Aufgaben:Exercise 1.10Z: Gaussian Band-Pass - Revision history

Guenter at 14:13, 7 May 2022

2022-05-07T14:13:35Z

Guenter at 14:00, 7 May 2022

2022-05-07T14:00:18Z

Hwang at 15:46, 29 March 2022

2022-03-29T15:46:42Z

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Guenter at 14:07, 19 March 2022

2022-03-19T14:07:30Z

Guenter: Guenter moved page Aufgaben:Exercise 1.10Z: Gaussian Band-pass to Aufgaben:Exercise 1.10Z: Gaussian Band-Pass

2022-03-19T14:07:06Z

Guenter moved page Exercise 1.10Z: Gaussian Band-pass to Exercise 1.10Z: Gaussian Band-Pass

Guenter: Guenter moved page Aufgaben:Aufgabe 1.10Z: Gauß-Bandpass to Aufgaben:Exercise 1.10Z: Gaussian Band-pass

2022-03-19T14:03:21Z

Guenter moved page Aufgabe 1.10Z: Gauß-Bandpass to Exercise 1.10Z: Gaussian Band-pass

Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:47:17Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

Javier: Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

2020-07-09T12:43:46Z

Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

Tasnad: Text replacement - "[File:" to "[File:"

2020-05-26T14:09:24Z

Text replacement - "[File:" to "[File:"

Guenter at 15:11, 7 February 2019

2019-02-07T15:11:49Z

@@ Line 72: / Line 72: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; For the bandpass frequency response $H_{\rm K}(f)$ we can write:
+'''(1)'''&nbsp; For the band-pass frequency response&nbsp; $H_{\rm K}(f)$&nbsp; we can write:
 :$$H_{\rm K}(f) = H_{\rm K,\hspace{0.04cm} TP}(f) \star \big [ \delta (f - f_{\rm M}) + \delta (f + f_{\rm M}) \big ] .$$
-*The Fourier inverse transform of the bracket expression yields a cosine function of frequency $f_{\rm M}$ with amplitude $2$.
+*The Fourier inverse transform of the bracket expression yields a cosine function of frequency&nbsp; $f_{\rm M}$&nbsp; with amplitude&nbsp; $2$.
 *Thus, according to the convolution theorem:
 :$$h_{\rm K}(t) = 2 \cdot \Delta f_{\rm K} \cdot {\rm exp} \left [ - \pi \cdot \left ( {\Delta f_{\rm K}} \cdot t \right )^2 \right ] \cdot \cos(2 \pi f_{\rm M} t ) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}h_{\rm K}(t = 0)/\Delta f_{\rm K} \hspace{0.1cm}\underline {= 2}.$$
-*This means: The TP impulse response $h_{\rm K,\hspace{0.04cm}TP}(t)$ is identical in shape to the envelope of the bp impulse response $h_{\rm K}(t)$, but twice as large.
+*This means:&nbsp; The low-pass impulse response&nbsp; $h_{\rm K,\hspace{0.04cm}TP}(t)$&nbsp; is identical in shape to the envelope of the band-pass impulse response&nbsp; $h_{\rm K}(t)$,&nbsp; but twice as large.
-'''(2)'''&nbsp;  <u>Statements 2, 3 and 4</u> are correct:
+*The same is true for&nbsp; $h_{\rm MKD}(t)$&nbsp; also considering the&nbsp; $\pm 2f_{\rm T}$&nbsp; parts,&nbsp; since&nbsp; $H_{\rm MKD}(f)$&nbsp; is an even function with respect&nbsp; to $f = 0$.
-*The first statement is false because $H_{\rm MKD}(f)$ also has components around $\pm 2f_{\rm T}$.
-*The time function $h_{\rm K,\hspace{0.04cm}TP}(t)$ is real according to the given equation.
+*The diagram shows&nbsp; $H_{\rm MKD}(f)$,&nbsp; which also has components around&nbsp; $\pm 2f_{\rm T}$.&nbsp; At low frequencies,&nbsp; $H_{\rm K,\hspace{0.04cm}TP}(f)$&nbsp; is identical to&nbsp; $H_{\rm MKD}(f)$.
 '''(3)'''&nbsp;  Only <u>solution 4</u> is correct:
 *Here $H_{\rm K,\hspace{0.04cm}TP}(f)$ and $H_{\rm MKD}(f)$ differ even at the low frequencies.
-*$H_{\rm K,\hspace{0.04cm}TP}(f)$ is a Gaussian function with the maximum at $f_{ε} = f_{\rm M} - f_{\rm T}$.
+*$H_{\rm K,\hspace{0.04cm}TP}(f)$&nbsp; is a Gaussian function with maximum at&nbsp; $f_{ε} = f_{\rm M} - f_{\rm T}$.
-*Because of this asymmetry, $h_{\rm K,\hspace{0.04cm}TP}(t)$ is complex.
+*Because of this asymmetry,&nbsp; $h_{\rm K,\hspace{0.04cm}TP}(t)$&nbsp; is complex.
-*In contrast, $H_{\rm MKD}(f)$ is still an even function with respect to $f = 0$ with real impulse response $h_{\rm MKD}(t)$.
+*In contrast,&nbsp; $H_{\rm MKD}(f)$&nbsp; is still an even function with respect to&nbsp; $f = 0$&nbsp; with real impulse response&nbsp; $h_{\rm MKD}(t)$.
-*$H_{\rm MKD}(f)$ is composed of two Gaussian functions at $± f_ε$.
+*$H_{\rm MKD}(f)$&nbsp; is composed of two Gaussian functions at&nbsp; $± f_ε$.
-'''(4)'''&nbsp; Correct is of course the <u>first answer.</u>
+'''(4)'''&nbsp; Correct is of course the&nbsp; <u>first answer.</u>

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1697__Dig_Z_4_3.png|right|frame|Gaussian bandpass channel]]
+[[File:P_ID1697__Dig_Z_4_3.png|right|frame|Gaussian band-pass channel]]
 For this exercise we assume:
-*Binary phase modulation (BPSK) is used for modulation.
+*Binary phase modulation&nbsp; $\rm (BPSK)$&nbsp; is used for modulation.
-*Demodulation is frequency and phase synchronous.
+*Demodulation is  synchronous in frequency and phase.
-For carrier frequency modulated transmission, the channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; must always be assumed to be a bandpass. The channel parameters are, for example, the center frequency &nbsp;$f_{\rm M}$&nbsp; and the bandwidth &nbsp;$\Delta f_{\rm K}$, where the center frequency &nbsp;$f_{\rm M}$&nbsp; often coincides with the carrier frequency &nbsp;$f_{\rm T}$.&nbsp;
+For carrier frequency modulated transmission,&nbsp; the channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; must always be assumed to be a band-pass.&nbsp; The channel parameters are&nbsp; e.g.&nbsp; the center frequency &nbsp;$f_{\rm M}$&nbsp; and the bandwidth &nbsp;$\Delta f_{\rm K}$,&nbsp; where the center frequency&nbsp; (German:&nbsp; "Mittenfrequenz" &nbsp; &rArr; &nbsp; subscipt:&nbsp; "M")&nbsp; $f_{\rm M}$&nbsp; often coincides with the carrier frequency&nbsp; (German:&nbsp; "Trägerfrequenz" &nbsp; &rArr; &nbsp; subscipt:&nbsp; "T")&nbsp; $f_{\rm T}$.&nbsp;
-In this exercise, in particular, we will assume a Gaussian bandpass according to the diagram. For its frequency response holds:
+In this exercise we will assume a Gaussian band-pass according to the diagram.&nbsp; For its frequency response holds:
 :$$H_{\rm K}(f) = {\rm exp} \left [ - \pi \cdot \left ( \frac {f - f_{\rm M} }{\Delta f_{\rm K}}\right )^2 \right ]
   +{\rm exp} \left [ - \pi \cdot \left ( \frac {f + f_{\rm M} }{\Delta f_{\rm K}}\right )^2 \right ]$$
-For a simpler description, one often uses the equivalent TP frequency response &nbsp;$H_{\rm K,TP}(f)$. This results from &nbsp;$H_{\rm K}(f)$&nbsp; by
+For a simpler description,&nbsp; one often uses the equivalent low-pass&nbsp; ("TP")&nbsp; frequency response &nbsp;$H_{\rm K,TP}(f)$.&nbsp; This results from &nbsp;$H_{\rm K}(f)$&nbsp; by
 *truncating the components at negative frequencies,
 *shifting the spectrum by &nbsp;$f_{\rm T}$&nbsp; to the left.
-In the considered example with &nbsp;$f_{\rm T} = f_{\rm M}$&nbsp; for the equivalent TP frequency response results:
+In the considered example with &nbsp;$f_{\rm T} = f_{\rm M}$&nbsp; for the equivalent low-pass frequency response results:
 :$$ H_{\rm K,\hspace{0.04cm} TP}(f) = {\rm e}^ { - \pi \hspace{0.04cm}\cdot \hspace{0.04cm}\left ( {f }/{\Delta f_{\rm K}}\right )^2 }.$$
-The corresponding time function (Fourier inverse transform) is:
+The corresponding time function&nbsp; ("inverse Fourier transform")&nbsp; is:
 :$$ h_{\rm K,\hspace{0.04cm} TP}(t) = \Delta f_{\rm K} \cdot {\rm e}^ { - \pi \hspace{0.04cm}\cdot \hspace{0.04cm}\left ( {\Delta f_{\rm K}} \cdot t \right )^2 }.$$
-However, the frequency response is also suitable for describing a phase-synchronous BPSK system in the low-pass range
+However,&nbsp; the frequency response is also suitable for describing a phase-synchronous BPSK system in the low-pass range
 :$$H_{\rm MKD}(f) = {1}/{2} \cdot \left [ H_{\rm K}(f-f_{\rm T}) + H_{\rm K}(f+f_{\rm T})\right ] ,$$
-where "MKD" stands for modulator – channel (Kanal) – demodulator. Often - but not always - &nbsp;$H_{\rm MKD}(f)$&nbsp; and &nbsp;$H_{\rm K,TP}(f)$&nbsp; are identical.
+where&nbsp; "MKD"&nbsp; stands for&nbsp; "modulator – channel (Kanal) – demodulator".&nbsp; Often - but not always - &nbsp;$H_{\rm MKD}(f)$&nbsp; and &nbsp;$H_{\rm K,TP}(f)$&nbsp; are identical.
-''Notes:''
+*Reference is made in particular to the section&nbsp; [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Baseband_model_for_ASK_and_BPSK|"Baseband model for ASK and BPSK"]].
@@ Line 42: / Line 43: @@
 <quiz display=simple>
-{Give the impulse response &nbsp;$h_{\rm K}(t)$&nbsp; of the Gaussian bandpass channel. What is the (normalized) value for time &nbsp;$t = 0$?
+{Give the impulse response &nbsp;$h_{\rm K}(t)$&nbsp; of the Gaussian band-pass channel.&nbsp; What is the&nbsp; (normalized)&nbsp; value for time &nbsp;$t = 0$?
 |type="{}"}
 $ h_{\rm K}(t)/\Delta f_{\rm K} \  =  \ $ { 2 3% }
@@ Line 54: / Line 55: @@
 +The time function &nbsp;$h_{\rm MKD}(t)$&nbsp; is real.
-{Which statements are true under the condition $f_{\rm T} \neq f_{\rm M}$?
+{Which statements are true under the condition&nbsp; $f_{\rm T} \neq f_{\rm M}$?
 |type="[]"}
 -$H_{\rm K,TP}(f)$&nbsp; and &nbsp;$H_{\rm MKD}(f)$&nbsp; coincide completely.

← Older revision		Revision as of 14:07, 19 March 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^1.5 ~~Lineare digitale~~ Modulation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^1.5 Linear Digital Modulation^]]

← Older revision		Revision as of 12:47, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^1.5 Lineare digitale Modulation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^1.5 Lineare digitale Modulation^]]

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 ''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digitalsignalübertragung/Lineare_digitale_Modulation_–_Kohärente_Demodulation|Lineare digitale Modulation – Kohärente Demodulation]].
+*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digital_Signal_Transmission/Lineare_digitale_Modulation_–_Kohärente_Demodulation|Lineare digitale Modulation – Kohärente Demodulation]].
-*Bezug genommen wird insbesondere auf die Seite&nbsp; [[Digitalsignalübertragung/Lineare_digitale_Modulation_–_Kohärente_Demodulation#Basisbandmodell_f.C3.BCr_ASK_und_BPSK|Basisbandmodell für ASK und BPSK]].
+*Bezug genommen wird insbesondere auf die Seite&nbsp; [[Digital_Signal_Transmission/Lineare_digitale_Modulation_–_Kohärente_Demodulation#Basisbandmodell_f.C3.BCr_ASK_und_BPSK|Basisbandmodell für ASK und BPSK]].

@@ Line 72: / Line 72: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; Für den Bandpass–Frequenzgang $H_{\rm K}(f)$ kann geschrieben werden:
-:$$H_{\rm K}(f) = H_{\rm K,\hspace{0.04cm} TP}(f) \star \left [ \delta (f - f_{\rm M}) + \delta (f + f_{\rm M}) \right ] .$$
+:$$H_{\rm K}(f) = H_{\rm K,\hspace{0.04cm} TP}(f) \star \big [ \delta (f - f_{\rm M}) + \delta (f + f_{\rm M}) \big ] .$$
-Die Fourierrücktransformierte des Klammerausdrucks liefert eine Cosinusfunktion der Frequenz $f_{\rm M}$ mit der Amplitude $2$. Nach dem Faltungssatz gilt somit:
+*Die Fourierrücktransformierte des Klammerausdrucks liefert eine Cosinusfunktion der Frequenz $f_{\rm M}$ mit der Amplitude $2$.
 :$$h_{\rm K}(t) = 2 \cdot \Delta f_{\rm K} \cdot {\rm exp} \left [ - \pi \cdot \left ( {\Delta f_{\rm K}} \cdot t \right )^2 \right ] \cdot \cos(2 \pi f_{\rm M} t ) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}h_{\rm K}(t = 0)/\Delta f_{\rm K} \hspace{0.1cm}\underline {= 2}.$$
-Das heißt: Die TP–Impulsantwort $h_{\rm K,\hspace{0.04cm}TP}(t)$ ist formgleich mit der Hüllkurve der BP–Impulsantwort $h_{\rm K}(t)$, aber doppelt so groß.
+*Das heißt: Die TP–Impulsantwort $h_{\rm K,\hspace{0.04cm}TP}(t)$ ist formgleich mit der Hüllkurve der BP–Impulsantwort $h_{\rm K}(t)$, aber doppelt so groß.
 '''(2)'''&nbsp;  Richtig sind die <u>Aussagen 2, 3 und 4:</u>
@@ Line 88: / Line 90: @@
 '''(3)'''&nbsp;  Richtig ist nur der <u>Lösungsvorschlag 4:</u>
 *Hier unterscheiden sich $H_{\rm K,\hspace{0.04cm}TP}(f)$ und $H_{\rm MKD}(f)$ auch bei den tiefen Frequenzen.
-*$H_{\rm K,\hspace{0.04cm}TP}(f)$ ist eine Gaußfunktion mit dem Maximum bei $f_{ε} = f_{\rm M} – f_{\rm T}$.
+*$H_{\rm K,\hspace{0.04cm}TP}(f)$ ist eine Gaußfunktion mit dem Maximum bei $f_{ε} = f_{\rm M} - f_{\rm T}$.
 *Aufgrund dieser Unsymmetrie ist $h_{\rm K,\hspace{0.04cm}TP}(t)$ komplex.
-*Dagegen ist $H_{\rm MKD}(f)$ weiterhin eine bezüglich $f = 0$ gerade Funktion mit reeller Impulsantwort $h_{\rm MKD}(t)$. $H_{\rm MKD}(f)$ setzt sich aus zwei Gaußfunktionen bei $± f_ε$ zusammen.
+*Dagegen ist $H_{\rm MKD}(f)$ weiterhin eine bezüglich $f = 0$ gerade Funktion mit reeller Impulsantwort $h_{\rm MKD}(t)$.