Aufgaben:Exercise 5.5Z: About the Rake Receiver - Revision history

Maintenance script: Add German interlanguage link

2026-03-16T12:38:20Z

Add German interlanguage link

← Older revision		Revision as of 14:38, 16 March 2026
Line 119:		Line 119:

	[[Category:Modulation Methods: Exercises\|^5.4 BER of the PN Modulation^]]		[[Category:Modulation Methods: Exercises\|^5.4 BER of the PN Modulation^]]
			[[de:Aufgaben:Aufgabe_5.5Z:_Zum_RAKE–Empfänger]]

Guenter at 15:03, 7 February 2023

2023-02-07T15:03:42Z

← Older revision		Revision as of 17:03, 7 February 2023
Line 3:		Line 3:
	}}		}}

	[[File:EN_Mod_Z_5_5.png\|right\|frame\|2-way channel <br>& rake receiver]]		[[File:EN_Mod_Z_5_5.png\|right\|frame\|Two-way channel <br>& rake receiver]]
	The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:		The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:
	:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$		:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$

Hwang at 08:25, 29 December 2021

2021-12-29T08:25:53Z

← Older revision		Revision as of 10:25, 29 December 2021
Line 3:		Line 3:
	}}		}}

	[[File:EN_Mod_Z_5_5.png\|right\|frame\|2-way channel <br>& ~~rake receiver '''KORREKTUR''': two-way channel,~~ rake receiver]]		[[File:EN_Mod_Z_5_5.png\|right\|frame\|2-way channel <br>& rake receiver]]
	The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:		The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:
	:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$		:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$

Guenter at 16:39, 20 December 2021

2021-12-20T16:39:34Z

← Older revision		Revision as of 18:39, 20 December 2021
Line 3:		Line 3:
	}}		}}

	[[File:~~P_ID1888__Mod_Z_5_5~~.png\|right\|frame\|2-way channel <br>& rake receiver '''KORREKTUR''': two-way channel, rake receiver]]		[[File:EN_Mod_Z_5_5.png\|right\|frame\|2-way channel <br>& rake receiver '''KORREKTUR''': two-way channel, rake receiver]]
	The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:		The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:
	:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$		:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$

Guenter at 15:09, 17 December 2021

2021-12-17T15:09:16Z

← Older revision		Revision as of 17:09, 17 December 2021
Line 68:		Line 68:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solution 1</u> is correct:		'''(1)'''  <u>Solution 1</u> is correct:
	*The impulse response  $h_{\rm K}(t)$  is obtained as the received signal  $r(t)$ when there is a ~~dirac~~ pulse at the input   ⇒   $s(t) = δ(t)$.  It follows that:		*The impulse response  $h_{\rm K}(t)$  is obtained as the received signal  $r(t)$  when there is a Dirac delta pulse at the input   ⇒   $s(t) = δ(t)$.  It follows that:
	:$$ h_(t) = 0.6 \cdot \delta (t ) + 0.4 \cdot \delta (t - \tau) \hspace{0.05cm}.$$		:$$ h_{\rm K}(t) = 0.6 \cdot \delta (t ) + 0.4 \cdot \delta (t - \tau) \hspace{0.05cm}.$$


	'''(2)'''  <u>Solutions 2 and 3</u> are correct:		'''(2)'''  <u>Solutions 2 and 3</u>  are correct:
	*By definition, the channel frequency response  $H_{\rm K}(f)$  is the Fourier transform of the impulse response  $h_{\rm K}(t)$.  With the shift theorem this results in:		*By definition,  the channel frequency response  $H_{\rm K}(f)$  is the Fourier transform of the impulse response  $h_{\rm K}(t)$.  With the shift theorem this results in:
	:$$H_{\rm K}(f) = 0.6 + 0.4 \cdot {\rm e}^{ \hspace{0.03cm}{\rm j} \hspace{0.03cm} \cdot \hspace{0.03cm}2 \pi f \tau}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_{\rm K}(f= 0) = 0.6 + 0.4 = 1 \hspace{0.05cm}.$$		:$$H_{\rm K}(f) = 0.6 + 0.4 \cdot {\rm e}^{ \hspace{0.03cm}{\rm j} \hspace{0.03cm} \cdot \hspace{0.03cm}2 \pi f \tau}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_{\rm K}(f= 0) = 0.6 + 0.4 = 1 \hspace{0.05cm}.$$
	*Accordingly, the first proposed solution is incorrect in contrast to the other two:   $H_{\rm K}(f)$ is complex-valued and the magnitude is periodic with  $1/τ$, as the following calculation shows:		*Accordingly,  the first proposed solution is incorrect in contrast to the other two:
	:$$\|H_{\rm K}(f)\|^2 = \left [0.6 + 0.4 \cdot \cos(2 \pi f \tau) \right ]^2 + \left [ 0.4 \cdot \sin(2 \pi f \tau) \right ]^2 = \left [0.6^2 + 0.4^2 \cdot \left ( \cos^2(2 \pi f \tau) + \sin^2(2 \pi f \tau)\right ) \right ] + 2 \cdot 0.6 \cdot 0.4 \cdot \cos(2 \pi f \tau)$$		#  $H_{\rm K}(f)$ is complex-valued and
			# the magnitude is periodic with  $1/τ$,  as the following calculation shows:
			:$$\|H_{\rm K}(f)\|^2 = \left [0.6 + 0.4 \cdot \cos(2 \pi f \tau) \right ]^2 + \left [ 0.4 \cdot \sin(2 \pi f \tau) \right ]^2 = \left [0.6^2 + 0.4^2 \cdot \left ( \cos^2(2 \pi f \tau) + \sin^2(2 \pi f \tau)\right ) \right ] + 2 \cdot 0.6 \cdot 0.4 \cdot \cos(2 \pi f \tau).$$
	*For  $f = 0$,   $\|H_{\rm K}(f)\| = 1$.  This value is repeated in the respective frequency spacing  $1/τ$.		*For  $f = 0$,   $\|H_{\rm K}(f)\| = 1$.  This value is repeated in the respective frequency spacing  $1/τ$.



Line 85:		Line 88:
	*To satisfy the given  $h_{\rm KR}(t)$ equation, either  $τ_0 = 0$  must hold or  $τ_1 = 0$.  With  $τ_0 = 0$  we obtain for the impulse response:		*To satisfy the given  $h_{\rm KR}(t)$ equation, either  $τ_0 = 0$  must hold or  $τ_1 = 0$.  With  $τ_0 = 0$  we obtain for the impulse response:
	:$$h_{\rm KR}(t) = 0.6 \cdot h_0 \cdot \delta (t ) + 0.4 \cdot h_0 \cdot \delta (t - \tau) + 0.6 \cdot h_1 \cdot \delta (t -\tau_1) + 0.4 \cdot h_1 \cdot \delta (t - \tau-\tau_1) \hspace{0.05cm}.$$		:$$h_{\rm KR}(t) = 0.6 \cdot h_0 \cdot \delta (t ) + 0.4 \cdot h_0 \cdot \delta (t - \tau) + 0.6 \cdot h_1 \cdot \delta (t -\tau_1) + 0.4 \cdot h_1 \cdot \delta (t - \tau-\tau_1) \hspace{0.05cm}.$$
	*To be able to focus the "main energy" on a time point,  $τ_1 = τ$  would ~~then~~ have to be chosen.  With  $h_0 = 0.6$  and  $h_1 = 0.4$,  we then obtain  $A_0 ≠ A_2$:		*To be able to focus the  "main energy"  at a certain time point,  $τ_1 = τ$  would have to be chosen.
			* With  $h_0 = 0.6$  and  $h_1 = 0.4$,  we then obtain  $A_0 ≠ A_2$:
	:$$h_{\rm KR}(t) = 0.36 \cdot \delta (t ) +0.48 \cdot \delta (t - \tau) + 0.16 \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$		:$$h_{\rm KR}(t) = 0.36 \cdot \delta (t ) +0.48 \cdot \delta (t - \tau) + 0.16 \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$
	*In contrast, with  $h_0 = 0.6$,  $h_1 = 0.4$,  $τ_0 = τ$  and  $τ_1 = 0$:		*In contrast, with  $h_0 = 0.6$,  $h_1 = 0.4$,  $τ_0 = τ$  and  $τ_1 = 0$:
	:$$h_{\rm KR}(t) = 0.6 \cdot h_0 \cdot \delta (t - \tau ) + 0.4 \cdot h_0 \cdot \delta (t - 2\tau) + 0.6 \cdot h_1 \cdot \delta (t) + 0.4 \cdot h_1 \cdot \delta (t - \tau)= 0.24 \cdot \delta (t ) +0.52 \cdot \delta (t - \tau) + 0.24 \cdot \delta (t - 2\tau) \hspace{0.05cm}.$$		:$$h_{\rm KR}(t) = 0.6 \cdot h_0 \cdot \delta (t - \tau ) + 0.4 \cdot h_0 \cdot \delta (t - 2\tau) + 0.6 \cdot h_1 \cdot \delta (t) + 0.4 \cdot h_1 \cdot \delta (t - \tau)= 0.24 \cdot \delta (t ) +0.52 \cdot \delta (t - \tau) + 0.24 \cdot \delta (t - 2\tau) \hspace{0.05cm}.$$
	*Here, the additional condition  $A_0 = A_2$  is satisfied.  Thus, the result we are looking for is:		*Here,  the additional condition  $A_0 = A_2$  is satisfied.  Thus,  the result we are looking for is:
	:$$ \underline{\tau_0 = \tau = 1\,{\rm µ s} \hspace{0.05cm},\hspace{0.2cm}\tau_1 =0} \hspace{0.05cm}.$$		:$$ \underline{\tau_0 = \tau = 1\,{\rm µ s} \hspace{0.05cm},\hspace{0.2cm}\tau_1 =0} \hspace{0.05cm}.$$

Line 96:		Line 100:
	'''(4)'''  The following must apply to the normalization factor:		'''(4)'''  The following must apply to the normalization factor:
	:$$K= \frac{1}{h_0^2 + h_1^2} = \frac{1}{0.6^2 + 0.4^2} = \frac{1}{0.52} \hspace{0.15cm}\underline {\approx 1.923} \hspace{0.05cm}.$$		:$$K= \frac{1}{h_0^2 + h_1^2} = \frac{1}{0.6^2 + 0.4^2} = \frac{1}{0.52} \hspace{0.15cm}\underline {\approx 1.923} \hspace{0.05cm}.$$
	*This gives for the common impulse response  $($it holds $0.24/0.52 = 6/13)$:		*This gives for the common impulse response  $($it holds  $0.24/0.52 = 6/13)$:
	:$$ h_{\rm KR}(t) = \frac{6}{13} \cdot \delta (t ) + 1.00 \cdot \delta (t - \tau) + \frac{6}{13} \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$		:$$ h_{\rm KR}(t) = \frac{6}{13} \cdot \delta (t ) + 1.00 \cdot \delta (t - \tau) + \frac{6}{13} \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$

Line 103:		Line 107:
	[[File:P_ID1902__Mod_Z_5_5e.png\|right\|frame\|Signals to illustrate the rake receiver]]		[[File:P_ID1902__Mod_Z_5_5e.png\|right\|frame\|Signals to illustrate the rake receiver]]
	'''(5)'''  <u>Statements 1 and 4</u> are correct, as shown in the diagram:		'''(5)'''  <u>Statements 1 and 4</u> are correct, as shown in the diagram:
	*For the received signal  $r(t)$  ~~and for the rake output signal  $b(t)$ ~~ holds:		*For the received signal  $r(t)$  holds:
	:$$r(t) = 0.6 \cdot s(t) + 0.4 \cdot s (t - 1\,{\rm µ s})\hspace{0.05cm},$$		:$$r(t) = 0.6 \cdot s(t) + 0.4 \cdot s (t - 1\,{\rm µ s})\hspace{0.05cm},$$
			*and for the rake output signal  $b(t)$:
	:$$b(t) = \frac{6}{13} \cdot s(t) + 1 \cdot s (t - 1\,{\rm µ s}) + \frac{6}{13} \cdot s (t - 2\,{\rm µ s}) \hspace{0.05cm}.$$		:$$b(t) = \frac{6}{13} \cdot s(t) + 1 \cdot s (t - 1\,{\rm µ s}) + \frac{6}{13} \cdot s (t - 2\,{\rm µ s}) \hspace{0.05cm}.$$
	*The overshoot of the output signal   ⇒   $b(t) > 1$  is due to the normalization factor  $K = 25/13$.		*The overshoot of the output signal   ⇒   $b(t) > 1$  is due to the normalization factor  $K = 25/13$.

Guenter at 14:49, 17 December 2021

2021-12-17T14:49:42Z

← Older revision		Revision as of 16:49, 17 December 2021
Line 23:		Line 23:


			Notes:



	''Notes:''
	*The exercise belongs to the chapter  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation\|Error Probability of Direct-Sequence Spread Spectrum Modulation]].		*The exercise belongs to the chapter  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation\|Error Probability of Direct-Sequence Spread Spectrum Modulation]].
	*Reference is made in particular to the section  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Principle_of_the_rake_receiver \|Principle of the rake receiver]].		*Reference is made in particular to the section  [[Modulation_Methods/Error_Probability_of_Direct-Sequence_Spread_Spectrum_Modulation#Principle_of_the_rake_receiver \|Principle of the rake receiver]].

Guenter at 14:48, 17 December 2021

2021-12-17T14:48:55Z

← Older revision		Revision as of 16:48, 17 December 2021
Line 6:		Line 6:
	The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:		The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:
	:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$		:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$
	Let the delay on the secondary path be  $τ = 1 \ \rm µ s$.  ~~Drawn below is the structure of a rake receiver  (green background)  with general coefficients  $K$,  $h_0$,  $h_1$,  $τ_0$  and  $τ_1$.~~		Let the delay on the secondary path be  $τ = 1 \ \rm µ s$.

	~~The purpose~~ of ~~the~~ rake receiver ~~is to combine the energy of the two signal paths~~,  ~~making the decision more reliable.~~		Drawn below is the structure of a rake receiver  (green background)  with general coefficients  $K$,  $h_0$,  $h_1$,  $τ_0$  and  $τ_1$.

	The combined impulse response of the channel  (German:  "Kanal"   ⇒   subscript "K")  and the rake receiver can be expressed in the form		*The purpose of the rake receiver is to combine the energy of the two signal paths,  making the decision more reliable.

			*The combined impulse response of the channel  (German:  "Kanal"   ⇒   subscript "K")  and the rake receiver can be expressed in the form
	:$$h_{\rm KR}(t) = A_0 \cdot \delta (t ) + A_1 \cdot \delta (t - \tau) + A_2 \cdot \delta (t - 2\tau)$$		:$$h_{\rm KR}(t) = A_0 \cdot \delta (t ) + A_1 \cdot \delta (t - \tau) + A_2 \cdot \delta (t - 2\tau)$$
	but only if the rake coefficients  $h_0$,  $h_1$,  $τ_0$  and  $τ_1$  are appropriately chosen.  The main part of  $h_{\rm KR}(t)$  is supposed to be at  $t = τ$.		:but only if the rake coefficients  $h_0$,  $h_1$,  $τ_0$  and  $τ_1$  are appropriately chosen.
			*The main part of  $h_{\rm KR}(t)$  is supposed to be at  $t = τ$.
	The constant  $K$  is to be chosen so that the amplitude of the main path  $A_1 = 1$ :		*The constant  $K$  is to be chosen so that the amplitude of the main path  $A_1 = 1$ :
	:$$K= \frac{1}{h_0^2 + h_1^2}.$$		:$$K= \frac{1}{h_0^2 + h_1^2}.$$
	Apart from the rake parameters,  the signals  $r(t)$  and  $b(t)$ are sought when $s(t)$  is a rectangle of height  $s_0 = 1$  and width  $T = \ \rm 5 µ s$.		Apart from the rake parameters,  the signals  $r(t)$  and  $b(t)$ are sought when $s(t)$  is a rectangle of height  $s_0 = 1$  and width  $T = \ \rm 5 µ s$.

Guenter at 14:46, 17 December 2021

2021-12-17T14:46:41Z

Hwang at 12:16, 13 December 2021

2021-12-13T12:16:40Z

← Older revision		Revision as of 14:16, 13 December 2021
Line 83:		Line 83:


	'''(3)'''  ~~Wir setzen zunächst vereinbarungsgemäß~~  $K = 1$.		'''(3)'''  We first set  $K = 1$ as agreed.
	*~~Insgesamt kommt man über vier Wege von~~  $s(t)$  ~~zum Ausgangssignal~~  $b(t)$.		*Altogether we get from  $s(t)$  to the output signal  $b(t)$ via four paths.
	*~~Um die vorgegebene~~  $h_{\rm KR}(t)$~~–Gleichung zu erfüllen~~, ~~muss entweder~~  $τ_0 = 0$  ~~gelten oder~~  $τ_1 = 0$.  ~~Mit~~  $τ_0 = 0$  ~~erhält man für die Impulsantwort~~:		*To satisfy the given  $h_{\rm KR}(t)$ equation, either  $τ_0 = 0$  must hold or  $τ_1 = 0$.  With  $τ_0 = 0$  we obtain for the impulse response:
	:$$h_{\rm KR}(t) = 0.6 \cdot h_0 \cdot \delta (t ) + 0.4 \cdot h_0 \cdot \delta (t - \tau) + 0.6 \cdot h_1 \cdot \delta (t -\tau_1) + 0.4 \cdot h_1 \cdot \delta (t - \tau-\tau_1) \hspace{0.05cm}.$$		:$$h_{\rm KR}(t) = 0.6 \cdot h_0 \cdot \delta (t ) + 0.4 \cdot h_0 \cdot \delta (t - \tau) + 0.6 \cdot h_1 \cdot \delta (t -\tau_1) + 0.4 \cdot h_1 \cdot \delta (t - \tau-\tau_1) \hspace{0.05cm}.$$
	*~~Um die „Hauptenergie” auf einen Zeitpunkt bündeln zu können~~, ~~müsste dann~~  $τ_1 = τ$  ~~gewählt werden~~.  ~~Mit~~  $h_0 = 0.6$  ~~und~~  $h_1 = 0.4$  ~~erhält man dann~~  $A_0 ≠ A_2$:		*To be able to focus the "main energy" on a time point,  $τ_1 = τ$  would then have to be chosen.  With  $h_0 = 0.6$  and  $h_1 = 0.4$,  we then obtain  $A_0 ≠ A_2$:
	:$$h_{\rm KR}(t) = 0.36 \cdot \delta (t ) +0.48 \cdot \delta (t - \tau) + 0.16 \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$		:$$h_{\rm KR}(t) = 0.36 \cdot \delta (t ) +0.48 \cdot \delta (t - \tau) + 0.16 \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$
	*~~Dagegen ergibt sich mit~~  $h_0 = 0.6$,  $h_1 = 0.4$,  $τ_0 = τ$  ~~und~~  $τ_1 = 0$:		*In contrast, with  $h_0 = 0.6$,  $h_1 = 0.4$,  $τ_0 = τ$  and  $τ_1 = 0$:
	:$$h_{\rm KR}(t) = 0.6 \cdot h_0 \cdot \delta (t - \tau ) + 0.4 \cdot h_0 \cdot \delta (t - 2\tau) + 0.6 \cdot h_1 \cdot \delta (t) + 0.4 \cdot h_1 \cdot \delta (t - \tau)= 0.24 \cdot \delta (t ) +0.52 \cdot \delta (t - \tau) + 0.24 \cdot \delta (t - 2\tau) \hspace{0.05cm}.$$		:$$h_{\rm KR}(t) = 0.6 \cdot h_0 \cdot \delta (t - \tau ) + 0.4 \cdot h_0 \cdot \delta (t - 2\tau) + 0.6 \cdot h_1 \cdot \delta (t) + 0.4 \cdot h_1 \cdot \delta (t - \tau)= 0.24 \cdot \delta (t ) +0.52 \cdot \delta (t - \tau) + 0.24 \cdot \delta (t - 2\tau) \hspace{0.05cm}.$$
	*~~Hier ist die Zusatzbedingung~~  $A_0 = A_2$  ~~erfüllt~~.  ~~Somit lautet das gesuchte Ergebnis~~:		*Here, the additional condition  $A_0 = A_2$  is satisfied.  Thus, the result we are looking for is:
	:$$ \underline{\tau_0 = \tau = 1\,{\rm µ s} \hspace{0.05cm},\hspace{0.2cm}\tau_1 =0} \hspace{0.05cm}.$$		:$$ \underline{\tau_0 = \tau = 1\,{\rm µ s} \hspace{0.05cm},\hspace{0.2cm}\tau_1 =0} \hspace{0.05cm}.$$



	'''(4)'''  ~~Für den Normierungsfaktor muss gelten~~:		'''(4)'''  The following must apply to the normalization factor:
	:$$K= \frac{1}{h_0^2 + h_1^2} = \frac{1}{0.6^2 + 0.4^2} = \frac{1}{0.52} \hspace{0.15cm}\underline {\approx 1.923} \hspace{0.05cm}.$$		:$$K= \frac{1}{h_0^2 + h_1^2} = \frac{1}{0.6^2 + 0.4^2} = \frac{1}{0.52} \hspace{0.15cm}\underline {\approx 1.923} \hspace{0.05cm}.$$
	*~~Damit erhält man für die gemeinsame Impulsantwort~~  $($~~es gilt~~ $0.24/0.52 = 6/13)$:		*This gives for the common impulse response  $($it holds $0.24/0.52 = 6/13)$:
	:$$ h_{\rm KR}(t) = \frac{6}{13} \cdot \delta (t ) + 1.00 \cdot \delta (t - \tau) + \frac{6}{13} \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$		:$$ h_{\rm KR}(t) = \frac{6}{13} \cdot \delta (t ) + 1.00 \cdot \delta (t - \tau) + \frac{6}{13} \cdot \delta (t - 2\tau)\hspace{0.05cm}.$$



	[[File:P_ID1902__Mod_Z_5_5e.png\|right\|frame\|~~Signale zur Verdeutlichung des Rake–Empfängers~~]]		[[File:P_ID1902__Mod_Z_5_5e.png\|right\|frame\|Signals to illustrate the rake receiver]]
	'''(5)'''  ~~Richtig sind die~~ <u>~~Aussagen~~ 1 ~~und~~ 4</u>, ~~wie die Grafik zeigt~~:		'''(5)'''  <u>Statements 1 and 4</u> are correct, as shown in the diagram:
	*~~Für das Empfangssignal~~  $r(t)$  ~~und für das Rake–Ausgangssignal~~  $b(t)$  ~~gilt~~:		*For the received signal  $r(t)$  and for the rake output signal  $b(t)$  holds:
	:$$r(t) = 0.6 \cdot s(t) + 0.4 \cdot s (t - 1\,{\rm µ s})\hspace{0.05cm},$$		:$$r(t) = 0.6 \cdot s(t) + 0.4 \cdot s (t - 1\,{\rm µ s})\hspace{0.05cm},$$
	:$$b(t) = \frac{6}{13} \cdot s(t) + 1 \cdot s (t - 1\,{\rm µ s}) + \frac{6}{13} \cdot s (t - 2\,{\rm µ s}) \hspace{0.05cm}.$$		:$$b(t) = \frac{6}{13} \cdot s(t) + 1 \cdot s (t - 1\,{\rm µ s}) + \frac{6}{13} \cdot s (t - 2\,{\rm µ s}) \hspace{0.05cm}.$$
	*~~Die Überhöhung des Ausgangssignals~~   ⇒   $b(t) > 1$  ~~ist auf den Normierungsfaktor~~  $K = 25/13$  ~~zurückzuführen.~~		*The overshoot of the output signal   ⇒   $b(t) > 1$  is due to the normalization factor  $K = 25/13$.
	*~~Mit~~  $K = 1$  ~~wäre der Maximalwert von~~  $b(t)$  ~~tatsächlich~~  $1$.		*With  $K = 1$,  the maximum value of  $b(t)$  would actually be  $1$.

	{{ML-Fuß}}		{{ML-Fuß}}

Hwang at 10:57, 13 December 2021

2021-12-13T10:57:34Z

Show changes

← Older revision		Revision as of 16:46, 17 December 2021
Line 3:		Line 3:
	}}		}}

	[[File:P_ID1888__Mod_Z_5_5.png\|right\|frame\|~~Two~~-way channel <br>& rake '''KORREKTUR''': two-way channel, rake receiver]]		[[File:P_ID1888__Mod_Z_5_5.png\|right\|frame\|2-way channel <br>& rake receiver '''KORREKTUR''': two-way channel, rake receiver]]
	The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:		The diagram shows a two-way channel  (yellow background).  The corresponding descriptive equation is:
	:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$		:$$ r(t) =0.6 \cdot s(t) + 0.4 \cdot s (t - \tau) \hspace{0.05cm}.$$
	Let the delay on the secondary path be  $τ = 1 \ \rm µ s$.  Drawn below is the structure of a rake receiver  (green background)  with general coefficients  $K$,  $h_0$,  $h_1$,  $τ_0$  and  $τ_1$.		Let the delay on the secondary path be  $τ = 1 \ \rm µ s$.  Drawn below is the structure of a rake receiver  (green background)  with general coefficients  $K$,  $h_0$,  $h_1$,  $τ_0$  and  $τ_1$.

	The purpose of the rake receiver is to combine the energy of the two signal paths, making the decision more reliable.		The purpose of the rake receiver is to combine the energy of the two signal paths,  making the decision more reliable.

	The combined impulse response of the channel and the rake receiver can be expressed in the form		The combined impulse response of the channel  (German:  "Kanal"   ⇒   subscript "K")  and the rake receiver can be expressed in the form
	:$$h_{\rm KR}(t) = A_0 \cdot \delta (t ) + A_1 \cdot \delta (t - \tau) + A_2 \cdot \delta (t - 2\tau)$$		:$$h_{\rm KR}(t) = A_0 \cdot \delta (t ) + A_1 \cdot \delta (t - \tau) + A_2 \cdot \delta (t - 2\tau)$$
	but only if the rake coefficients  $h_0$,  $h_1$,  $τ_0$  and  $τ_1$  are appropriately chosen.  The main part of  $h_{\rm KR}(t)$  is supposed to be at  $t = τ$.		but only if the rake coefficients  $h_0$,  $h_1$,  $τ_0$  and  $τ_1$  are appropriately chosen.  The main part of  $h_{\rm KR}(t)$  is supposed to be at  $t = τ$.
Line 16:		Line 16:
	The constant  $K$  is to be chosen so that the amplitude of the main path  $A_1 = 1$ :		The constant  $K$  is to be chosen so that the amplitude of the main path  $A_1 = 1$ :
	:$$K= \frac{1}{h_0^2 + h_1^2}.$$		:$$K= \frac{1}{h_0^2 + h_1^2}.$$
	Apart from the rake parameters, the signals  $r(t)$  and  $b(t)$ are sought when $s(t)$  is a rectangle of height  $s_0 = 1$  and width  $T = \ \rm 5 µ s$.		Apart from the rake parameters,  the signals  $r(t)$  and  $b(t)$ are sought when $s(t)$  is a rectangle of height  $s_0 = 1$  and width  $T = \ \rm 5 µ s$.


Line 36:		Line 36:
	{Which statements are valid for the channel impulse response  $h_{\rm K}(t)$?		{Which statements are valid for the channel impulse response  $h_{\rm K}(t)$?
	\|type="[]"}		\|type="[]"}
	+ $h_{\rm K}(t)$  consists of two Dirac functions.		+ $h_{\rm K}(t)$  consists of two Dirac delta functions.
	- $h_{\rm K}(t)$  is complex-valued.		- $h_{\rm K}(t)$  is complex-valued.
	- $h_{\rm K}(t)$  is a function periodic with delay time  $\tau$.		- $h_{\rm K}(t)$  is a function periodic with delay time  $\tau$.
Line 42:		Line 42:
	{Which statements are true for the channel frequency response  $H_{\rm K}(f)$?		{Which statements are true for the channel frequency response  $H_{\rm K}(f)$?
	\|type="[]"}		\|type="[]"}
	- ~~ ~~$H_{\rm K}(f = 0) = 2$ is true.		- $H_{\rm K}(f = 0) = 2$  is true.
	+ $H_{\rm K}(f)$  is complex-valued.		+ $H_{\rm K}(f)$  is complex-valued.
	+ $\|H_{\rm K}(f)\|$  is a function periodic with frequency  $1/τ$.		+ $\|H_{\rm K}(f)\|$  is a function periodic with frequency  $1/τ$.