Aufgaben:Exercise 4.3: Different Frequencies - Revision history

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	[[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.~~3_Unterschiedliche_Frequenzen~~]]		[[de:Aufgaben:Aufgabe 4.3: Unterschiedliche Frequenzen]]

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	'''(3)'''  The energy of the signal  $s_0(t)$,  which is constant within  $T$,  is equal to		'''(3)'''  The energy of the signal  $s_0(t)$,  which is constant within  $T$,  is equal to
	:$$E_0 = \|\|s_0(t)\|\|^2 = A^2 \cdot T		:$$E_0 = \|\|s_0(t)\|\|^2 = A^2 \cdot T\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|\|s_0(t)\|\| = A \cdot \sqrt{T} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \varphi_0 (t) = \frac{s_0(t)}{\|\|s_0(t)\|\|} =\left\{ \begin{array}{c} 1/\sqrt{T} \\0 \end{array} \right.\quad\begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm},\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|\|s_0(t)\|\| = A \cdot \sqrt{T} \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} \varphi_0 (t) = \frac{s_0(t)}{\|\|s_0(t)\|\|} =
	\left\{ \begin{array}{c} 1/\sqrt{T} \\
	0 \end{array} \right.\quad
	\begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm},
	\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

	Therefore,  <u>solution 2</u>  is correct.		Therefore,  <u>solution 2</u>  is correct.
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	'''(4)'''  The  <u>last solution</u>  is correct because of		'''(4)'''  The  <u>last solution</u>  is correct because of
	:$$E_1 = \|\|s_1(t)\|\|^2 = \frac{A^2 \cdot T}{2}		:$$E_1 = \|\|s_1(t)\|\|^2 = \frac{A^2 \cdot T}{2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|\|s_1(t)\|\| = A \cdot \sqrt{{T}/{2}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \varphi_1 (t) = \frac{s_1(t)}{\|\|s_1(t)\|\|} =\left\{ \begin{array}{c} \sqrt{2/T} \cdot \cos(2\pi t/T) \\0 \end{array} \right.\quad\begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm},\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|\|s_1(t)\|\| = A \cdot \sqrt{{T}/{2}} \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} \varphi_1 (t) = \frac{s_1(t)}{\|\|s_1(t)\|\|} =
	\left\{ \begin{array}{c} \sqrt{2/T} \cdot \cos(2\pi t/T) \\
	0 \end{array} \right.\quad
	\begin{array}{*{1}c} 0 \le t < T \hspace{0.05cm},
	\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	'''(2)'''  The energy-limited signals   $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$   are orthogonal to each other   ⇒   the inner product of two signals  $s_i(t)$,  $s_k(t)$  with  $i ≠ k$  is always  $0$:		'''(2)'''  The energy-limited signals   $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$   are orthogonal to each other   ⇒   the inner product of two signals  $s_i(t)$,  $s_k(t)$  with  $i ≠ k$  is always  $0$:
	:$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A^2 \cdot \int_{0}^{T}\cos(2\pi \cdot i \cdot t/T) \cdot \cos(2\pi \cdot k \cdot t/T)\,{\rm d} t $$		:$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A^2 \cdot \int_{0}^{T}\cos(2\pi \cdot i \cdot t/T) \cdot \cos(2\pi \cdot k \cdot t/T)\,{\rm d} t $$
	:$$ \Rightarrow \hspace{0.3cm} < \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} {A^2}/{2} \cdot \int_{0}^{T}\cos(2\pi (i-k) t/T) \,{\rm d} t +		:$$ \Rightarrow \hspace{0.3cm} < \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} {A^2}/{2} \cdot \int_{0}^{T}\cos(2\pi (i-k) t/T) \,{\rm d} t +\frac{A^2}{2} \cdot \int_{0}^{T}\cos(2\pi (i+k) t/T) \,{\rm d} t\hspace{0.05cm}.$$
	\frac{A^2}{2} \cdot \int_{0}^{T}\cos(2\pi (i+k) t/T) \,{\rm d} t
	\hspace{0.05cm}.$$

	*With  $i ∈ \{0, \ \text{...} \ , 4\}$  and  $k ∈ \{0, \ \text{...}\ , 4\}$  as well as  $i ≠ j$,  both  $i \, - k$  is integer  $\ne0$,  as is the  sum $i + k$.		*With  $i ∈ \{0, \ \text{...} \ , 4\}$  and  $k ∈ \{0, \ \text{...}\ , 4\}$  as well as  $i ≠ j$,  both  $i \, - k$  is integer  $\ne0$,  as is the  sum $i + k$.
	*Thus,  both integrals yield the result zero:		*Thus,  both integrals yield the result zero:
	:$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm}= 0		:$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm}= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \hspace{0.05cm}\hspace{0.15cm}\underline {N = M = 5}\hspace{0.05cm}.$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \hspace{0.05cm}\hspace{0.15cm}\underline {N = M = 5}
	\hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]		[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]
			[[de:Aufgaben:4.3_Unterschiedliche_Frequenzen]]

Guenter at 14:56, 13 July 2022

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct is the <u>solution 2</u>:		'''(1)'''  Correct is the  <u>solution 2</u>:
	* This takes into account the different frequencies and the limitation to the range $0 ≤ t < T$.		* This takes into account the different frequencies and the limitation to the range  $0 ≤ t < T$.
	*The signals $s_i(t)$ according to suggestion 3, on the other hand, do not differ with respect to frequency, but have different phase positions.
			*The signals  $s_i(t)$  according to suggestion 3,  on the other hand,  do not differ with respect to frequency,  but have different phase positions.



	'''(2)'''  The energy-limited signals $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$ are ~~all~~ orthogonal to each other~~, that is,~~ the inner product of two signals $s_i(t)$ ~~and~~ $s_k(t)$ with $i ≠ k$ is always $0$:		'''(2)'''  The energy-limited signals   $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$   are orthogonal to each other   ⇒   the inner product of two signals  $s_i(t)$,  $s_k(t)$  with  $i ≠ k$  is always  $0$:
	:$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A^2 \cdot \int_{0}^{T}\cos(2\pi \cdot i \cdot t/T) \cdot \cos(2\pi \cdot k \cdot t/T)\,{\rm d} t $$		:$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} A^2 \cdot \int_{0}^{T}\cos(2\pi \cdot i \cdot t/T) \cdot \cos(2\pi \cdot k \cdot t/T)\,{\rm d} t $$
	:$$ \Rightarrow \hspace{0.3cm} < \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} {A^2}/{2} \cdot \int_{0}^{T}\cos(2\pi (i-k) t/T) \,{\rm d} t +		:$$ \Rightarrow \hspace{0.3cm} < \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm} {A^2}/{2} \cdot \int_{0}^{T}\cos(2\pi (i-k) t/T) \,{\rm d} t +
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*With $i ∈ \{0, \ \text{...} \ , 4\}$ and $k ∈ \{0, \ \text{...}\ , 4\}$ as well as $i ≠ j$, both $i \, &~~ndash~~; k$ is integer ~~not equal to~~ $0$, as is the sum $i + k$.		*With  $i ∈ \{0, \ \text{...} \ , 4\}$  and  $k ∈ \{0, \ \text{...}\ , 4\}$  as well as  $i ≠ j$,  both  $i \, - k$  is integer  $\ne0$,  as is the  sum $i + k$.
	*Thus, both integrals yield the result zero:		*Thus,  both integrals yield the result zero:
	:$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm}= 0		:$$< \hspace{-0.1cm}s_i(t), \hspace{0.1cm} s_k(t)\hspace{-0.1cm} > \hspace{0.1cm} \hspace{-0.1cm}= 0
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \hspace{0.05cm}\hspace{0.15cm}\underline {N = M = 5}		\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \hspace{0.05cm}\hspace{0.15cm}\underline {N = M = 5}
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	'''(3)'''  The energy of the signal $s_0(t)$, which is constant within $T$, is equal to		'''(3)'''  The energy of the signal  $s_0(t)$,  which is constant within  $T$,  is equal to
	:$$E_0 = \|\|s_0(t)\|\|^2 = A^2 \cdot T		:$$E_0 = \|\|s_0(t)\|\|^2 = A^2 \cdot T
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|\|s_0(t)\|\| = A \cdot \sqrt{T} \hspace{0.3cm}		\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|\|s_0(t)\|\| = A \cdot \sqrt{T} \hspace{0.3cm}
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	\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$		\\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

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


	'''(4)'''  The <u>last solution</u> is correct because of		'''(4)'''  The  <u>last solution</u>  is correct because of
	:$$E_1 = \|\|s_1(t)\|\|^2 = \frac{A^2 \cdot T}{2}		:$$E_1 = \|\|s_1(t)\|\|^2 = \frac{A^2 \cdot T}{2}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|\|s_1(t)\|\| = A \cdot \sqrt{{T}/{2}} \hspace{0.3cm}		\hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|\|s_1(t)\|\| = A \cdot \sqrt{{T}/{2}} \hspace{0.3cm}

Guenter at 14:48, 13 July 2022

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	<quiz display=simple>		<quiz display=simple>
	{Describe the signal set  $\{s_i(t)\}$  with  $0 ≤ i ≤ 4$  as compactly as possible. <br>Which description form is correct?		{Describe the signal set  $\{s_i(t)\}$  with  $0 ≤ i ≤ 4$  as compactly as possible. <br>Which description form is correct?
	\|type="[]"}		\|type="()"}
	- $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$.		- $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$.
	+ $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$  for  $0 ≤ t < T$,  otherwise $0$.		+ $s_i(t) = A \cdot \cos {(2\pi \cdot i \cdot t/T)}$  for  $0 ≤ t < T$,  otherwise $0$.

Guenter at 14:47, 13 July 2022

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	[[File:P_ID1999__Dig_A_4_3.png\|right\|frame\|Given signal set  $\{s_i(t)\}$]]		[[File:P_ID1999__Dig_A_4_3.png\|right\|frame\|Given signal set  $\{s_i(t)\}$]]
	In the diagram  $M = 5$  different signals  $s_i(t)$  are shown. Contrary to the nomenclature in the theory section, the indexing variable  $i$  can have the values  $0, \ \text{...} \ , M-1$.		In the diagram  $M = 5$  different signals  $s_i(t)$  are shown.  Contrary to the nomenclature in the theory section,  the indexing variable  $i$  can have the values  $0, \ \text{...} \ , M-1$.

	To be noted:		To be noted:
	* All signals are time-limited to  $0$  ~~to  $T$; thus the energy of all signals is also finite~~.		* All signals are time-limited to  $0$  ...   $T$;  thus the energy of all signals is finite.
	* The signal  $s_1(t)$  has the period  $T_0 = T$. ~~The frequency is therefore equal to~~  $~~f_0 = 1/~~T$.
	* The signals  ~~$s_i(t)$~~  ~~with  $i ≠ 0$  are cosine oscillations with frequency  $i \cdot f_0$.~~
	*In contrast,  $s_0(t)$  is constant between  $0$  and  $T$.
	* The maximum value of all signals is~~  $A$  and  $\|s_i(t)\| ≤ A$ holds~~.

			* The signal  $s_1(t)$  has the period  $T_0 = T$.  The frequency is therefore  $f_0 = 1/T$.

	~~In this exercise we are looking for the~~  $N$  ~~basis functions, which~~ are ~~numbered here~~ with  $~~j = 0,~~ \ ~~\text{~~.~~..} \~~ , ~~N-1~~$.		* The signals  $s_i(t)$  with  $i ≠ 0$  are cosine oscillations with frequency  $i \cdot f_0$.

			*In contrast,  $s_0(t)$  is constant between  $0$  and  $T$.

			* The maximum value of all signals is  $A$  and  $\|s_i(t)\| ≤ A$ holds.


			In this exercise we are looking for the  $N$  basis functions,  which are numbered here with  $j = 0, \ \text{...} \ , N-1$.



	''Note:''		Note:  The exercise belongs to the chapter  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].


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	{What is the basis function  $\varphi_0(t)$ that is equal in form to  $s_0(t)$?		{What is the basis function  $\varphi_0(t)$ that is equal in form to  $s_0(t)$?
	\|type="[]"}		\|type="()"}
	- $\varphi_0(t) = s_0(t)$,		- $\varphi_0(t) = s_0(t)$,
	+ $\varphi_0(t) = \sqrt{1/T}$ for $0 ≤ t < T$,  outside  $0$.		+ $\varphi_0(t) = \sqrt{1/T}$  for  $0 ≤ t < T$,  outside  $0$.
	- $\varphi_0(t) = \sqrt{2/T}$ for $0 ≤ t < T$,  outside  $0$.		- $\varphi_0(t) = \sqrt{2/T}$  for  $0 ≤ t < T$,  outside  $0$.

	{What is the basis function  $\varphi_1(t)$ that is equal in form to  $s_1(t)$?		{What is the basis function  $\varphi_1(t)$  that is equal in form to  $s_1(t)$?
	\|type="[]"}		\|type="()"}
	- $\varphi_1(t) = s_1(t)$,		- $\varphi_1(t) = s_1(t)$,
	- $\varphi_1(t) = \sqrt{1/T} \cdot \cos {(2\pi t/T)}$ for $0 ≤ t < T$,  outside $0$.		- $\varphi_1(t) = \sqrt{1/T} \cdot \cos {(2\pi t/T)}$ for $0 ≤ t < T$,  outside $0$.

Hwang at 20:43, 14 June 2022

2022-06-14T20:43:21Z

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Guenter at 11:55, 8 June 2022

2022-06-08T11:55:58Z

← Older revision		Revision as of 13:55, 8 June 2022
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	[[Category:Digital Signal Transmission: Exercises\|^4.1 ~~Basisfunktionen~~ & ~~Vektorräume~~^]]		[[Category:Digital Signal Transmission: Exercises\|^4.1 Basis Functions & Vector Spaces^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 4.3: Unterschiedliche Frequenzen to Aufgaben:Exercise 4.3: Different Frequencies

2022-06-08T11:51:06Z

Guenter moved page Aufgaben:Aufgabe 4.3: Unterschiedliche Frequenzen to Aufgaben:Exercise 4.3: Different Frequencies

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