Difference between revisions of "Aufgaben:Exercise 3.11: Viterbi Receiver and Trellis Diagram"
(19 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Digital_Signal_Transmission/Viterbi_Receiver}} |
− | [[File:P_ID1475__Dig_A_3_11.png|right|frame| | + | [[File:P_ID1475__Dig_A_3_11.png|right|frame|Trellis diagram <br>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 [[Aufgaben:Exercise_3.11Z:_Maximum_Likelihood_Error_Variables|Exercise 3.11Z]]. | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | === | + | |
+ | Notes: | ||
+ | *The exercise belongs to the chapter [[Digital_Signal_Transmission/Viterbi_Receiver|"Viterbi Receiver"]]. | ||
+ | |||
+ | *Reference is also made to the section [[Digital_Signal_Transmission/Optimal_Receiver_Strategies#Maximum-a-posteriori_and_maximum.E2.80.93likelihood_decision_rule|"Maximum-a-posteriori and Maximum–Likelihood decision rule"]]. | ||
+ | |||
+ | * All quantities here are to be understood normalized. Also assume unipolar and equal probability amplitude coefficients: | ||
+ | :$${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$$ | ||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which of the following statements are true? |
|type="[]"} | |type="[]"} | ||
− | + | + | + 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 $H_{\rm DF}(f)$. |
− | { | + | {At what times $\nu$ can we finally decide the current symbol $a_{\rm \nu}$? |
|type="[]"} | |type="[]"} | ||
+ $\nu = 1,$ | + $\nu = 1,$ | ||
Line 38: | Line 43: | ||
+ $\nu = 5.$ | + $\nu = 5.$ | ||
− | { | + | {What is the total sequence decided by the Viterbi receiver? |
|type="{}"} | |type="{}"} | ||
$a_1 \ = \ $ { 0. } | $a_1 \ = \ $ { 0. } | ||
Line 46: | Line 51: | ||
$a_5 \ = \ $ { 0. } | $a_5 \ = \ $ { 0. } | ||
− | { | + | {Which of the following statements are true? |
|type="[]"} | |type="[]"} | ||
− | - | + | - 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. |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' The <u>first two solutions</u> are correct: |
+ | *The signal $m(t)$ after the matched filter $H_{\rm MF}(f)$ has the largest possible signal-to-interference power ratio $\rm (SNR)$. | ||
+ | * 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 bindings, 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 times $\underline {\nu = 3}$ 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)''' | + | '''(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_{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} | a_{2}\hspace{0.15cm}\underline { =0} \hspace{0.05cm},\hspace{0.2cm}a_{3}\hspace{0.15cm}\underline {=1} | ||
Line 70: | Line 84: | ||
− | '''(4)''' | + | '''(4)''' Only the <u>second statement</u> 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 BER 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 false. Indeed, this would mean that the Viterbi receiver can have error probability $0$ in the presence of AWGN noise. <br>This is not possible for information-theoretic reasons. | |
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Digital Signal Transmission: Exercises|^3.8 Viterbi Receiver^]] |
Latest revision as of 14:00, 5 July 2022
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:
- $${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$$
Questions
Solution
- The signal $m(t)$ after the matched filter $H_{\rm MF}(f)$ has the largest possible signal-to-interference power ratio $\rm (SNR)$.
- 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 bindings, 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 times $\underline {\nu = 3}$ 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 BER 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 false. Indeed, this would mean that the Viterbi receiver can have error probability $0$ in the presence of AWGN noise.
This is not possible for information-theoretic reasons.