Aufgaben:Exercise 1.1: Music Signals: Difference between revisions

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{{quiz-Header|Buchseite=Signaldarstellung/Prinzip der Nachrichtenübertragung}}
{{quiz-Header|Buchseite=Signaldarstellung/Prinzip der Nachrichtenübertragung}}


[[File:P_ID339__Sig_A_1_1.png|right|frame|Music signals, original and <br> noisy and/or distorted?]]
[[File:P_ID339__Sig_A_1_1.png|right|frame|Music signals, original, <br> noisy and/or distorted?]]
On the right you see a ca.&nbsp; $\text{30 ms}$&nbsp; long section of a music signal&nbsp; <math>q(t)</math>. It is the piece &bdquo;For Elise&rdquo; by Ludwig van Beethoven.
On the right you see a ca.&nbsp; $\text{30 ms}$&nbsp; long section of a music signal&nbsp; <math>q(t)</math>. It is the piece &bdquo;For Elise&rdquo; by Ludwig van Beethoven.


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*The signal&nbsp; <math>v_1(t)</math>&nbsp; is undistorted compared to the original signal <math>q(t)</math>. The following applies: &nbsp; $v_1(t)=\alpha \cdot q(t-\tau) .$
*The signal&nbsp; <math>v_1(t)</math>&nbsp; is undistorted compared to the original signal <math>q(t)</math>. The following applies: &nbsp; $v_1(t)=\alpha \cdot q(t-\tau) .$


*Eine Dämpfung&nbsp; <math>\alpha</math>&nbsp; und eine Laufzeit&nbsp; <math>\tau</math>&nbsp; führen nicht zu Verzerrungen, sondern das Signal ist dann nur leiser und es kommt später als das Original.
*An attenuation&nbsp; <math>\alpha</math>&nbsp; and a delay&nbsp; <math>\tau</math>&nbsp; do not cause distortion, but the signal is then only quieter and comes later than the original.






'''(3)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
'''(3)'''&nbsp; Correct are the <u>solutions 1 and 3</u>:
*Man erkennt sowohl im dargestellten Signalverlauf&nbsp; <math>v_2(t)</math>&nbsp; als auch im Audiosignal&nbsp; ''additives Rauschen'' &nbsp; ⇒ &nbsp;   <u>Lösungsvorschlag 3</u>.  
*One can recognize both in the displayed signal&nbsp; <math>v_2(t)</math>&nbsp; and in the audio signal&nbsp; ''additive noise'' &nbsp; ⇒ &nbsp; <u>solution 3</u>.  
*Der Signalrauschabstand beträgt dabei ca.&nbsp; $\text{30 dB}$; dies ist aber aus dieser Darstellung nicht erkennbar.  
*The signal-to-noise ratio is approx. &nbsp; $\text{30 dB}$; but this cannot be seen from this representation.  
*Richtig ist aber auch der <u>Lösungsvorschlag 1</u>: &nbsp; Ohne diesen Rauschanteil wäre&nbsp; <math>v_2(t)</math>&nbsp; identisch mit&nbsp; <math>q(t)</math>.
*Correct is also the <u>solution 1</u>: &nbsp; Without this noise component&nbsp; <math>v_2(t)</math>&nbsp; identical with&nbsp; <math>q(t)</math>.




'''(4)'''&nbsp;  Das Signal&nbsp; <math>v_1(t)</math>&nbsp; ist formgleich mit dem Originalsignal&nbsp; <math>q(t)</math>&nbsp; und unterscheidet sich von diesem lediglich
'''(4)'''&nbsp;  The signal&nbsp; <math>v_1(t)</math>&nbsp; is identical in form to the original signal&nbsp; <math>q(t)</math>&nbsp; and differs from it only
*durch den Amplitudenfaktor&nbsp; $\alpha = \underline{\text{0.3}}$&nbsp;  (dies entspricht etwa&nbsp; $\text{–10 dB)}$  
*by the attenuation factor&nbsp; $\alpha = \underline{\text{0.3}}$&nbsp;  (dies entspricht etwa&nbsp; $\text{–10 dB)}$  
*und die Laufzeit&nbsp;  $\tau = \underline{10\,\text{ms}}$.
*and the delay&nbsp;  $\tau = \underline{10\,\text{ms}}$.
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[[Category:Aufgaben zu Signaldarstellung|^1. Grundbegriffe der Nachrichtentechnik^]]
[[Category:Aufgaben zu Signaldarstellung|^1. Grundbegriffe der Nachrichtentechnik^]]

Revision as of 14:15, 8 August 2020

Music signals, original,
noisy and/or distorted?

On the right you see a ca.  $\text{30 ms}$  long section of a music signal  [math]\displaystyle{ q(t) }[/math]. It is the piece „For Elise” by Ludwig van Beethoven.

  • Underneath are drawn two sink signals  [math]\displaystyle{ v_1(t) }[/math]  and  [math]\displaystyle{ v_2(t) }[/math], which were recorded after the transmission of the music signal  [math]\displaystyle{ q(t) }[/math]  over two different channels.
  • The following controls allow you to listen to the first fourteen seconds of each of the three audio signals  [math]\displaystyle{ q(t) }[/math][math]\displaystyle{ v_1(t) }[/math]  and  [math]\displaystyle{ v_2(t) }[/math].


Originalsignal  [math]\displaystyle{ q(t) }[/math]

Sinkensignal  [math]\displaystyle{ v_1(t) }[/math]

Sinkensignal  [math]\displaystyle{ v_2(t) }[/math]



Notes:



Questions

1 Estimate the signal frequency of  [math]\displaystyle{ q(t) }[/math]  in the displayed section.

The signal frequency is approximately  [math]\displaystyle{ f = 250\,\text{Hz} }[/math].
The signal frequency is approximately  [math]\displaystyle{ f = 500\,\text{Hz} }[/math].
The signal frequency is about  [math]\displaystyle{ f = 1\,\text{kHz} }[/math].

2 Which statements are true for the signal  [math]\displaystyle{ v_1(t) }[/math] ?

The signal  [math]\displaystyle{ v_1(t) }[/math]  is undistorted compared to [math]\displaystyle{ q(t) }[/math].
The signal  [math]\displaystyle{ v_1(t) }[/math]  shows distortions compared to  [math]\displaystyle{ q(t) }[/math] .
The signal  [math]\displaystyle{ v_1(t) }[/math]  is noisy compared to  [math]\displaystyle{ q(t) }[/math] .

3 Which statements are true for the signal  [math]\displaystyle{ v_2(t) }[/math] ?

The signal  [math]\displaystyle{ v_2(t) }[/math]  is undistorted compared to  [math]\displaystyle{ q(t) }[/math] .
The signal  [math]\displaystyle{ v_2(t) }[/math]  shows distortions compared to  [math]\displaystyle{ q(t) }[/math] .
The signal  [math]\displaystyle{ v_2(t) }[/math]  is noisy compared to  [math]\displaystyle{ q(t) }[/math] .

4 One of the signals is opposite the original  [math]\displaystyle{ q(t) }[/math]  undistorted and not noisy.
Estimate the attenuation factor and the running time for this.

[math]\displaystyle{ \alpha \ = \ }[/math]
[math]\displaystyle{ \tau \ = \ }[/math] $\ \text{ms}$


Solutions

(1)  Correct is the solution 2:

  • In the marked range of $20$ milliseconds approx.   $10$  oscillations can be detected.
  • From this the result  follows approximately for the signal frequency; $f = {10}/(20 \,\text{ms}) = 500 \,\text{Hz}$.


(2)  Correct is the solution 1:

  • The signal  [math]\displaystyle{ v_1(t) }[/math]  is undistorted compared to the original signal [math]\displaystyle{ q(t) }[/math]. The following applies:   $v_1(t)=\alpha \cdot q(t-\tau) .$
  • An attenuation  [math]\displaystyle{ \alpha }[/math]  and a delay  [math]\displaystyle{ \tau }[/math]  do not cause distortion, but the signal is then only quieter and comes later than the original.


(3)  Correct are the solutions 1 and 3:

  • One can recognize both in the displayed signal  [math]\displaystyle{ v_2(t) }[/math]  and in the audio signal  additive noise   ⇒   solution 3.
  • The signal-to-noise ratio is approx.   $\text{30 dB}$; but this cannot be seen from this representation.
  • Correct is also the solution 1:   Without this noise component  [math]\displaystyle{ v_2(t) }[/math]  identical with  [math]\displaystyle{ q(t) }[/math].


(4)  The signal  [math]\displaystyle{ v_1(t) }[/math]  is identical in form to the original signal  [math]\displaystyle{ q(t) }[/math]  and differs from it only

  • by the attenuation factor  $\alpha = \underline{\text{0.3}}$  (dies entspricht etwa  $\text{–10 dB)}$
  • and the delay  $\tau = \underline{10\,\text{ms}}$.