Difference between revisions of "Aufgaben:Exercise 1.1: Music Signals"
From LNTwww
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 09:32, 8 April 2021
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:
- 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 \(v_1(t)\) 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 $\tau = \underline{10\,\text{ms}}$.