Aufgaben:Exercise 1.1: Music Signals: Difference between revisions
From LNTwww
No edit summary |
No edit summary |
||
| Line 51: | Line 51: | ||
+ The signal <math>v_2(t)</math> is noisy compared to <math>q(t)</math> . | + The signal <math>v_2(t)</math> is noisy compared to <math>q(t)</math> . | ||
{One of the signals is undistorted and not noisy compared to the original <math>q(t)</math> . <br> Estimate the attenuation factor and the | {One of the signals is undistorted and not noisy compared to the original <math>q(t)</math> . <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 } | ||
| Line 62: | Line 62: | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' Correct is the <u>solution 2</u>: | '''(1)''' Correct is the <u>solution 2</u>: | ||
*In the marked range of $20$ milliseconds approx. $10$ oscillations can be detected. | *In the marked range of $20$ milliseconds approx. $10$ oscillations can be detected. | ||
*From this the result follows approximately for the signal frequency; $f = {10}/(20 \,\text{ms}) = 500 \,\text{Hz}$. | *From this the result follows approximately for the signal frequency $f = {10}/(20 \,\text{ms}) = 500 \,\text{Hz}$. | ||
'''(2)''' Correct is the <u>solution 1</u>: | '''(2)''' Correct is the <u>solution 1</u>: | ||
*The signal <math>v_1(t)</math> is undistorted compared to the original signal <math>q(t)</math>. The following applies: $v_1(t)=\alpha \cdot q(t-\tau) . | *The signal <math>v_1(t)</math> is undistorted compared to the original signal <math>q(t)</math>. The following applies: $v_1(t)=\alpha \cdot q(t-\tau)$. | ||
*An attenuation <math>\alpha</math> and a delay <math>\tau</math> do not cause distortion, but the signal is then only quieter and delayed in time, compared to the original. | *An attenuation <math>\alpha</math> and a delay <math>\tau</math> do not cause distortion, but the signal is then only quieter and delayed in time, compared to the original. | ||
| Line 75: | Line 75: | ||
'''(3)''' Correct are the <u>solutions 1 and 3</u>: | '''(3)''' Correct are the <u>solutions 1 and 3</u>: | ||
*One can recognize both in the displayed signal <math>v_2(t)</math> and in the audio signal | *One can recognize additive noise both in the displayed signal <math>v_2(t)</math> and in the audio signal ⇒ <u>solution 3</u>. | ||
*The signal-to-noise ratio is approx. $\text{30 dB}$; but this cannot be seen from this representation. | *The signal-to-noise ratio is approx. $\text{30 dB}$; but this cannot be seen from this representation. | ||
*Correct is also the <u>solution 1</u>: Without this noise component <math>v_2(t)</math> identical with <math>q(t)</math>. | *Correct is also the <u>solution 1</u>: Without this noise component <math>v_2(t)</math> would be identical with <math>q(t)</math>. | ||
'''(4)''' The signal <math>v_1(t)</math> is identical in form to the original signal <math>q(t)</math> and differs from it only | '''(4)''' The signal <math>v_1(t)</math> is identical in form to the original signal <math>q(t)</math> and differs from it only | ||
*by the attenuation factor $\alpha = \underline{\text{0.3}}$ ( | *by the attenuation factor $\alpha = \underline{\text{0.3}}$ $($this corresponds to about $\text{–10 dB)}$ | ||
*and the delay $\tau = \underline{10\,\text{ms}}$. | *and the delay $\tau = \underline{10\,\text{ms}}$. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 10:32, 8 April 2021

noisy and/or distorted?
On the right you see a $\text{30 ms}$ 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 operating elements 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].
Original signal [math]\displaystyle{ q(t) }[/math]:
Sink signal [math]\displaystyle{ v_1(t) }[/math]:
Sink signal [math]\displaystyle{ v_2(t) }[/math]:
Notes:
- The task belongs to the chapter Principles of Communication.
Questions
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 [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 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 [math]\displaystyle{ v_2(t) }[/math] 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 [math]\displaystyle{ v_2(t) }[/math] would be 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 $\alpha = \underline{\text{0.3}}$ $($this corresponds to about $\text{–10 dB)}$
- and the delay $\tau = \underline{10\,\text{ms}}$.