Aufgaben:Exercise 1.1: Music Signals: Difference between revisions
From LNTwww
Add German interlanguage link |
Fix interlanguage link: resolve redirect chain |
||
| Line 90: | Line 90: | ||
__NOEDITSECTION__ | __NOEDITSECTION__ | ||
[[Category:Signal Representation: Exercises|^1.1 Principles of Communication^]] | [[Category:Signal Representation: Exercises|^1.1 Principles of Communication^]] | ||
[[de:Aufgaben:1. | [[de:Aufgaben:Aufgabe 1.1: Musiksignale]] | ||
Latest revision as of 17:53, 16 March 2026

original, 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 exercise belongs to the chapter »Principles of Communication«.
Questions
Solution
(1) Correct is 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 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 time [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 the mentioned data$)$.
- Correct is also 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 shape 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 time $\tau = \underline{10\,\text{ms}}$.