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
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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.
x ^ ( t ) = x ( t ) ∗ h ( t ) = ∫ − ∞ ∞ x ( τ ) π ( t − τ ) ∂ τ {\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)
F [ sign ( t ) ] = 1 j π f {\displaystyle \mathbb {F} [\operatorname {sign} (t)]={\frac {1}{j\pi f}}} Por propiedades: F [ sign ( f ) ] = 1 − j π t {\displaystyle \mathbb {F} [\operatorname {sign} (f)]={\frac {1}{-j\pi t}}}
X ^ ( f ) = X ( f ) ⋅ H ( f ) = X ( f ) ( − j sign ( f ) ) {\displaystyle {\hat {X}}(f)=X(f)\cdot H(f)=X(f)\left(-j\operatorname {sign} (f)\right)}
E j : x ( t ) = cos ( 2 π f 0 t ) x ( t ) → X ( f ) = 1 2 δ ( f − f 0 ) + 1 2 δ ( f + f 0 ) X ^ ( f ) = X ( f ) ( − j sign ( f ) ) = − j 2 δ ( f − f 0 ) − − j 2 δ ( f + f 0 ) F [ sin ( 2 π f 0 t ) ] = 1 2 j δ ( f − f 0 ) − 1 2 j δ ( f + f 0 ) → j = 1 − j x ^ ( t ) = sin ( 2 π f 0 t ) {\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:
H ( f ) = − j sign ( f ) | H ( f ) | = 1 ∀ f ∡ H ( f ) = { π ╱ 2 , f > 0 − π ╱ 2 , f < 0 {\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}}}
F [ x ^ ( t ) ] = x ^ ^ ( t ) = − x ( t ) X ^ ^ ( f ) = X ( f ) ⋅ H 2 ( f ) = X ( f ) ⋅ [ − j sign ( f ) ] 2 = X ( f ) ⋅ ( − sign 2 ( f ) ) ⏟ sign 2 ( f ) = 1 = X ( f ) ⋅ ( − 1 ) = − X ( f ) {\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
G x ^ ( f ) = G x ( f ) ⋅ | H ( f ) | 2 ⏟ H ( f ) = − j sign ( f ) = G x ( f ) → R x ^ ( τ ) = R x ( τ ) {\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
R y x ( τ ) = R x y ∗ ( τ ) ⇒ R x ( τ ) = R x ∗ ( τ ) R y x ( τ ) = R x ( τ ) ∗ h ( τ ) → R x ^ x ( τ ) = R x ( τ ) ∗ 1 π t R x x ^ ( τ ) = R x ^ x ∗ ( − τ ) = R x ∗ ( − τ ) ∗ 1 π ( − t ) = R x ( τ ) ∗ − 1 π t R x ^ x ( τ ) = − R x x ^ ( τ ) {\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)
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Sign(f) 1+sign(f)
x + ( t ) = x ( t ) + j x ^ ( t ) x − ( t ) = x ( t ) − j x ^ ( t ) x + ( t ) + x − ( t ) = 2 x ( t ) X + ( f ) = X ( f ) + j ( − j sign ( f ) X ( f ) ) = X ( f ) + X ( f ) ⋅ sign ( f ) = X ( f ) ( 1 + sign ( f ) ) X + ( f ) = X ( f ) ( 1 + sign ( f ) ) X − ( f ) = X ( f ) ( 1 − sign ( f ) ) X + ( f ) = { 2 X ( f ) , f > 0 0 , f < 0 X − ( f ) = { 0 , f > 0 2 X ( f ) , f < 0 X + ( f ) + X − ( f ) = 2 X ( f ) G x ( f ) = lim T → ∞ | X ( f ) | 2 T G x + ( f ) = { 4 G x ( f ) , f > 0 0 , f < 0 = 2 G x ( f ) ( 1 + sign ( f ) ) G x − ( f ) = { 0 , f > 0 4 G x ( f ) , f < 0 = 2 G x ( f ) ( 1 − sign ( f ) ) G x + ( f ) + G x − ( f ) = 4 G x ( f ) S x + = 2 S x S x − = 2 S x R x + x − ( τ ) = 0 → G x + x − ( f ) = 0 {\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}}}
Envolvente compleja
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La envolvente compleja se define como (se toma x(t) como señal paso-banda):
x ~ ( t ) = x + ( t ) e − j ω c t X ~ ( f ) = X + ( f + f c ) R x ~ ( τ ) = lim T → ∞ 1 T ⋅ ∫ − ∞ ∞ x ~ ( t + τ ) ⋅ x ~ ∗ ( t ) ∂ t = lim T → ∞ 1 T ⋅ ∫ − ∞ ∞ [ x + ( t + τ ) e − j ω c ( t + τ ) ] ⋅ [ x + ( t ) e − j ω c t ] ∗ ∂ t = lim T → ∞ 1 T ⋅ ∫ − ∞ ∞ [ x + ( t + τ ) e − j ω c t e − j ω c τ ] ⋅ x + ∗ ( t ) e + j ω c t ∂ t = lim T → ∞ 1 T ⋅ ∫ − ∞ ∞ x + ( t + τ ) x + ∗ ( t ) e − j ω c τ ∂ t = R x ~ ( τ ) = R x + ( τ ) e − j ω c τ → G x ~ ( f ) = G x + ( f + f c ) G x ~ ( f ) = G x − ( f − f c ) S x ~ = ∫ − ∞ ∞ G x + ( f + f c ) ∂ f = 2 S x {\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}}}
x ~ ( t ) = x + ( t ) e j ω c t x ~ ( t ) = x I ( t ) + j x Q ( t ) = e ( t ) e j φ ( t ) x I ( t ) = Componente en fase de x ( t ) ( I : In phase) x Q ( t ) = Componente en cuadratura de x ( t ) ( Q : Quadrature) e ( t ) = Envolvente (el modulo) de x ( t ) φ ( t ) = Fase de x ( t ) x + ( t ) = x ( t ) + j x ^ ( t ) x ( t ) = ℜ { x + ( t ) } = ℜ { x ~ ( t ) e + j ω c t } = ℜ { [ x I ( t ) + j x Q ( t ) ] e + j ω c t } = ℜ { [ x I ( t ) + j x Q ( t ) ] [ cos ( ω c t ) + j sin ( ω c t ) ] } = x I ( t ) cos ( ω c t ) − x Q ( t ) sin ( ω c t ) x ( t ) = ℜ { x ~ ( t ) e + j ω c t } = ℜ { e ( t ) ⏟ valor absoluto e j φ ( t ) e + j ω c t } = e ( t ) cos ( ω c t + φ ( t ) ) x ( t ) = x I ( t ) cos ( ω c t ) − x Q ( t ) sin ( ω c t ) = e ( t ) cos ( ω c t + φ ( t ) ) {\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
x ~ ( t ) = x + ( t ) e − j ω c t ; x + ( t ) = x ( t ) + j x ^ ( t ) x ~ ( t ) = x I ( t ) + j x Q ( t ) = e ( t ) e j φ ( t ) e ( t ) = x I 2 ( t ) + x Q 2 ( t ) = | x ~ ( t ) | = | x + ( t ) e − j ω c t | = | x + ( t ) | = x 2 ( t ) + x ^ 2 ( t ) φ ( t ) = arctan ( x Q ( t ) x I ( t ) ) ℜ { x ~ ( t ) } = x I ( t ) = ℜ { e ( t ) e j φ ( t ) } = e ( t ) cos ( φ ( t ) ) → x Q ( t ) = e ( t ) sin ( φ ( t ) ) {\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 ( t ) : 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
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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}}}
R x I x Q ( τ ) = R x Q x I ( τ ) → R x m ( τ ) = R x I ( τ ) cos ( ω c τ ) 2 + R x Q ( τ ) cos ( ω c τ ) 2 {\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
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Y ( f ) = X ( f ) ⋅ H ( f ) Y ~ ( f ) ∝ X ( f ) ⋅ H ( f ) ? X + ( f ) = { 2 X ( f ) , f > 0 0 , f < 0 Y ( f ) = X ( f ) ⋅ H ( f ) x ( t ) = ℜ { x + ( t ) } = ℜ { x ~ ( t ) e + j ω c t } = x + ( t ) + x + ∗ ( t ) 2 = x ~ ( t ) e + j ω c t + x ~ ∗ ( t ) e − j ω c t 2 X + ( f ) = X ~ ( f − f c ) X ( f ) = X + ( f ) + X + ∗ ( − f ) 2 = X ~ ( f − f c ) + X ~ ∗ ( − f − f c ) 2 = X ~ ( f − f c ) + X ~ ∗ ( − ( f + f c ) ) 2 X ( f ) ⋅ H ( f ) = X ~ ( f − f c ) + X ~ ∗ ( − ( f + f c ) ) 2 ⋅ H ~ ( f − f c ) + H ~ ∗ ( − ( f + f c ) ) 2 Suponiendo se n ~ al paso-banda Y ( f ) = X ( f ) ⋅ H ( f ) Y ~ ( f − f c ) + Y ~ ∗ ( − ( f + f c ) ) 2 = X ~ ( f − f c ) + X ~ ∗ ( − ( f + f c ) ) 2 ⋅ H ~ ( f − f c ) + H ~ ∗ ( − ( f + f c ) ) 2 Y ~ ( f − f c ) + Y ~ ∗ ( − ( f + f c ) ) 2 = X ~ ( f − f c ) ⋅ H ~ ( f − f c ) + X ~ ∗ ( − ( f + f c ) ) ⋅ H ~ ∗ ( − ( f + f c ) ) 4 → Y ~ ( f − f c ) 2 = X ~ ( f − f c ) ⋅ H ~ ( f − f c ) 4 → Y ~ ( f ) 2 = X ~ ( f ) ⋅ H ~ ( f ) 4 Y ~ ( f ) = X ~ ( f ) ⋅ H ~ ( f ) 2 {\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}}}