Exercise 3.10: Maximum Likelihood Tree Diagram

Signals and tree diagram

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 transmitted. 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

$I_0/E_{\rm B} \ = \ $

$I_2/E_{\rm B} \ = \ $

$I_4/E_{\rm B} \ = \ $

$I_6/E_{\rm B} \ = \ $

	The tree diagram can be further described by straight line segments.
	If $I_3$ is the maximum $I_i$ value, the receiver decides correctly.
	$I_0 = I_6$ is valid independent of the strength of the noise.

	The tree diagram can be further described by straight line segments.
	The signal energies $E_i(i = 0, \ \text{...} \ , 7$) are different.
	Both the decision variables $I_i$ and $W_i$ are suitable.

	Without intersymbol interference (blue), $t_1 = 0$ and $t_2 = 3T$ are best possible.
	With intersymbol interference (red), $t_1 = 0$ and $t_2 = 3T$ are best possible.

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.

Tree diagram of the correlation receiver

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 (noise) disturbances, 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, for example, 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 pure 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 $0 \ \text{...} \ 3T$ is further reduced.