# Exercise 3.10: Maximum Likelihood Tree Diagram

From LNTwww

As in "Exercise 3.9" we consider the joint decision of three binary symbols ("bits") by means of the correlation receiver.

- The possible transmitted signals $s_0(t), \ \text{...} \ , \ s_7(t)$ are bipolar.

- In the graphic the functions $s_0(t)$, $s_1(t)$, $s_2(t)$ and $s_3(t)$ are shown.

- The blue curves are valid for rectangular NRZ transmission pulses.

Below is drawn the so-called "tree diagram" for this constellation under the condition that the signal $s_3(t)$ was sent. Shown here in the range from $0$ to $3T$ are the functions

- $$i_i(t) = \int_{0}^{t} s_3(\tau) \cdot s_i(\tau) \,{\rm d} \tau \hspace{0.3cm}( i = 0, \ \text{...} \ , 7)\hspace{0.05cm}.$$

- The correlation receiver compares the final values $I_i = i_i(3T)$ with each other and searches for the largest possible value $I_j$.

- The corresponding signal $s_j(t)$ is then the one most likely to have been sent according to the maximum likelihood criterion.

- Note that the correlation receiver generally makes the decision based on the corrected quantities

- $$W_i = I_i \ - E_i/2.$$

- But since for bipolar rectangles all transmitted signals $(i = 0, \ \text{...} \ , \ 7)$ have exactly the same energy

- $$E_i = \int_{0}^{3T} s_i^2(t) \,{\rm d} t,$$

- the integrals $I_i$ provide exactly the same maximum likelihood information as the corrected quantities $W_i$.

The red signal waveforms $s_i(t)$ are obtained from the blue ones by convolution with the impulse response $h_{\rm G}(t)$ of a Gaussian low-pass filter with cutoff frequency $f_{\rm G} \cdot T = 0.35$.

- Each individual rectangular pulse is broadened.

- The red signal waveforms lead to "intersymbol interference" in case of threshold decision.

Note: The exercise belongs to the chapter "Optimal Receiver Strategies".

### Questions

### Solution

**(1)**The left graph shows the tree diagram (without noise) with all final values. Highlighted in green is the curve $i_0(t)/E_{\rm B}$ with the final result $I_0/E_{\rm B} = \ -1$, which first rises linearly to $+1$ $($the first bit of $s_0(t)$ and $s_3(t)$ in each case coincide$)$ and then falls off over two bit durations.

- The correct results are thus:

- $$I_0/E_{\rm B}\hspace{0.15cm}\underline { = -1},$$
- $$I_2/E_{\rm B} \hspace{0.15cm}\underline {= +1}, $$
- $$I_4/E_{\rm B} \hspace{0.15cm}\underline {= -3}, $$
- $$I_6/E_{\rm B}\hspace{0.15cm}\underline { = -1} \hspace{0.05cm}.$$

**(2)** Only the __second solution__ is correct:

- In the presence of noises, the functions $i_i(t)$ no longer increase or decrease linearly, but have a curve as shown in the right graph.
- As long as $I_3 > I_{\it i≠3}$, the correlation receiver decides correctly.
- In the presence of noise, $I_0 ≠ I_6$ always holds, in contrast to the noise-free tree diagram.

**(3)** Only the __second statement__ is true:

- Since now the possible transmitted signals $s_i(t)$ can no longer be composed of isolated horizontal sections, also the tree diagram without noise does not consist of straight line segments.
- Since the energies $E_i$ are different – this can be seen e.g. by comparing the (red) signals $s_0(t)$ and $s_2(t)$ – it is essential to use the corrected quantities $W_i$ for the decision.
- The use of the uncorrected correlation values $I_i$ can already lead to wrong decisions without noise disturbances.

**(4)** __Answer 1__ is correct:

- In the case
__without intersymbol interference__(blue rectangular signals), all signals are limited to the range $0 \ ... \ 3T$. - Outside this range the received signal $r(t)$ is pure noise. Therefore in this case also the integration over the range $0 \ \text{...} \ 3T$.
- In contrast, when intersymbol interference (red signals) is taken into account, the integrands $s_3(t) \cdot s_i(t)$ also differ outside this range.
- Therefore, if $t_1 = \ -T$ and $t_2 = +4T$ are chosen, the error probability of the correlation receiver is further reduced compared to the integration range $0 \ \text{...} \ 3T$.