s 4 − A S K ( t ) = A c ∑ k = − ∞ ∞ a I k p ( t − k T s ) cos ( ω c t ) a I k = { 0 , A , + 2 A , + 3 A } {\displaystyle {\begin{aligned}&s_{4-ASK}(t)=A_{c}\sum \limits _{k=-\infty }^{\infty }{a_{I_{k}}p\left(t-kT_{s}\right)\cos \left(\omega _{c}t\right)}\\&a_{I_{k}}=\left\{0,A,+2A,+3A\right\}\\\end{aligned}}}
La constelación de una señal ASK de 4 niveles es:
y R ( t ) = A R ∑ k = − ∞ ∞ a I k p ( t − k T s ) + n I ( t ) a I k = { a ′ 00 ′ = 0 a ′ 01 ′ = + A a ′ 10 ′ = + 2 A a ′ 11 ′ = + 3 A } → { m ′ 00 ′ = ∫ 0 T s s ′ 00 ′ ( t ) k p ∗ ( t ) ∂ t = ∫ 0 T s 0 p ∗ ( t ) ∂ t = 0 m ′ 01 ′ = ∫ 0 T s s ′ 01 ′ ( t ) k p ∗ ( t ) ∂ t = ∫ 0 T s + A p ( t ) p ∗ ( t ) ∂ t = + A T s m ′ 10 ′ = ∫ 0 T s s ′ 10 ′ ( t ) k p ∗ ( t ) ∂ t = ∫ 0 T s + 2 A p ( t ) p ∗ ( t ) ∂ t = + 2 A T s m ′ 11 ′ = ∫ 0 T s s ′ 00 ′ ( t ) k p ∗ ( t ) ∂ t = ∫ 0 T s + 3 A p ( t ) p ∗ ( t ) ∂ t = + 3 A T s V T = { A T s 2 , 3 A T s 2 , 5 A T s 2 } σ 2 = η 2 T s P e ≃ 1 4 ⋅ Q ( | V T − m | σ 2 ) + 1 4 ⋅ 2 Q ( | V T − m | σ 2 ) + 1 4 ⋅ 2 Q ( | V T − m | σ 2 ) + 1 4 ⋅ Q ( | V T − m | σ 2 ) = P e ≃ 1 4 ⋅ Q ( | A T s 2 − 0 | σ 2 ) + 1 4 ⋅ 2 Q ( | A T s 2 − ( A T s ) | σ 2 ) + 1 4 ⋅ 2 Q ( | 3 A T s 2 − 2 A T s | σ 2 ) + 1 4 ⋅ Q ( | 5 A T s 2 − 3 A T s | σ 2 ) = P e ≃ 3 2 Q ( A T s 2 σ 2 ) = 3 2 Q ( A T s 2 η 2 T s ) = 3 2 Q ( A 2 T s 2 4 η 2 T s ) = 3 2 Q ( A 2 T s 2 η ) {\displaystyle {\begin{aligned}&y_{R}(t)=A_{R}\sum \limits _{k=-\infty }^{\infty }{a_{I_{k}}p\left(t-kT_{s}\right)+n_{I}(t)}\\&a_{I_{k}}=\left\{{\begin{aligned}&a_{'00'}=0\\&a_{'01'}=+A\\&a_{'10'}=+2A\\&a_{'11'}=+3A\\\end{aligned}}\right\}\to \left\{{\begin{aligned}&m_{'00'}=\int _{0}^{T_{s}}{s_{'00'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{0p^{*}\left(t\right)\partial t}=0\\&m_{'01'}=\int _{0}^{T_{s}}{s_{'01'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{+Ap(t)p^{*}\left(t\right)\partial t}=+AT_{s}\\&m_{'10'}=\int _{0}^{T_{s}}{s_{'10'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{+2Ap(t)p^{*}\left(t\right)\partial t}=+2AT_{s}\\&m_{'11'}=\int _{0}^{T_{s}}{s_{'00'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{+3Ap(t)p^{*}\left(t\right)\partial t}=+3AT_{s}\\\end{aligned}}\right.\\&V_{T}=\left\{{\frac {AT_{s}}{2}},{\frac {3AT_{s}}{2}},{\frac {5AT_{s}}{2}}\right\}\\&{\sqrt {\sigma ^{2}}}={\sqrt {{\frac {\eta }{2}}T_{s}}}\\&P_{e}\simeq {\frac {1}{4}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=\\&P_{e}\simeq {\frac {1}{4}}\cdot Q\left({\frac {\left|{\frac {AT_{s}}{2}}-0\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|{\frac {AT_{s}}{2}}-\left(AT_{s}\right)\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|{\frac {3AT_{s}}{2}}-2AT_{s}\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {\left|{\frac {5AT_{s}}{2}}-3AT_{s}\right|}{\sqrt {\sigma ^{2}}}}\right)=\\&P_{e}\simeq {\frac {3}{2}}Q\left({\frac {\frac {AT_{s}}{2}}{\sqrt {\sigma ^{2}}}}\right)={\frac {3}{2}}Q\left({\frac {\frac {AT_{s}}{2}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {\frac {A^{2}T_{s}^{2}}{4}}{{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)\\\end{aligned}}}
E s c = 1 4 ⋅ E ′ 00 ′ + 1 4 ⋅ E ′ 01 ′ + 1 4 ⋅ E ′ 10 ′ + 1 4 ⋅ E ′ 11 ′ E ′ 00 ′ = 0 E ′ 01 ′ = ∫ 0 T s s ′ 01 ′ 2 ( t ) ∂ t = ( + A ) 2 T s = A 2 T s E ′ 10 ′ = ∫ 0 T s s ′ 10 ′ 2 ( t ) ∂ t = ( + 2 A ) 2 T s = 4 A 2 T s E ′ 11 ′ = ∫ 0 T s s ′ 11 ′ 2 ( t ) ∂ t = ( + 3 A ) 2 T s = 9 A 2 T s E s c = 14 A 2 T s 4 = 7 A 2 T s 2 E s c = 7 A 2 T s 2 ↔ A 2 T s 2 = E s c 7 P s c ≃ 3 2 Q ( A 2 T s 2 η ) = 3 2 Q ( E s c 7 η ) E s = E s c 1 + E s c 2 ⏟ 0 = E s c 1 E s = log 2 ( M = 4 ) E b = 2 E b P s ≃ 3 2 Q ( E s 7 η ) = 3 2 Q ( 2 E b 7 η ) {\displaystyle {\begin{aligned}&E_{sc}={\frac {1}{4}}\cdot E_{'00'}+{\frac {1}{4}}\cdot E_{'01'}+{\frac {1}{4}}\cdot E_{'10'}+{\frac {1}{4}}\cdot E_{'11'}\\&E_{'00'}=0\\&E_{'01'}=\int _{0}^{T_{s}}{s_{'01'}^{2}(t)\partial t}=\left(+A\right)^{2}T_{s}=A^{2}T_{s}\\&E_{'10'}=\int _{0}^{T_{s}}{s_{'10'}^{2}(t)\partial t}=\left(+2A\right)^{2}T_{s}=4A^{2}T_{s}\\&E_{'11'}=\int _{0}^{T_{s}}{s_{'11'}^{2}(t)\partial t}=\left(+3A\right)^{2}T_{s}=9A^{2}T_{s}\\&E_{sc}={\frac {14A^{2}T_{s}}{4}}={\frac {7A^{2}T_{s}}{2}}\\&E_{sc}={\frac {7A^{2}T_{s}}{2}}\leftrightarrow {\frac {A^{2}T_{s}}{2}}={\frac {E_{sc}}{7}}\\&P_{sc}\simeq {\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {E_{sc}}{7\eta }}}\right)\\&E_{s}=E_{sc_{1}}+\underbrace {E_{sc_{2}}} _{0}=E_{sc_{1}}\\&E_{s}=\log _{2}\left(M=4\right)E_{b}=2E_{b}\\&P_{s}\simeq {\frac {3}{2}}Q\left({\sqrt {\frac {E_{s}}{7\eta }}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {2E_{b}}{7\eta }}}\right)\\\end{aligned}}}
4-ASK
