Exercise 3.11: Viterbi Receiver and Trellis Diagram

Trellis diagram
for one precursor

The Viterbi receiver allows a low-effort realization of the maximum likelihood decision rule. It contains the system components listed below:

A matched filter adapted to the basic transmission pulse with frequency response $H_{\rm MF}(f)$ and output signal $m(t)$,

a sampler spaced at the symbol duration $T$, which converts the continuous-time signal $m(t)$ into the discrete-time sequence $〈m_{\rm \nu}〉$,

a decorrelation filter with frequency response $H_{\rm DF}(f)$ for removing statistical ties between noise components of the sequence $〈d_{\rm \nu}〉$,

the Viterbi decision, which uses a trellis-based algorithm to obtain the sink symbol sequence $〈v_{\rm \nu}〉$.

The graph shows the simplified trellis diagram of the two states "$0$" and "$1$" for time points $\nu ≤ 5$. This diagram is obtained as a result of evaluating the two minimum accumulated metrics ${\it \Gamma}_{\rm \nu}(0)$ and ${\it \Gamma}_{\rm \nu}(1)$ corresponding to Exercise 3.11Z.

Notes:

The exercise belongs to the chapter "Viterbi Receiver".

Reference is also made to the section "Maximum-a-posteriori and Maximum–Likelihood decision rule".

All quantities here are to be understood normalized. Also assume unipolar and equal probability amplitude coefficients:

$${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$$

The topic is also covered with the interactive SWF applet "Properties of the Viterbi Receiver".

Questions

Solution

(1) The first two solutions are correct:

The signal $m(t)$ after the matched filter $H_{\rm MF}(f)$ has the largest possible signal-to-interference power ratio.
However, the noise components of the sequence $〈m_{\rm \nu}〉$ are (strongly) correlated due to the spectral shaping.
The task of the discrete-time decorrelation filter with the frequency response $H_{\rm DF}(f)$ is to dissolve these bonds, which is why the name "whitening filter" is also used for $H_{\rm DF}(f)$.
However, this is possible only at the cost of increased noise power ⇒ consequently, the last proposed solution does not apply.

(2) The two arrows arriving at $\underline {\nu = 1}$ are each drawn in blue and indicate the symbol $a_1 = 0$. Thus, the initial symbol $a_1$ is already fixed at this point. Similarly, the symbols $a_3 = 1$ and $a_5 = 0$ are already fixed at the timesn $\underline {\nu = 3}$ and $\underline {\nu = 5}$, respectively.

In contrast, at time $\nu = 2$, a decision regarding symbol $a_2$ is not possible.

Under the hypothesis that the following symbol $a_3 = 0$ would result in symbol $a_2 = 1$ (at "$0$" a red path arrives, thus coming from "$1$").
In contrast, the hypothesis $a_3 = 1$ leads to the result $a_2 = 0$ (the path arriving at "$1$" is blue).

The situation is similar at time $\nu = 4$.

(3) From the continuous paths at $\nu = 5$ it can be seen:

$$a_{1}\hspace{0.15cm}\underline {=0} \hspace{0.05cm},\hspace{0.2cm} a_{2}\hspace{0.15cm}\underline { =0} \hspace{0.05cm},\hspace{0.2cm}a_{3}\hspace{0.15cm}\underline {=1} \hspace{0.05cm},\hspace{0.2cm} a_{4}\hspace{0.15cm}\underline {=0} \hspace{0.05cm},\hspace{0.2cm} a_{5}\hspace{0.15cm}\underline {=0} \hspace{0.05cm}.$$

(4) Only the second statement is correct:

Since the source symbols "$0$" and "$1$" were assumed to be equally probable, the ML receiver (Viterbi) is identical to the MAP receiver.
A threshold decision (which makes a symbol-by-symbol decision at each clock) has the same error probability as the Viterbi receiver only if there is no intersymbol interference.
This is obviously not the case here, otherwise it should be possible to make a final decision at every time $\nu$.
The first statement is also not true. Indeed, this would mean that the Viterbi receiver can have error probability $0$ in the presence of statistical noise. This is not possible for information-theoretic reasons.

	$\nu = 1,$
	$\nu = 2,$
	$\nu = 3,$
	$\nu = 4,$
	$\nu = 5.$

	The matched filter $H_{\rm MF}(f)$ is mainly used for noise power limitation.
	The decorrelation filter removes ties between samples.
	The noise power is only affected by $H_{\rm MF}(f)$, not by $H_{\rm DF}(f)$.

	It is certain that the detected sequence was also sent.
	A MAP receiver would have the same error probability.
	Threshold decision is the same as this maximum likelihood receiver.