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.
Señal 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.
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{\displaystyle {\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):
Sign(f)
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{\displaystyle \mathbb {F} [\operatorname {sign} (t)]={\frac {1}{j\pi f}}}
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{\displaystyle \mathbb {F} [\operatorname {sign} (f)]={\frac {1}{-j\pi t}}}
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{\displaystyle {\hat {X}}(f)=X(f)\cdot H(f)=X(f)\left(-j\operatorname {sign} (f)\right)}
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{\displaystyle {\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:
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{\displaystyle {\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}}}
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{\displaystyle {\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
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{\displaystyle 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
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{\displaystyle {\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)[ editar ]
Sign(f)
1+sign(f)
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{\displaystyle {\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}}}
La envolvente compleja se define como (se toma x(t) como señal paso-banda):
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{\displaystyle {\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}}}
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Componente en fase de
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Componente en cuadratura de
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{\displaystyle {\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
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{\displaystyle {\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:
x
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Se
n
~
al paso-banda
x
~
(
t
)
,
x
I
(
t
)
,
x
Q
(
t
)
,
e
(
t
)
: Se
n
~
ales paso-bajo
{\displaystyle {\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:
x
+
(
t
)
=
x
−
∗
(
t
)
x
I
(
t
)
=
ℜ
{
x
~
(
t
)
}
=
x
~
(
t
)
+
x
~
∗
(
t
)
2
=
x
+
(
t
)
e
−
j
ω
c
t
+
x
+
∗
(
t
)
e
+
j
ω
c
t
2
R
x
I
(
τ
)
=
lim
T
→
∞
1
T
⋅
∫
−
∞
∞
x
I
(
t
+
τ
)
⋅
x
I
∗
(
t
)
∂
t
=
lim
T
→
∞
1
T
⋅
∫
−
∞
∞
[
x
+
(
t
+
τ
)
e
−
j
ω
c
(
t
+
τ
)
+
x
+
∗
(
t
+
τ
)
e
+
j
ω
c
(
t
+
τ
)
2
]
⋅
[
x
+
(
t
)
e
−
j
ω
c
t
+
x
+
∗
(
t
)
e
+
j
ω
c
t
2
]
∗
∂
t
=
lim
T
→
∞
1
T
⋅
1
4
∫
−
∞
∞
[
x
+
(
t
+
τ
)
e
−
j
ω
c
(
t
+
τ
)
+
x
+
∗
(
t
+
τ
)
e
+
j
ω
c
(
t
+
τ
)
]
⋅
[
x
+
∗
(
t
)
e
+
j
ω
c
t
+
x
+
(
t
)
e
−
j
ω
c
t
]
∂
t
=
lim
T
→
∞
1
T
⋅
1
4
∫
−
∞
∞
x
+
(
t
+
τ
)
x
+
∗
(
t
)
e
−
j
ω
c
τ
+
x
+
(
t
+
τ
)
x
+
(
t
)
e
−
j
ω
c
(
2
t
+
τ
)
⏞
no entra en area de integracion
+
.
.
.
x
+
∗
(
t
+
τ
)
x
+
∗
(
t
)
e
+
j
ω
c
(
2
t
+
τ
)
⏞
i
^
dem
+
x
+
∗
(
t
+
τ
)
x
+
(
t
)
e
+
j
ω
c
τ
∂
t
=
R
x
I
(
τ
)
=
1
╱
4
(
R
x
+
(
τ
)
e
−
j
ω
c
τ
+
R
x
−
(
τ
)
e
+
j
ω
c
τ
)
G
x
I
(
f
)
=
1
╱
4
(
G
x
+
(
f
+
f
c
)
+
G
x
−
(
f
−
f
c
)
)
G
x
+
(
f
)
+
G
x
−
(
f
)
=
4
G
x
(
f
)
S
x
I
=
∫
−
∞
∞
G
x
I
(
f
)
∂
f
=
S
x
x
Q
(
t
)
=
ℑ
{
x
~
(
t
)
}
=
x
~
(
t
)
−
x
~
∗
(
t
)
2
j
→
G
x
Q
(
f
)
=
1
╱
4
(
G
x
+
(
f
+
f
c
)
+
G
x
−
(
f
−
f
c
)
)
{\displaystyle {\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:
R
x
Q
x
I
(
τ
)
=
lim
T
→
∞
1
T
⋅
∫
−
∞
∞
x
Q
(
t
+
τ
)
⋅
x
I
∗
(
t
)
∂
t
=
lim
T
→
∞
1
T
⋅
∫
−
∞
∞
[
x
+
(
t
+
τ
)
e
−
j
ω
c
(
t
+
τ
)
−
x
+
∗
(
t
+
τ
)
e
+
j
ω
c
(
t
+
τ
)
2
j
]
⋅
[
x
+
(
t
)
e
−
j
ω
c
t
+
x
+
∗
(
t
)
e
+
j
ω
c
t
2
]
∗
∂
t
=
lim
T
→
∞
1
T
⋅
1
4
j
∫
−
∞
∞
[
x
+
(
t
+
τ
)
e
−
j
ω
c
(
t
+
τ
)
−
x
+
∗
(
t
+
τ
)
e
+
j
ω
c
(
t
+
τ
)
]
⋅
[
x
+
∗
(
t
)
e
+
j
ω
c
t
+
x
+
(
t
)
e
−
j
ω
c
t
]
∂
t
=
lim
T
→
∞
1
T
⋅
1
4
j
∫
−
∞
∞
x
+
(
t
+
τ
)
x
+
∗
(
t
)
e
−
j
ω
c
τ
+
x
+
(
t
+
τ
)
x
+
(
t
)
e
−
j
ω
c
(
2
t
+
τ
)
⏞
no entra en area de integracion
+
.
.
.
−
x
+
∗
(
t
+
τ
)
x
+
∗
(
t
)
e
+
j
ω
c
(
2
t
+
τ
)
⏞
i
^
dem
−
x
+
∗
(
t
+
τ
)
x
+
(
t
)
⏞
=
x
−
(
t
+
τ
)
x
−
∗
(
t
)
e
+
j
ω
c
τ
∂
t
=
R
x
Q
x
I
(
τ
)
=
1
4
j
(
R
x
+
(
τ
)
e
−
j
ω
c
τ
−
R
x
−
(
τ
)
e
+
j
ω
c
τ
)
R
x
I
x
Q
(
τ
)
=
R
x
Q
x
I
∗
(
−
τ
)
=
1
4
(
−
j
)
∗
(
R
x
+
(
−
τ
)
e
+
j
ω
c
τ
−
R
x
−
(
−
τ
)
e
−
j
ω
c
τ
)
∗
=
1
4
j
(
R
x
+
∗
(
−
τ
)
⏞
R
x
+
(
τ
)
e
−
j
ω
c
τ
−
R
x
−
∗
(
−
τ
)
⏞
R
x
−
(
τ
)
e
+
j
ω
c
τ
)
⇒
R
x
I
x
Q
(
τ
)
=
R
x
Q
x
I
(
τ
)
{\displaystyle {\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[ editar ]
Con la representación de una señal paso-banda en sus componentes de fase y cuadratura, tenemos que, si obtenemos su correlación:
R
x
m
(
τ
)
=
lim
T
→
∞
1
T
⋅
∫
−
∞
∞
x
m
∗
(
t
)
⋅
x
m
(
t
+
τ
)
∂
t
=
x
m
(
t
)
=
x
I
(
t
)
cos
(
ω
c
t
)
−
x
Q
(
t
)
sin
(
ω
c
t
)
lim
T
→
∞
1
T
∫
−
∞
∞
[
x
I
(
t
)
cos
(
ω
c
t
)
−
x
Q
(
t
)
sin
(
ω
c
t
)
]
∗
⋅
[
x
I
(
t
+
τ
)
cos
(
ω
c
(
t
+
τ
)
)
−
x
Q
(
t
)
sin
(
ω
c
(
t
+
τ
)
)
]
∂
t
=
lim
T
→
∞
1
T
∫
−
∞
∞
x
I
∗
(
t
)
cos
(
ω
c
t
)
x
I
(
t
+
τ
)
cos
(
ω
c
(
t
+
τ
)
)
+
−
x
Q
∗
(
t
)
sin
(
ω
c
t
)
x
I
(
t
+
τ
)
cos
(
ω
c
(
t
+
τ
)
)
+
x
I
∗
(
t
)
cos
(
ω
c
t
)
(
−
x
Q
(
t
+
τ
)
sin
(
ω
c
(
t
+
τ
)
)
)
+
−
x
Q
∗
(
t
)
sin
(
ω
c
t
)
(
−
x
Q
(
t
+
τ
)
sin
(
ω
c
(
t
+
τ
)
)
)
∂
t
=
{
Relaciones Trigonometricas
}
lim
T
→
∞
1
T
∫
−
∞
∞
x
I
∗
(
t
)
x
I
(
t
+
τ
)
(
cos
(
ω
c
τ
)
+
cos
(
ω
c
(
2
t
+
τ
)
)
⏞
fuera del area de
∫
2
)
+
←
{
cos
a
cos
b
=
cos
(
a
+
b
)
+
cos
(
a
−
b
)
2
}
−
x
Q
∗
(
t
)
x
I
(
t
+
τ
)
(
sin
(
−
ω
c
τ
)
+
sin
(
ω
c
(
2
t
+
τ
)
)
⏞
fuera del area de
∫
2
)
←
{
sin
a
cos
b
=
sin
(
a
+
b
)
+
sin
(
a
−
b
)
2
}
−
x
I
∗
(
t
)
x
Q
(
t
+
τ
)
(
sin
(
ω
c
τ
)
+
sin
(
ω
c
(
2
t
+
τ
)
)
⏞
fuera del area de
∫
2
)
+
←
{
sin
a
cos
b
=
sin
(
a
+
b
)
+
sin
(
a
−
b
)
2
}
x
Q
∗
(
t
)
x
Q
(
t
+
τ
)
(
cos
(
ω
c
τ
)
−
cos
(
ω
c
(
2
t
+
τ
)
)
⏞
fuera del area de
∫
2
)
∂
t
←
{
sin
a
sin
b
=
cos
(
a
−
b
)
−
cos
(
a
+
b
)
2
}
=
lim
T
→
∞
1
T
∫
−
∞
∞
x
I
∗
(
t
)
x
I
(
t
+
τ
)
cos
(
ω
c
τ
)
2
+
x
Q
∗
(
t
)
x
I
(
t
+
τ
)
sin
(
ω
c
τ
)
2
−
x
I
∗
(
t
)
x
Q
(
t
+
τ
)
sin
(
ω
c
τ
)
2
+
x
Q
∗
(
t
)
x
Q
(
t
+
τ
)
cos
(
ω
c
τ
)
2
=
R
x
I
(
τ
)
cos
(
ω
c
τ
)
2
+
R
x
I
x
Q
(
τ
)
sin
(
ω
c
τ
)
2
−
R
x
Q
x
I
(
τ
)
sin
(
ω
c
τ
)
2
+
R
x
Q
(
τ
)
cos
(
ω
c
τ
)
2
{\displaystyle {\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}}}
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{\displaystyle {\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[ editar ]
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{\displaystyle {\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}}}