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

Revision as of 13:15, 8 August 2020

Music signals, original,
noisy and/or distorted?

On the right you see a ca. $\text{30 ms}$ 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 controls allow you to listen to the first fourteen seconds of each of the three audio signals $q(t)$, $v_1(t)$ and $v_2(t)$.

Originalsignal $q(t)$

Sinkensignal $v_1(t)$

Sinkensignal $v_2(t)$

Notes:

The task belongs to chapter Prinzip der Nachrichtenübertragung.

Questions

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

	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 $q(t)$ .

	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 $q(t)$ .

$ \alpha \ = \ $

$ \tau \ = \ $

$\ \text{ms}$

Solutions

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 comes later than the original.

(3) Correct are the solutions 1 and 3:

One can recognize both in the displayed signal $v_2(t)$ and in the audio signal additive noise ⇒ 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 $v_2(t)$ 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 $\alpha = \underline{\text{0.3}}$ (dies entspricht etwa $\text{–10 dB)}$
and the delay $\tau = \underline{10\,\text{ms}}$.

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Signaldarstellung/Prinzip der Nachrichtenübertragung}}
-[[File:P_ID339__Sig_A_1_1.png|right|frame|Music signals, original and <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 ca.&nbsp; $\text{30 ms}$&nbsp; long section of a music signal&nbsp; <math>q(t)</math>. It is the piece &bdquo;For Elise&rdquo; by Ludwig van Beethoven.
@@ Line 70: / Line 70: @@
 *The signal&nbsp; <math>v_1(t)</math>&nbsp; is undistorted compared to the original signal <math>q(t)</math>. The following applies: &nbsp; $v_1(t)=\alpha \cdot q(t-\tau) .$
-*Eine Dämpfung&nbsp; <math>\alpha</math>&nbsp; und eine Laufzeit&nbsp; <math>\tau</math>&nbsp; führen nicht zu Verzerrungen, sondern das Signal ist dann nur leiser und es kommt später als das 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 comes later than the original.
-'''(3)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
+'''(3)'''&nbsp; Correct are the <u>solutions 1 and 3</u>:
-*Man erkennt sowohl im dargestellten Signalverlauf&nbsp; <math>v_2(t)</math>&nbsp; als auch im Audiosignal&nbsp; ''additives Rauschen''  &nbsp; ⇒ &nbsp;   <u>Lösungsvorschlag 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>.
-*Der Signalrauschabstand beträgt dabei ca.&nbsp; $\text{30 dB}$; dies ist aber aus dieser Darstellung nicht erkennbar.
+*The signal-to-noise ratio is approx. &nbsp; $\text{30 dB}$; but this cannot be seen from this representation.
-*Richtig ist aber auch der <u>Lösungsvorschlag 1</u>: &nbsp; Ohne diesen Rauschanteil wäre&nbsp; <math>v_2(t)</math>&nbsp; identisch mit&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; identical with&nbsp; <math>q(t)</math>.
-'''(4)'''&nbsp;  Das Signal&nbsp; <math>v_1(t)</math>&nbsp; ist formgleich mit dem Originalsignal&nbsp; <math>q(t)</math>&nbsp; und unterscheidet sich von diesem lediglich
+'''(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
-*durch den Amplitudenfaktor&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;   (dies entspricht etwa&nbsp; $\text{–10 dB)}$
-*und die Laufzeit&nbsp;  $\tau = \underline{10\,\text{ms}}$.
+*and the delay&nbsp;  $\tau = \underline{10\,\text{ms}}$.
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 [[Category:Aufgaben zu Signaldarstellung|^1. Grundbegriffe der Nachrichtentechnik^]]

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

	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 \(q(t)\) .

	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 \(q(t)\) .