Aufgaben:Exercise 3.10: Maximum Likelihood Tree Diagram

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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

1

Give the following normalized final values  $I_i/E_{\rm B}$  for rectangular signals (without noise).

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

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

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

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

2

Which statements are valid when considering a noise term?

	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.

3

Which statements are valid for the red signal waveforms (with intersymbol interference)?

	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.

4

How should the intergration range  $(t_1 \ \text{...} \ t_2)$  be chosen?

	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

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.