Error Probability of Binary Transmission Systems MCQ Quiz in বাংলা - Objective Question with Answer for Error Probability of Binary Transmission Systems - বিনামূল্যে ডাউনলোড করুন [PDF]

The probability of error for a baseband signal receiver is

$P_e=\frac{1}{2}erfc\left[\sqrt{\frac{E\mathrm{s}}{\eta }}\right]$

The transmitted signal is of form

Energy of signal E_s = V².T

(T = Pulse duration)

E_s = 25 × 72 μs

= 1.8 × 10^-3 Joule

Two – sided noise PSD given 10^-4 V²/Hz ⇒ η = 2 × 10^-4

Error probability $P_e=\frac{1}{2}erfc\left[\sqrt{\frac{1.8\times {10}^{-3}}{2\times {10}^{-4}}}\right]$

$\begin{array}{l}=\frac{1}{2}erfc\left(3\right)\\ =\frac{1}{2}\frac{e^{-9}}{\sqrt{\pi }\times 3}\end{array}$

P_e = 1.16 × 10^-5

$\mathrm{Q}\left(\mathrm{x}\right)$ is given by the definition, $\mathrm{Q}\left(\mathrm{x}\right)=\frac{1}{\sqrt{2\pi }}\underset{\mathrm{x}}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{u}}^2/2}\mathrm{d}\mathrm{u}$, and $\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{c}\left(\mathrm{x}\right)$ is given by, $\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{c}\left(\mathrm{x}\right)=\frac{2}{\sqrt{\pi }}\underset{\mathrm{x}}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{v}}^2}\mathrm{d}\mathrm{v}$. Now, changing variable $\mathrm{v}=\frac{\mathrm{u}}{\sqrt{2}}$, we get,

when $\mathrm{u}=\mathrm{x}\Rightarrow \mathrm{v}=\mathrm{x}/\surd 2$

when, $\mathrm{u}=\mathrm{\infty }\Rightarrow \mathrm{v}=\mathrm{\infty }$

and$\mathrm{d}\mathrm{v}=\frac{\mathrm{d}\mathrm{u}}{\sqrt{2}}\Rightarrow \mathrm{d}\mathrm{u}=\sqrt{2}\mathrm{d}\mathrm{v}$

Thus, $\mathrm{Q}\left(\mathrm{x}\right)=\frac{1}{\sqrt{2\pi }}\underset{\frac{\mathrm{x}}{\sqrt{2}}}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{v}}^2}\sqrt{2}\mathrm{d}\mathrm{v}$

$\begin{array}{l}\Rightarrow \mathrm{Q}\left(\mathrm{x}\right)=\frac{1}{2}\frac{2}{\sqrt{\pi }}\underset{\frac{\mathrm{x}}{\sqrt{2}}}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{v}}^2}\mathrm{d}\mathrm{v}\\ \mathrm{Q}\left(\mathrm{x}\right)=\frac{1}{2}\mathrm{e}\mathrm{r}\mathrm{f}\mathrm{c}\left(\frac{\mathrm{x}}{\sqrt{2}}\right)\end{array}$ 

We see, from the pdf of AWGN, that the mean is zero. Since, the noise is additive, the pdf simply shifts by the value of the voltage of the transmitted bit. Thus, we have the following of the transmitted bit. Thus, we have the following pdfs

Drawing the two pdfs together, we have

Let the decision threshold be at x. we assume for now that – 0.3 /, x < 0.4.

Thus, we have

Probability of error when zero is transmitted

$P_e\left(1/0\right)=1/{}_2(0.4-x)(0.4-x)$

Similarly,

$P_e\left(0/1\right)=1/{}_2(0.3+x)(0.3+x)$

Overall probability of error is

$P_e=0.4P_e\left(1/0\right)+0.6P_e\left(0/1\right)$

For minimum probability of error

$\frac{d{P}_e}{dx}=0$

$\Rightarrow \frac{d}{dx}[\frac{2}{5}\cdot \frac{1}{2}{(x-0.4)}^2+\frac{3}{5}\cdot \frac{1}{2}{(x+0.3)}^2]$

$=\frac{2}{5}(x-0.4)+\frac{3}{5}(x+0.3)$

$=(\frac{2}{5}+\frac{3}{5})x+(-\frac{0.8}{5}+\frac{0.9}{5})$

$\Rightarrow x+\left(\frac{0.1}{5}\right)=0$

Using $\frac{d{P}_e}{dx}=0$, we have

$x=\frac{-0.1}{5}=-0.02V$

Thus, minimum probability of error occurs for $x=-20mV$.

It can shown that for $x>0.4$ or $x<-0.3$, the probability of error will be always more than that obtained for threshold $x=20mV$.

The probability of error increases as the bit rate increases

For a bipolar signaling based PCM transmission

${\mathrm{P}}_{\mathrm{e}}=\mathrm{Q}\left(\sqrt{\frac{2{\mathrm{A}}^2{\mathrm{T}}_{\mathrm{b}}}{2\eta }}\right)$

Now as bit rate increase, bit duration decreases. Now complementary error function is a decreasing function. So as argument $\mathrm{x}=\sqrt{\frac{2{\mathrm{A}}^2{\mathrm{T}}_{\mathrm{b}}}{2\eta }}$ decrease $\mathrm{Q}(\mathrm{x})$ increase and hence probability of error increases.

We have, for bipolar signal transmission

${\mathrm{P}}_{\mathrm{e}}=\mathrm{Q}\left(\sqrt{\frac{{\mathrm{E}}_{\mathrm{d}}}{2\eta }}\right)$

Bit duration $=\frac{1}{7.26\mathrm{k}}=0.1377\mathrm{m}\mathrm{s}\mathrm{e}\mathrm{c}$

Now, the difference signal energy

${\mathrm{E}}_{\mathrm{d}}=\underset{0}{\overset{0.1377\mathrm{m}}{\int }}{\left[\mathrm{A}-\left(-\mathrm{A}\right)\right]}^2\mathrm{d}\mathrm{t}={\left(2\mathrm{A}\right)}^2\left(0.1377\times {10}^{-3}\right)]$

Now,${\mathrm{P}}_{\mathrm{e}}={10}^{-4}$ when $\sqrt{\frac{{\mathrm{E}}_{\mathrm{d}}}{2\eta }}=3.71$

 $\Rightarrow \sqrt{\frac{{\left(2\mathrm{A}\right)}^2\left(0.1377\times {10}^{-3}\right)}{4\times {10}^{-5}}}=3.71$

$\Rightarrow \mathrm{A}=\sqrt{\frac{{\left(3.71\right)}^2\times 1\times {10}^{-5}}{\left(0.1377\times {10}^{-3}\right)}}=0.999\mathrm{V}$

Probability density function of $0$ is

Since, it is uniformly distributed between $-0.25$ to $0.25\mathrm{V}$. So, we have

$\mathrm{k}\times 0.5=1$

$\Rightarrow \mathrm{k}=2$

Similarly, the pdf of $1$ can be plotted. It is shown below,

Now, probability of error when $0$ is transmitted $\mathrm{P}\left(\frac{1}{0}\right)$ is shown as shaded area.

Similarly, probability of error when $1$ is transmitted $\mathrm{P}\left(\frac{0}{1}\right)$ is shown in shaded region below.

Now probability of error

${\mathrm{P}}_{\mathrm{e}}=\mathrm{P}\left(0\right)\mathrm{P}\left(\frac{1}{0}\right)+\mathrm{P}\left(1\right)\mathrm{P}\left(\frac{0}{1}\right)$

$=0.5(0.05\times 2)+0.5(0.2\times 1)$

$=0.05+0.1$

$\Rightarrow {\mathrm{P}}_{\mathrm{e}}=0.15$