Aufgaben:Exercise 4.11: C Program "acf1" - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]
	[[de:Aufgaben:4.~~11_C~~-~~Programm_„akf1”~~]]		[[de:Aufgaben:Aufgabe 4.11: C-Programm „akf1”]]

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.4 Auto-Correlation Function^]]
			[[de:Aufgaben:4.11_C-Programm_„akf1”]]

Guenter at 15:28, 28 April 2022

2022-04-28T15:28:39Z

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	}}		}}

	[[File:~~EN_Sto_A_4_11~~.png\|right\|frame\|C program  $1$  for the ACF calculation ~~'''correction'''~~]]		[[File:EN_Sto_A_4_11_neu2.png\|right\|frame\|C program  $1$  for the ACF calculation]]
	See the C–program "acf1" for calculating the discrete ACF values  $\varphi_x(k)$  with index  $k = 0$, ... , $l$.  The following should be noted about this:		See the C–program "acf1" for calculating the discrete ACF values  $\varphi_x(k)$  with index  $k = 0$, ... , $l$.  The following should be noted about this:

Guenter at 16:39, 31 March 2022

2022-03-31T16:39:04Z

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	[[File:~~P_ID391__Sto_A_4_11~~.png\|right\|frame\|C program  $1$  for the ACF calculation '''correction''']]		[[File:EN_Sto_A_4_11.png\|right\|frame\|C program  $1$  for the ACF calculation '''correction''']]
	See the C–program "acf1" for calculating the discrete ACF values  $\varphi_x(k)$  with index  $k = 0$, ... , $l$.  The following should be noted about this:		See the C–program "acf1" for calculating the discrete ACF values  $\varphi_x(k)$  with index  $k = 0$, ... , $l$.  The following should be noted about this:

Guenter at 15:53, 21 March 2022

2022-03-21T15:53:01Z

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	<quiz display=simple>		<quiz display=simple>
	{Which elements  $i$  and  $j$  of the auxiliary field  ${\rm H}\big[ \ \big]$  are used <u>on the first loop pass</u>  $(z=0)$  to calculate the ACF–value  $\varphi(k=6)$ ? <br>What random values  $x_\nu$  are in these memory cells?		{Which elements  $i$  and  $j$  of the auxiliary field  ${\rm H}\big[ \ \big]$  are used  <u>on the first loop pass</u>   $(z=0)$  to calculate the ACF–value  $\varphi(k=6)$ ? <br>What random values  $x_\nu$  are in these memory cells?
	\|type="{}"}		\|type="{}"}
	$i \ = \ $ { 0. }		$i \ = \ $ { 0. }
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	{Which memory cell  ${\rm H}\big[\hspace{0.03cm} i \hspace{0.03cm} \big]$  will be occupied <u>after the first loop pass</u>  $(z=0)$  with a new random variable  $x_\nu$ ? <br> Which index   $\nu$  is entered in the process?		{Which memory cell  ${\rm H}\big[\hspace{0.03cm} i \hspace{0.03cm} \big]$  will be occupied  <u>after the first loop pass</u>   $(z=0)$  with a new random variable  $x_\nu$ ? <br> Which index   $\nu$  is entered in the process?
	\|type="{}"}		\|type="{}"}
	$i \ = \ $ { 0. }		$i \ = \ $ { 0. }

Guenter at 15:47, 21 March 2022

2022-03-21T15:47:05Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID417__Sto_A_4_11_b.png\|right\|frame\|For numerical ACF calculation]]		[[File:P_ID417__Sto_A_4_11_b.png\|right\|frame\|For the exemplary numerical ACF calculation]]
	<br>		<br>
	'''(1)'''  With  $z= 0$  and  $k=6$  results according to the program:   $\underline{i= 0}$  and  $\underline{j= 6}$.		'''(1)'''  With  $z= 0$  and  $k=6$  results according to the program:   $\underline{i= 0}$  and  $\underline{j= 6}$.
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	'''(3)'''  The graph shows the allocation of the auxiliary field with the random values  $x_\nu$.		'''(3)'''  The graph shows the allocation of the auxiliary field with the random values  $x_\nu$.

	*In each case, the memory cell ~~is highlighted in green~~  ${\rm H}\big[\hspace{0.03cm} i \hspace{0.03cm}\big]$.  The new random variable is entered into this memory location at the end of each loop (line 18).		*In each case,  the memory cell  ${\rm H}\big[\hspace{0.03cm} i \hspace{0.03cm}\big]$  is highlighted in green.  The new random variable is entered into this memory location at the end of each loop  (line 18).
	*For  $z= 83$  and  $K=6$  this results in.		*For  $z= 83$  and  $K=6$  this results in
	:$$\underline{i= 83 \hspace{-0.2cm}\mod \hspace{-0.15cm} \ 11 = 6},\hspace{1cm} \underline{j= (i+k)\hspace{-0.2cm}\mod \hspace{-0.15cm} \ 11 = 1}.$$		:$$\underline{i= 83 \hspace{-0.2cm}\mod \hspace{-0.15cm} \ 11 = 6},\hspace{1cm} \underline{j= (i+k)\hspace{-0.2cm}\mod \hspace{-0.15cm} \ 11 = 1}.$$
	*Loop pass  $z= 83$:   In memory cell  ${\rm H}\big[\hspace{0.03cm} 6 \hspace{0.03cm}\big]$  is the random variable  $x_{84}$  and in the memory cell  ${\rm H}\big[\hspace{0.03cm} 1 \hspace{0.03cm}\big]$  is the random variable  $x_{90}$.		*Loop pass  $z= 83$:   In memory cell  ${\rm H}\big[\hspace{0.03cm} 6 \hspace{0.03cm}\big]$  is the random variable  $x_{84}$  and in the memory cell  ${\rm H}\big[\hspace{0.03cm} 1 \hspace{0.03cm}\big]$  is the random variable  $x_{90}$.
	*At the end of ~~the~~ loop pass  $z= 83$  in  ${\rm H}\big[\hspace{0.03cm} 6 \hspace{0.03cm}\big]$  ~~the~~ content  $x_{84}$  is replaced by  $x_{95}$ .		*At the end of loop pass  $z= 83$  in  ${\rm H}\big[\hspace{0.03cm} 6 \hspace{0.03cm}\big]$:  The content  $x_{84}$  is replaced by  $x_{95}$ .
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:39, 21 March 2022

2022-03-21T15:39:41Z

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	}}		}}

	[[File:P_ID391__Sto_A_4_11.png\|right\|frame\|C program  $1$  for ACF~~–~~calculation '''correction''']]		[[File:P_ID391__Sto_A_4_11.png\|right\|frame\|C program  $1$  for the ACF calculation '''correction''']]
	See the C–program "acf1" for calculating the discrete ACF values  $\varphi_x(k)$  with index  $k = 0$, ... , $l$.  The following should be noted about this:		See the C–program "acf1" for calculating the discrete ACF values  $\varphi_x(k)$  with index  $k = 0$, ... , $l$.  The following should be noted about this:

	* ~~Let the~~ long–value passed to the program  $l = 10$.  The ACF values  $\varphi_x(0)$, ... , $\varphi_x(10)$  are returned to the calling program with the float field  $\rm ACF\big[ \ \big]$  . In lines 7 and 8 of the program given on the right, this field is pre-populated with zeros.		* The long–value passed to the program is  $l = 10$.  The ACF values  $\varphi_x(0)$, ... , $\varphi_x(10)$  are returned to the calling program with the float field  $\rm ACF\big[ \ \big]$.  In lines 7 and 8 of the program given on the right,  this field is pre-populated with zeros.

	* The random variables ~~to be analyzed~~  $x_\nu$  are generated with the float function  $x( \ )$  (see line 4).  This function is called a total of  $N + l + 1 = 10011$  times (lines 9 and 18).		* The random variables  $x_\nu$  to be analyzed are generated with the float function  $x( \ )$,  see line 4.  This function is called a total of  $N + l + 1 = 10011$  times  $x_\nu$  (lines 9 and 18).

	* In contrast to the algorithm given in the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Numerical_ACF_determination\|Theory section]]  program  "acf2"  of  [[Aufgaben:Exercise_4.11Z:_C_Program_"acf2"\|Task 4.11Z]]  ~~directly implemented,~~ one needs here an auxiliary field  ${\rm H}\big[ \ \big]$  with only  $l + 1 = 11$  memory elements.		* In contrast to the algorithm given in the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Numerical_ACF_determination\|Theory section]]  directly implemented in the program  "acf2"  of  [[Aufgaben:Exercise_4.11Z:_C_Program_"acf2"\|Task 4.11Z]],  one needs here an auxiliary field  ${\rm H}\big[ \ \big]$  with only  $l + 1 = 11$  memory elements.

			* Before starting the actual calculation algorithm (lines 11 to 21),  the eleven memory cells of  ${\rm H}\big[ \ \big]$  contain the random values  $x_1$, ... ,  $x_{11}$.  The outer loop with the index  $z$   (marked in red)  is run  $N$  times.
			*In the inner loop   (marked white)  with the index  $k = 0$, ... ,  $l$  all memory cells of the field  ${\rm ACF}\big[\hspace{0.03cm} k \hspace{0.03cm} \big]$  are increased by the contribution  $x_\nu \cdot x_{\nu+k}$.

	* ~~Before starting the actual calculation algorithm (lines 11 to 21), the eleven memory cells of  ${\rm H}\big[ \ \big]$  contain the random values  $x_1$, ...~~ ,  ~~$x_{11}$.~~		* Finally,  in lines 22 and 23,  all ACF values are divided by the number  $N$  of analyzed data.
	*The outer loop with the index  $z$  (marked in red) is run  $N$ times.
	*In the inner loop  (marked white)  with the index  $k = 0$, ... ,  $l$  all memory cells of the field  ${\rm ACF}\big[\hspace{0.03cm} k \hspace{0.03cm} \big]$  are increased by the contribution  $x_\nu \cdot x_{\nu+k}$ .

	* Finally, in lines 22 and 23, all ACF~~–~~values are divided by the number  $N$  of data ~~analyzed~~.


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	<quiz display=simple>		<quiz display=simple>
	{Which elements  $i$  and  $j$  of the auxiliary field  ${\rm H}\big[ \ \big]$  are used <u>on the first pass</u>  $(z=0)$  to calculate the ACF–value  $\varphi(k=6)$ ? <br>What random values  $x_\nu$  are in these memory cells?		{Which elements  $i$  and  $j$  of the auxiliary field  ${\rm H}\big[ \ \big]$  are used <u>on the first loop pass</u>  $(z=0)$  to calculate the ACF–value  $\varphi(k=6)$ ? <br>What random values  $x_\nu$  are in these memory cells?
	\|type="{}"}		\|type="{}"}
	$i \ = \ $ { 0. }		$i \ = \ $ { 0. }

Noah at 12:43, 6 March 2022

2022-03-06T12:43:07Z

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	[[File:P_ID391__Sto_A_4_11.png\|right\|frame\|C program  $1$  for ACF–calculation '''correction''']]		[[File:P_ID391__Sto_A_4_11.png\|right\|frame\|C program  $1$  for ACF–calculation '''correction''']]
	See the C–program "~~akf1~~" for calculating the discrete ACF values  $\varphi_x(k)$  with index  $k = 0$, ... , $l$.  The following should be noted about this:		See the C–program "acf1" for calculating the discrete ACF values  $\varphi_x(k)$  with index  $k = 0$, ... , $l$.  The following should be noted about this:

	* Let the long–value passed to the program  $l = 10$.  The ACF values  $\varphi_x(0)$, ... , $\varphi_x(10)$  are returned to the calling program with the float field  $\rm ACF\big[ \ \big]$  . In lines 7 and 8 of the program given on the right, this field is pre-populated with zeros.		* Let the long–value passed to the program  $l = 10$.  The ACF values  $\varphi_x(0)$, ... , $\varphi_x(10)$  are returned to the calling program with the float field  $\rm ACF\big[ \ \big]$  . In lines 7 and 8 of the program given on the right, this field is pre-populated with zeros.
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	* The random variables to be analyzed  $x_\nu$  are generated with the float function  $x( \ )$  (see line 4).  This function is called a total of  $N + l + 1 = 10011$  times (lines 9 and 18).		* The random variables to be analyzed  $x_\nu$  are generated with the float function  $x( \ )$  (see line 4).  This function is called a total of  $N + l + 1 = 10011$  times (lines 9 and 18).

	* In contrast to the algorithm given in the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Numerical_ACF_determination\|Theory section]]  program  "~~akf2~~"  of  [[Aufgaben:Exercise_4.11Z:_C_Program_"acf2"\|Task 4.11Z]]  directly implemented, one needs here an auxiliary field  ${\rm H}\big[ \ \big]$  with only  $l + 1 = 11$  memory elements.		* In contrast to the algorithm given in the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Numerical_ACF_determination\|Theory section]]  program  "acf2"  of  [[Aufgaben:Exercise_4.11Z:_C_Program_"acf2"\|Task 4.11Z]]  directly implemented, one needs here an auxiliary field  ${\rm H}\big[ \ \big]$  with only  $l + 1 = 11$  memory elements.

Noah at 15:25, 4 March 2022

2022-03-04T15:25:07Z

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	*For  $z= 83$  and  $K=6$  this results in.		*For  $z= 83$  and  $K=6$  this results in.
	:$$\underline{i= 83 \hspace{-0.2cm}\mod \hspace{-0.15cm} \ 11 = 6},\hspace{1cm} \underline{j= (i+k)\hspace{-0.2cm}\mod \hspace{-0.15cm} \ 11 = 1}.$$		:$$\underline{i= 83 \hspace{-0.2cm}\mod \hspace{-0.15cm} \ 11 = 6},\hspace{1cm} \underline{j= (i+k)\hspace{-0.2cm}\mod \hspace{-0.15cm} \ 11 = 1}.$$
	*~~loop~~ pass  $z= 83$:   In memory cell  ${\rm H}\big[\hspace{0.03cm} 6 \hspace{0. 03cm}\big]$  is the random variable  $x_{84}$  and in the memory cell  ${\rm H}\big[\hspace{0.03cm} 1 \hspace{0.03cm}\big]$  is the random variable  $x_{90}$.		*Loop pass  $z= 83$:   In memory cell  ${\rm H}\big[\hspace{0.03cm} 6 \hspace{0.03cm}\big]$  is the random variable  $x_{84}$  and in the memory cell  ${\rm H}\big[\hspace{0.03cm} 1 \hspace{0.03cm}\big]$  is the random variable  $x_{90}$.
	*At the end of the loop pass  $z= 83$  in  ${\rm H}\big[\hspace{0.03cm} 6 \hspace{0.03cm}\big]$  the content  $x_{84}$  is replaced by  $x_{95}$ .		*At the end of the loop pass  $z= 83$  in  ${\rm H}\big[\hspace{0.03cm} 6 \hspace{0.03cm}\big]$  the content  $x_{84}$  is replaced by  $x_{95}$ .
	{{ML-Fuß}}		{{ML-Fuß}}

Noah at 00:00, 1 March 2022

2022-03-01T00:00:27Z

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	* Before starting the actual calculation algorithm (lines 11 to 21), the eleven memory cells of  ${\rm H}\big[ \ \big]$  contain the random values  $x_1$, ... ,  $x_{11}$.		* Before starting the actual calculation algorithm (lines 11 to 21), the eleven memory cells of  ${\rm H}\big[ \ \big]$  contain the random values  $x_1$, ... ,  $x_{11}$.
	*The outer loop with the ~~run variable~~  $z$  (marked in red) is run  $N$ times.		*The outer loop with the index  $z$  (marked in red) is run  $N$ times.
	*In the inner loop  (marked white)  with the index  $k = 0$, ... ,  $l$  all memory cells of the field  ${\rm ACF}\big[\hspace{0.03cm} k \hspace{0.03cm} \big]$  are increased by the contribution  $x_\nu \cdot x_{\nu+k}$ .		*In the inner loop  (marked white)  with the index  $k = 0$, ... ,  $l$  all memory cells of the field  ${\rm ACF}\big[\hspace{0.03cm} k \hspace{0.03cm} \big]$  are increased by the contribution  $x_\nu \cdot x_{\nu+k}$ .