Exercise 1.08Z: BPSK Error Probability

Numerical values of function ${\rm Q}(x)$

We assume the optimal baseband transmission system for binary signals with

bipolar amplitude coefficients $a_{\nu} \in \{–1, +1\}$,

rectangular transmitted signal with signal values $\pm s_{0}$ and bit duration $T_{\rm B}$,

AWGN noise with noise power density $N_{0}$,

receiver filter according to the matched filter principle,

decision with the optimal threshold $E = 0$.

Unless otherwise specified, you should also assume the following numerical values:

$$ s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$$

The bit error probability of this "baseband system" has already been given in the chapter "Error Probability for Baseband Transmission" $($Index: $\rm BB)$:

$$p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}.$$

Here, $\sigma_{d}$ denotes the noise rms value at the decision device and ${\rm Q}(x)$ denotes the complementary Gaussian error function, which is given here in tabular form. This error probability can also be expressed in the form

$$p_{\rm BB} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ),$$

where $E_{\rm B}$ denotes "energy per bit."

The error probability of a comparable transmission system with "Binary Phase Shift Keying" is $($Index: $\rm BPSK)$:

$$p_{\rm BPSK} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{{N_0}/{T_{\rm B}}}.$$

Notes:

The exercise belongs to the chapter Linear Digital Modulation - Coherent Demodulation.

Reference is also made to the chapter "Error Probability for Baseband Transmission".

You can check the results with the HTML5/JavaScript applet "Complementary Gaussian Error Functions".

Since the signal value $s_{0}$ is specified here in "volts" and no specification is made for the reference resistance, $E_{\rm B}$ has the unit "$\rm V^{2}/Hz$".

Questions

$p_{\rm BB} \ = \ $

$\ \% $

$E_{\rm B} \ = \ $

$\ \cdot 10^{-8}\ \rm V^{2}s $

$p_{\rm BB} \ = \ $

$\ \% $

	$p_{\rm BPSK} = {\rm Q}\big [(E_{\rm B}/N_{0})^{1/2}\big ]$,
	$p_{\rm BPSK} = {\rm Q}\big [(2 \cdot E_{\rm B}/N_{0})^{1/2}\big ]$,
	$p_{\rm BPSK} = {\rm Q}\big [(4\cdot E_{\rm B}/N_{0})^{1/2}\big ]$.

$E_{\rm B}/N_{0} = 8\text{:}\hspace{0.4cm} p_{\rm BPSK} \ = \ $

$\ \% $

$E_{\rm B}/N_{0} = 2\text{:}\hspace{0.4cm} p_{\rm BPSK} \ = \ $

$\ \% $

Solution

(1) The noise rms value is given here by

$$\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}= \sqrt{\frac{2 \cdot 10^{-9}\,{\rm V^2/Hz}}{2 \cdot 1\,{\rm ns}}}= 1\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}p_{\rm BB} = {\rm Q}\left ({s_0}/{\sigma_d } \right )= {\rm Q}(4)= 0.317 \cdot 10^{-4}\hspace{0.1cm}\underline {= 0.00317 \%}.$$

(2) For the baseband system:

$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot 10^{-9}\,{\rm s}\hspace{0.1cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$

Of course, the additional given equation gives the exact same error probability:

$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 16 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$

A comparison with question (4) of Exercise 1.8 shows that $E_{\rm B}/N_{0} = 8$ is not (exactly) equal to $10 \cdot \lg E_{\rm B}/N_{0} = 9 \ \rm dB$.
In the first case $p_{\rm BB} = 0.317 \cdot 10^{–4}$ is obtained, in the second $p_{\rm BB} = 0.336 \cdot 10^{-4}$.

(3) At half the transmission amplitude $s_{0} = 2 \ \rm V$, the energy per bit decreases to a quarter and the following equations apply:

$$p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )\hspace{0.1cm}\underline {= {\rm Q}(2)= 2.27 \%},$$

$$p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)= 2.27 \%.$$

(4) Considering only half the energy $E_{\rm B} = s^{2}_{0} \cdot T_{\rm B}/2$, we obtain with $\sigma^{2}_{d} = N_{0}/T_{\rm B}$ and

$$p_{\rm BPSK} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )= {\rm Q}\left ( \sqrt{{s_0^2 \cdot T_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }}\hspace{0.1cm}\right )$$

the exact same result as for the optimal baseband system ⇒ solution 2.

(5) Of course, this also gives the exact same results as for the baseband transmission:

$${ E_{\rm B}}/{N_0 }= 8{\rm :} \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{16}) = {\rm Q}(4)\hspace{0.1cm}\underline {= 0.00317 \%},$$

$${ E_{\rm B}}/{N_0 }= 2{\rm :} \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{4}) = {\rm Q}(2) \hspace{0.1cm}\underline {= 2.27 \%}.$$