Aufgaben:Exercise 1.7: Coding for Broadband ISDN - Revision history

Guenter at 11:14, 10 November 2022

2022-11-10T11:14:34Z

Guenter at 14:54, 8 November 2022

2022-11-08T14:54:25Z

Guenter at 16:39, 4 November 2022

2022-11-04T16:39:48Z

Guenter at 16:38, 4 November 2022

2022-11-04T16:38:28Z

Guenter at 16:26, 4 November 2022

2022-11-04T16:26:18Z

Guenter: Guenter moved page Aufgaben:Exercise 1.7: Coding for B-ISDN to Aufgaben:Exercise 1.7: Coding for Broadband ISDN

2022-11-04T15:48:57Z

Guenter moved page Exercise 1.7: Coding for B-ISDN to Exercise 1.7: Coding for Broadband ISDN

Hwang at 16:33, 30 October 2022

2022-10-30T16:33:52Z

Show changes

Guenter at 15:12, 27 October 2022

2022-10-27T15:12:57Z

Guenter at 12:45, 10 September 2022

2022-09-10T12:45:17Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.7: Codierung bei B–ISDN to Aufgaben:Exercise 1.7: Coding for B-ISDN

2022-09-10T12:43:33Z

Guenter moved page Aufgabe 1.7: Codierung bei B–ISDN to Exercise 1.7: Coding for B-ISDN

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Bei_A_1_7_neu.png|right|frame|HDB3 and 1T2B coding '''Korrektur''']]
+[[File:EN_Bei_A_1_7_neu.png|right|frame|HDB3 and 1T2B coding]]
 For conventional ISDN over copper lines the&nbsp; $\rm HDB3$&nbsp; ("High Definition Bipolar")&nbsp; code is used &ndash; see&nbsp; [[Aufgaben:Exercise_1.5:_HDB3_Coding|$\text{Exercise 1.5}$]]:

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Bei_A_1_7.png|right|frame|HDB3 and 1T2B coding '''Korrektur''']]
+[[File:EN_Bei_A_1_7_neu.png|right|frame|HDB3 and 1T2B coding '''Korrektur''']]
 For conventional ISDN over copper lines the&nbsp; $\rm HDB3$&nbsp; ("High Definition Bipolar")&nbsp; code is used &ndash; see&nbsp; [[Aufgaben:Exercise_1.5:_HDB3_Coding|$\text{Exercise 1.5}$]]:

@@ Line 95: / Line 95: @@
-'''(5)'''&nbsp; The resulting redundancy of both codes is obtained by relating the given equation to the input signal&nbsp; $q(t)$&nbsp; and the output signal&nbsp; $c(t)$.&nbsp; *With&nbsp; $M_{q} = M_{b} = 2$&nbsp; and&nbsp; $T_{b} = T_{q}/2$&nbsp; it follows:
+'''(5)'''&nbsp; The resulting redundancy of both codes is obtained by relating the given equation to the input signal&nbsp; $q(t)$&nbsp; and the output signal&nbsp; $c(t)$.
 :$$r_{\rm HDB3+1T2B} = 1 - \frac{T_b \cdot {\rm log_2}\hspace{0.1cm}(M_q)}{T_q \cdot {\rm log_2}\hspace{0.1cm}(M_b)} = 1 - \frac{T_b}{T_q} \hspace{0.15cm}\underline{= 50\,\%} \hspace{0.05cm}.$$
 *The same result is obtained by the calculation

@@ Line 76: / Line 76: @@
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Solution 2</u> is correct, as a comparison of the signal characteristics $c(t)$ and $b(t)$ shows.
+'''(1)'''&nbsp; <u>Solution 2</u>&nbsp; is correct,&nbsp; as a comparison of the signal characteristics&nbsp; $c(t)$&nbsp; and&nbsp; $b(t)$&nbsp; shows.
-'''(2)'''&nbsp; The symbol duration (bit duration) of $q(t)$ is &nbsp; $T_{q} \hspace{0.15cm}\underline{ = 1/R_{q} = 0.488 \ \rm &micro; s}$.
+'''(2)'''&nbsp; The symbol duration&nbsp; $(=$ bit duration$)$&nbsp; of&nbsp; $q(t)$&nbsp; is &nbsp; $T_{q} \hspace{0.15cm}\underline{ = 1/R_{q} = 0.488 \ \rm &micro; s}$.
 '''(3)'''&nbsp; Using the given equation, with $M_{q} = 2, M_{c} = 3$ and $T_{c} = T_{q}$, we get:
 :$$r_{\rm HDB3} = 1 - \frac{T_c \cdot {\rm log_2}\hspace{0.1cm}(M_q)}{T_q \cdot {\rm log_2}\hspace{0.1cm}(M_c)} = 1 - \frac{1}{{\rm log_2}\hspace{0.1cm}(3)} \hspace{0.15cm}\underline{= 36.9\,\%} \hspace{0.05cm}.$$
-'''(4)'''&nbsp; Fitting the equation to the 1T2B code, we obtain with $M_{c} = 3, M_{b} = 2, T_{b} = T_{c}/2$:
+'''(4)'''&nbsp; Fitting the equation to the 1T2B code,&nbsp; we obtain with&nbsp; $M_{c} = 3, \hspace{0.15cm} M_{b} = 2,&nbsp; T_{b} = T_{c}/2$:
 :$$r_{\rm 1T2B} = 1 - \frac{T_b \cdot {\rm log_2}\hspace{0.1cm}(M_c)}{T_c \cdot {\rm log_2}\hspace{0.1cm}(M_b)} = 1 - \frac{{\rm log_2}\hspace{0.1cm}(3)}{2} \hspace{0.15cm}\underline{= 20.7\,\%} \hspace{0.05cm}.$$
-'''(5)'''&nbsp; The resulting redundancy of both codes is obtained by relating the given equation to the input signal $q(t)$ and the output signal $c(t)$. With $M_{q} = M_{b} = 2$ and $T_{b} = T_{q}/2$ it follows:
+'''(5)'''&nbsp; The resulting redundancy of both codes is obtained by relating the given equation to the input signal&nbsp; $q(t)$&nbsp; and the output signal&nbsp; $c(t)$.&nbsp; *With&nbsp; $M_{q} = M_{b} = 2$&nbsp; and&nbsp; $T_{b} = T_{q}/2$&nbsp; it follows:
 :$$r_{\rm HDB3+1T2B} = 1 - \frac{T_b \cdot {\rm log_2}\hspace{0.1cm}(M_q)}{T_q \cdot {\rm log_2}\hspace{0.1cm}(M_b)} = 1 - \frac{T_b}{T_q} \hspace{0.15cm}\underline{= 50\,\%} \hspace{0.05cm}.$$
-The same result is obtained by the calculation
+*The same result is obtained by the calculation
 :$$1-r_{\rm HDB3+1T2B} \ = \ (1-r_{\rm HDB3}) \cdot (1-r_{\rm 1T2B}) =(1- 1 +\frac{1}{{\rm log_2}\hspace{0.1cm}(3)}) \cdot (1-1+ \frac{{\rm log_2}\hspace{0.1cm}(3)}{2}) = 50\,\% \hspace{0.05cm}.$$
 :$$\Rightarrow \hspace{0.3cm}r_{\rm HDB3+1T2B}= 50\,\% \hspace{0.05cm}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Bei_A_1_7.png|right|frame|HDB3 and 1T2B coding]]
+[[File:EN_Bei_A_1_7.png|right|frame|HDB3 and 1T2B coding '''Korrektur''']]
-For conventional ISDN over copper lines the HDB3 code is used &ndash; see&nbsp; [[Aufgaben:Exercise_1.5:_HDB3_Coding|"Exercise 1.5"]]:
+For conventional ISDN over copper lines the&nbsp; $\rm HDB3$&nbsp; ("High Definition Bipolar")&nbsp; code is used &ndash; see&nbsp; [[Aufgaben:Exercise_1.5:_HDB3_Coding|$\text{Exercise 1.5}$]]:
-This was derived from the so-called AMI code,
+This was derived from the so-called&nbsp; "AMI code",
 *is like the latter a pseudo-ternary code,
 *but avoids more than three consecutive&nbsp; "$0$" symbols,
 *by deliberately violating the stricter AMI coding rule for longer zero sequences.
-The graph shows the HDB3 encoded signal&nbsp; $c(t)$ resulting from the binary redundancy-free source signal&nbsp; $q(t)$.&nbsp; Since there are no more than three consecutive zeros in the source signal,&nbsp; $c(t)$&nbsp; is identical to the AMI-encoded signal.
+The graph shows the HDB3 encoded signal&nbsp; $c(t)$&nbsp; resulting from the binary redundancy-free source signal&nbsp; $q(t)$.&nbsp; Since there are no more than three consecutive zeros in the source signal,&nbsp; $c(t)$&nbsp; is identical to the AMI-encoded signal.
-However, since the HDB3-encoded signal&nbsp; $c(t) ∈ \{–1, \ 0, +1\}$&nbsp; cannot be transmitted by means of light, a second encoding was required. The&nbsp; '''1T2B code'''&nbsp; provided for this purpose replaces each ternary symbol with two binary symbols. The diagram below shows an example of the binary signal&nbsp; $b(t) ∈ \{0, 1\}$, which results from the signal&nbsp; $c(t)$&nbsp; after this 1T2B coding.
+*secondly,&nbsp; the transmission medium had to be changed from copper to fiber optics.
-The equivalent bit rate of the pseudo-ternary signal&nbsp; $c(t)$&nbsp; is&nbsp; $R_{c} = {\rm log_2}(3)/T_{c}$, from which the bit rate&nbsp; $R_{q} = 1/T_{q}$&nbsp; of the source signal can be used to calculate the relative redundancy of the AMI or HDB3 code:
+However,&nbsp; since the HDB3-encoded signal&nbsp; $c(t) ∈ \{–1, \ 0, +1\}$&nbsp; cannot be transmitted by means of light,&nbsp; a second encoding was required.&nbsp;
 :$$r_{\rm HDB3} = \frac{R_c - R_q}{R_c}= 1 - \frac{T_c \cdot {\rm log_2}\hspace{0.1cm}(M_q)}{T_q \cdot {\rm log_2}\hspace{0.1cm}(M_c)} \hspace{0.05cm}.$$
-A similar equation can be established for the 1T2B code, as well as for the two codes in combination.
+A similar equation can be established for the 1T2B code,&nbsp; as well as for the two codes in combination.
 * Redundancy is defined and illustrated with examples in the chapter&nbsp; [[Digital_Signal_Transmission/Basics_of_Coded_Transmission|"Basics of Coded Transmission"]]&nbsp; of the book "Digital Signal Transmission".
@@ Line 50: / Line 53: @@
 - $c(t) = +1 \Rightarrow b(t) = 01, \hspace{1cm}c(t) = 0 \Rightarrow b(t) = 11, \hspace{1cm}c(t) = -1 \Rightarrow b(t) = 10.$
-{What are the symbol durations of&nbsp; $q(t),&nbsp; c(t)$&nbsp; and&nbsp; $b(t)$?
+{What are the symbol durations of&nbsp; $q(t), &nbsp; c(t)$&nbsp; and &nbsp; $b(t)$?
 |type="{}"}
 $T_{q} \ = \ $ { 0.488 3% } $\ \rm &micro; s$

← Older revision	Revision as of 15:48, 4 November 2022
(No difference)

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1630__Bei_A_1_7.png|right|frame|HDB3- und 1T2B-Codierung]]
+[[File:EN_Bei_A_1_7.png|right|frame|HDB3- und 1T2B-Codierung]]
 Bei herkömmlichem ISDN über Kupferleitungen wird der HDB3–Code verwendet &ndash; siehe&nbsp; [[Aufgaben:Aufgabe_1.5:_HDB3–Codierung|Aufgabe 1.5]]:

← Older revision		Revision as of 12:45, 10 September 2022
Line 100:		Line 100:


−	[[Category:Examples of Communication Systems: Exercises\|^1.4 ~~Weiterentwicklungen von~~ ISDN^]]	+	[[Category:Examples of Communication Systems: Exercises\|^1.4 Further Developments of ISDN^]]