# Exercise 3.09: Correlation Receiver for Unipolar Signaling

The joint decision of  $N = 3$  binary symbols  ("bits")  by means of the correlation receiver is considered.

The  $M = 8$  possible source symbol sequences  $Q_i$  all have the same probability and they are defined by the following unipolar amplitude coefficients:

$$Q_0 = 000, \hspace{0.15cm}Q_1 = 001,\hspace{0.15cm}Q_2 = 010,\hspace{0.15cm}Q_3 = 011 \hspace{0.05cm},\hspace{0.15cm} Q_4 = 100, \hspace{0.15cm}Q_5 = 101,\hspace{0.15cm}Q_6 = 110,\hspace{0.15cm}Q_7 = 111 \hspace{0.05cm}.$$

Further applies:

• The possible transmitted signals  $s_i(t)$  – each with duration  $3T$  – are all rectangular with the exception of  $s_0(t) \equiv 0$.
• The signals  $s_1(t)$,  $s_2(t)$  and  $s_4(t)$  with only one  "$1$"  each have the signal energy  $E_{\rm B}$  (stands for "energy per bit"),  while for example the energy of  $s_7(t)=3E_{\rm B}$.

The correlation receiver forms from the noisy received signal  $r(t) = s(t) + n(t)$  a total of  $2^3 = 8$  decision variables (metrics)

$$W_i = I_i - {E_i}/{2 }\hspace{0.3cm}{\rm with}\hspace{0.3cm} I_i =\int_{0}^{3T} r(t) \cdot s_i(t) \,{\rm d} t \hspace{0.3cm}( i = 0,\text{...} , 7)$$

and sets the sink symbol sequence  $V = Q_j$,  if  $W_j$  is larger than all other  $W_{i \ne j}$.  Thus,  it makes an optimal decision in the sense of  "maximum likelihood".

In the table,  the  (uncorrected)  correlation values  $I_0, \ \text{...} \ , I_7$  for three different systems differing in terms of noise  $n(t)$  and labeled  $\rm A$,  $\rm B$  or  $\rm C$.

• One of these columns stands for  "no noise",
• one for  "minor noise",  and
• another one for  "strong noise".

Note:

• The same source symbol sequence was always sent to determine the metrics for the three system variants.

### Question

1

For which system is there  "no noise"   ⇒   $n(t)=0$? At

 $\rm System \ A$, $\rm System \ B$, $\rm System \ C$.

2

Which source symbol sequence  $Q_k ∈ {Q_0, \ \text{...} \ , Q_7}$  was actually sent?

 $k \ = \$

3

Which decision value  $W_j$  is largest for system  $\rm A$?

 ${\rm System \ A} \text{:} \hspace{0.2cm} j \ = \$

4

Which decision value  $W_j$  is largest for system  $\rm C$?

 ${\rm System \ C} \text{:} \hspace{0.2cm} j \ = \$

5

For which system do the largest noise occur? At

 $\rm System \ A$, $\rm System \ B$, $\rm System \ C$.

6

Which statements are valid under the assumption that  $Q_2$  was sent and the correlation receiver normally chooses  $Q_2$  as well?

 The difference between  $W_2$  and the next largest value  $W_{i \ne 2}$  is smaller the stronger the noise is. When falsification occurs,  the receiver is most likely to decide in favor of the symbol sequence  $Q_6$. The probabilities for erroneous decisions in favor of  $Q_0$,  $Q_3$  and  $Q_6$,  respectively, are equal.

### Solution

#### Solution

(1)  Solution 2  is correct:

• For system  $\rm B$,  metrics  "$0$"  occur four times and metrics  "$1$"  occur four times.
• This points to  $n(t) = 0$,  otherwise – as in systems  $\rm A$  and  $\rm C$  – all  $I_i$  would have to differ.

(2)  For system  $\rm B$,  the decision values  $W_i = I_i \ - E_i/2$,  each normalized to  $E_{\rm B}$,  are as follows:

$$W_0 = 0 - 0 = 0, \hspace{0.2cm}W_1 = 0 - 0.5 = -0.5 \hspace{0.05cm},$$
$$W_2 = 1 - 0.5 = 0.5, \hspace{0.2cm}W_3 = 1 - 1 = 0 \hspace{0.05cm},$$
$$W_4 = 0 - 0.5 = -0.5, \hspace{0.2cm}W_5 = 0 - 1 = -1 \hspace{0.05cm}.$$
$$W_6 = 1 - 1 = 0, \hspace{0.2cm}W_7 = 1 - 1.5 = -0.5 \hspace{0.05cm}.$$
• The maximum value  $W_2 = 0.5$   ⇒   $i = 2$.
• Thus,  the correlation receiver decides to use  $V = Q_2$.
• Since there is no noise,  $Q_2 =$ "$\rm 010$"  was indeed also sent   ⇒   $\underline { k= 2}$.

(3)  For the decision values of system  $\rm A$  holds:

$$W_0 = 0.00 - 0.00 = 0.00, \hspace{0.2cm}W_1 = -0.07 - 0.50 = -0.57,$$
$$W_2 = 1.13 - 0.50 = 0.63, \hspace{0.2cm}W_3 = 1.06 - 1.00 = 0.06 \hspace{0.05cm},$$
$$W_4 = 0.05 - 0.50 = -0.45, \hspace{0.2cm}W_5 = -0.02 - 1.00 = -1.02\hspace{0.05cm},$$
$$W_6 = 1.18 - 1.00 = 0.18, \hspace{0.2cm}W_7 = 1.11 - 1.50 = -0.39 \hspace{0.05cm}.$$
• The maximum is  = $W_j = W_2$   ⇒   $\underline { j= 2}$.
• This means that the correlation receiver also makes the correct decision  $V = Q_2$  for system  $\rm A$.
• However,  without the correction term  $(– E_i/2)$,  the receiver would have made the wrong decision  $V = Q_6$.

(4)  The correlation receiver  $\rm C$  has to compare the following values:

$$W_0 = 0.00 - 0.00 = 0.00, \hspace{0.2cm}W_1 = -1.31 - 0.50 = -1.81 \hspace{0.05cm},$$
$$W_2 = 3.59 - 0.50 = 3.09, \hspace{0.2cm}W_3 = 2.28 - 1.00 = 1.28 \hspace{0.05cm},$$
$$W_4 = 0.97 - 0.50 = 0.47, \hspace{0.2cm}W_5 = -0.34 - 1.00 = -1.34 \hspace{0.05cm},$$
$$W_6 = 4.56 - 1.00 = 3.56, \hspace{0.2cm}W_7 = 3.25 - 1.50 = 1.75 \hspace{0.05cm}.$$

The maximization here gives  $\underline {j = 6}$   ⇒   $V = Q_6$.

• But since  $Q_2$  was sent,  the correlation receiver decides wrong here.
• The noise is too strong.

(5)  Solution 3  is correct:

• The noise is greatest for system  $\rm C$  and is even so great for the current received values that the correlation receiver makes an incorrect decision.

(6)  Statements 1 and 3 are correct:

• In the error-free case  $($system  $\rm B)$,  the difference between  $W_2 = 0.5$  and the next largest values  $W_0 = W_3 = W_6 = 0$  is equal to  $D_{\hspace{0.02cm}\rm min} =0.5$ in each case.
• In system  $\rm A$  (light noise),  the difference between  $W_2 = 0.63$  and the next largest value  $W_6 = 0.18$  is still  $D_{\hspace{0.02cm}\rm min} = 0.45$.
• If the noise power is increased by factor  $50$,  the correlation receiver still decides correctly,  but then the minimum difference  $D_{\hspace{0.02cm}\rm min} = 0.16$  is significantly smaller.
• For system  $\rm C$,  where the correlation receiver is overcharged   ⇒   subtask  (4),  a noise power larger by a factor of  $400$  compared to system  $\rm A$  was used as a basis.
• If the correlation receiver decides the transmitted sequence  $Q_2$  incorrectly,  a falsification to the sequences  $Q_0$,  $Q_3$  resp.  $Q_6$  is most likely,
since all these three sequences differ from  $Q_2$  only in one bit each.
• The fact that  $W_6$  is always larger than  $W_0$  or  $W_3$  in the described simulation is "coincidence" and should not be overinterpreted.