[[File:P_ID1683__Dig_A_4_3.png|right|frame|Unbalanced channel frequency response]]
Wir betrachten in dieser Aufgabe ein BPSK–System mit kohärenter Demodulation, das heißt, es gilt
In this exercise, we consider a BPSK system with coherent demodulation, i.e.
:$$s(t) \ = \ z(t) \cdot q(t),$$
:$$s(t) \ = \ z(t) \cdot q(t),$$
:$$b(t) \ = \ 2 \cdot z(t) \cdot r(t) .$$
:$$b(t) \ = \ 2 \cdot z(t) \cdot r(t) .$$
Die hier gewählten Bezeichnungen lehnen sich an das [[Digitalsignalübertragung/Lineare_digitale_Modulation_–_Kohärente_Demodulation#Gemeinsames_Blockschaltbild_f.C3.BCr_ASK_und_BPSK|Blockschaltbild]] im Theorieteil an.
The designations chosen here are based on the [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Common_block_diagram_for_ASK_and_BPSK|"block diagram"]] in the theory section.
Der Einfluss eines Kanalfrequenzgangs $H_{\rm K}(f)$ lässt sich in einfacher Weise berücksichtigen, wenn man diesen zusammen mit Modulator und Demodulator durch einen gemeinsamen Basisbandfrequenzgang beschreibt:
*Modulator und Demodulator quasi gegeneinander gekürzt,
*der Bandpasskanal $H_{\rm K}(f)$ in den Tiefpassbereich transformiert.
The influence of a channel frequency response $H_{\rm K}(f)$ can be taken into account in a simple way if it is described together with modulator and demodulator by a common baseband frequency response:
Die resultierende Übertragungsfunktion $H_{\rm MKD}(f)$ sollte man nicht mit der Tiefpass–Übertragungsfunktion $H_{\rm K,TP}(f)$ gemäß der Beschreibung in [[Signaldarstellung/Äquivalentes_Tiefpass-Signal_und_zugehörige_Spektralfunktion|Äquivalentes Tiefpass-Signal und zugehörige Spektralfunktion]] des Buches „Signaldarstellung” verwechseln, die sich aus $H_{\rm K}(f)$ durch Abschneiden der Anteile bei negativen Frequenzen sowie einer Frequenzverschiebung um $f_{\rm T}$ nach links ergibt.
*Thus the modulator and demodulator are virtually shortened against each other, and
''Hinweis:''
*the band-pass channel $H_{\rm K}(f)$ is transformed into the low-pass range.
Die Aufgabe gehört zum Themengebiet von [[Digitalsignalübertragung/Lineare_digitale_Modulation_–_Kohärente_Demodulation|Lineare digitale Modulation – Kohärente Demodulation]].
===Fragebogen===
The resulting transmission function $H_{\rm MKD}(f)$ should not be confused with the low-pass transmission function $H_{\rm K, \, TP}(f)$ as described in the chapter [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function|"Equivalent Low-Pass Signal and its Spectral Function"]] of the book "Signal Representation", which results from $H_{\rm K}(f)$ by truncating the components at negative frequencies as well as a frequency shift by the carrier frequency $f_{\rm T}$ to the left.
For frequency responses, in contrast to spectral functions, the doubling of the components at positive frequencies must be omitted.
Notes:
*The exercise belongs to the chapter [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation|"Linear Digital Modulation - Coherent Demodulation"]].
*Reference is made in particular to the section [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Baseband_model_for_ASK_and_BPSK|"Baseband model for ASK and BPSK"]].
-$h_{\rm MKD}(t)$ has equidistant zero crossings at distance $1/\Delta f_{\rm K}$.
+$h_{\rm MKD}(t)$ has equidistant zero crossings at distance $2/\Delta f_{\rm K}$.
Line 38:
Line 65:
</quiz>
</quiz>
===Musterlösung===
===Solution===
{{ML-Kopf}}
{{ML-Kopf}}
'''(1)'''
'''(1)''' <u>Statements 2, 3 and 4</u> are correct:
'''(2)'''
*$H_{\rm K,TP}(f)$ results from $H_{\rm K}(f)$ by cutting off the negative frequency components and shifting $f_{\rm T}$ to the left.
'''(3)'''
'''(4)'''
* For frequency responses – in contrast to spectra – the doubling of the components at positive frequencies is omitted. Therefore:
*Because of the real and asymmetrical spectral functions $H_{\rm K,\hspace{0.04cm}TP}(f),$ the corresponding time function (inverse Fourier transform) $h_{\rm K,\hspace{0.04cm}TP}(t)$ is complex according to the "Allocation Theorem".
'''(2)''' Here only the <u>third proposed solution</u> is correct:
*The spectral function $H_{\rm MKD}(f)$ always has an even real part and no imaginary part. Consequently $h_{\rm MKD}(t)$ is always real.
*If $H_{\rm K}(f)$ had additionally an imaginary part odd by $f= f_{\rm T}$, $H_{\rm MKD}(f)$ would have an imaginary part odd by $f = 0$. Thus $h_{\rm MKD}(t)$ would still be a real function.
The diagram illustrates the differences between $H_{\rm K,\hspace{0.04cm}TP}(f)$ and $H_{\rm MKD}(f)$. The parts of $H_{\rm MKD}(f)$ in the range around $\pm 2f_{\rm T}$ need not be considered further.
'''(3)''' $H_{\rm MKD}(f)$ is additively composed of a rectangle and a triangle, each with width $\Delta f_{\rm K}$ and height $0.5$. It follows:
In this exercise, we consider a BPSK system with coherent demodulation, i.e.
$$s(t) \ = \ z(t) \cdot q(t),$$
$$b(t) \ = \ 2 \cdot z(t) \cdot r(t) .$$
The designations chosen here are based on the "block diagram" in the theory section.
The influence of a channel frequency response $H_{\rm K}(f)$ can be taken into account in a simple way if it is described together with modulator and demodulator by a common baseband frequency response:
Thus the modulator and demodulator are virtually shortened against each other, and
the band-pass channel $H_{\rm K}(f)$ is transformed into the low-pass range.
The resulting transmission function $H_{\rm MKD}(f)$ should not be confused with the low-pass transmission function $H_{\rm K, \, TP}(f)$ as described in the chapter "Equivalent Low-Pass Signal and its Spectral Function" of the book "Signal Representation", which results from $H_{\rm K}(f)$ by truncating the components at negative frequencies as well as a frequency shift by the carrier frequency $f_{\rm T}$ to the left.
For frequency responses, in contrast to spectral functions, the doubling of the components at positive frequencies must be omitted.
Because of the real and asymmetrical spectral functions $H_{\rm K,\hspace{0.04cm}TP}(f),$ the corresponding time function (inverse Fourier transform) $h_{\rm K,\hspace{0.04cm}TP}(t)$ is complex according to the "Allocation Theorem".
Low-pass functions for $H_{\rm K}(f)$
(2) Here only the third proposed solution is correct:
The spectral function $H_{\rm MKD}(f)$ always has an even real part and no imaginary part. Consequently $h_{\rm MKD}(t)$ is always real.
If $H_{\rm K}(f)$ had additionally an imaginary part odd by $f= f_{\rm T}$, $H_{\rm MKD}(f)$ would have an imaginary part odd by $f = 0$. Thus $h_{\rm MKD}(t)$ would still be a real function.
The diagram illustrates the differences between $H_{\rm K,\hspace{0.04cm}TP}(f)$ and $H_{\rm MKD}(f)$. The parts of $H_{\rm MKD}(f)$ in the range around $\pm 2f_{\rm T}$ need not be considered further.
(3) $H_{\rm MKD}(f)$ is additively composed of a rectangle and a triangle, each with width $\Delta f_{\rm K}$ and height $0.5$. It follows:
