Aufgaben:Exercise 3.10: Maximum Likelihood Tree Diagram - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^3.7 Optimal Receiver Strategies^]]		[[Category:Digital Signal Transmission: Exercises\|^3.7 Optimal Receiver Strategies^]]
	[[de:Aufgaben:3.~~10_Baumdiagramm_bei_Maximum~~-Likelihood]]		[[de:Aufgaben:Aufgabe 3.10: Baumdiagramm bei Maximum-Likelihood]]

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

	*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 correlation receiver compares the final values  $I_i = i_i(3T)$  with each other and searches for the largest possible value  $I_j$.
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	:$$I_2/E_{\rm B} \hspace{0.15cm}\underline {= +1}, $$		:$$I_2/E_{\rm B} \hspace{0.15cm}\underline {= +1}, $$
	:$$I_4/E_{\rm B} \hspace{0.15cm}\underline {= -3}, $$		:$$I_4/E_{\rm B} \hspace{0.15cm}\underline {= -3}, $$
	:$$I_6/E_{\rm B}\hspace{0.15cm}\underline { = -1}		:$$I_6/E_{\rm B}\hspace{0.15cm}\underline { = -1}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^3.7 Optimal Receiver Strategies^]]		[[Category:Digital Signal Transmission: Exercises\|^3.7 Optimal Receiver Strategies^]]
			[[de:Aufgaben:3.10_Baumdiagramm_bei_Maximum-Likelihood]]

Guenter at 16:24, 30 June 2022

2022-06-30T16:24:26Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(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 $+1$ &~~ndash~~; the first bit of $s_0(t)$ and $s_3(t)$ in each case coincide &~~ndash~~; and then falls off over two bit durations.		'''(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.
	[[File:EN_Dig_A_3_10_ML.png\|right\|frame\|Tree diagram of the correlation receiver]]		[[File:EN_Dig_A_3_10_ML.png\|right\|frame\|Tree diagram of the correlation receiver]]
	The correct results are thus:
			*The correct results are thus:
	:$$I_0/E_{\rm B}\hspace{0.15cm}\underline { = -1},$$		:$$I_0/E_{\rm B}\hspace{0.15cm}\underline { = -1},$$
	:$$I_2/E_{\rm B} \hspace{0.15cm}\underline {= +1}, $$		:$$I_2/E_{\rm B} \hspace{0.15cm}\underline {= +1}, $$
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	:$$I_6/E_{\rm B}\hspace{0.15cm}\underline { = -1}		:$$I_6/E_{\rm B}\hspace{0.15cm}\underline { = -1}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	~~<br clear=all>~~
	~~'''(2)'''  Only the <u>second solution</u> 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 <u>second statement</u> is true:		'''(2)'''  Only the  <u>second solution</u>  is correct:
	*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.		*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.
	*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.		*As long as  $I_3 > I_{\it i≠3}$,  the correlation receiver decides correctly.
	*The use of the ~~pure~~ correlation values $I_i$ can already lead to wrong decisions without noise disturbances.		*In the presence of noise,  $I_0 ≠ I_6$  always holds,  in contrast to the noise-free tree diagram.


			'''(3)'''  Only the  <u>second statement</u>  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)'''  <u>Answer 1</u> is correct:		'''(4)'''  <u>Answer 1</u>  is correct:
	*In the case <u>without intersymbol interference</u> (blue rectangular signals), all signals are limited to the range $0 \ ... \ 3T$.		*In the case  <u>without intersymbol interference</u>  (blue rectangular signals),  all signals are limited to the range  $0 \ ... \ 3T$.
	*Outside this range the received signal $r(t)$ is pure noise.		*Outside this range the received signal  $r(t)$  is pure noise.  Therefore in this case also the integration over the range  $0 \ \text{...} \ 3T$.
	*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.
	*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  $0 \ \text{...} \ 3T$.
	*Therefore, if $t_1 = \ &~~ndash~~;T$ 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~~.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 16:07, 30 June 2022

2022-06-30T16:07:46Z

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	[[File:P_ID1465__Dig_A_3_10_95.png\|right\|frame\|Signals and tree diagram]]		[[File:P_ID1465__Dig_A_3_10_95.png\|right\|frame\|Signals and tree diagram]]
	As in  [[Aufgaben:Exercise_3.09:_Correlation_Receiver_for_Unipolar_Signaling\|"Exercise 3.9"]]  we consider the joint decision of three binary symbols (bits) by means of the correlation receiver.		As in  [[Aufgaben:Exercise_3.09:_Correlation_Receiver_for_Unipolar_Signaling\|"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.		*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.		*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.		*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		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
	:$$i_i(t) = \int_{0}^{t} s_3(\tau) \cdot s_i(\tau) \,{\rm d}		:$$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}.$$		\tau \hspace{0.3cm}( i = 0, \ \text{...} \ , 7)\hspace{0.05cm}.$$

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

	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		:the integrals  $I_i$  provide exactly the same maximum likelihood information as the corrected quantities  $W_i$.
	:~~$$E_i = \int_{0}^{3T} s_i^2(t) \,{\rm d} t$$~~

	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  $f_{\rm G} \cdot T = 0.35$.		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.		*Each individual rectangular pulse is broadened.
	*The red signal waveforms lead to intersymbol interference in case of threshold decision.




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


	''Note:''		Note:  The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimal_Receiver_Strategies\|"Optimal Receiver Strategies"]].
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Optimal_Receiver_Strategies\|"Optimal Receiver Strategies"]].


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Give the following normalized final values  $I_i/E_{\rm B}$  for rectangular signals (without noise).		{Give the following normalized final values  $I_i/E_{\rm B}$  for rectangular signals  $($without noise$)$.
	\|type="{}"}		\|type="{}"}
	$I_0/E_{\rm B} \ = \ $ { -1.03--0.97 }		$I_0/E_{\rm B} \ = \ $ { -1.03--0.97 }
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	\|type="[]"}		\|type="[]"}
	- The tree diagram can be further described by straight line segments.		- The tree diagram can be further described by straight line segments.
	+ If  $I_3$  is the maximum $I_i$ value, the receiver decides correctly.		+ 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.		-  $I_0 = I_6$  is valid independent of the strength of the noise.

	{Which statements are valid for the red signal waveforms (with intersymbol interference)?		{Which statements are valid for the red signal waveforms  $($with intersymbol interference$)$?
	\|type="[]"}		\|type="[]"}
	- The tree diagram can be further described by straight line segments.		- The tree diagram can be further described by straight line segments.

Guenter at 15:09, 28 June 2022

2022-06-28T15:09:25Z

← Older revision		Revision as of 17:09, 28 June 2022
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(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.		'''(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.
	[[File:~~P_ID1466__Dig_A_3_10~~.png\|right\|frame\|Tree diagram of the correlation receiver]]		[[File:EN_Dig_A_3_10_ML.png\|right\|frame\|Tree diagram of the correlation receiver]]
	The correct results are thus:		The correct results are thus:
	:$$I_0/E_{\rm B}\hspace{0.15cm}\underline { = -1},$$		:$$I_0/E_{\rm B}\hspace{0.15cm}\underline { = -1},$$

Hwang at 11:32, 31 May 2022

2022-05-31T11:32:47Z

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Guenter at 08:56, 18 May 2022

2022-05-18T08:56:46Z

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	[[Category:Digital Signal Transmission: Exercises\|^3.7 ~~Optimale Empfängerstrategien~~^]]		[[Category:Digital Signal Transmission: Exercises\|^3.7 Optimal Receiver Strategies^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 3.10: Baumdiagramm bei Maximum-Likelihood to Aufgaben:Exercise 3.10: Maximum Likelihood Tree Diagram

2022-05-18T08:55:08Z

Guenter moved page Aufgaben:Aufgabe 3.10: Baumdiagramm bei Maximum-Likelihood to Aufgaben:Exercise 3.10: Maximum Likelihood Tree Diagram

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Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

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	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^3.7 Optimale Empfängerstrategien^]]		[[Category:Digital Signal Transmission: Exercises\|^3.7 Optimale Empfängerstrategien^]]