Exercise 4.8Z: BPSK Error Probability

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Table of the Complementary Gaussian Error Function  ${\rm Q}(x)$

We assume the optimal baseband transmission system for binary signals with

• bipolar amplitude coefficients  $a_ν ∈ \{-1, +1\}$,
• rectangular transmitted signal  $s(t)$  with signal values  $±s_0$  and bit duration  $T_{\rm B}$,
• AWGN noise with the (one-sided) noise power density  $N_0$,
• receiver filter according to the matched filter principle,
• decision with optimal threshold value  $E = 0$.

If nothing else is specified,  you should assume the following numerical values:

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

Using the noise rms value  $σ_d$  at the decision and the complementary Gaussian error function  ${\rm Q}(x)$,  the bit error probability of this baseband  $\rm (BB)$  system is   ⇒   see table:

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

This bit 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  $E_{\rm B}$  indicates the  "signal energy per bit".

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

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

Notes:

• The exercise belongs to the chapter  "Linear Digital Modulation".
• Reference is made in particular to the section  "Error probabilities - a brief overview".
• The derivations can be found in the chapter  "Linear Digital Modulation - Coherent Demodulation"  of the book "Digital Signal Transmission".
• The specification of a power in  $\rm V^2$  or an energy in  $\rm V^2 s$  means a conversion to the reference resistance  $1 \ \rm \Omega$.

Questions

1

Let  $s_0 = 4\,{\rm V},\hspace{0.2cm} T_{\rm B} = 1\,{\rm ns},\hspace{0.2cm}N_0 = 2 \cdot 10^{-9}\, {\rm V^2/Hz} \hspace{0.05cm}.$  What is the error probability  $p_{\rm BB}$  of the baseband system?

$p_{\rm BB} \ = \ $ $\ \cdot 10^{-4}$

2

For this parameter set,  what is the energy per bit   ⇒    $E_{\rm B}$  for the baseband system?

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

3

What is the error probability at half the transmitted amplitude   ⇒    $s_0 = 2\,{\rm V}$?

$p_{\rm BB} \ = \ $ $\ \cdot 10^{-4}$

4

Give the error probability of the BPSK depending on the quotient  $E_{\rm B}/N_0$.  Which result is correct?

	$p_{\rm BPSK} = {\rm Q}\big[(E_{\rm B}/N_0)^{1/2}\big],$
	$p_{\rm BPSK} = {\rm Q}\big[(2E_{\rm B}/N_0)^{1/2}\big],$
	$p_{\rm BPSK} = {\rm Q}\big[(4E_{\rm B}/N_0)^{1/2}\big].$

5

What are the error probabilities for BPSK with  $E_{\rm B}/N_0 = 8$  and  $E_{\rm B}/N_0 = 2$?

$E_{\rm B}/N_0 = 8\text{:} \ \ \ \ p_{\rm BPSK} \ = \ $ $\ \cdot 10^{-4}$

$E_{\rm B}/N_0 = 2\text{:} \ \ \ \ p_{\rm BPSK} \ = \ $ $\ \cdot 10^{-4}$

Solution

Solution

(1)  The noise rms value here is given by

$$\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}= \sqrt{\frac{2 \cdot 10^{-9}\,{\rm V^2/Hz}}{2 \cdot 1\,{\rm ns}}}= 1\,{\rm V} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm BB} = {\rm Q}\left ( {s_0}/{\sigma_d } \right )= {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}}.$$

(2)  For the baseband system:

$$E_{\rm B} = s_0^2 \cdot T_{\rm B}= (4\,{\rm V})^2 \cdot 10^{-9}\,{\rm s}\hspace{0.15cm}\underline {= 1.6 \cdot 10^{-8}\,{\rm V^2s}}.$$

• Of course,  the second equation gives the exact same error probability

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

(3)  When the transmitted amplitude is half   ⇒   $s_0 = 2\,{\rm V}$,  the energy per bit decreases to one-fourth and the following equations apply:

$$ p_{\rm BB} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )= {\rm Q}\left ( \frac{2\,{\rm V}}{1\,{\rm V}} \right )= {\rm Q}(2)= 227 \cdot 10^{-4},$$

$$ p_{\rm BB} = {\rm Q}\left ( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}\left ( \sqrt{\frac{2 \cdot 4 \cdot 10^{-9}\,{\rm V^2s}}{2 \cdot 10^{-9}\, {\rm V^2/Hz} }} \hspace{0.1cm}\right ) = {\rm Q}(2)\hspace{0.15cm}\underline {= 227 \cdot 10^{-4}}.$$

(4)  Answer 2  is correct:

• Considering the energy  $E_{\rm B} = s_0^2 · T_{\rm B}/2$  we obtain

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

• Thus,  the same result is obtained as for the optimal baseband transmission system.

(5)  Exactly the same results are obtained as for the baseband transmission in questions  (1)  and  (3):

$${ E_{\rm B}}/{N_0 }= 8: \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{16}) = {\rm Q}(4)\hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4}},$$

$$ { E_{\rm B}}/{N_0 }= 2: \hspace{0.2cm}p_{\rm BPSK} = {\rm Q}(\sqrt{4}) = {\rm Q}(2)\hspace{0.15cm}\underline {= 227 \cdot 10^{-4}}.$$