Aufgaben:Exercise 5.5: Multi-User Interference - Revision history

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:11:50Z

Text replacement - "rms value" to "standard deviation"

Hwang at 14:50, 20 December 2021

2021-12-20T14:50:25Z

Hwang at 14:49, 20 December 2021

2021-12-20T14:49:35Z

Guenter at 16:26, 17 December 2021

2021-12-17T16:26:52Z

Guenter at 14:14, 17 December 2021

2021-12-17T14:14:18Z

Guenter at 13:53, 17 December 2021

2021-12-17T13:53:46Z

Guenter at 13:53, 17 December 2021

2021-12-17T13:53:18Z

Guenter at 13:51, 17 December 2021

2021-12-17T13:51:46Z

Guenter at 13:50, 17 December 2021

2021-12-17T13:50:28Z

Guenter at 13:48, 17 December 2021

2021-12-17T13:48:02Z

@@ Line 10: / Line 10: @@
 * 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&nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system").&nbsp; As a result, the bit error probability for noise rms value &nbsp;$σ_d$&nbsp; before the decision stage $($ originating from the AWGN noise$)$&nbsp; can be stated as follows: &nbsp;
+* The value of the signal at sampling time, without noise or interfering subscribers, is equal to&nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system").&nbsp; As a result, the bit error probability for noise standard deviation &nbsp;$σ_d$&nbsp; before the decision stage $($ originating from the AWGN noise$)$&nbsp; can be stated as follows: &nbsp;
 :$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$
@@ Line 47: / Line 47: @@
 <quiz display=simple>
-{What is the&nbsp; (normalized)&nbsp; noise rms value at the decision maker?
+{What is the&nbsp; (normalized)&nbsp; noise standard deviation at the decision maker?
 |type="{}"}
 $σ_d/s_0 \ = \ $  { 0.4 3% }

@@ Line 10: / Line 10: @@
 * 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&nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system").&nbsp; As a result, the bit error probability for noise rms value &nbsp;$σ_d$&nbsp; before the decision stage $($&nbsp; originating from the AWGN noise$)$&nbsp; can be stated as follows: &nbsp;
+* The value of the signal at sampling time, without noise or interfering subscribers, is equal to&nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system").&nbsp; As a result, the bit error probability for noise rms value &nbsp;$σ_d$&nbsp; before the decision stage $($ originating from the AWGN noise$)$&nbsp; can be stated as follows: &nbsp;
 :$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$

@@ Line 10: / Line 10: @@
 * The bit error probability without interfering subscribers in the same frequency band is:
 :$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$
-* Since in the absence of interfering subscribers all detection signal samples&nbsp; (without input noise)&nbsp; are equal to &nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system"),&nbsp; the bit error probability with the noise rms value &nbsp;$σ_d$&nbsp; $($before the decision maker,&nbsp; originating from the AWGN noise$)$&nbsp; is given as follows: &nbsp;
+* The value of the signal at sampling time, without noise or interfering subscribers, is equal to&nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system").&nbsp; As a result, the bit error probability for noise rms value &nbsp;$σ_d$&nbsp; before the decision stage $($&nbsp; originating from the AWGN noise$)$&nbsp; can be stated as follows: &nbsp;
 :$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$

@@ Line 10: / Line 10: @@
 * The bit error probability without interfering subscribers in the same frequency band is:
 :$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$
-* Since in the absence of interfering subscribers all payload samples are equal to &nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system"),&nbsp; the bit error probability with the noise effective value &nbsp;$σ_d$&nbsp; before the decision maker&nbsp; $($originating from the AWGN noise$)$&nbsp; is given as follows: &nbsp;
+* Since in the absence of interfering subscribers all detection signal samples&nbsp; (without input noise)&nbsp; are equal to &nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system"),&nbsp; the bit error probability with the noise rms value &nbsp;$σ_d$&nbsp; $($before the decision maker,&nbsp; originating from the AWGN noise$)$&nbsp; is given as follows: &nbsp;
 :$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$

@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  From the two equations given above, it follows directly:
+'''(1)'''&nbsp;  From the two equations given above,&nbsp; 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
+*However,&nbsp; one could also calculate this quantity using the more general equation
-:$$ \sigma_d^2 = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty} |H_{\rm I}(f) |^2 \,\,{\rm d} {\it f}\hspace{0.05cm} = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty}{\rm si}^2(\pi f T)\,\,{\rm d} {\it f} = \frac{N_0}{2T } \hspace{0.05cm}.$$
+:$$ \sigma_d^2 = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty} |H_{\rm I}(f) |^2 \,\,{\rm d} {\it f}\hspace{0.05cm} = \frac{N_0}{2 }\cdot\int^{+\infty}_{-\infty}{\rm sinc}^2(f T)\,\,{\rm d} {\it f} = \frac{N_0}{2T } \hspace{0.05cm}.$$
-:&nbsp; Here $H_{\rm I}(f)$ describes the integrator in the frequency domain.
+:&nbsp; Here&nbsp; $H_{\rm I}(f)$&nbsp; describes the integrator in the frequency domain.
-*With $E_{\rm B}= s_0^2 · T$ the same result is obtained:
+*With&nbsp; $E_{\rm B}= s_0^2 · T$&nbsp; 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)'''&nbsp; If the interfering participant uses the same M-sequence&nbsp; $(45)$&nbsp; as the considered user, <br>&nbsp; &nbsp; &nbsp; &nbsp; then the (normalized) detection utility samples are equal to&nbsp; $+2$&nbsp; $($at $25\%)$,&nbsp; $-2$&nbsp; $($at $25\%)$&nbsp; and&nbsp; $0$&nbsp; $($at $50\%)$.
+'''(2)'''&nbsp; If the interfering participant uses the same M-sequence&nbsp; $(45)$&nbsp; as the considered user, <br>&nbsp; &nbsp; &nbsp; &nbsp; then the&nbsp; (normalized)&nbsp; detection samples&nbsp; (without noise)&nbsp; are equal to&nbsp; $+2$&nbsp; $($at&nbsp; $25\%)$,&nbsp; $-2$&nbsp; $($at&nbsp; $25\%)$&nbsp; and&nbsp; $0$&nbsp; $($at&nbsp; $50\%)$.
-*When &nbsp;$d(νT) = ±2$,&nbsp; the error probability for the considered user is significantly reduced.&nbsp; In this case, both users transmit the same bit $($"$+1$" or "$-1$"$)$ and the distance from the threshold is doubled:
+*When &nbsp;$d(νT) = ±2$,&nbsp; the error probability for the considered user is significantly reduced.&nbsp;
 :$$ p_{\rm B}\,\,\big [{\rm if}\,\, d (\nu T) = \pm 2s_0 \big ] = {\rm Q} \left ( 2 \cdot 2.515 \right ) = {\rm Q} \left ( 5.03 \right ) \approx 2.45 \cdot 10^{-7} \approx 0 \hspace{0.05cm}.$$
-*On the other hand, if &nbsp;$d(νT) = 0$&nbsp; (for example, if&nbsp; $a_\text{1(s)} = +1$&nbsp; and&nbsp; $a_\text{1(i)} = -1$&nbsp; holds or vice versa), the signals cancel completely and we obtain
+*On the other hand,&nbsp; if &nbsp;$d(νT) = 0$&nbsp; (for example, if&nbsp; $a_\text{1(s)} = +1$ &nbsp; and&nbsp; $a_\text{1(i)} = -1$ &nbsp; holds or vice versa),&nbsp; the signals cancel completely and we obtain
-:$$p_{\rm B}\,\,\big[{\rm if}\,\, d (\nu T) = 0 \big] = {\rm Q} \left ( 0 \right ) = 0.5 \hspace{0.05cm}.$$
+:$$p_{\rm B}\big[{\rm if}\,\, d (\nu T) = 0 \big] = {\rm Q} \left ( 0 \right ) = 0.5 \hspace{0.05cm}.$$
-*Averaging over these two equally probable possibilities, we thus obtain for the mean bit error probability:
+*Averaging over these two equally probable possibilities,&nbsp; 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)'''&nbsp;  We first consider only the useful part &nbsp; ⇒ &nbsp; $n(t) = 0$,&nbsp; restricting ourselves to the first data symbol and assuming the coefficient $a_\text{1(s)} = +1$.
+'''(3)'''&nbsp;  We first consider only the useful part &nbsp; ⇒ &nbsp; $n(t) = 0$,&nbsp; restricting ourselves to the first data symbol and assuming the coefficient&nbsp; $a_\text{1(s)} = +1$.
 *Then within this data bit&nbsp; $s(t) = c_{45}(t)$ holds.
-*If the coefficient&nbsp; $a_\text{1(i)} $&nbsp; of the interfering participant is also&nbsp; $+1$, then for the signals specified in front in the time interval from&nbsp; $0$&nbsp; to&nbsp; $T$ we obtain:
+*If the coefficient&nbsp; $a_\text{1(i)} $&nbsp; of the interfering participant is also&nbsp; $+1$,&nbsp; then for the signals specified in front in the time interval from&nbsp; $0$&nbsp; to&nbsp; $T$ 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, &nbsp; $φ_\text{45, 75}(τ)$&nbsp; denotes the PCCF between the spreading sequences with octal identifiers&nbsp; $(45)$&nbsp; and&nbsp; $(75)$, which can be found in the table on the data page.
+*Here, &nbsp; $φ_\text{45, 75}(τ)$&nbsp; denotes the PCCF between the spreading sequences with octal identifiers&nbsp; $(45)$&nbsp; and&nbsp; $(75)$,&nbsp; which can be found in the table on the data page.
+*Correspondingly,&nbsp; for the detection sample without noise,&nbsp; given&nbsp; $a_\text{1(s)} = +1$&nbsp; and&nbsp; $a_\text{1(i)} =-1$:
 :$$d (T) = 1 - {\it \varphi}_{45,\hspace{0.05cm}75}(\lambda = 0) \hspace{0.05cm}.$$
-*For symmetry reasons, the coefficients&nbsp; $a_\text{1(s)}&nbsp; = -1$,&nbsp; $a_\text{1(i)} = -1$&nbsp; as well as&nbsp; $a_\text{1(s)} = -1$,&nbsp; $a_\text{1(i)} = +1$&nbsp; provide exactly the same contributions for the bit error probability as&nbsp; $a_\text{1(s)} = +1$,&nbsp; $a_\text{1(i)} = +1$&nbsp; and&nbsp; $a_{1(s)} = +1$,&nbsp; $a_{1(i)} = –1$ respectively, if we also consider the AWGN noise.
+*For symmetry reasons,&nbsp; the coefficients&nbsp; $a_\text{1(s)}&nbsp; = -1$,&nbsp; $a_\text{1(i)} = -1$&nbsp; as well as&nbsp; $a_\text{1(s)} = -1$,&nbsp; $a_\text{1(i)} = +1$&nbsp; provide exactly the same contributions for the bit error probability as&nbsp; $a_\text{1(s)} = +1$,&nbsp; $a_\text{1(i)} = +1$&nbsp; and&nbsp; $a_{1(s)} = +1$,&nbsp; $a_{1(i)} = -1$ respectively,&nbsp; if we also consider the AWGN noise.
-*Thus, using the result of subtask&nbsp; '''(1)'''&nbsp; and with&nbsp; $φ_\text{45, 75}(λ = 0) = 7/31$,&nbsp; we obtain approximately:
+*Thus,&nbsp; using the result of subtask&nbsp; '''(1)'''&nbsp; and with&nbsp; $φ_\text{45, 75}(λ = 0) = 7/31$,&nbsp; 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)'''&nbsp; Possible solutions are <u>2 and 3</u>:
 * The PCCF value&nbsp; $φ_\text{45, 57}(λ = 0)$&nbsp; is only&nbsp; $1/31$&nbsp; in magnitude and thus the error probability is only slightly larger than&nbsp; $0.6\%$.
-*In contrast, the sequence with octal identifiers&nbsp; $(67)$&nbsp; leads to the same PCCF as sequence&nbsp; $(75)$.
+*In contrast,&nbsp; the sequence with octal identifiers&nbsp; $(67)$&nbsp; leads to the same PCCF as sequence&nbsp; $(75)$.
-*Without interfering participants, the following applies according to the data sheet: &nbsp; $p_{\rm B} = 0.6\%$.
+*Without interfering participants,&nbsp; the following applies according to the data sheet: &nbsp; $p_{\rm B} = 0.6\%$.
-*With interference, this value cannot be undercut &nbsp; ⇒  &nbsp; solution 1 is not possible.
+*With interference,&nbsp; this value cannot be undercut &nbsp; ⇒  &nbsp; solution 1 is not possible.
 {{ML-Fuß}}

@@ Line 40: / Line 40: @@
 *The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation|Error Probability of Direct-Sequence Spread Spectrum Modulation]].
 *Reference is made in particular to the section&nbsp; [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Two_users_with_M-sequence_spreading|Two users with M-sequence spreading]].
-*For the so-called [[Applets:Complementary_Gaussian_Error_Functions|Q-function]],&nbsp; the following approximations can be assumed:
+*For the so-called&nbsp; [[Applets:Complementary_Gaussian_Error_Functions|Q-function]],&nbsp; the following approximations can be assumed:
 :$$ {\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$

@@ Line 4: / Line 4: @@
 [[File:P_ID1887__Mod_A_5_5.png|right|frame|$\rm PACF$&nbsp; and&nbsp; $\rm PCCF$&nbsp; of M-sequences with &nbsp;$P = 31$]]
-We consider PN modulation with the following parameters:
+We consider&nbsp; "Direct-Sequence Spread Spectrum Modulation"&nbsp; with the following parameters:
 * The spreading is done with the M-sequence &nbsp;$(45)_{\rm oct}$,&nbsp; starting from the degree &nbsp;$G = 5$.&nbsp; The period length is thus &nbsp;
 :$$P = 2^5 –1 = 31.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1887__Mod_A_5_5.png|right|frame|$\rm PACF$&nbsp; and&nbsp; $\rm PCCF$&nbsp; of M-sequence with &nbsp;$P = 31$]]
+[[File:P_ID1887__Mod_A_5_5.png|right|frame|$\rm PACF$&nbsp; and&nbsp; $\rm PCCF$&nbsp; of M-sequences with &nbsp;$P = 31$]]
 We consider PN modulation with the following parameters:
-* The spreading is done with the M-sequence with the octal identifier &nbsp;$(45)$, starting from the degree &nbsp;$G = 5$.&nbsp; The period length is thus &nbsp;
+* The spreading is done with the M-sequence &nbsp;$(45)_{\rm oct}$,&nbsp; starting from the degree &nbsp;$G = 5$.&nbsp; The period length is thus &nbsp;
 :$$P = 2^5 –1 = 31.$$
 * The AWGN parameter is set as &nbsp;$10 · \lg \ (E_{\rm B}/N_0) = 5 \ \rm  dB$&nbsp; &nbsp; &rArr; &nbsp;  $E_{\rm B}/N_0 = 3.162 = 1/0.316$.
 * The bit error probability without interfering subscribers in the same frequency band is:
 :$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2\cdot E_{\rm B}}/{N_{\rm 0}}}\right ) \approx {\rm Q} \left ( \sqrt{2 \cdot 3.162}\right ) = {\rm Q} \left ( 2.515 \right ) \approx 6 \cdot 10^{-3} \hspace{0.05cm}.$$
-* Since in the absence of interfering subscribers all payload samples are equal to &nbsp;$±s_0$&nbsp;&nbsp; (Nyquist system), the bit error probability with the noise effective value &nbsp;$σ_d$&nbsp; before the decision maker&nbsp; $($originating from the AWGN noise$)$&nbsp; is given as follows: &nbsp;
+* Since in the absence of interfering subscribers all payload samples are equal to &nbsp;$±s_0$&nbsp;&nbsp; ("Nyquist system"),&nbsp; the bit error probability with the noise effective value &nbsp;$σ_d$&nbsp; before the decision maker&nbsp; $($originating from the AWGN noise$)$&nbsp; is given as follows: &nbsp;
 :$$p_{\rm B} = {\rm Q} \left ( {s_0}/{\sigma_d}\right ) \hspace{0.05cm}.$$
-In this task, we want to investigate how the bit error probability is changed by an additional participant.
+In this exercise,&nbsp; 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 $P = 31$. &nbsp; The PN generators with octal identifiers &nbsp;$(45)$, &nbsp;$(51)$, &nbsp;$(57)$, &nbsp;$(67)$, &nbsp;$(73)$ and &nbsp;$(75)$ are available.
+*The possible spreading sequences of the interfering participant are also defined by&nbsp; $P = 31$. &nbsp;
+*The PN generators with octal identifiers &nbsp;$(45)$, &nbsp;$(51)$, &nbsp;$(57)$, &nbsp;$(67)$, &nbsp;$(73)$ and &nbsp;$(75)$ are available.
-*In the table the PCCF values for &nbsp;$λ = 0$&nbsp; are given, furthermore also the respective maximum value for another initial phase:
+*In the first column of the table the PCCF values for &nbsp;$λ = 0$&nbsp; 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 &nbsp;$φ_\text{45, 45}(λ = 0)$&nbsp; gives the PACF value of the spreading sequence with the octal identifier &nbsp;&nbsp;$(45)$.&nbsp;
@@ Line 23: / Line 23: @@
 In the course of this exercise and in the sample solution the following signals are mentioned:
-:&nbsp;$q(t)$: &nbsp; binary bipolar source signal, symbol duration &nbsp;$T$,
+:&nbsp;$q(t)$: &nbsp; binary bipolar source signal,&nbsp; symbol duration &nbsp;$T$,
-:&nbsp;$c(t)$:  &nbsp; $±1$ spreading signal, chip duration &nbsp;$T_c$,
+:&nbsp;$c(t)$:  &nbsp; $±1$ spreading signal,&nbsp; chip duration &nbsp;$T_c$,
-:&nbsp;$s(t)$:  &nbsp; band spreading transmit signal; it holds that &nbsp;$s(t) = q(t) · c(t)$, amplitude &nbsp;$±s_0$, chip duration &nbsp;$T_c$,
+:&nbsp;$s(t)$:  &nbsp; band-spread transmission signal;&nbsp; it holds that &nbsp;$s(t) = q(t) · c(t)$,&nbsp; amplitude &nbsp;$±s_0$,&nbsp; chip duration &nbsp;$T_c$,
-:&nbsp;$n(t)$:  &nbsp; AWGN noise, characterized by the quotient &nbsp;$E_{\rm B}/N_0$,
+:&nbsp;$n(t)$:  &nbsp; AWGN noise,&nbsp; characterized by the quotient &nbsp;$E_{\rm B}/N_0$,
 :&nbsp;$i(t)$:  &nbsp; interference signal of the interfering subscriber,
 :&nbsp;$r(t)$:  &nbsp; received signal; it holds that &nbsp;$r(t) = s(t) + n(t) + i(t)$,
-:&nbsp;$b(t)$:  &nbsp; bandpassed signal; it holds that &nbsp;$b(t)= r(t) · c(t)$,
+:&nbsp;$b(t)$:  &nbsp; band-compressed signal;&nbsp; it holds that &nbsp;$b(t)= r(t) · c(t)$,
 :&nbsp;$d(t)$:  &nbsp; detection signal after integration of &nbsp;$b(t)$&nbsp; over the symbol duration &nbsp;$T$,
-:&nbsp;$v(t)$:  &nbsp; sink signal, comparison with &nbsp;$q(t)$&nbsp; gives the error probability.
+:&nbsp;$v(t)$:  &nbsp; sink signal,&nbsp; comparison with &nbsp;$q(t)$&nbsp; gives the error probability.
@@ Line 40: / Line 37: @@
-''Notes:''
+Notes:
 *The exercise belongs to the chapter&nbsp; [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation|Error Probability of Direct-Sequence Spread Spectrum Modulation]].
 *Reference is made in particular to the section&nbsp; [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Two_participants_with_M-sequence_spreading|Two participants with M-sequence spreading]].
+*For the so-called Q-function,&nbsp; the following approximations can be assumed:
 :$$ {\rm Q} (2) \approx 0.02275, \hspace{0.2cm}{\rm Q} (3) \approx 0.00135, \hspace{0.2cm}{\rm Q} (5) \approx 2.45 \cdot 10^{-7} \hspace{0.05cm}.$$
@@ Line 51: / Line 47: @@
 <quiz display=simple>
-{What is the (normalized) noise effective value at the decision maker?
+{What is the&nbsp; (normalized)&nbsp; noise rms value at the decision maker?
 |type="{}"}
 $σ_d/s_0 \ = \ $  { 0.4 3% }
@@ Line 65: / Line 61: @@
 {What statements could possibly be made for a different spreading sequence of the interfering participant?
 |type="[]"}
-- With the octal identifier &nbsp;$(51)$,&nbsp; &nbsp; $p_{\rm B} = 0.1\%$ &nbsp; is possible.
+- With octal identifier &nbsp;$(51)$,&nbsp; &nbsp; $p_{\rm B} = 0.1\%$ &nbsp; is possible.
-+ With the octal identifier &nbsp;$(57)$,&nbsp; &nbsp;  $p_{\rm B} = 0.7\%$ &nbsp; is possible.
++ With octal identifier &nbsp;$(57)$,&nbsp; &nbsp;  $p_{\rm B} = 0.7\%$ &nbsp; is possible.
-+ With the octal identifier &nbsp;$(67)$,&nbsp; &nbsp; $p_{\rm B} = 1.2\%$ &nbsp; is possible.
++ With octal identifier &nbsp;$(67)$,&nbsp; &nbsp; $p_{\rm B} = 1.2\%$ &nbsp; is possible.
 </quiz>