# Exercise 5.5: Multi-User Interference

$\rm PACF$  and  $\rm PCCF$  of M-sequences with  $P = 31$

We consider  "Direct-Sequence Spread Spectrum Modulation"  with the following parameters:

• The spreading is done with the M-sequence  $(45)_{\rm oct}$,  starting from the degree  $G = 5$.  The period length is thus
$$P = 2^5 –1 = 31.$$
• The AWGN parameter is set as  $10 · \lg \ (E_{\rm B}/N_0) = 5 \ \rm dB$    ⇒   $E_{\rm B}/N_0 = 3.162 = 1/0.316$.
• The bit error probability without interfering subscribers in the same frequency band is:
$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$
• The value of the signal at sampling time, without noise or interfering subscribers, is equal to $±s_0$   ("Nyquist system").  As a result, the bit error probability for noise standard deviation  $σ_d$  before the decision stage $($ originating from the AWGN noise$)$  can be stated as follows:
$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$

In this exercise,  we want to investigate how the bit error probability is changed by an additional participant.

• The possible spreading sequences of the interfering participant are also defined by  $P = 31$.
• The PN generators with octal identifiers  $(45)$,  $(51)$,  $(57)$,  $(67)$,  $(73)$ and  $(75)$ are available.
• In the first column of the table the PCCF values for  $λ = 0$  are given, furthermore also the respective maximum value for another initial phase:
$${\rm Max}\,\,|{\it \varphi}_{45,\hspace{0.05cm}i}| = \max_{\lambda} \,\,|{\it \varphi}_{45,\hspace{0.05cm}i}(\lambda)| \hspace{0.05cm}.$$
• The special case  $φ_\text{45, 45}(λ = 0)$  gives the PACF value of the spreading sequence with the octal identifier   $(45)$.

In the course of this exercise and in the sample solution the following signals are mentioned:

$q(t)$:   binary bipolar source signal,  symbol duration  $T$,
$c(t)$:   $±1$ spreading signal,  chip duration  $T_c$,
$s(t)$:   band-spread transmission signal;  it holds that  $s(t) = q(t) · c(t)$,  amplitude  $±s_0$,  chip duration  $T_c$,
$n(t)$:   AWGN noise,  characterized by the quotient  $E_{\rm B}/N_0$,
$i(t)$:   interference signal of the interfering subscriber,
$r(t)$:   received signal; it holds that  $r(t) = s(t) + n(t) + i(t)$,
$b(t)$:   band-compressed signal;  it holds that  $b(t)= r(t) · c(t)$,
$d(t)$:   detection signal after integration of  $b(t)$  over the symbol duration  $T$,
$v(t)$:   sink signal,  comparison with  $q(t)$  gives the error probability.

Notes:

$${\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$

### Questions

1

What is the  (normalized)  noise standard deviation at the decision maker?

 $σ_d/s_0 \ = \$

2

What is the bit error probability  $p_{\rm B}$ if the interfering participant  $i(t)$  uses the same M-sequence with octal identifier  $(45)$  as the participant under consideration?

 $p_{\rm B}\ = \$ $\ \%$

3

What is the approximate bit error probability  $p_{\rm B}$  if the interfering subscriber $i(t)$  uses the M-sequence with octal identifier  $(75)$?

 $p_{\rm B}\ = \$ $\ \%$

4

What statements could possibly be made for a different spreading sequence of the interfering participant?

 With octal identifier  $(51)$,    $p_{\rm B} = 0.1\%$   is possible. With octal identifier  $(57)$,    $p_{\rm B} = 0.7\%$   is possible. With octal identifier  $(67)$,    $p_{\rm B} = 1.2\%$   is possible.

### Solution

#### Solution

(1)  From the two equations given above,  it follows directly:

$$p_{\rm B} = {\rm Q}(2.515) = {\rm Q}({s_0}/{\sigma_d}) \hspace{0.3cm}\Rightarrow\hspace{0.3cm} \frac{\sigma_d}{s_0} = \frac{1}{2.515} = 0.398 \hspace{0.15cm}\underline {\approx 0.4} \hspace{0.05cm}.$$
• However,  one could also calculate this quantity using the more general equation
$$\sigma_d^2 = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty} |H_{\rm I}(f) |^2 \,\,{\rm d} {\it f}\hspace{0.05cm} = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty}{\rm sinc}^2(f T)\,\,{\rm d} {\it f} = \frac{N_0}{2T } \hspace{0.05cm}.$$
Here  $H_{\rm I}(f)$  describes the integrator in the frequency domain.
• With  $E_{\rm B}= s_0^2 · T$  the same result is obtained:
$$\frac{\sigma_d^2}{s_0^2} = \frac{N_0}{2 \cdot s_0^2 \cdot T } = \frac{N_0}{2 E_{\rm B} } = \frac{0.316}{2 } = 0.158\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\sigma_d}/{s_0} = 0.398 \approx 0.4 \hspace{0.05cm}.$$

(2)  If the interfering participant uses the same M-sequence  $(45)$  as the considered user,
then the  (normalized)  detection samples  (without noise)  are equal to  $+2$  $($at  $25\%)$,  $-2$  $($at  $25\%)$  and  $0$  $($at  $50\%)$.

• When  $d(νT) = ±2$,  the error probability for the considered user is significantly reduced.
• In this case,  both users transmit the same bit  $($"$+1$"  or  "$-1$"$)$  and the distance from the threshold is doubled:
$$p_{\rm B}\,\,\big [{\rm if}\,\, d (\nu T) = \pm 2s_0 \big ] = {\rm Q} \left ( 2 \cdot 2.515 \right ) = {\rm Q} \left ( 5.03 \right ) \approx 2.45 \cdot 10^{-7} \approx 0 \hspace{0.05cm}.$$
• On the other hand,  if  $d(νT) = 0$  (for example, if  $a_\text{1(s)} = +1$   and  $a_\text{1(i)} = -1$   holds or vice versa),  the signals cancel completely and we obtain
$$p_{\rm B}\big[{\rm if}\,\, d (\nu T) = 0 \big] = {\rm Q} \left ( 0 \right ) = 0.5 \hspace{0.05cm}.$$
• Averaging over these two equally probable possibilities,  we thus obtain for the mean bit error probability:
$$p_{\rm B}= 0.5 \cdot 2.45 \cdot 10^{-7}+ 0.5 \cdot 0.5 \hspace{0.15cm}\underline {\approx 25\%} \hspace{0.05cm}.$$

(3)  We first consider only the useful part   ⇒   $n(t) = 0$,  restricting ourselves to the first data symbol and assuming the coefficient  $a_\text{1(s)} = +1$.

• Then within this data bit  $s(t) = c_{45}(t)$ holds.
• If the coefficient  $a_\text{1(i)}$  of the interfering participant is also  $+1$,  then for the signals specified in front in the time interval from  $0$  to  $T$ we obtain:
$$r(t) = c_{45}(t) + c_{75}(t)\hspace{0.05cm},$$
$$b(t) = r(t) \cdot c_{45}(t) = \left [c_{45}(t) + c_{75}(t) \right ] \cdot c_{45}(t) = 1+ c_{45}(t) \cdot c_{75}(t) \hspace{0.05cm},$$
$$d (T) = \frac{1}{T} \cdot \int_{0 }^{ T} b (t )\hspace{0.1cm} {\rm d}t = 1 + {\it \varphi}_{45,\hspace{0.05cm}75}(\lambda = 0) \hspace{0.05cm}.$$
• Here,   $φ_\text{45, 75}(τ)$  denotes the PCCF between the spreading sequences with octal identifiers  $(45)$  and  $(75)$,  which can be found in the table on the data page.
• Correspondingly,  for the detection sample without noise,  given  $a_\text{1(s)} = +1$  and  $a_\text{1(i)} =-1$:
$$d (T) = 1 - {\it \varphi}_{45,\hspace{0.05cm}75}(\lambda = 0) \hspace{0.05cm}.$$
• For symmetry reasons,  the coefficients  $a_\text{1(s)} = -1$,  $a_\text{1(i)} = -1$  as well as  $a_\text{1(s)} = -1$,  $a_\text{1(i)} = +1$  provide exactly the same contributions for the bit error probability as  $a_\text{1(s)} = +1$,  $a_\text{1(i)} = +1$  and  $a_{1(s)} = +1$,  $a_{1(i)} = -1$ respectively,  if we also consider the AWGN noise.
• Thus,  using the result of subtask  (1)  and with  $φ_\text{45, 75}(λ = 0) = 7/31$,  we obtain approximately:
$$p_{\rm B} = \frac{1}{2} \cdot {\rm Q} \left ( \frac{1+ 7/31}{0.4} \right ) + \frac{1}{2} \cdot {\rm Q} \left ( \frac{1- 7/31}{0.4} \right ) = \frac{1}{2} \cdot {\rm Q} \left ( \frac{1.225}{0.4} \right ) + \frac{1}{2} \cdot {\rm Q} \left ( \frac{0.775}{0.4} \right ) = \frac{1}{2} \cdot {\rm Q} \left ( 3.06 \right ) + \frac{1}{2} \cdot {\rm Q} \left ( 1.94 \right )$$
$$\Rightarrow \hspace{0.3cm} p_{\rm B}\approx \frac{1}{2} \cdot \left [{\rm Q} \left ( 3 \right ) + {\rm Q} \left ( 2 \right ) \right ] = \frac{1}{2} \cdot \left [0.00135 + 0.02275 \right ] \hspace{0.15cm}\underline {= 1.2\%}\hspace{0.05cm}.$$

(4)  Possible solutions are  2 and 3:

• The PCCF value  $φ_\text{45, 57}(λ = 0)$  is only  $1/31$  in magnitude and thus the error probability is only slightly larger than  $0.6\%$.
• In contrast,  the sequence with octal identifiers  $(67)$  leads to the same PCCF as sequence  $(75)$.
• Without interfering participants,  the following applies according to the data sheet:   $p_{\rm B} = 0.6\%$.
• With interference,  this value cannot be undercut   ⇒   solution 1 is not possible.