QAM

QAM (Quaternary Amplitude Modulation)

QAM = ASK + PSK

$\begin{array}{cc}& s_{QAM}(t)=A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{k}p(t-k{T}_s)\cos ({\omega }_{c}t+{\phi }_k)=\\ & A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\underset{a_{I_k}}{\underbrace{a_k\underset{I_k}{\underbrace{\cos \left({\phi }_k\right)}}}}p(t-k{T}_s)\cos \left({\omega }_{c}t\right)-A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\underset{a_{Q_k}}{\underbrace{a_k\underset{Q_k}{\underbrace{\sin \left({\phi }_k\right)}}}}p(t-k{T}_s)\sin \left({\omega }_{c}t\right)=\\ & A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{I_k}p(t-k{T}_s)\cos \left({\omega }_{c}t\right)-A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{Q_k}p(t-k{T}_s)\sin \left({\omega }_{c}t\right)\\ \end{array}$

16-QAM

Constellation diagram for rectangular 16-QAM with Gray coding. Each adjacent symbol only differs by one bit.

$\begin{array}{cc}& a_{I_k}=\{-\frac{3}{\sqrt{2}},-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}}\}\\ & a_{Q_k}=\{-\frac{3}{\sqrt{2}},-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}}\}\\ \end{array}$

$\begin{array}{cc}& s_I(t)=A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{I_k}\cdot p(t-k{T}_s)\\ & s_Q(t)=A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{Q_k}\cdot p(t-k{T}_s)\to \\ & {\overline{G}}_{I/Q}(f)={\sigma }_{a_k}^2\cdot R_s{|P(f)|}^2+m_{a_k}^2\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & {|P(f)|}^2=T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)\\ \end{array}$

$\begin{array}{cc}& m_{a_{I_k}}=m_{a_{Q_k}}=0\\ & P_x={(-\frac{3}{\sqrt{2}})}^2\cdot \frac{4}{16}+{(-\frac{1}{\sqrt{2}})}^2\cdot \frac{4}{16}+{\left(\frac{1}{\sqrt{2}}\right)}^2\cdot \frac{4}{16}+{\left(\frac{3}{\sqrt{2}}\right)}^2\cdot \frac{4}{16}=\\ & \frac{1}{4}(\frac{9}{2}+\frac{1}{2}+\frac{1}{2}+\frac{9}{2})=\frac{1}{4}\cdot 10=2.5\\ & {\sigma }_{a_{I_k}}^2={\sigma }_{a_{Q_k}}^2=2.5\\ & {\overline{G}}_x(f)={\sigma }_{a_k}^2\cdot R_s{|P(f)|}^2+\underset{0}{\underbrace{m_{a_k}^2}}\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)={\sigma }_{a_k}^2\cdot R_s\cdot T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)=\\ & {\overline{G}}_I(f)=G_{polar}(f)=A_c^2{\sigma }_{a_k}^2{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)=A_c^{2}2.5\cdot T_s{\mathrm{sinc}}^2\left(T_{s}f\right)\\ \end{array}$

$\begin{array}{cc}& m_{I_k}=m_{Q_k}=0\\ & {\sigma }_{I_k}^2={\sigma }_{Q_k}^2\\ & {\overline{G}}_I(f)={\overline{G}}_Q(f)=A_c^2{\sigma }_{a_k}^2{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)=A_c^{2}2.5\cdot T_s{\mathrm{sinc}}^2\left(T_{s}f\right)\\ & G_x(f)=\frac{G_I(f-f_c)+G_I(f+f_c)}{4}+\frac{G_Q(f-f_c)+G_Q(f+f_c)}{4}\to \\ & G_{16-QAM}(f)=2\frac{G_{I/Q}(f-f_c)+G_{I/Q}(f+f_c)}{4}=\frac{G_{I/Q}(f\pm f_c)}{2}=\frac{A_c^2}{2}2.5\cdot T_s{\mathrm{sinc}}^2\left(T_{s}f\right)\\ \end{array}$

Para la probabilidad de error BER:

(en proceso) BER de 16-QAM