ASK

ASK (Amplitude Shift Keying)

ASK, también llamado OOK(On-Off Keying).

Para calcular su DEP nos basamos en que es una señal unipolar NRZ modulada por un coseno:

Constelación ASK

$\begin{array}{cc}& s_{ASK}(t)=A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{k}p(t-k{T}_s)\cos \left({\omega }_{c}t\right)\\ & x_I(t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{k}p(t-k{T}_s),x_Q(t)=0\\ & {\overline{G}}_{x_I}(f)=G_{NRZ}(f)=\frac{A^2}{4}{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)+\frac{A^2}{4}\delta (f-R_s)\\ & G_x(f)=\frac{G_I(f-f_c)+G_I(f+f_c)}{4}+\frac{G_Q(f-f_c)+G_Q(f+f_c)}{4}\to \\ & G_x(f)=\frac{G_I(f\pm f_c)}{4}\\ & G_{x_{ASK}}(f)=\frac{A^2}{4^2}{T}_s{\mathrm{sinc}}^2\left(T_s(f\pm f_c)\right)+\frac{A^2}{4^2}\delta (f\pm f_c-R_s)\\ \end{array}$

Para la probabilidad de error (BER):

BER de ASK

4-ASK

4-ASK

$\begin{array}{cc}& s_{4-ASK}(t)=A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{k}p(t-k{T}_s)\cos \left({\omega }_{c}t\right)\\ & s_{4-AS{K}_I}(t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{k}p(t-k{T}_s)\\ & a_k=\{0,A,2A,3A\}\\ & {\overline{G}}_x(f)={\sigma }_{a_k}^2\cdot R_s{|P(f)|}^2+m_{a_k}^2\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & {|P(f)|}^2=T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)\\ & m_{I_k}=0\cdot \frac{1}{4}+A\cdot \frac{1}{4}+2A\cdot \frac{1}{4}+3A\cdot \frac{1}{4}=\frac{3A}{2}\\ & P_{I_k}={\left(0\right)}^2\cdot \frac{1}{4}+{\left(A\right)}^2\cdot \frac{1}{4}+{\left(2A\right)}^2\cdot \frac{1}{4}+{\left(3A\right)}^2\cdot \frac{1}{4}=\frac{14A^2}{4}\\ & {\sigma }_{I_k}^2=P_{I_k}-m_{I_k}^2=\frac{14A^2}{4}-{\left(\frac{3A}{2}\right)}^2=\frac{14A^2}{4}-\frac{9A^2}{4}=\frac{5A^2}{4}\\ & {\overline{G}}_x(f)=\frac{5A^2}{4}\cdot R_s{|P(f)|}^2+\frac{9A^2}{4}\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & {|P(f)|}^2=T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)\\ & {\overline{G}}_I(f)=\frac{5A^2}{4}{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)+\frac{9A^2}{4}\cdot \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|{\mathrm{sinc}}^2\left(k\right)\right|}^2\delta (f-k{R}_s)=\\ & {\overline{G}}_I(f)=\frac{5A^2}{4}{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)+\frac{9A^2}{4}\cdot \delta (f-R_s)\\ \end{array}$

$\begin{array}{cc}& G_x(f)=\frac{G_I(f-f_c)+G_I(f+f_c)}{4}+\frac{G_Q(f-f_c)+G_Q(f+f_c)}{4}\to G_Q(f)=0\\ & G_x(f)=\frac{G_I(f\pm f_c)}{4}\to \\ & G_{4-ASK}(f)=\frac{5A^2}{4^2}{T}_s{\mathrm{sinc}}^2\left(T_s(f\pm f_c)\right)+\frac{9A^2}{4^2}\cdot \delta ((f\pm f_c)-R_s)\\ \end{array}$

Para la probabilidad de error (BER):

BER de 4-ASK

M-ASK

$\begin{array}{cc}& m_{I_k}=\sum _{i=0}^{M-1}\frac{1}{M}\cdot i\Rightarrow \sum _{i=1}^{n}i=\frac{n(n+1)}{2}\Rightarrow \frac{1}{M}\cdot \left(\frac{(M-1)M}{2}\right)=\frac{M-1}{2}\\ & P_{I_k}=\sum _{i=0}^{M-1}\frac{1}{M}\cdot i^2\Rightarrow \sum _{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}\Rightarrow \frac{1}{M}\cdot \left(\frac{(M-1)M(2(M-1)+1)}{6}\right)=\\ & P_{I_k}=\frac{(M-1)(2M-1)}{6}\\ & {\sigma }_{I_k}^2=P_{I_k}-m_{I_k}^2=\frac{(M-1)(2M-1)}{6}-{\left(\frac{M-1}{2}\right)}^2=\frac{2M^2-M-2M+1}{6}-\frac{M^2-2M+1}{4}=\\ & {\sigma }_{I_k}^2=\frac{4\cdot (2M^2-3M+1)}{4\cdot 6}-\frac{6\cdot (M^2-2M+1)}{4\cdot 6}=\frac{2M^2-2}{4\cdot 6}=\frac{M^2-1}{2\cdot 6}\\ & {\overline{G}}_I(f)={\sigma }_{a_k}^2\cdot R_s{|P(f)|}^2+m_{a_k}^2\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & G_{M-ASK}(f)=\frac{M^2-1}{12}\cdot \frac{1}{4}{T}_s{\mathrm{sinc}}^2\left(T_s(f\pm f_c)\right)+{\left(\frac{M-1}{2}\right)}^2\frac{1}{4}\cdot \delta ((f\pm f_c)-R_s)\\ \end{array}$

Para la probabilidad de error (BER):

BER de M-ASK