Aufgaben:Exercise 2.7: AMI Code - Revision history

Hwang at 15:11, 25 May 2022

2022-05-25T15:11:32Z

Guenter at 16:19, 19 May 2022

2022-05-19T16:19:56Z

Guenter at 15:54, 19 May 2022

2022-05-19T15:54:52Z

Guenter at 15:54, 19 May 2022

2022-05-19T15:54:07Z

Hwang at 15:51, 16 May 2022

2022-05-16T15:51:40Z

Hwang at 15:50, 16 May 2022

2022-05-16T15:50:03Z

Guenter at 15:13, 9 May 2022

2022-05-09T15:13:17Z

Guenter at 14:20, 5 May 2022

2022-05-05T14:20:10Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.7: AMI-Code to Aufgaben:Exercise 2.7: AMI Code

2022-05-05T14:14:15Z

Guenter moved page Aufgabe 2.7: AMI-Code to Exercise 2.7: AMI Code

Hwang at 19:03, 22 April 2022

2022-04-22T19:03:03Z

Show changes

← Older revision		Revision as of 15:11, 25 May 2022
Line 1:		Line 1:

−	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/~~Symbol~~-~~Wise_Coding_with_Pseudo_Ternary_Codes~~	+	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes
	}}		}}

@@ Line 80: / Line 80: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The modulo-2 addition can also be taken as an antivalence.
+'''(1)'''&nbsp; The modulo-2 addition can also be taken as an&nbsp; "antivalence".
-*It is $b_{\nu} = +1$ if $q_{\nu}$ and $b_{\nu – 1}$ differ, otherwise set $b_{\nu} = -1$.
+*It is&nbsp; $b_{\nu} = +1$&nbsp; if&nbsp; $q_{\nu}$&nbsp; and&nbsp; $b_{\nu – 1}$&nbsp; differ,&nbsp; otherwise set&nbsp; $b_{\nu} = -1$.
-*With the initial value $b_{0} = -1$ we obtain:
+*With the initial value&nbsp; $b_{0} = -1$&nbsp; we obtain:
 :$$b_1\hspace{0.15cm}\underline { = +1}, \hspace{0.2cm} b_2 = +1, \hspace{0.2cm}b_3 = -1, \hspace{0.2cm}b_4 = +1, \hspace{0.2cm}b_5 = +1, \hspace{0.2cm}b_6 = -1\hspace{0.05cm},$$
 :$$b_7 = +1, \hspace{0.2cm} b_8 = +1, \hspace{0.2cm}b_9 = +1, \hspace{0.2cm}b_{10} = +1, \hspace{0.2cm}b_{11} \hspace{0.15cm}\underline {= +1}, \hspace{0.2cm}b_{12} \hspace{0.15cm}\underline {= -1}\hspace{0.05cm}.$$
@@ Line 90: / Line 90: @@
 :$$a_1\hspace{0.15cm}\underline { = +1}, \hspace{0.2cm} a_2 = 0, \hspace{0.2cm}a_3 = -1, \hspace{0.2cm}a_4 = +1, \hspace{0.2cm}a_5 = 0, \hspace{0.2cm}a_6 = -1\hspace{0.05cm},$$
 :$$a_7 = +1, \hspace{0.2cm} a_8 = 0, \hspace{0.2cm}a_9 = 0, \hspace{0.2cm}a_{10} = 0, \hspace{0.2cm}a_{11}\hspace{0.15cm}\underline { = 0}, \hspace{0.2cm}a_{12} \hspace{0.15cm}\underline {= -1}\hspace{0.05cm}.$$
-This result is obtained by the equation $a_{\nu} = (b_{\nu} - b_{\nu –1})/2$ or by direct application of the AMI coding rule:
+This result is obtained by the equation&nbsp; $a_{\nu} = (b_{\nu} - b_{\nu –1})/2$&nbsp; or by direct application of the AMI coding rule:
-*A source symbol $q_{\nu} = -1$ always leads to $a_{\nu} = 0$.
+*A source symbol&nbsp; $q_{\nu} = -1$&nbsp; always leads to&nbsp; $a_{\nu} = 0$.
-*Source symbols $q_{\nu} = +1$ lead alternately to $a_{\nu} = +1$ and $a_{\nu} = -1$.
+*A source symbols&nbsp; $q_{\nu} = +1$&nbsp; lead alternately to&nbsp; $a_{\nu} = +1$&nbsp; and&nbsp; $a_{\nu} = -1$.
 '''(4)'''&nbsp; Assuming equally probable binary values $±1$, we obtain ${\Pr}(a_{\nu} = 0) = {\Pr}(q_{\nu} = -1)\hspace{0.15cm}\underline{ = 1/2}$ and for symmetry reasons
 : ${\Pr}(a_{\nu} = +1) = {\Pr}(a_{\nu} = -1) \hspace{0.15cm}\underline{ = 1/4}.$
 '''(5)'''&nbsp; Using the probabilities calculated in '''(4)''', we obtain:
 :$${\rm E}\big[a_\nu \big] = \ {1}/{4} \cdot (+1) +{1}/{2} \cdot 0+ {1}/{4} \cdot (-1)\hspace{0.15cm}\underline {= 0}\hspace{0.05cm},$$
 :$$ {\rm E}\big[a_\nu^2 \big] = \ {1}/{4} \cdot (+1)^2 +{1}/{2} \cdot 0^2 + {1}/{4} \cdot (-1)^2 \hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$
-'''(6)'''&nbsp; The ACF value at $\lambda = 0$ is equal to the root mean square of the amplitude coefficients:
 :$$ \varphi_a(\lambda = 0) = {\rm E}[a_\nu^2] \hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$
-*Since the order of the AMI code is $N = 1$, for $\lambda > 1$: &nbsp; $\varphi_a(\lambda > 1) = {\rm E}[a_\nu^2] \hspace{0.15cm}\underline {= 0} \hspace{0.05cm}.$
+*Since the order of the AMI code is&nbsp; $N = 1$, &nbsp; for $\lambda > 1$: &nbsp; $\varphi_a(\lambda > 1) = {\rm E}[a_\nu^2] \hspace{0.15cm}\underline {= 0} \hspace{0.05cm}.$
-*The ACF value $\varphi_{a}(\lambda = 1)$ must be determined by averaging:  &nbsp; $\varphi_a(\lambda = 1) = {\rm E}[a_\nu \cdot a_{\nu+1} \cdot {\rm Pr}(a_\nu \cap a_{\nu+1})] \hspace{0.05cm}.$
+*The ACF value&nbsp; $\varphi_{a}(\lambda = 1)$&nbsp; must be determined by averaging:  &nbsp; $\varphi_a(\lambda = 1) = {\rm E}[a_\nu \cdot a_{\nu+1} \cdot {\rm Pr}(a_\nu \cap a_{\nu+1})] \hspace{0.05cm}.$
-*Of the nine possible combinations for $a_{\nu} \cdot a_{\nu +1}$, only four yield a nonzero value. In the other cases, either $a_{\nu} = 0$ or $a_{\nu +1} = 0$.
+*Of the nine possible combinations for&nbsp; $a_{\nu} \cdot a_{\nu +1}$,&nbsp; only four yield a non-zero value.&nbsp; In the other cases,&nbsp; either&nbsp; $a_{\nu} = 0$&nbsp; or&nbsp; $a_{\nu +1} = 0$.
+*However,&nbsp; since in AMI code also
 :$${\rm Pr}[(a_\nu = +1) \cap (a_{\nu+1}= +1)]  = \ 0 \hspace{0.05cm},$$
 :$$ {\rm Pr}[(a_\nu = -1) \cap (a_{\nu+1}= -1)]  = \ 0$$
-is true,
+:is true,&nbsp; one obtains as the final result&nbsp; (since the ACF is always an even function):
 :$${\rm Pr}[(a_\nu = +1) \cap (a_{\nu+1}= -1)] = \ {\rm Pr}(a_\nu = +1)\cdot {\rm Pr}(a_{\nu+1} = -1 | a_\nu = +1) = {1}/{4}\cdot{1}/{2} ={1}/{8} \hspace{0.05cm},$$
 :$${\rm Pr}[(a_\nu = -1) \cap (a_{\nu+1}= +1)] = \ {\rm Pr}(a_\nu = -1)\cdot {\rm Pr}(a_{\nu+1} = +1 | a_\nu = -1) =  {1}/{4}\cdot {1}/{2} = {1}/{8}$$
 [[File:P_ID1353__Dig_A_2_7f.png|right|frame|Auto-correlation functions of the AMI code]]
-is obtained as the final result (since the ACF is always an even function):
-:$$\varphi_{a}(\lambda = +1) = \varphi_{a}(\lambda = –1) = –0.25.$$
+:$$\Rightarrow \hspace{0.3cm} \varphi_{a}(\lambda = +1) = \varphi_{a}(\lambda = -1) = -0.25.$$
-*This takes into account that $a_{\nu} = +1$ is followed by $a_{\nu +1} = +1$ and $a_{\nu +1} = -1$ with equal probability.
+*This takes into account that&nbsp; $a_{\nu} = +1$&nbsp; is followed by&nbsp; $a_{\nu +1} = +1$&nbsp; and&nbsp; $a_{\nu +1} = -1$&nbsp; with equal probability.&nbsp; Thus,&nbsp; the result is:
 :$$\varphi_a(\lambda = 0)\hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}, $$
 :$$\varphi_a(\lambda = 1)\hspace{0.15cm}\underline {= -0.25} \hspace{0.05cm},$$
 :$$\varphi_a(\lambda = 2)\hspace{0.15cm}\underline {= 0}.$$
-<br clear=all>
 The graph shows
-*the discrete ACF $\varphi_{a}(\lambda)$ of the amplitude coefficients and
+#the discrete ACF&nbsp; $\varphi_{a}(\lambda)$&nbsp; of the amplitude coefficients,
-*the ACF $\varphi_{s}(\tau)$ of the transmitted signal under the condition of NRZ rectangular pulses and AMI coding.
+#the ACF&nbsp; $\varphi_{s}(\tau)$&nbsp; of the transmitted signal under the condition of NRZ rectangular pulses and AMI coding.
-'''(7)'''&nbsp; From the given equation, taking into account the discrete ACF values calculated in '''(6)'''
+'''(7)'''&nbsp; From the given equation,&nbsp; taking into account the discrete ACF values calculated in&nbsp; '''(6)''',
 :$$\varphi_{a}(\lambda = 0) = 1/2,$$
 :$$\varphi_{a}(|\lambda| = 1) = -1/4,$$
-:$$\varphi_{a}(|\lambda| > 1) = 0$$
+:$$\varphi_{a}(|\lambda| > 1) = 0,$$
 we obtain the following result:
 :$${\it \Phi}_a(f) = \ \sum_{\lambda = -\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda T} = \varphi_a(\lambda = 0) + 2 \cdot \varphi_a(\lambda = 1 )\cdot\cos ( 2 \pi f \hspace{0.02cm} \lambda T) = \ {1}/{2} \cdot \left [ 1 - \cos ( 2 \pi f \hspace{0.02cm} T)\right ] = \sin^2 ( \pi f \hspace{0.02cm} T) \hspace{0.05cm}.$$
-In particular, holds:
+In particular holds:
 :$${\it \Phi}_a(f = 0) \hspace{0.15cm}\underline {= 0},$$
 :$${\it \Phi}_a(f = {1}/({2T})) = \sin^2 ({\pi}/{2})\hspace{0.15cm}\underline {= 1} \hspace{0.05cm}.$$

@@ Line 27: / Line 27: @@
 Notes:
-*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Symbol-Wise_Coding_with_Pseudo_Ternary_Codes|"Symbolwise Coding with Pseudo-Ternary Codes"]].
+*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes|"Symbolwise Coding with Pseudo-Ternary Codes"]].
 *You can check the results with the&nbsp; (German language)&nbsp; SWF applet &nbsp;[[Applets:Pseudoternaercodierung|"Signals, ACF and PSD of pseudo-ternary codes"]].&nbsp;

@@ Line 4: / Line 4: @@
-[[File:EN_Dig_A_2_7.png|right|frame|Block diagram of a pseudo ternary encoder]]
+[[File:EN_Dig_A_2_7.png|right|frame|Block diagram of a <br>pseudo-ternary encoder]]
-The diagram shows the block diagram for AMI coding, assuming binary bipolar amplitude coefficients &nbsp;$q_{\nu} ∈ \{–1, +1\}$&nbsp; at the input. This recoding is done in two stages:
+The diagram shows the block diagram for AMI coding,&nbsp; assuming binary bipolar amplitude coefficients &nbsp;$q_{\nu} ∈ \{–1, +1\}$&nbsp; at the input.&nbsp; This encoding is done in two stages:
-*In the first part of the block diagram, a binary precoded symbol &nbsp;$b_{\nu}$&nbsp; is generated at each clock step from the modulo-2 addition of &nbsp;$q_{\nu}$&nbsp; and &nbsp;$b_{\nu -1}$.&nbsp; It holds &nbsp;$b_{\nu} ∈ \{–1, +1\}.$
+#In the first part of the block diagram,&nbsp; a binary pre-encoded symbol &nbsp;$b_{\nu}$&nbsp; is generated at each clock from the modulo-2 addition of &nbsp;$q_{\nu}$&nbsp; and &nbsp;$b_{\nu -1}$.&nbsp; It holds &nbsp;$b_{\nu} ∈ \{–1, +1\}.$<br>
-*Then, the current amplitude coefficient of the ternary transmitted signal &nbsp;$s(t)$&nbsp; is determined by a conventional subtraction. Thereby holds:
+#Then,&nbsp; the current amplitude coefficient of the ternary transmitted signal &nbsp;$s(t)$&nbsp; is determined by a conventional subtraction.&nbsp; Thereby holds:
-:$$a_\nu = {1}/{2} \cdot \left [ b_\nu - b_{\nu-1} \right ] \hspace{0.05cm}.$$
+::$$a_\nu = {1}/{2} \cdot \big [ b_\nu - b_{\nu-1} \big ] \hspace{0.05cm}.$$
-Due to AMI coding, it is ensured that no long "$+1$"– or "$–1$" sequences are created. Modified AMI codes have also been developed to avoid long zero sequences:
+Due to AMI coding,&nbsp; it is ensured that the AMI encoder signal does not contain these two "long sequences"<br>
-*In the HDB3 code, four consecutive zeros each are marked by a specific violation of the AMI coding rule.
+&nbsp; &nbsp;  &nbsp; &nbsp; $ \langle c_\nu \rangle =  \langle \text{...}, +1, +1, +1, +1, +1, \text{...}\rangle$ &nbsp; resp.
 *In the B6ZS code, six consecutive zeros are marked by a targeted violation of the AMI coding rule.
@@ Line 20: / Line 26: @@
 Notes:
+*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Symbol-Wise_Coding_with_Pseudo_Ternary_Codes|"Symbolwise Coding with Pseudo-Ternary Codes"]].
-*You can check the results with the interactive applet &nbsp;[[Applets:Pseudoternaercodierung|"Signals, ACF and PSD of pseudo ternary codes"]].&nbsp;
+*You can check the results with the&nbsp; (German language)&nbsp; SWF applet &nbsp;[[Applets:Pseudoternaercodierung|"Signals, ACF and PSD of pseudo-ternary codes"]].&nbsp;
@@ Line 31: / Line 35: @@
 <quiz display=simple>
-{At the input, &nbsp;$\langle q_{\nu} \rangle = \langle +1, –1, +1, +1, –1, +1, +1, –1, –1, –1, –1, +1 \rangle$&nbsp; is applied. Determine the binary precoded sequence &nbsp;$\langle b_{\nu} \rangle$&nbsp; with the default $b_{0} = \hspace{0.05cm}–1$. <br>Enter the following values as a check:
+{At the input, &nbsp;$\langle q_{\nu} \rangle = \langle +1, –1, +1, +1, –1, +1, +1, –1, –1, –1, –1, +1 \rangle$&nbsp; is applied. <br>Determine the binary pre-encoded sequence &nbsp;$\langle b_{\nu} \rangle$&nbsp; with the default $b_{0} = \hspace{0.05cm}-1$.&nbsp; Enter the following values as a check:
 |type="{}"}
 $b_{1} \hspace{0.26cm} = \ $ { 1 3% }
@@ Line 37: / Line 41: @@
 $b_{12} \ = \ $ { -1.03--0.97 }
-{Furthermore, determine the sequence &nbsp;$\langle a_{\nu} \rangle$&nbsp; of the amplitude coefficients of the AMI-coded transmitted signal &nbsp;$s(t)$. <br>Enter the following values to check the results:
+{Furthermore, determine the sequence &nbsp;$\langle a_{\nu} \rangle$&nbsp; of the amplitude coefficients of the AMI-encoded transmitted signal &nbsp;$s(t)$. <br>Enter the following values to check the results:
 |type="{}"}
 $a_{1} \hspace{0.28cm} = \ $ { 1 3% }
@@ Line 60: / Line 64: @@
 $\E\big[a_{\nu}^{2}\big] \ = \ $ { 0.5 3% }
-{Calculate the auto-correlation function &nbsp;$\varphi_{a}(\lambda)$, in particular the following ACF values:
+{Calculate the auto-correlation function &nbsp;$\varphi_{a}(\lambda)$,&nbsp; in particular the following ACF values:
 |type="{}"}
 $\varphi_{a}(\lambda = 0) \ = \ $ { 0.5 3% }
@@ Line 66: / Line 70: @@
 $\varphi_{a}(\lambda = 2) \ = \ $ { 0. }
-{What is the power-spectral density &nbsp;${\it \Phi}_{a}(f)$? What are the values for &nbsp;$f = 0$&nbsp; and &nbsp;$f = 1/(2T)$?
+{What is the power-spectral density &nbsp;${\it \Phi}_{a}(f)$?&nbsp; What are the values for &nbsp;$f = 0$&nbsp; and &nbsp;$f = 1/(2T)$?
 |type="{}"}
 ${\it \Phi}_{a}(f = 0) \ = \ $ { 0. }

@@ Line 23: / Line 23: @@
 ''Notes:''
-*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Symbol-Wise_Coding_with_Pseudo_Ternary_Codes|"Symbol-Wise Coding with Pseudo Ternary Codes"]].
+*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Symbol-Wise_Coding_with_Pseudo_Ternary_Codes|"Symbolwise Coding with Pseudo Ternary Codes"]].
 *You can check the results with the interactive applet &nbsp;[[Applets:Pseudoternaercodierung|"Signals, ACF and PSD of pseudo ternary codes"]].&nbsp;

← Older revision		Revision as of 14:20, 5 May 2022
Line 153:		Line 153:


−	[[Category:Digital Signal Transmission: Exercises\|^2.4 ~~Pseudoternärcodes~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^2.4 Pseudo-Ternary Codes^]]