t
)
=
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
)
+
j
⋅
n
R
Q
(
t
)
y
~
R
(
t
)
=
(
A
R
(
1
+
m
x
(
t
)
)
+
n
R
I
(
t
)
)
+
n
R
Q
(
t
)
⋅
e
j
π
2
{\displaystyle {\begin{aligned}&{\tilde {x}}(t)=x_{I}(t)+jx_{Q}(t)=e(t)\cdot e^{j\phi (t)}\\&{\tilde {y}}_{R}(t)=\left(A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)\right)+j\cdot n_{R_{Q}}(t)\\&{\tilde {y}}_{R}(t)=\left(A_{R}\left(1+mx(t)\right)+n_{R_{I}}(t)\right)+n_{R_{Q}}(t)\cdot e^{j{\frac {\pi }{2}}}\\\end{aligned}}}
A
R
(
1
+
m
x
(
t
)
)
≪
n
R
I
(
t
)
{\displaystyle A_{R}\left(1+mx(t)\right)\ll n_{R_{I}}(t)}
No podemos sacar
y
D
(
t
)
=
e
y
R
(
t
)
{\displaystyle y_{D}(t)=e_{y_{R}}(t)}
directamente, por lo que tendremos que usar la siguiente aproximación:
ϕ
y
R
(
t
)
≈
ϕ
n
(
t
)
{\displaystyle \phi _{y_{R}}(t)\approx \phi _{n}(t)}
Por lo que:
b
(
t
)
≈
e
n
(
t
)
{\displaystyle b(t)\approx e_{n}(t)}
y sabiendo
cos
x
=
cateto adyacente
hipotenusa
{\displaystyle \cos x={\frac {\text{cateto adyacente}}{\text{hipotenusa}}}}
cos
ϕ
y
R
(
t
)
=
a
(
t
)
A
R
(
1
+
m
x
(
t
)
)
a
(
t
)
=
A
R
(
1
+
m
x
(
t
)
)
.
cos
ϕ
y
R
(
t
)
→
ϕ
y
R
(
t
)
≈
ϕ
n
(
t
)
a
(
t
)
=
A
R
(
1
+
m
x
(
t
)
)
.
cos
ϕ
n
(
t
)
e
y
R
(
t
)
=
a
(
t
)
+
b
(
t
)
=
A
R
(
1
+
m
x
(
t
)
)
.
cos
ϕ
n
(
t
)
+
e
n
(
t
)
{\displaystyle {\begin{aligned}&\cos \phi _{y_{R}}(t)={\frac {a(t)}{A_{R}\left(1+mx(t)\right)}}\\&a(t)=A_{R}\left(1+mx(t)\right).\cos \phi _{y_{R}}(t){\xrightarrow[{\phi _{y_{R}}(t)\approx \phi _{n}(t)}]{}}a(t)=A_{R}\left(1+mx(t)\right).\cos \phi _{n}(t)\\&e_{y_{R}}(t)=a(t)+b(t)=A_{R}\left(1+mx(t)\right).\cos \phi _{n}(t)+e_{n}(t)\\\end{aligned}}}
Como se ve la demodulación no funcionará bien, habrá mucho ruido.
Por estos motivos se suele poner un limite inferior o threshold como valor mínimo de la relación señal a ruido para considerar que podemos demodular correctamente.
(
S
╱
N
)
R
≥
(
S
╱
N
)
R
threshold
{\displaystyle \left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{R}\geq \left({}^{S}\!\!\diagup \!\!{}_{N}\;\right)_{R_{\text{threshold}}}}
El valor de threshold no es fijo y dependerá de la calidad que queramos, pero podemos considerar 10 como un valor adecuado. (Además, el decibelios 10 vale también 10).