Consider a binary, digital communication system which used pulses $\mathrm{g}\left(\mathrm{t}\right)$ and $-\mathrm{g}\left(\mathrm{t}\right)$ for transmitting bits over an AWGN channel. If the receiver uses a matched filter, which one of the following pulses will give the minimum probability of bit error?

Concept:

Probability of error for PSK is given as:

$P_e=Q\left(\sqrt{\frac{2E}{N_0}}\right)$ 

Complementary error function Q(x) is defined as:

$Q\left(x\right)={\int }_x^{\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{t^2}{2}}dt$ 

Q(x) decreases as x increases.

Calculation:

Since:

$P_e=Q\left(\sqrt{\frac{2E}{N_0}}\right)$ 

$Q\left(\sqrt{\frac{2E}{N_0}}\right)$ decreases as $\sqrt{\frac{2E}{N_0}}$ increases.

Therefore, to minimize the probability of error, maximize the energy of the signal E.

Energy of the signal for option A = E_A = 1 × 1 = 1

Energy of the signal for option B:

$E_B=2\times {\int }_0^{1/2}{\left(2t\right)}^{2}dt=2\times \frac{4t^3}{3}|\begin{array}{c}1/2\\ 0\end{array}=\frac{8}{3}\times \frac{1}{8}=\frac{1}{3}$ 

The energy of the signal for option C:

$E_C={\int }_0^1{\left(1-t\right)}^{2}dt={\int }_0^1\left(1-2t+t^2\right)dt$ 

$=\left(t-t^2+\frac{t^3}{3}\right)\begin{array}{c}1\\ 0\end{array}$ 

$E_C=1-1+\frac{1}{3}$ 

The energy of the signal for option D:

$E_D={\int }_0^1{\left(t\right)}^{2}dt={\int }_0^1{t}^{2}dt$ 

$=\left(\frac{t^3}{3}\right)\begin{array}{c}1\\ 0\end{array}=\frac{1}{3}$ 

Since E_A > E_B = E_C = E_D

Therefore, the probability of error will be minimum for option A.