Difference between revisions of "Aufgaben:Exercise 1.1: Music Signals"

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+ The signal&nbsp; <math>v_2(t)</math>&nbsp; is noisy compared to&nbsp; <math>q(t)</math>&nbsp;.
 
+ The signal&nbsp; <math>v_2(t)</math>&nbsp; is noisy compared to&nbsp; <math>q(t)</math>&nbsp;.
  
{One of the signals is undistorted and not noisy compared to the original &nbsp; <math>q(t)</math>&nbsp;. <br> Estimate the attenuation factor and the running time for this.
+
{One of the signals is undistorted and not noisy compared to the original &nbsp; <math>q(t)</math>&nbsp;. <br> Estimate the attenuation factor and the delay time for this.
 
|type="{}"}
 
|type="{}"}
 
<math> \alpha \ = \ </math> { 0.2-0.4 }
 
<math> \alpha \ = \ </math> { 0.2-0.4 }
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{{ML-Kopf}}
 
{{ML-Kopf}}
 
'''(1)'''&nbsp;  Correct is the <u>solution 2</u>:
 
'''(1)'''&nbsp;  Correct is the <u>solution 2</u>:
*In the marked range of $20$ milliseconds approx. &nbsp; $10$&nbsp; oscillations can be detected.  
+
*In the marked range of&nbsp; $20$&nbsp; milliseconds approx.&nbsp; $10$&nbsp; oscillations can be detected.  
*From this the result&nbsp; follows approximately for the signal frequency; $f = {10}/(20 \,\text{ms}) =  500 \,\text{Hz}$.
+
*From this the result&nbsp; follows approximately for the signal frequency&nbsp; $f = {10}/(20 \,\text{ms}) =  500 \,\text{Hz}$.
  
  
  
 
'''(2)'''&nbsp; Correct is the <u>solution 1</u>:
 
'''(2)'''&nbsp; Correct is the <u>solution 1</u>:
*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>.&nbsp; The following applies: &nbsp; $v_1(t)=\alpha \cdot q(t-\tau)$.
  
 
*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 delayed in time, compared to the 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 delayed in time, compared to the original.
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'''(3)'''&nbsp; Correct are the <u>solutions 1 and 3</u>:
 
'''(3)'''&nbsp; Correct are the <u>solutions 1 and 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>.  
+
*One can recognize additive noise both in the displayed signal&nbsp; <math>v_2(t)</math>&nbsp; and in the audio signal&nbsp; &nbsp; ⇒ &nbsp; <u>solution 3</u>.  
*The signal-to-noise ratio is approx. &nbsp; $\text{30 dB}$; but this cannot be seen from this representation.  
+
*The signal-to-noise ratio is approx.&nbsp; $\text{30 dB}$;&nbsp; but this cannot be seen from this representation.  
*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>.
+
*Correct is also the <u>solution 1</u>: &nbsp; Without this noise component&nbsp; <math>v_2(t)</math>&nbsp; would be identical with&nbsp; <math>q(t)</math>.
  
  
 
'''(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  
 
'''(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  
*by the attenuation factor&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;  $($this corresponds to about&nbsp; $\text{–10 dB)}$  
 
*and the delay&nbsp;  $\tau = \underline{10\,\text{ms}}$.
 
*and the delay&nbsp;  $\tau = \underline{10\,\text{ms}}$.
 
{{ML-Fuß}}
 
{{ML-Fuß}}

Revision as of 10:32, 8 April 2021

Music signals, original,
noisy and/or distorted?

On the right you see a  $\text{30 ms}$  long section of a music signal  \(q(t)\).  It is the piece  "For Elise"  by Ludwig van Beethoven.

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


Original signal  \(q(t)\):

Sink signal  \(v_1(t)\):

Sink signal  \(v_2(t)\):



Notes:



Questions

1

Estimate the signal frequency of  \(q(t)\)  in the displayed section.

The signal frequency is approximately  \(f = 250\,\text{Hz}\).
The signal frequency is approximately  \(f = 500\,\text{Hz}\).
The signal frequency is about  \(f = 1\,\text{kHz}\).

2

Which statements are true for the signal  \(v_1(t)\) ?

The signal  \(v_1(t)\)  is undistorted compared to  \(q(t)\).
The signal  \(v_1(t)\)  shows distortions compared to  \(q(t)\) .
The signal  \(v_1(t)\)  is noisy compared to  \(q(t)\) .

3

Which statements are true for the signal  \(v_2(t)\) ?

The signal  \(v_2(t)\)  is undistorted compared to  \(q(t)\) .
The signal  \(v_2(t)\)  shows distortions compared to  \(q(t)\) .
The signal  \(v_2(t)\)  is noisy compared to  \(q(t)\) .

4

One of the signals is undistorted and not noisy compared to the original   \(q(t)\) .
Estimate the attenuation factor and the delay time for this.

\( \alpha \ = \ \)

\( \tau \ = \ \)

$\ \text{ms}$


Solution

(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  \(v_1(t)\)  is undistorted compared to the original signal \(q(t)\).  The following applies:   $v_1(t)=\alpha \cdot q(t-\tau)$.
  • An attenuation  \(\alpha\)  and a delay  \(\tau\)  do not cause distortion, but the signal is then only quieter and delayed in time, compared to the original.


(3)  Correct are the solutions 1 and 3:

  • One can recognize additive noise both in the displayed signal  \(v_2(t)\)  and in the audio signal    ⇒   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  \(v_2(t)\)  would be identical with  \(q(t)\).


(4)  The signal  \(v_1(t)\)  is identical in form to the original signal  \(q(t)\)  and differs from it only

  • by the attenuation factor  $\alpha = \underline{\text{0.3}}$   $($this corresponds to about  $\text{–10 dB)}$
  • and the delay  $\tau = \underline{10\,\text{ms}}$.