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:

The exercise belongs to the chapter Error Probability of Direct-Sequence Spread Spectrum Modulation.
Reference is made in particular to the section Two users with M-sequence spreading.
For the so-called Q-function, the following approximations can be assumed:

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

Questions

$σ_d/s_0 \ = \ $

$p_{\rm B}\ = \ $

$\ \%$

$p_{\rm B}\ = \ $

$\ \%$

	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

(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.