Aufgaben:Exercise 4.6: AWGN Channel Capacity - Revision history

Guenter at 12:18, 3 November 2021

2021-11-03T12:18:56Z

Guenter at 12:17, 3 November 2021

2021-11-03T12:17:55Z

Noah at 11:49, 1 November 2021

2021-11-01T11:49:36Z

Javier: Text replacement - "continuous-value" to "continuous"

2021-10-11T08:28:22Z

Text replacement - "continuous-value" to "continuous"

Guenter at 15:57, 9 October 2021

2021-10-09T15:57:00Z

Guenter at 10:08, 4 October 2021

2021-10-04T10:08:57Z

Guenter at 09:13, 4 October 2021

2021-10-04T09:13:39Z

Guenter at 09:12, 4 October 2021

2021-10-04T09:12:15Z

Guenter at 12:31, 2 October 2021

2021-10-02T12:31:56Z

Noah at 13:11, 1 October 2021

2021-10-01T13:11:38Z

Show changes

@@ Line 136: / Line 136: @@
 *However,&nbsp; according to the adjacent graph,&nbsp; also holds:
 :$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) = h(Y) - I(X;Y) = 4.047 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm}. $$
 *From this,&nbsp; the differential equivocation can be calculated as follows:
 :$$h(X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y) = h(X) - I(X;Y) = 4.000 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 2.000\,{\rm bit}}\hspace{0.05cm}.$$
 *Finally,&nbsp; the differential composite entropy is also given,&nbsp; which cannot be read directly from the above diagram:
 :$$h(XY) = h(X) + h(Y) - I(X;Y) = 4.000 \,{\rm bit} + 4.047 \,{\rm bit}  - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 6.047\,{\rm bit}}\hspace{0.05cm}.$$

@@ Line 101: / Line 101: @@
 '''(1)'''&nbsp; The equation for the AWGN channel capacity in&nbsp; "bit"&nbsp; is:
 :$$C_{\rm bit} = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$
-With&nbsp; $C_{\rm bit} = 2$&nbsp; this results in:
+:With&nbsp; $C_{\rm bit} = 2$&nbsp; this results in:
 :$$4 \stackrel{!}{=} {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )
 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} 1 + {P_X}/{P_N} \stackrel {!}{=} 2^4 = 16
@@ Line 113: / Line 113: @@
 * With &nbsp;$P_X = 15 \ \rm mW$&nbsp; the maximum mutual information &nbsp;$I(X; Y) = 2$&nbsp; bit is only achievable if the input quantity&nbsp; $X$&nbsp; is Gaussian distributed.&nbsp; <br>The output quantity&nbsp; $Y$&nbsp; is then also Gaussian distributed.
 * If the random variable&nbsp; $X$&nbsp; has a constant proportion &nbsp;$m_X$&nbsp; then the variance &nbsp;$\sigma_X^2 = P_X - m_X^2 $&nbsp;  for given&nbsp; $P_X$ &nbsp;is smaller, and it holds &nbsp; <br>$I(X; Y) = 1/2 &middot; \log_2 \ (1 + \sigma_X^2/P_N) < 2$&nbsp; bit.
-* The precondition for the given channel capacity equation is that&nbsp; $X$ &nbsp;and&nbsp; $N$&nbsp; are uncorrelated.&nbsp; On the other hand, if the random variables&nbsp; $X$ &nbsp;and&nbsp; $N$&nbsp; were uncorrelated, then &nbsp;$I(X; Y) = 0$ would result.
+* The precondition for the given channel capacity equation is that&nbsp; $X$ &nbsp;and&nbsp; $N$&nbsp; are uncorrelated.&nbsp; On the other hand, if the random variables&nbsp; $X$ &nbsp;and&nbsp; $N$&nbsp; were uncorrelated, then &nbsp;$I(X; Y) = 0$&nbsp; would result.
 '''(3)'''&nbsp; The given equation for differential entropy makes sense only for dimensionless power.&nbsp; With the proposed normalization, one obtains:
 * For &nbsp;$P_N = 1 \ \rm mW$&nbsp; &nbsp;&#8658;&nbsp; &nbsp;$P_N\hspace{0.05cm}' = 1$:
 :$$h(N) \  =  \ {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e} \cdot 1 \right )
@@ Line 126: / Line 126: @@
 {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left (15 \right )
 \hspace{0.15cm}\underline{= 4.000\,{\rm bit}}\hspace{0.05cm}, $$
-*  Für &nbsp;$P_Y = P_X  + P_N = 16 \ \rm mW$&nbsp; &nbsp;&#8658;&nbsp; $P_Y\hspace{0.01cm}' = 16$:
+*  For &nbsp;$P_Y = P_X  + P_N = 16 \ \rm mW$&nbsp; &nbsp;&#8658;&nbsp; $P_Y\hspace{0.01cm}' = 16$:
 :$$h(Y) = 2.047\,{\rm bit} + 2.000\,{\rm bit}
 \hspace{0.15cm}\underline{= 4.047\,{\rm bit}}\hspace{0.05cm}.$$
@@ Line 132: / Line 132: @@
-'''(4)'''&nbsp; For differential irrelevance, for the AWGN channel:
+'''(4)'''&nbsp; The differential irrelevance for the AWGN channel:
 :$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) = h(N) \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm}.$$
-*However, according to the adjacent graph, also holds:
+*However,&nbsp; according to the adjacent graph,&nbsp; also holds:
 :$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) = h(Y) - I(X;Y) = 4.047 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm}. $$
-*From this, the differential equivocation can be calculated as follows:
+[[File: P_ID2900__Inf_A_4_6e.png |right|frame|Information-theoretical values with the ideal channel]]
 :$$h(X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y) = h(X) - I(X;Y) = 4.000 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 2.000\,{\rm bit}}\hspace{0.05cm}.$$
+*Finally,&nbsp; the differential composite entropy is also given,&nbsp; which cannot be read directly from the above diagram:
 :$$h(XY) = h(X) + h(Y) - I(X;Y) = 4.000 \,{\rm bit} + 4.047 \,{\rm bit}  - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 6.047\,{\rm bit}}\hspace{0.05cm}.$$
-[[File: P_ID2900__Inf_A_4_6e.png |right|frame|Information-theoretical quantities with the ideal channel]]
+'''(5)'''&nbsp; For the ideal channel with&nbsp; $h(X)\hspace{0.15cm}\underline{=  4.000 \,{\rm bit}}$:
 :$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) \  =  \ h(N) \hspace{0.15cm}\underline{= 0\,{\rm (bit)}}\hspace{0.05cm},$$
 :$$h(Y) \  =  \ h(X) \hspace{0.15cm}\underline{= 4\,{\rm bit}}\hspace{0.05cm},$$
@@ Line 151: / Line 150: @@
 h(X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y) \  =  \ h(X) - I(X;Y)\hspace{0.15cm}\underline{= 0\,{\rm (bit)}}\hspace{0.05cm}.$$
-*The graph shows these quantities in a flowchart.
+*The graph shows these quantities in a flowchart.&nbsp; The same diagram would result in the discrete value case with&nbsp; $M = 16$&nbsp; equally probable symbols &nbsp; &#8658; &nbsp; $H(X)=  4.000 \,{\rm bit}$.
+*One only would have to replace each&nbsp; $h$&nbsp; by an&nbsp; $H$.
 {{ML-Fuß}}

@@ Line 116: / Line 116: @@
-[[File: P_ID2901__Inf_A_4_6c.png |right|frame|Informationstheoretische Größen für den AWGN-Kanal]]
+[[File: P_ID2901__Inf_A_4_6c.png |right|frame|Information-theoretical quantities for the AWGN channel]]
-'''(3)'''&nbsp; Die angegebene Gleichung für die differentielle Entropie macht nur bei dimensionsloser Leistung Sinn.&nbsp; Mit der vorgeschlagenen Normierung erhält man:
+'''(3)'''&nbsp; The given equation for differential entropy makes sense only for dimensionless power.&nbsp; With the proposed normalization, one obtains:
-* Für &nbsp;$P_N = 1 \ \rm mW$&nbsp; &nbsp;&#8658;&nbsp; &nbsp;$P_N\hspace{0.05cm}' = 1$:
+* For &nbsp;$P_N = 1 \ \rm mW$&nbsp; &nbsp;&#8658;&nbsp; &nbsp;$P_N\hspace{0.05cm}' = 1$:
 :$$h(N) \  =  \ {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e} \cdot 1 \right )
    =  \ {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 17.08 \right )
 \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm},$$
-* Für &nbsp;$P_X = 15 \ \rm mW$&nbsp; &nbsp;&#8658;&nbsp; &nbsp;$P_X\hspace{0.01cm}' = 15$:
+* For &nbsp;$P_X = 15 \ \rm mW$&nbsp; &nbsp;&#8658;&nbsp; &nbsp;$P_X\hspace{0.01cm}' = 15$:
 :$$h(X) \  =  \ {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e} \cdot 15 \right )   =  {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e}  \right ) +
 {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left (15 \right )
@@ Line 132: / Line 132: @@
-'''(4)'''&nbsp; Für die differentielle Irrelevanz gilt beim AWGN&ndash;Kanal:
+'''(4)'''&nbsp; For differential irrelevance, for the AWGN channel:
 :$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) = h(N) \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm}.$$
-*Entsprechend nebenstehender Grafik gilt aber auch:
+*However, according to the adjacent graph, also holds:
 :$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) = h(Y) - I(X;Y) = 4.047 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 2.047\,{\rm bit}}\hspace{0.05cm}. $$
-*Daraus kann die differentielle Äquivokation wie folgt berechnet werden:
+*From this, the differential equivocation can be calculated as follows:
 :$$h(X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y) = h(X) - I(X;Y) = 4.000 \,{\rm bit} - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 2.000\,{\rm bit}}\hspace{0.05cm}.$$
-*Abschließend wird auch noch die differentielle Verbundentropie angegeben, die aus obigem Schaubild nicht direkt ablesbar ist:
+*Finally, the differential composite entropy is also given, which cannot be read directly from the above diagram:
 :$$h(XY) = h(X) + h(Y) - I(X;Y) = 4.000 \,{\rm bit} + 4.047 \,{\rm bit}  - 2 \,{\rm bit} \hspace{0.15cm}\underline{= 6.047\,{\rm bit}}\hspace{0.05cm}.$$
-[[File: P_ID2900__Inf_A_4_6e.png |right|frame|Informationstheoretische Größen beim idealen Kanal]]
+[[File: P_ID2900__Inf_A_4_6e.png |right|frame|Information-theoretical quantities with the ideal channel]]
-'''(5)'''&nbsp; Beim idealen Kanal erhält man mit $h(X)\hspace{0.15cm}\underline{=  4.000 \,{\rm bit}}$:
+'''(5)'''&nbsp; In the ideal channel, $h(X)\hspace{0.15cm}\underline{=  4.000 \,{\rm bit}}$ is obtained::
 :$$h(Y \hspace{-0.05cm}\mid \hspace{-0.05cm} X) \  =  \ h(N) \hspace{0.15cm}\underline{= 0\,{\rm (bit)}}\hspace{0.05cm},$$
 :$$h(Y) \  =  \ h(X) \hspace{0.15cm}\underline{= 4\,{\rm bit}}\hspace{0.05cm},$$
@@ Line 151: / Line 151: @@
 h(X \hspace{-0.05cm}\mid \hspace{-0.05cm} Y) \  =  \ h(X) - I(X;Y)\hspace{0.15cm}\underline{= 0\,{\rm (bit)}}\hspace{0.05cm}.$$
-*Die Grafik zeigt diese Größen in einem Flussdiagramm.
+*The graph shows these quantities in a flowchart.
-*Das gleiche Diagramm ergäbe sich auch im wertdiskreten Fall mit&nbsp; $M = 16$&nbsp; gleichwahrscheinlichen Symbolen &nbsp; &#8658; &nbsp; $H(X)=  4.000 \,{\rm bit}$.
+*The same diagram would result in the discrete value case with&nbsp; $M = 16$&nbsp; equally probable symbols &nbsp; &#8658; &nbsp; $H(X)=  4.000 \,{\rm bit}$.
-*Man müsste nur jedes&nbsp; $h$&nbsp; durch ein&nbsp; $H$&nbsp; ersetzen.
+*One would only have to replace each&nbsp; $h$&nbsp; by an&nbsp; $H$&nbsp;.
 {{ML-Fuß}}

@@ Line 47: / Line 47: @@
 Hints:
-*The exercise belongs to the chapter&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang|AWGN channel capacity with continuous-value input]].
+*The exercise belongs to the chapter&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang|AWGN channel capacity with continuous input]].
 *Since the results are to be given in&nbsp; "bit",&nbsp; "log" &nbsp;&#8658;&nbsp; "log<sub>2</sub>"&nbsp; is used in the equations.

@@ Line 109: / Line 109: @@
+'''(2)'''&nbsp; Correct are&nbsp; <u>proposed solutions 1 through 4</u>.&nbsp; Justification:
-'''(2)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 bis 4</u>. Begründung:
+* For &nbsp;$P_X < 15 \ \rm  mW$&nbsp; the mutual information &nbsp;$I(X; Y)$&nbsp; will always be less than&nbsp; $2$&nbsp; bit,&nbsp; regardless of all other conditions.
-* Für &nbsp;$P_X < 15 \ \rm mW$&nbsp; wird die Transinformation &nbsp;$I(X; Y)$&nbsp; stets kleiner als&nbsp; $2$&nbsp; bit sein, unabhängig von allen anderen Gegebenheiten.
+* With &nbsp;$P_X = 15 \ \rm mW$&nbsp; the maximum mutual information &nbsp;$I(X; Y) = 2$&nbsp; bit is only achievable if the input quantity&nbsp; $X$&nbsp; is Gaussian distributed.&nbsp; <br>The output quantity&nbsp; $Y$&nbsp; is then also Gaussian distributed.
-* Mit &nbsp;$P_X = 15 \ \rm mW$&nbsp; ist die maximale Transinformation &nbsp;$I(X; Y) = 2$&nbsp; bit nur  erreichbar, wenn die Eingangsgröße&nbsp; $X$&nbsp; gaußverteilt ist.&nbsp;  <br>Die Ausgangsgröße&nbsp; $Y$&nbsp; ist dann ebenfalls gaußverteilt.
+* If the random variable&nbsp; $X$&nbsp; has a constant proportion &nbsp;$m_X$&nbsp; then the variance &nbsp;$\sigma_X^2 = P_X - m_X^2 $&nbsp;  for given&nbsp; $P_X$ &nbsp;is smaller, and it holds &nbsp; <br>$I(X; Y) = 1/2 &middot; \log_2 \ (1 + \sigma_X^2/P_N) < 2$&nbsp; bit.
-* Weist die Zufallsgröße&nbsp; $X$&nbsp; einen Gleichanteil &nbsp;$m_X$&nbsp; auf, so ist die Varianz &nbsp;$\sigma_X^2 = P_X - m_X^2 $&nbsp;  bei gegebenem&nbsp; $P_X$ &nbsp;kleiner, und es gilt &nbsp;$I(X; Y) = 1/2 &middot; \log_2 \ (1 + \sigma_X^2/P_N) < 2$&nbsp; bit.
+* The precondition for the given channel capacity equation is that&nbsp; $X$ &nbsp;and&nbsp; $N$&nbsp; are uncorrelated.&nbsp; On the other hand, if the random variables&nbsp; $X$ &nbsp;and&nbsp; $N$&nbsp; were uncorrelated, then &nbsp;$I(X; Y) = 0$ would result.

@@ Line 99: / Line 99: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The equation for the AWGN channel capacity in "bit" is:
+'''(1)'''&nbsp; The equation for the AWGN channel capacity in&nbsp; "bit"&nbsp; is:
 :$$C_{\rm bit} = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + {P_X}/{P_N} \right )\hspace{0.05cm}.$$
 With&nbsp; $C_{\rm bit} = 2$&nbsp; this results in:

@@ Line 48: / Line 48: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/AWGN–Kanalkapazität_bei_wertkontinuierlichem_Eingang|AWGN channel capacity with continuous-value input]].
-*Since the results are to be given in "bit", "log" &nbsp;&#8658;&nbsp; "log<sub>2</sub>" is used in the equations.
+*Since the results are to be given in&nbsp; "bit",&nbsp; "log" &nbsp;&#8658;&nbsp; "log<sub>2</sub>"&nbsp; is used in the equations.
@@ Line 56: / Line 56: @@
 <quiz display=simple>
-{What transmit power is required for &nbsp;$C = 2 \ \rm bit$?
+{What transmission power is required for &nbsp;$C = 2 \ \rm bit$?
 |type="{}"}
 $P_X \ = \ $ { 15 3% } $\ \rm mW$

@@ Line 9: / Line 9: @@
 * $Y = X +N$&nbsp; describes the output (receiver) in case of additive noise.
 For the probability density function of the noise, let hold:
 :$$f_N(n) = \frac{1}{\sqrt{2\pi \hspace{0.03cm}\sigma_{\hspace{-0.05cm}N}^2}} \cdot {\rm e}^{
 - \hspace{0.05cm}{n^2}\hspace{-0.05cm}/{(2 \hspace{0.03cm} \sigma_{\hspace{-0.05cm}N}^2) }} \hspace{0.05cm}.$$
-Since the random variable&nbsp; $N$&nbsp; is zero mean &nbsp; &#8658; &nbsp; $m_{N} = 0$, , we can equate the variance&nbsp;$\sigma_{\hspace{-0.05cm}N}^2$&nbsp; with the power &nbsp;$P_N$&nbsp;.&nbsp; In this case, the differential entropy of the random variable&nbsp; $N$&nbsp;  is specifiable (with the pseudo&ndash;unit "bit") as follows:
+Since the random variable&nbsp; $N$&nbsp; is zero mean &nbsp; &#8658; &nbsp; $m_{N} = 0$,&nbsp; we can equate the variance&nbsp;$\sigma_{\hspace{-0.05cm}N}^2$&nbsp; with the power &nbsp;$P_N$&nbsp;.&nbsp; In this case, the differential entropy of the random variable&nbsp; $N$&nbsp;  is specifiable&nbsp; (with the pseudo&ndash;unit "bit")&nbsp; as follows:
 :$$h(N) = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 2\pi {\rm e} \cdot P_N \right )\hspace{0.05cm}.$$
-In this exercise, &nbsp;$P_N = 1 \rm mW$&nbsp; is given.&nbsp; It should be noted:
+In this exercise, &nbsp;$P_N = 1 \hspace{0.15cm} \rm mW$&nbsp; is given.&nbsp; It should be noted:
 * The power&nbsp; $P_N$&nbsp; in the above equation, like the variance &nbsp;$\sigma_{\hspace{-0.05cm}N}^2$&nbsp;, must be dimensionless.
-* To work with this equation, the physical quantity &nbsp;$P_N$&nbsp; must be suitably normalized, for example corresponding to&nbsp; $P_N = 1 \rm mW$ &nbsp;&nbsp; &#8658; &nbsp;&nbsp; $P_N\hspace{0.01cm}' = 1$.
+* To work with this equation, the physical quantity &nbsp;$P_N$&nbsp; must be suitably normalized, for example corresponding to&nbsp; $P_N = 1 \hspace{0.15cm} \rm mW$ &nbsp;&nbsp; &#8658; &nbsp;&nbsp; $P_N\hspace{0.01cm}' = 1$.
-* With other normalization, for example &nbsp;$P_N = 1 \rm mW$&nbsp; &nbsp;&nbsp; &#8658; &nbsp;&nbsp; &nbsp;$P_N\hspace{0.01cm}' = 0.001$&nbsp; a completely different numerical value would result fo &nbsp;$h(N)$&nbsp;.
+* With other normalization, for example &nbsp;$P_N = 1 \hspace{0.15cm} \rm mW$&nbsp; &nbsp;&nbsp; &#8658; &nbsp;&nbsp; &nbsp;$P_N\hspace{0.01cm}' = 0.001$&nbsp; a completely different numerical value would result fo &nbsp;$h(N)$&nbsp;.
-Further, you can consider for the solution of this exercise:
+Further,&nbsp; you can consider for the solution of this exercise:
-* The channel capacitance is defined as the maximum transinformation between input&nbsp; $X$&nbsp; and output&nbsp; $Y$&nbsp; with the best possible input distribution:
+* The channel capacity is defined as the maximum mutual information between input&nbsp; $X$&nbsp; and output&nbsp; $Y$&nbsp; with the best possible input distribution:
 :$$C = \max_{\hspace{-0.15cm}f_X:\hspace{0.05cm} {\rm E}[X^2] \le P_X} \hspace{-0.2cm}  I(X;Y)
 \hspace{0.05cm}.$$
@@ Line 32: / Line 33: @@
 :$$C_{\rm AWGN} = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{P_X}{P_N} \right )
 = {1}/{2} \cdot {\rm log}_2\hspace{0.05cm}\left ( 1 + \frac{P_{\hspace{-0.05cm}X}\hspace{0.01cm}'}{P_{\hspace{-0.05cm}N}\hspace{0.01cm}'} \right )\hspace{0.05cm}.$$
-:It can be seen:&nbsp; the channel capacity&nbsp; $C$&nbsp; and also the mutual information &nbsp;$I(X; Y)$&nbsp; are independent of the above normalization, in contrast to the differential entropies.
+:It can be seen:&nbsp; The channel capacity&nbsp; $C$&nbsp; and also the mutual information &nbsp;$I(X; Y)$&nbsp; are independent of the above normalization, in contrast to the differential entropies.
-* With Gaussian noise PDF &nbsp;$f_N(n)$&nbsp;, an input PDF $f_X(x)$&nbsp; , also Gaussian, leads to the maximum transinformation and thus to the channel capacity.
+* With Gaussian noise PDF &nbsp;$f_N(n)$,&nbsp; an Gaussian input PDF $f_X(x)$&nbsp;  leads to the maximum mutual information and thus to the channel capacity.