Aufgaben:Exercise 4.5Z: Tangent Hyperbolic and Inverse - Revision history

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Noah at 23:18, 30 October 2022

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Guenter: Guenter moved page Aufgaben:Aufgabe 4.5Z: Tangens Hyperbolikus und Inverse to Aufgaben:Exercise 4.5Z: Tangent Hyperbolic and Inverse

2022-10-27T13:30:14Z

Guenter moved page Aufgabe 4.5Z: Tangens Hyperbolikus und Inverse to Exercise 4.5Z: Tangent Hyperbolic and Inverse

@@ Line 104: / Line 104: @@
-'''(2)'''&nbsp; <u>The correct solutions are 1, 2, 3, and 5</u>:
+'''(2)'''&nbsp;Correct are the&nbsp; <u>solutions 1, 2, 3, and 5</u>:&nbsp; The function
 :$$y ={\rm tanh}(x) = \frac{{\rm e}^{x}-{\rm e}^{-x}}{{\rm e}^{x}+{\rm e}^{-x}}
   = \frac{1-{\rm e}^{-2x}}{1+{\rm e}^{-2x}}$$
+*By rearranging the given equation
 :$$x ={\rm tanh}^{-1}(y) = 1/2 \cdot {\rm ln} \hspace{0.2cm} \frac{1+y}{1-y}$$
-one obtains:
+:one obtains:
 :$${\rm e}^{2x} =  \frac{1+y}{1-y} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 {\rm e}^{-2x} =  \frac{1-y}{1+y} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
@@ Line 127: / Line 125: @@
 {\rm tanh}(x) \hspace{0.05cm}.$$
-This means:
+*This means:
-* The equation given in the proposed solution 2 is correct.
+# The equation given in the&nbsp; <u>proposed solution 2</u>&nbsp; is correct.
-* In the limiting case $y &#8594; 1$, $x = \tanh^{-1}(y) &#8594; &#8734;$ holds.
+# In the limiting case&nbsp; $y &#8594; 1$, &nbsp; $x = \tanh^{-1}(y) &#8594; &#8734;$&nbsp; holds.
-* Also the inverse function is odd &nbsp;&#8658;&nbsp; in the limiting case $y &#8594; -1$ goes $x &#8594; -&#8734;$.
+# Also the inverse function is odd &nbsp; &#8658; &nbsp; in the limiting case&nbsp; $y &#8594; -1$&nbsp; goes&nbsp; $x &#8594; -&#8734;$.
-Accordingly, the <u>proposed solutions 2 and 4</u> are correct.
+*Accordingly,&nbsp; the&nbsp; <u>proposed solutions 2 and 4</u>&nbsp; are correct.
@@ Line 140: / Line 137: @@
 :$$L_{\rm E}(i) = {\rm ln} \hspace{0.2cm}  \frac{1 + \pi}{1 - \pi}$$
-one arrives with the result of '''(3)''' at the equivalent equation corresponding to <u>proposed solution 2</u>:
+one arrives with the result of&nbsp; '''(3)'''&nbsp; at the equivalent equation corresponding to&nbsp; <u>proposed solution 2</u>:
 :$$L_{\rm E}(i) = 2 \cdot {\rm tanh}^{-1}(\pi)\hspace{0.05cm}.$$
-'''(5)'''&nbsp; With the result of the subtask '''(1)''' we get.
+'''(5)'''&nbsp; With the result of the subtask&nbsp; '''(1)'''&nbsp; we get
-* for the first extrinsic $L$ value, since $\pi_1 = -0.0912$:
+* for the first extrinsic&nbsp; $L$&nbsp; value,&nbsp; since&nbsp; $\pi_1 = -0.0912$:
 :$$L_{\rm E}(1) = 2 \cdot {\rm tanh}^{-1}(-0.0912)= -2 \cdot {\rm tanh}^{-1}(0.0912)
 = -2 \cdot 0.0915\hspace{0.15cm}\underline{=-0.1830}
 \hspace{0.05cm}.$$
-* for the second extrinsic $L$ value, since $\pi_2 = -0.2135$:
+* for the second extrinsic&nbsp; $L$&nbsp; value,&nbsp; since&nbsp; $\pi_2 = -0.2135$:
 :$$L_{\rm E}(2) =  -2 \cdot {\rm tanh}^{-1}(0.2135)
 = -2 \cdot 0.2168\hspace{0.15cm}\underline{=-0.4336}
 \hspace{0.05cm}.$$
-* for the third extrinsic $L$ value, since $\pi_3 = +0.0912 = -\pi_1$:
+* for the third extrinsic&nbsp; $L$ value,&nbsp; since&nbsp; $\pi_3 = +0.0912 = -\pi_1$:
 :$$L_{\rm E}(3) = -L_{\rm E}(1) \hspace{0.15cm}\underline{=+0.1830}
 \hspace{0.05cm}.$$
-The result was determined using the red table entries on the information section and, except for rounding errors (multiplication/division by $2$), agrees with the results of subtask '''(1)'''.
+<u>Note:</u>
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^4.1 Soft–in Soft–out Decoder^]]

@@ Line 80: / Line 80: @@
   \hspace{0.05cm}.$$
-From the table on the specification section can be read:
+*From the table on the specification section can be read:
 :$$\tanh {(L_1/2)} = \tanh {(0.5)} = 0.4621,$$
 :$$\tanh {(L_2/2)} = \tanh {(0.2)} = 0.1974.$$
-Since the hyperbolic tangent is an odd function, the following applies further
+*Since the hyperbolic tangent is an odd function,&nbsp; the following applies further
 :$$\tanh {(L_3/2)} = -\tanh {(0.5)} = -0.4621.$$
-* Calculation of $L_{\rm E}(1)$:
+* Calculation of&nbsp; $L_{\rm E}(1)$:
 :$$\pi = {\rm tanh}(L_2/2) \cdot {\rm tanh}(L_3/2) = (+0.1974) \cdot (-0.4621) = - 0.0912\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} L_{\rm E}(1) = {\rm ln} \hspace{0.2cm}  \frac{1 -0.0912}{1 +0.0912}\hspace{0.15cm}\underline{=-0.1829}
   \hspace{0.05cm}.$$
-* Calculation of $L_{\rm E}(2)$:
+* Calculation of&nbsp; $L_{\rm E}(2)$:
 :$$\pi = {\rm tanh}(L_1/2) \cdot {\rm tanh}(L_3/2) = (+0.4621) \cdot (-0.4621) = - 0.2135\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} L_{\rm E}(2) = {\rm ln} \hspace{0.2cm}  \frac{1 -0.2135}{1 +0.2135}\hspace{0.15cm}\underline{=-0.4337}
   \hspace{0.05cm}.$$
-* Calculation of $L_{\rm E}(3)$:
+* Calculation of&nbsp; $L_{\rm E}(3)$:
 :$$\pi = {\rm tanh}(L_1/2) \cdot {\rm tanh}(L_2/2) = (+0.4621) \cdot (+0.1974) = + 0.0912\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} L_{\rm E}(3) = {\rm ln} \hspace{0.2cm}  \frac{1 +0.0912}{1 -0.0912}\hspace{0.15cm}\underline{=+0.1829}= - L_{\rm E}(1)

@@ Line 39: / Line 39: @@
 ===Questions===
 <quiz display=simple>
-{It holds&nbsp; $\underline{L}_{\rm APP} = (+1.0, +0.4, -1.0)$. Calculate the extrinsic&nbsp; $L$ values &nbsp; &#8658; &nbsp; $\underline{L}_E = \big (L_{\rm E}(1), \ L_{\rm E}(2), \ L_{\rm E}(3) \big)$&nbsp; according to the second equation given:
+{It holds &nbsp; $\underline{L}_{\rm APP} = (+1.0, +0.4, -1.0)$. &nbsp; Calculate the extrinsic&nbsp; $L$ values &nbsp; &#8658; &nbsp; $\underline{L}_{\rm E} = \big (L_{\rm E}(1), \ L_{\rm E}(2), \ L_{\rm E}(3) \big)$ &nbsp; according to the second equation given:
 |type="{}"}
 $L_{\rm E}(1) \ = \ ${ -0.188387--0.177413 }
@@ Line 45: / Line 45: @@
 $L_{\rm E}(3) \ = \ ${ 0.1829 3% }
-{Which of the properties does the function&nbsp; $y = \tanh\hspace{-0.05cm}{(x)}$&nbsp; exhibit?
+{Which of the properties does the function &nbsp; $y = \tanh\hspace{-0.05cm}{(x)}$ &nbsp; exhibit?
 |type="[]"}
-+ $\tanh\hspace{-0.05cm} {(x)} = ({\rm e}^x - {\rm e}^{-x}) \ / \ ({\rm e}^x + {\rm e}^{-x})$ is valid.
++ $\tanh\hspace{-0.05cm} {(x)} = ({\rm e}^x - {\rm e}^{-x}) \ / \ ({\rm e}^x + {\rm e}^{-x})$&nbsp; is valid.
-+ $\tanh\hspace{-0.05cm} {(x)} = (1 - {\rm e}^{-2x}) \ / \ (1 + {\rm e}^{-2x})$ is valid.
++ $\tanh\hspace{-0.05cm} {(x)} = (1 - {\rm e}^{-2x}) \ / \ (1 + {\rm e}^{-2x})$&nbsp; is valid.
-+ The function&nbsp; $y = \tanh\hspace{-0.05cm} {(x)}$&nbsp; is defined for all&nbsp; $x$ values.
++ The function &nbsp; $y = \tanh\hspace{-0.05cm} {(x)}$ &nbsp; is defined for all&nbsp; $x$ values.
-- $y_{\rm min} = 0$&nbsp;and&nbsp; $y_{\rm max} &#8594; &#8734;$ is valid.
+- $y_{\rm min} = 0$ &nbsp;and &nbsp; $y_{\rm max} &#8594; &#8734;$&nbsp; is valid.
-+ $y_{\rm min} = -1$ &nbsp;and&nbsp; $y_{\rm max} = +1$ is valid.
++ $y_{\rm min} = -1$ &nbsp;and &nbsp; $y_{\rm max} = +1$ &nbsp; is valid.
-{What are the properties of the inverse function&nbsp; $x = \tanh^{-1}\hspace{-0.08cm} {(y)}$&nbsp;?
+{What are the properties of the inverse function&nbsp; $x = \tanh^{-1}\hspace{-0.08cm} {(y)}$?
 |type="[]"}
-- The function&nbsp; $x = \tanh^{-1}\hspace{-0.05cm} (y)$&nbsp; is defined for all&nbsp; $y$ values.
+- The function &nbsp; $x = \tanh^{-1}\hspace{-0.05cm} (y)$ &nbsp; is defined for all&nbsp; $y$&nbsp; values.
-+ $x = \tanh^{-1}\hspace{-0.08cm} {(y)} = 1/2 \cdot \ln {[(1 + y) \ / \ (1 - y)]}$ is valid.
++ $x = \tanh^{-1}\hspace{-0.08cm} {(y)} = 1/2 \cdot \ln {[(1 + y) \ / \ (1 - y)]}$&nbsp; is valid.
-- $x_{\rm min} = -1$ &nbsp;and&nbsp; $x_{\rm max} = +1$.
+- $x_{\rm min} = -1$ &nbsp;and &nbsp; $x_{\rm max} = +1$&nbsp; is valid.
-+ $x_{\rm min} &#8594; -&#8734;$ &nbsp;and&nbsp; $x_{\rm max} &#8594; +&#8734;$ is valid.
++ $x_{\rm min} &#8594; -&#8734;$ &nbsp;and &nbsp; $x_{\rm max} &#8594; +&#8734;$&nbsp; is valid.
-{How can&nbsp; $L_{\rm E}(i)$&nbsp; also be represented? Let&nbsp; $\pi$&nbsp; be defined as in the specification section.
+{How can&nbsp; $L_{\rm E}(i)$&nbsp; also be represented?&nbsp; Let&nbsp; $\pi$&nbsp; be defined as in the specification section.
 |type="[]"}
-- $L_{\rm E}(i) = \tanh^{-1}\hspace{-0.08cm} {(\pi)}$ is valid.
+- $L_{\rm E}(i) = \tanh^{-1}\hspace{-0.08cm} {(\pi)}$&nbsp; is valid.
-+ $L_{\rm E}(i) = 2 \cdot \tanh^{-1}\hspace{-0.08cm} {(\pi)}$ is valid.
++ $L_{\rm E}(i) = 2 \cdot \tanh^{-1}\hspace{-0.08cm} {(\pi)}$&nbsp; is valid.
-- $L_{\rm E}(i) = 2 \cdot \tanh^{-1}\hspace{-0.05cm}\big [ {\ln {[(1 + \pi) \ / \ (1 - \pi)]}}\big ]$ is valid.
+- $L_{\rm E}(i) = 2 \cdot \tanh^{-1}\hspace{-0.05cm}\big [ {\ln {[(1 + \pi) \ / \ (1 - \pi)]}}\big ]$&nbsp; is valid.
-{Calculate the extrinsic&nbsp; $L$ values using the equation given in exercise '''(4)'''. Use the table in the information section for this purpose.
+{Calculate the extrinsic&nbsp; $L$ values using the equation given in exercise&nbsp; '''(4)'''.&nbsp; Use the table in the information section for this purpose.
 |type="{}"}
 $L_{\rm E}(1) \ = \ ${ -0.18849--0.17751 }

@@ Line 28: / Line 28: @@
 *This exercise belongs to the chapter&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder|"Soft&ndash;in Soft&ndash;out Decoder"]].
-*Reference is made in particular to the&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"Calculations of extrinsic LLRs"]] section.
+*Reference is made in particular to the&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_log_likelihood_ratios|"Calculations of extrinsic log likelihood ratios"]] section.
 * Above you can see a table with the numerical values of the function&nbsp; $y = \tanh(x)$ &nbsp; &#8658; &nbsp; "hyperbolic tangent".

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}
-[[File:P_ID3025__KC_Z_4_5_v1.png|right|frame|$y = \tanh {(x)}$&nbsp; represented on a table]]
+[[File:P_ID3025__KC_Z_4_5_v1.png|right|frame|$y = \tanh {(x)}$&nbsp; tabularly]]
-In&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"Theory Part"]]&nbsp; it was shown, using the example of&nbsp; <i>single parity&ndash;check code</i>&nbsp; that the extrinsic&nbsp; $L$ value with respect to the&nbsp; $i$th symbol is defined as follows:
+In the&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"Theory Part"]]&nbsp; it was shown,&nbsp; using the example of a&nbsp; "single parity&ndash;check code"&nbsp; that the extrinsic&nbsp; $L$ value with respect to the&nbsp; $i^{th}$&nbsp; symbol is defined as follows:
 :$$L_{\rm E}(i) = {\rm ln} \hspace{0.2cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
   \hspace{0.05cm}.$$
-This equation is also applicable to many other channel codes. The code word&nbsp; $\underline{x}^{(-i)}$&nbsp; in this definition includes all symbols except&nbsp; $x_i$&nbsp; and thus has length&nbsp; $n-1$ only.
+*This equation is also applicable to many other channel codes.
-In the&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC|"Exercise 4.4"]]&nbsp; it was shown that the extrinsic&nbsp; $L$ value can also be written as follows:
+*The code word&nbsp; $\underline{x}^{(-i)}$&nbsp; in this definition includes all symbols except&nbsp; $x_i$&nbsp; and has thus only length&nbsp; $n-1$.
 :$$L_{\rm E}(i) = {\rm ln} \hspace{0.2cm}  \frac{1 + \pi}{1 - \pi}\hspace{0.05cm}, \hspace{0.3cm}
-{\rm mit} \hspace{0.3cm} \pi = \prod\limits_{j \ne i}^{n} \hspace{0.15cm}{\rm tanh}(L_j/2)
+{\rm with} \hspace{0.3cm} \pi = \prod\limits_{j \ne i}^{n} \hspace{0.15cm}{\rm tanh}(L_j/2)
   \hspace{0.05cm}.$$
-In this exercise, we will now look for another calculation possibility.
+In this exercise,&nbsp; we will now look for another calculation possibility.
@@ Line 24: / Line 27: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder|"Soft&ndash;in Soft&ndash;out Decoder"]].
 *Reference is made in particular to the&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"Calculations of extrinsic LLRs"]] section.
-* Above you can see a table with the numerical values of the function&nbsp; $y = \tanh(x)$ &nbsp; &#8658; &nbsp; <i>hyperbolic tangent</i>.
+*Reference is made in particular to the&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"Calculations of extrinsic LLRs"]] section.
-*With the rows highlighted in red you can read the values of the inverse function&nbsp; $x = \tanh^{-1}(y)$&nbsp; needed for the subtask '''(5)'''.

@@ Line 100: / Line 100: @@
 '''(2)'''&nbsp; <u>The correct solutions are 1, 2, 3, and 5</u>:
 *The function
-:$$y ={\rm tanh}(x) = \frac{\rm e}^{x}-{\rm e}^{-x}}{\rm e}^{x}+{\rm e}^{-x}}
+:$$y ={\rm tanh}(x) = \frac{{\rm e}^{x}-{\rm e}^{-x}}{{\rm e}^{x}+{\rm e}^{-x}}
   = \frac{1-{\rm e}^{-2x}}{1+{\rm e}^{-2x}}$$
 is computable for all $x$ values and $\tanh(-x) = -\tanh(x)$ holds.

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