Difference between revisions of "Aufgaben:Exercise 1.1: Music Signals"
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[[File:P_ID339__Sig_A_1_1.png|right|frame|Music signals, original, <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 $\text{30 ms}$ long section of a music signal <math>q(t)</math>. It is the piece | + | On the right you see a $\text{30 ms}$ long section of a music signal <math>q(t)</math>. It is the piece »For Elise« by Ludwig van Beethoven. |
*Underneath are drawn two sink signals <math>v_1(t)</math> and <math>v_2(t)</math>, which were recorded after the transmission of the music signal <math>q(t)</math> over two different channels. | *Underneath are drawn two sink signals <math>v_1(t)</math> and <math>v_2(t)</math>, which were recorded after the transmission of the music signal <math>q(t)</math> over two different channels. | ||
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''Notes:'' | ''Notes:'' | ||
− | *The | + | *The exercise belongs to the chapter [[Signal_Representation/Principles_of_Communication|»Principles of Communication«]]. |
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- The signal frequency is approximately <math>f = 250\,\text{Hz}</math>. | - The signal frequency is approximately <math>f = 250\,\text{Hz}</math>. | ||
+ The signal frequency is approximately <math>f = 500\,\text{Hz}</math>. | + The signal frequency is approximately <math>f = 500\,\text{Hz}</math>. | ||
− | - The signal frequency is | + | - The signal frequency is approximately <math>f = 1\,\text{kHz}</math>. |
{Which statements are true for the signal <math>v_1(t)</math> ? | {Which statements are true for the signal <math>v_1(t)</math> ? | ||
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'''(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: $f = {10}/(20 \,\text{ms}) = 500 \,\text{Hz}$. |
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*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 time <math>\tau</math> do not cause distortion, but the signal is then only quieter and delayed in time, compared to the original. |
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'''(3)''' Correct are the <u>solutions 1 and 3</u>: | '''(3)''' Correct are the <u>solutions 1 and 3</u>: | ||
*One can recognize additive noise both in the displayed signal <math>v_2(t)</math> and in the audio signal ⇒ <u>solution 3</u>. | *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. $\text{30 dB}$ | + | *The signal-to-noise ratio is approx. $\text{30 dB}$ $($but this cannot be seen from this representation$)$. |
− | *Correct is also | + | *Correct is also <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 $\alpha = \underline{\text{0.3}}$ $($this corresponds to about $\text{–10 dB)}$ | *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 time $\tau = \underline{10\,\text{ms}}$. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
__NOEDITSECTION__ | __NOEDITSECTION__ | ||
[[Category:Signal Representation: Exercises|^1.1 Principles of Communication^]] | [[Category:Signal Representation: Exercises|^1.1 Principles of Communication^]] |
Revision as of 16:42, 9 January 2024
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 exercise 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 time \(\tau\) 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 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 time $\tau = \underline{10\,\text{ms}}$.