Aufgaben:Exercise 4.2: AM/PM Oscillations - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:58:36Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:58, 16 March 2026
Line 152:		Line 152:

	[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]		[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]
	[[de:Aufgaben:4.~~2_AM~~/PM-Schwingungen]]		[[de:Aufgaben:Aufgabe 4.2: AM/PM-Schwingungen]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:32Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 29:		Line 29:

	The basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$  are to be found here by the   [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#The_Gram-Schmidt_process\|"Gram–Schmidt process"]],   which was described in detail in the theory section.  The required equations are summarized here again:		The basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$  are to be found here by the   [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#The_Gram-Schmidt_process\|"Gram–Schmidt process"]],   which was described in detail in the theory section.  The required equations are summarized here again:
	:$$\varphi_1(t) = \frac{s_1(t)}{\|\|s_1(t)\|\|}\hspace{0.4cm}{\rm with}\hspace{0.4cm}		:$$\varphi_1(t) = \frac{s_1(t)}{\|\|s_1(t)\|\|}\hspace{0.4cm}{\rm with}\hspace{0.4cm}s_{11} = \|\|s_1(t)\|\| = \sqrt{\int_{0}^{T}s_1^2(t) \, {\rm d} t}\hspace{0.05cm},\hspace{0.4cm}s_{21} = \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_1(t) \hspace{-0.1cm} > \hspace{0.1cm} =\int_{0}^{T}s_2(t) \cdot \varphi_1(t)\, {\rm d} t\hspace{0.05cm},$$
	s_{11} = \|\|s_1(t)\|\| = \sqrt{\int_{0}^{T}s_1^2(t) \, {\rm d} t}
	\hspace{0.05cm},\hspace{0.4cm}
	s_{21} = \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_1(t) \hspace{-0.1cm} > \hspace{0.1cm} =
	\int_{0}^{T}s_2(t) \cdot \varphi_1(t)\, {\rm d} t
	\hspace{0.05cm},$$
	:$$\theta_2(t) = s_2(t) - s_{21} \cdot \varphi_1(t)\hspace{0.05cm}, \hspace{0.2cm}\varphi_2(t) = \frac{\theta_2(t)}{\|\|\theta_2(t)\|\|}\hspace{0.05cm}.$$		:$$\theta_2(t) = s_2(t) - s_{21} \cdot \varphi_1(t)\hspace{0.05cm}, \hspace{0.2cm}\varphi_2(t) = \frac{\theta_2(t)}{\|\|\theta_2(t)\|\|}\hspace{0.05cm}.$$

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:31:07Z

Join split $$ formula lines for correct rendering

Show changes

Maintenance script: Add German interlanguage link

2026-03-16T12:33:25Z

Add German interlanguage link

← Older revision		Revision as of 14:33, 16 March 2026
Line 192:		Line 192:

	[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]		[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]
			[[de:Aufgaben:4.2_AM/PM-Schwingungen]]

Guenter at 16:11, 13 July 2022

2022-07-13T16:11:59Z

← Older revision		Revision as of 18:11, 13 July 2022
Line 17:		Line 17:


	For the ~~first~~ two signals  (shown in the graph)  holds:		For the two signals  (shown in the graph)  holds:
	:$$s_1(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos(2\pi f_{\rm T}t ) \hspace{0.05cm},$$		:$$s_1(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos(2\pi f_{\rm T}t ) \hspace{0.05cm},$$
	:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2A \cdot \cos(2\pi f_{\rm T}t + \pi/4) \hspace{0.05cm}. $$		:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2A \cdot \cos(2\pi f_{\rm T}t + \pi/4) \hspace{0.05cm}. $$

Guenter at 15:43, 13 July 2022

2022-07-13T15:43:51Z

← Older revision		Revision as of 17:43, 13 July 2022
Line 2:		Line 2:
	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces}}

	[[File:P_ID1997__Dig_A_4_2.png\|right\|frame\|~~Considered~~ AM/PM oscillations]]		[[File:P_ID1997__Dig_A_4_2.png\|right\|frame\|Two possible AM/PM oscillations]]
	We consider the signal set   $\{s_i(t)\}$   with the indexing variable  $i = 1, \ \text{...} \, M$.  All signals  $s_i(t)$  can be represented in the same way:		We consider the signal set   $\{s_i(t)\}$   with the indexing variable  $i = 1, \ \text{...} \, M$.  All signals  $s_i(t)$  can be represented in the same way:
	:$$s_i(t) =		:$$s_i(t) =

Guenter at 14:02, 13 July 2022

2022-07-13T14:02:00Z

← Older revision		Revision as of 16:02, 13 July 2022
Line 104:		Line 104:
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$

	*Here it is considered that $T$ is an even multiple of $1/f_{\rm T}$, so the second integral vanishes. Further:		*Here it is considered that  $T$  is an even multiple of  $1/f_{\rm T}$,  so the second integral vanishes.

			*Further:
	:$$\|\|s_1(t)\|\| = \sqrt{E_1} = \sqrt{E} = \hspace{0.1cm}\hspace{0.15cm}\underline{1 \cdot\sqrt{E}}		:$$\|\|s_1(t)\|\| = \sqrt{E_1} = \sqrt{E} = \hspace{0.1cm}\hspace{0.15cm}\underline{1 \cdot\sqrt{E}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	'''(2)'''  <u>Solution 3</u> is correct: The basis function $\varphi_1(t)$ is equal in form to $s_1(t)$, where holds:		'''(2)'''  <u>Solution 3</u>  is correct:  The basis function  $\varphi_1(t)$  is equal in form to  $s_1(t)$,  where holds:
	:$$\varphi_1(t) = \frac{s_1(t)}{\|\|s_1(t)\|\|}= \frac{A \cdot \cos(2\pi f_{\rm T}t )}{\sqrt{E}}= \frac{A \cdot \cos(2\pi f_{\rm T}t )}{\sqrt{1/2 \cdot A^2 \cdot T}} = \sqrt{{2}/{T}}		:$$\varphi_1(t) = \frac{s_1(t)}{\|\|s_1(t)\|\|}= \frac{A \cdot \cos(2\pi f_{\rm T}t )}{\sqrt{E}}= \frac{A \cdot \cos(2\pi f_{\rm T}t )}{\sqrt{1/2 \cdot A^2 \cdot T}} = \sqrt{{2}/{T}}
	\cdot \cos(2\pi f_{\rm T}t )		\cdot \cos(2\pi f_{\rm T}t )
Line 115:		Line 117:


	'''(3)'''  <u>Solution 1</u> is correct, since according to the equation given in '''(2)''':		'''(3)'''  <u>Solution 1</u>  is correct  since according to the equation given in  '''(2)''':
	:$$s_1(t) = \|\|s_1(t)\|\| \cdot \varphi_1(t) = \sqrt{E} \cdot \varphi_1(t)		:$$s_1(t) = \|\|s_1(t)\|\| \cdot \varphi_1(t) = \sqrt{E} \cdot \varphi_1(t)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	'''(4)'''  Using the signal $s_2(t)$ according to the given information, the basis function $\varphi_1(t)$ according to subtask '''(2)''' and the given trigonometric relation we get:		'''(4)'''  Using the signal  $s_2(t)$  according to the given information,  the basis function  $\varphi_1(t)$  according to subtask  '''(2)'''  and the given trigonometric relation we get:
	:$$s_{21} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_1(t) \hspace{-0.1cm} > \hspace{0.1cm} =		:$$s_{21} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \hspace{0.1cm} < \hspace{-0.1cm} s_2(t), \hspace{0.1cm}\varphi_1(t) \hspace{-0.1cm} > \hspace{0.1cm} =
	\int_{0}^{T}2A \cdot \cos(2\pi f_{\rm T}t + {\pi}/{4}) \cdot \sqrt{{2}/{T}}		\int_{0}^{T}2A \cdot \cos(2\pi f_{\rm T}t + {\pi}/{4}) \cdot \sqrt{{2}/{T}}
Line 130:		Line 132:
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$

	*The second component yields the value $0$ (orthogonality). The first component yields:		*The second component yields the value  $0$  (orthogonality).  The first component yields:
	:$$s_{21} = \sqrt{\frac{8A^2}{T}}\cdot \frac{1}{\sqrt{2}}\cdot \frac{T}{2} = \sqrt{A^2 \cdot T} = \sqrt{2E} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1.414 \cdot \sqrt{E}}		:$$s_{21} = \sqrt{\frac{8A^2}{T}}\cdot \frac{1}{\sqrt{2}}\cdot \frac{T}{2} = \sqrt{A^2 \cdot T} = \sqrt{2E} \hspace{0.1cm}\hspace{0.15cm}\underline { = 1.414 \cdot \sqrt{E}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	'''(5)'''  According to the Gram–Schmidt process, we obtain		'''(5)'''  According to the Gram–Schmidt process,  we obtain
	:$$\theta_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_2(t) - s_{21} \cdot \varphi_1(t)\hspace{0.05cm} =		:$$\theta_2(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_2(t) - s_{21} \cdot \varphi_1(t)\hspace{0.05cm} =
	2A \cdot \cos(2\pi f_{\rm T}t + {\pi}/{4}) - \sqrt{A^2 \cdot T}		2A \cdot \cos(2\pi f_{\rm T}t + {\pi}/{4}) - \sqrt{A^2 \cdot T}
Line 145:		Line 147:
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$

	*With $\cos {(\pi/4)} = \sin (\pi/4) =\sqrt{0.5}$ it follows:		*With  $\cos {(\pi/4)} = \sin (\pi/4) =\sqrt{0.5}$  it follows:
	:$$\theta_2(t) = - \sqrt{2} \cdot A \cdot \sin(2\pi f_{\rm T}t )		:$$\theta_2(t) = - \sqrt{2} \cdot A \cdot \sin(2\pi f_{\rm T}t )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*Therefore, <u>solution 2</u> is correct.		*Therefore,  <u>solution 2</u> is correct.



	'''(6)'''  Analogous to subtask '''(2)''', the orthonormal basis function $\varphi_2(t)$ is given by		'''(6)'''  Analogous to subtask  '''(2)''',  the orthonormal basis function  $\varphi_2(t)$  is given by
	:$$\varphi_2(t) = \frac{\theta_2(t)}{\|\|\theta_2(t)\|\|} = - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )		:$$\varphi_2(t) = \frac{\theta_2(t)}{\|\|\theta_2(t)\|\|} = - \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*Thus, the signal $s_2(t)$ can be represented by $s_{21}$ according to subtask '''(4)''' as follows:		*Thus,  the signal  $s_2(t)$  can be represented by  $s_{21}$  according to subtask  '''(4)'''  as follows:
	:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{21} \cdot \varphi_1(t) + s_{22} \cdot \varphi_2(t) \hspace{0.05cm}, \hspace{0.2cm}s_{21} = \underline{ = 1.414 \cdot \sqrt {E}}\hspace{0.05cm},$$		:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{21} \cdot \varphi_1(t) + s_{22} \cdot \varphi_2(t) \hspace{0.05cm}, \hspace{0.2cm}s_{21} = \underline{ = 1.414 \cdot \sqrt {E}}\hspace{0.05cm},$$
	:$$s_{22}\hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{\theta_2(t)}{\varphi_2(t)} = \frac{-\sqrt{2} \cdot A \cdot \sin(2\pi f_{\rm T}t )}		:$$s_{22}\hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{\theta_2(t)}{\varphi_2(t)} = \frac{-\sqrt{2} \cdot A \cdot \sin(2\pi f_{\rm T}t )}
Line 163:		Line 165:


	'''(7)'''  We consider very many energy-limited signals ($M \gg 2$) of the following form:		'''(7)'''  We consider very many energy-limited signals  $(M \gg 2)$  of the following form:
	:$$s_i(t)=		:$$s_i(t)=
	\left\{ \begin{array}{c} A_i \cdot \cos(2\pi f_{\rm T}t + \phi_i) \\		\left\{ \begin{array}{c} A_i \cdot \cos(2\pi f_{\rm T}t + \phi_i) \\
Line 170:		Line 172:
	\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$		\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

	The indexing variable can take the values $i = 1, 2, \ \text{...} \ , M$. Then holds:		The indexing variable can take the values  $i = 1, 2, \ \text{...} \ , M$.  Then holds:
	* All $M$ signals can be completely described by only $N = 2$ basis functions:		* All  $M$  signals can be completely described by only  $N = 2$  basis functions:
	:$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}. $$		:$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}. $$
	* If one proceeds according to the Gram–Schmidt process, one obtains for the two basis functions
			* If one proceeds according to the Gram–Schmidt process,  one obtains for the two basis functions
	:$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t + \phi_1)\hspace{0.05cm},\hspace{0.5cm}		:$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t + \phi_1)\hspace{0.05cm},\hspace{0.5cm}
	\varphi_2(t) = \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t + \phi_1 \pm {\pi}/{2})\hspace{0.05cm}.$$		\varphi_2(t) = \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t + \phi_1 \pm {\pi}/{2})\hspace{0.05cm}.$$
	* The sign in the argument of the second cosine function ($± \pi/2$) is not unique. Rather, the sign of $s_{i 2}$ also depends on whether the plus sign or the minus sign was used for $\varphi_2(t)$.
	* However, possible basis functions that then lead to other coefficients are also:		* The sign in the argument of the second cosine function  $(± \pi/2)$  is not unique.  Rather,  the sign of  $s_{i 2}$  also depends on whether the plus sign or the minus sign was used for  $\varphi_2(t)$.

			* However,  possible basis functions that then lead to other coefficients are also:
	:$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm},\hspace{0.5cm}		:$$\varphi_1(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{{2}/{T}} \cdot \cos(2\pi f_{\rm T}t )\hspace{0.05cm},\hspace{0.5cm}
	\varphi_2(t) \pm \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}.$$		\varphi_2(t) \pm \sqrt{{2}/{T}} \cdot \sin(2\pi f_{\rm T}t )\hspace{0.05cm}.$$

	So <u>solutions 2 and 3</u> are correct.		⇒   So the  <u>solutions 2 and 3</u>  are correct.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 13:47, 13 July 2022

2022-07-13T13:47:08Z

← Older revision		Revision as of 15:47, 13 July 2022
Line 2:		Line 2:
	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces}}

	[[File:P_ID1997__Dig_A_4_2.png\|right\|frame\|AM/PM oscillations]]		[[File:P_ID1997__Dig_A_4_2.png\|right\|frame\|Considered AM/PM oscillations]]
	We consider the signal set  $\{s_i(t)\}$  with the indexing variable  $i = 1, \ \text{...} \, M$. All signals  $s_i(t)$  can be represented in the same way:		We consider the signal set   $\{s_i(t)\}$   with the indexing variable  $i = 1, \ \text{...} \, M$.  All signals  $s_i(t)$  can be represented in the same way:
	:$$s_i(t) =		:$$s_i(t) =
	\left\{ \begin{array}{c} A_i \cdot \cos(2\pi f_{\rm T}t + \phi_i) \\		\left\{ \begin{array}{c} A_i \cdot \cos(2\pi f_{\rm T}t + \phi_i) \\
Line 10:		Line 10:
	\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$		\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

	The signal duration  $T$  is an integer multiple of  $1/f_{\rm T}$, where  $f_{\rm T}$  is the signal frequency (carrier frequency).		The signal duration  $T$  is an integer multiple of  $1/f_{\rm T}$,  where  $f_{\rm T}$  is the signal frequency  ("carrier frequency").

			*For the sketch,  the duration of the energy-limited signals is  $T = 4/f_{\rm T}$,  i.e. exactly four oscillations are recognized within  $T$  in each case.

	*For the sketch, the duration of the energy-limited signals is  $T = 4/f_{\rm T}$ in each case, i.e. exactly four oscillations are recognized within  $T$ in each case.
	*The individual signals  $s_i(t)$  differ in amplitude  $(A_i)$  and/or phase  $(\phi_i)$.		*The individual signals  $s_i(t)$  differ in amplitude  $(A_i)$  and/or phase  $(\phi_i)$.


	For the first two signals (shown in the graph) holds:		For the first two signals  (shown in the graph)  holds:
	:$$s_1(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos(2\pi f_{\rm T}t ) \hspace{0.05cm},$$		:$$s_1(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} A \cdot \cos(2\pi f_{\rm T}t ) \hspace{0.05cm},$$
	:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2A \cdot \cos(2\pi f_{\rm T}t + \pi/4) \hspace{0.05cm}. $$		:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 2A \cdot \cos(2\pi f_{\rm T}t + \pi/4) \hspace{0.05cm}. $$

	If we first restrict ourselves to these two signals  $s_1(t)$  and $s_2(t)$, they can be completely described by the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.  These are orthonormal to each other, that is, taking into account the time constraint on  $T$  holds:		If we first restrict ourselves to these two signals  $s_1(t)$  and  $s_2(t)$,  they can be completely described by the basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$.  These are orthonormal to each other,  that is,  taking into account the time constraint on  $T$  holds:
	:$$\int_{0}^{T}\varphi_1^2(t) \, {\rm d} t = \int_{0}^{T}\varphi_2^2(t) \, {\rm d} t = 1 \hspace{0.05cm}, ~~\hspace{0.2cm}~~		:$$\int_{0}^{T}\varphi_1^2(t) \, {\rm d} t = \int_{0}^{T}\varphi_2^2(t) \, {\rm d} t = 1 \hspace{0.05cm},$$
	\int_{0}^{T}\varphi_1(t) \cdot \varphi_2(t)\, {\rm d} t = 0		:$$ \int_{0}^{T}\varphi_1(t) \cdot \varphi_2(t)\, {\rm d} t = 0
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	With these basis functions, the two signals can be represented as follows:		With these basis functions,  the two signals can be represented as follows:
	:$$s_1(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{11} \cdot \varphi_1(t) \hspace{0.05cm},$$		:$$s_1(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{11} \cdot \varphi_1(t) \hspace{0.05cm},$$
	:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{21} \cdot \varphi_1(t) + s_{22} \cdot \varphi_2(t) \hspace{0.05cm}. $$		:$$s_2(t)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} s_{21} \cdot \varphi_1(t) + s_{22} \cdot \varphi_2(t) \hspace{0.05cm}. $$

	In subtask '''(7)''' we want to check whether all signals  $s_i(t)$  according to the above definition $($with arbitrary amplitude  $A_i$  and arbitrary phase  $\phi_i)$  can be described by the following equation:		In subtask  '''(7)'''  we want to check whether all signals  $s_i(t)$  according to the above definition   $($with arbitrary amplitude  $A_i$  and arbitrary phase  $\phi_i)$   can be described by the following equation:
	:$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}. $$		:$$s_i(t)= s_{i1} \cdot \varphi_1(t) + s_{i2} \cdot \varphi_2(t) \hspace{0.05cm}. $$

	The basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$  are to be found here by the  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#The_Gram-Schmidt_process\|"Gram–Schmidt process"]],  which was described in detail in the theory section. The required equations are summarized here again:		The basis functions  $\varphi_1(t)$  and  $\varphi_2(t)$  are to be found here by the   [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#The_Gram-Schmidt_process\|"Gram–Schmidt process"]],   which was described in detail in the theory section.  The required equations are summarized here again:
	:$$\varphi_1(t) = \frac{s_1(t)}{\|\|s_1(t)\|\|}\hspace{0.4cm}{\rm with}\hspace{0.4cm}		:$$\varphi_1(t) = \frac{s_1(t)}{\|\|s_1(t)\|\|}\hspace{0.4cm}{\rm with}\hspace{0.4cm}
	s_{11} = \|\|s_1(t)\|\| = \sqrt{\int_{0}^{T}s_1^2(t) \, {\rm d} t}		s_{11} = \|\|s_1(t)\|\| = \sqrt{\int_{0}^{T}s_1^2(t) \, {\rm d} t}
Line 45:		Line 46:


			Notes:
	''Notes:''		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].
	*~~Die Aufgabe gehört zum Kapitel~~  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].
			* For abbreviation,  use the energy  $E = 1/2 \cdot A^2 \cdot T$.

	* ~~For abbreviation~~, ~~use the energy~~  ~~$E = 1/2 \cdot A^2 \cdot T$.~~		* Furthermore,  the following trigonometric relation is given:
	* Furthermore, the following trigonometric relation is given:
	:$$\cos(\alpha \pm \beta) = \cos(\alpha )\cdot \cos(\beta) \mp \sin(\alpha )\cdot \sin(\beta)\hspace{0.05cm}.$$		:$$\cos(\alpha \pm \beta) = \cos(\alpha )\cdot \cos(\beta) \mp \sin(\alpha )\cdot \sin(\beta)\hspace{0.05cm}.$$

Line 56:		Line 57:
	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{What is the energy and 2–norm of the signal  $s_1(t)$, expressed by $E$?		{What is the energy and the  "2–norm"  of the signal  $s_1(t)$,  expressed by  $E$?
	\|type="{}"}		\|type="{}"}
	$E_1\ = \ $ { 1 3% } $\ \cdot E$		$E_1\ = \ $ { 1 3% } $\ \cdot E$
Line 73:		Line 74:
	- $s_1(t) = \sqrt{2/T} \cdot \varphi_1(t)$.		- $s_1(t) = \sqrt{2/T} \cdot \varphi_1(t)$.

	{What is the inner product  $s_{\rm 21} = 〈 s_2(t) \cdot \varphi_1(t)〉$?		{What is the  "inner product"  $s_{\rm 21} = 〈 s_2(t) \cdot \varphi_1(t)〉$?
	\|type="{}"}		\|type="{}"}
	$s_{\rm 21} \ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$		$s_{\rm 21} \ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$
Line 83:		Line 84:
	- $\theta_2(t) = \sqrt{2/T} \cdot {\rm sin}(2\pi f_{\rm T}t)$.		- $\theta_2(t) = \sqrt{2/T} \cdot {\rm sin}(2\pi f_{\rm T}t)$.

	{Give the coefficients of  $s_2(t) = s_{\rm 21} \cdot \varphi_1(t) + s_{\rm 22} \cdot \varphi_2(t)$.		{Give the coefficients of   $s_2(t) = s_{\rm 21} \cdot \varphi_1(t) + s_{\rm 22} \cdot \varphi_2(t)$.
	\|type="{}"}		\|type="{}"}
	$s_{\rm 21}\ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$		$s_{\rm 21}\ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$
	$s_{\rm 22}\ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$		$s_{\rm 22}\ = \ $ { 1.414 3% } $\ \cdot \sqrt{E}$

	{Which of the statements are generally true for the basis functions of the signal set  $\{s_i(t)\}$ with $i = 1, \ \text{ ...} \ , M$, ~~falls~~  $M \gg 2$?		{Which of the statements are generally true for the basis functions of the signal set  $\{s_i(t)\}$  with  $i = 1, \ \text{ ...} \ , M$,  if  $M \gg 2$?
	\|type="[]"}		\|type="[]"}
	- The number of basis functions is always  $N = M$.		- The number of basis functions is always  $N = M$.

Hwang at 13:03, 14 June 2022

2022-06-14T13:03:56Z

Show changes

Guenter at 11:54, 8 June 2022

2022-06-08T11:54:56Z

← Older revision		Revision as of 13:54, 8 June 2022
Line 185:		Line 185:


	[[Category:Digital Signal Transmission: Exercises\|^4.1 ~~Basisfunktionen~~ & ~~Vektorräume~~^]]		[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]