Probability of error of ASK

Probability of error of ASK using coherent Detection:-

The equation of ASK from it’s basic definition

S_{ASK}(t)=\left\{\begin{matrix} \sqrt{2P_{s}}\cos 2\pi f_{c}t\ \rightarrow \ symbol\ '1'\\ 0\ \rightarrow \ symbol\ '0' \end{matrix}\right. .

The inputs to the optimum filter are x_{1}(t) \ and \ x_{2}(t)  when the symbols 1 and 0 are being transmitted at the transmitter.

\therefore x_{1}(t) = \sqrt{2P_{s}}\cos 2\pi f_{c}t .

x_{2}(t)=0.

we know that P_{e} of an optimum filter is P_{e} = \frac{1}{2}\ erfc(\frac{x_{o1}(t)-x_{o2}(t)}{2\sqrt{2}\sigma })

now chose the ratio \rho _{max}^{2} =(\frac{x_{o1}(t)-x_{o2}(t)}{\sigma })^{2}

from the Matched filter concept (\frac{x_{o1}(t)-x_{o2}(t)}{\sigma })^{2}=\frac{2}{N_{o}}\int_{-\infty }^{\infty }\left | X(f) \right |^{2}df-----EQN(1)

from Parsevel’s relation  \int_{-\infty }^{\infty }\left | X(f) \right |^{2}df =\int_{-\infty }^{\infty }\left | x(t) \right |^{2}dt----EQN(2)

from equations (1) and (2)

\rho _{max}^{2}=\int_{-\infty }^{\infty }\left | x(t) \right |^{2}dt

Here x(t) is the signal x(t) =x_{1}(t)-x_{2}(t) .

x(t) =\sqrt{2P_{s}}\cos 2\pi f_{c}t .

over a duration of [0, T_{b}] the symbols are transmitted

\int_{0}^{T_{b}}\left | x(t) \right |^{2}dt=2P_{s}\int_{0}^{T_{b}}\cos^{2} 2\pi f_{c}t \ dt .

                          =P_{s}T_{b}.

from equation(1)      \rho _{max}^{2} = \frac{2}{N_{o}} P_{s}T_{b}.

\rho _{max} = \sqrt{\frac{2P_{s}T_{b}}{N_{o}}} .

\therefore P_{e} = \frac{1}{2} \ erfc(\sqrt{\frac{2P_{s}T_{b}}{N_{o}}}).\frac{1}{2\sqrt{2}})

\therefore P_{e} = \frac{1}{2} \ erfc(\sqrt{\frac{P_{s}T_{b}}{4N_{o}}}) .

we know that carrier signal power is   P_{s} = \frac{A^{2}}{2}.

A=\sqrt{2P_{s}} .

\therefore P_{e} = \frac{1}{2} \ erfc(\sqrt{\frac{A^{2}T_{b}}{8N_{o}}}) .

\therefore P_{e}= \frac{1}{2} \ erfc(\sqrt{\frac{E_{b}}{4N_{o}}})  . since A^{2}T_{b} = P_{s}T_{b} =E_{b}.

probability of error  of coherent ASK is  P_{e}= \frac{1}{2} \ erfc(\sqrt{\frac{E_{b}}{4N_{o}}})   (or)  in terms of Q function as P_{e}= Q(\sqrt{\frac{E_{b}}{2N_{o}}}) .

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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.