Aufgaben:Exercise 5.3Z: Zero-Padding - Revision history

Javier: Text replacement - "analyse" to "analyze"

2021-09-22T11:47:27Z

Text replacement - "analyse" to "analyze"

Guenter at 15:55, 17 May 2021

2021-05-17T15:55:06Z

Guenter at 15:47, 17 May 2021

2021-05-17T15:47:11Z

Guenter at 15:41, 17 May 2021

2021-05-17T15:41:54Z

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:39:07Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

Noah at 14:59, 22 March 2021

2021-03-22T14:59:36Z

Noah at 14:01, 22 March 2021

2021-03-22T14:01:10Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 5.3Z: Zero-Padding to Aufgaben:Exercise 5.3Z: Zero-Padding

2020-11-29T18:53:40Z

Oezdemir moved page Aufgabe 5.3Z: Zero-Padding to Exercise 5.3Z: Zero-Padding

Oezdemir at 18:53, 29 November 2020

2020-11-29T18:53:04Z

Javier: Text replacement - "Signaldarstellung/Fehlermöglichkeiten bei Anwendung der DFT" to "Signal_Representation/Possible_Errors_When_Using_DFT"

2020-09-01T13:32:20Z

Text replacement - "Signaldarstellung/Fehlermöglichkeiten bei Anwendung der DFT" to "Signal_Representation/Possible_Errors_When_Using_DFT"

@@ Line 6: / Line 6: @@
 We consider the DFT of a rectangular pulse&nbsp; $x(t)$&nbsp; of height&nbsp; $A =1$&nbsp; and duration&nbsp; $T$.&nbsp; Thus the spectral function&nbsp; $X(f)$&nbsp; has a&nbsp; $\sin(f)/f$–shaped course.
-For this special case the influence of the DFT parameter&nbsp; $N$&nbsp; is to be analysed, whereby the interpolation point distance in the time domain should always be&nbsp; $T_{\rm A} = 0.01T$&nbsp; or&nbsp; $T_{\rm A} = 0.05T$.
+For this special case the influence of the DFT parameter&nbsp; $N$&nbsp; is to be analyzed, whereby the interpolation point distance in the time domain should always be&nbsp; $T_{\rm A} = 0.01T$&nbsp; or&nbsp; $T_{\rm A} = 0.05T$.
 The resulting values for the "mean square error"&nbsp; $\rm (MSE)$&nbsp; of the grid values in the frequency domain are given opposite for different values of &nbsp; $N$.&nbsp; Here, we use instead of&nbsp; $\rm MSE$&nbsp; the designation&nbsp; $\rm MQF$ &nbsp; &rArr; &nbsp; (German:&nbsp; "Mittlerer Quadratischer Fehler"):

@@ Line 78: / Line 78: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp;  <u>Proposed solutions 1 and 3</u> are correct:
-*Already with&nbsp; $N = 128$&nbsp;,&nbsp; $T_{\rm P} = 1.28 \cdot T$, i.e. larger than the width of the rectangle.
+*Already with&nbsp; $N = 128$,&nbsp; $T_{\rm P} = 1.28 \cdot T$, i.e. larger than the width of the rectangle.
-*Thus the termination error plays no role at all here.
+*Thus the truncation error plays no role at all here.
 *The&nbsp; $\rm MQF$ value is determined solely by the aliasing error.
-*The numerical values clearly confirm that&nbsp; $\rm MQF$&nbsp;  is (almost) independent of&nbsp; $N$&nbsp; ist.
+*The numerical values clearly confirm that&nbsp; $\rm MQF$&nbsp;  is (almost) independent of&nbsp; $N$.
@@ Line 94: / Line 94: @@
 '''(3)'''&nbsp;  The <u>first statement</u> is correct:
-*For&nbsp; $N = 128$&nbsp;, the product is&nbsp; $\text{MQF} \cdot f_{\rm A} \approx 4.7 \cdot 10^{-6}/T$. For&nbsp; $N = 512$&nbsp;, the product is smaller by a factor of about&nbsp; $4$&nbsp;.
+*For&nbsp; $N = 128$&nbsp;, the product is&nbsp; $\text{MQF} \cdot f_{\rm A} \approx 4.7 \cdot 10^{-6}/T$.&nbsp; For&nbsp; $N = 512$&nbsp;, the product is smaller by a factor of about&nbsp; $4$&nbsp;.
-*This means that&nbsp;„zero padding” does not achieve greater DFT accuracy, but a finer „resolution” of the frequency range.
+*This means that&nbsp;„zero padding” does not achieve greater DFT accuracy, but a finer "resolution" of the frequency range.
 *The product&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; takes this fact into account; it should always be as small as possible.
@@ Line 101: / Line 101: @@
 '''(4)'''&nbsp;   <u>Proposed solutions 1 and 4</u> are correct:
-*Because of&nbsp; $T_{\rm A} \cdot f_{\rm A} \cdot N = 1$&nbsp;, a constant&nbsp; $N$&nbsp; always results in a smaller&nbsp; $f_{\rm A}$ value when $T_{\rm A}$ is increased.
+*Because of&nbsp; $T_{\rm A} \cdot f_{\rm A} \cdot N = 1$&nbsp;, a constant&nbsp; $N$&nbsp; always results in a smaller&nbsp; $f_{\rm A}$&nbsp; value when&nbsp; $T_{\rm A}$&nbsp; is increased.
-*From the table on the information page, one can see that the mean square error&nbsp; $\rm (MQF)$&nbsp; is significantly increased&nbsp; $($by a factor of about&nbsp; $400)$&nbsp;.
+*From the table on the information page, one can see that the mean square error&nbsp; $\rm (MQF)$&nbsp; is significantly increased&nbsp; $($by a factor of about&nbsp; $400)$.
 *The effect is due to the aliasing error, since the transition from&nbsp; $T_{\rm A}/T = 0.01$&nbsp; auf&nbsp; $T_{\rm A}/T = 0.05$&nbsp; reduces the frequency period by a factor of&nbsp; $5$&nbsp;.
-*The termination error, on the other hand, continues to play no role with the square pulse as long as&nbsp; $T_{\rm P} = N \cdot T_{\rm A}$&nbsp; is greater than the pulse duration&nbsp; $T$.
+*The truncation error, on the other hand, continues to play no role with the rectangular pulse as long as&nbsp; $T_{\rm P} = N \cdot T_{\rm A}$&nbsp; is greater than the pulse duration&nbsp; $T$.
 '''(5)'''&nbsp;  <u>All statements are true</u>:
-*With the parameter values&nbsp; $N = 64$&nbsp; and&nbsp; $T_{\rm A}/T = 0.01$&nbsp;, an extremely large termination error occurs.
+*With the parameter values&nbsp; $N = 64$&nbsp; and&nbsp; $T_{\rm A}/T = 0.01$&nbsp;, an extremely large truncation error occurs.
-*All time coefficients are&nbsp; $1$, so the DFT incorrectly interprets a DC signal instead of the square wave function.
+*All time coefficients are&nbsp; $1$, so the DFT incorrectly interprets a DC signal instead of the rectangular function.
 {{ML-Fuß}}

@@ Line 38: / Line 38: @@
 <quiz display=simple>
-{Which statements can be derived from the given MQF values&nbsp; $($valid for&nbsp; $T_{\rm A}/T = 0.01$&nbsp; and&nbsp; $N ≥ 128)$&nbsp; abgeleitet werden?
+{Which statements can be derived from the given MQF values&nbsp; $($valid for&nbsp; $T_{\rm A}/T = 0.01$&nbsp; and&nbsp; $N ≥ 128)$?
 |type="[]"}
-+ The&nbsp; $\rm MQF$ value here is almost independent o&nbsp; $N$.
++ The&nbsp; $\rm MQF$ value here is almost independent of&nbsp; $N$.
-- The&nbsp; $\rm MQF$ value is determined by the termination error.
+- The&nbsp; $\rm MQF$ value is determined by the truncation error.
 + The&nbsp; $\rm MQF$ value is determined by the aliasing error.
-{Let&nbsp; $T_{\rm A}/T = 0.01$. What is the distance&nbsp; $f_{\rm A}$&nbsp; of adjacent samples in the frequency domain for&nbsp; $N = 128$&nbsp; and&nbsp; $N = 512$?
+{Let&nbsp; $T_{\rm A}/T = 0.01$.&nbsp; What is the distance&nbsp; $f_{\rm A}$&nbsp; of adjacent samples in the frequency domain for&nbsp; $N = 128$&nbsp; and&nbsp; $N = 512$?
 |type="{}"}
 $N = 128$: &nbsp; &nbsp;  $f_{\rm A} \cdot T \ = \ $   { 0.781 3% }
@@ Line 56: / Line 56: @@
 - The product&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; should be as large as possible.
-{&nbsp; $N = 128$&nbsp; is now fixed. Which statements apply to the comparison of the DFT results with&nbsp; $T_{\rm A}/T = 0.01$&nbsp; und&nbsp; $T_{\rm A}/T = 0.05$ ?
+{&nbsp; $N = 128$&nbsp; is now fixed.&nbsp; Which statements apply to the comparison of the DFT results with&nbsp; $T_{\rm A}/T = 0.01$&nbsp; und&nbsp; $T_{\rm A}/T = 0.05$ ?
 |type="[]"}
-+ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; you get a finer frequency resolution.
++ With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; you get a finer frequency resolution.
-- Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the&nbsp; $\rm MQF$ value is smaller.
+- With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the&nbsp; $\rm MQF$ value is smaller.
-- Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the termination error decreases.
+- With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the truncation error decreases.
-+ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the aliasing error increases.
++ With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the aliasing error increases.
-{Now&nbsp; $N = 64$.Which statements are true for the comparison of the DFT results with&nbsp; $T_{\rm A}/T = 0.01$&nbsp; und&nbsp; $T_{\rm A}/T = 0.05$&nbsp;?
+{Now&nbsp; $N = 64$.&nbsp; Which statements are true for the comparison of the DFT results with&nbsp; $T_{\rm A}/T = 0.01$&nbsp; und&nbsp; $T_{\rm A}/T = 0.05$&nbsp;?
 |type="[]"}
 + With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; you get a finer frequency resolution.
 + With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the&nbsp; $\rm MQF$ value is smaller.
-+ With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the termination error decreases.
++ With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the truncation error decreases.
 + With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the aliasing error increases.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1146__Sig_Z_5_3_neu.png|right|frame|$\rm MQF$&ndash; values as a fuction of&nbsp; $T_{\rm A} /T$&nbsp; and&nbsp; $N$]]
+[[File:P_ID1146__Sig_Z_5_3_neu.png|right|frame|$\rm MQF$ values as a function <br>of&nbsp; $T_{\rm A} /T$&nbsp; and&nbsp; $N$]]
-We consider the DFT of a rectangular pulse&nbsp; $x(t)$&nbsp; of height&nbsp; $A =1$&nbsp; and duration&nbsp; $T$. Thus the spectral function&nbsp; $X(f)$&nbsp; a&nbsp; $\sin(f)/f$–shaped course.
+We consider the DFT of a rectangular pulse&nbsp; $x(t)$&nbsp; of height&nbsp; $A =1$&nbsp; and duration&nbsp; $T$.&nbsp; Thus the spectral function&nbsp; $X(f)$&nbsp; has a&nbsp; $\sin(f)/f$–shaped course.
-For this special case the influence of the DFT parameter&nbsp; $N$&nbsp; is to be analysed, whereby the interpolation point distance in the time domain should always be&nbsp; $T_{\rm A} = 0.01T$&nbsp; bzw.&nbsp; $T_{\rm A} = 0.05T$&nbsp;.
+For this special case the influence of the DFT parameter&nbsp; $N$&nbsp; is to be analysed, whereby the interpolation point distance in the time domain should always be&nbsp; $T_{\rm A} = 0.01T$&nbsp; or&nbsp; $T_{\rm A} = 0.05T$.
 :$${\rm MQF} =  \frac{1}{N}\cdot \sum_{\mu = 0 }^{N-1}
   \left|X(\mu \cdot f_{\rm A})-\frac{D(\mu)}{f_{\rm A}}\right|^2 \hspace{0.05cm}.$$
-Thus, for&nbsp; $T_A/T = 0.01$&nbsp;,&nbsp; $101$&nbsp; of the DFT coefficients&nbsp; $d(ν)$&nbsp; are always different from zero.
+Thus, for&nbsp; $T_{\rm A}/T = 0.01$&nbsp;,&nbsp; $101$&nbsp; of the DFT coefficients&nbsp; $d(ν)$&nbsp; are always different from zero.
 :* Of these, &nbsp; $99$&nbsp;  have the value&nbsp; $1$&nbsp; and the two marginal coefficients are each equal to&nbsp; $0.5$.
-:* If&nbsp; $N$, is increased, the DFT coefficient field is filled with zeros.
+:* If&nbsp; $N$&nbsp; is increased, the DFT coefficient field is filled with zeros.
@@ Line 28: / Line 26: @@
 ''Hints:''
-*This task belongs to the chapter&nbsp; [[Signal_Representation/Possible_Errors_When_Using_DFT|Possible Errors when Using DFT]].
+*This task belongs to the chapter&nbsp; [[Signal_Representation/Possible_Errors_When_Using_DFT|Possible errors when using DFT]].
-*The theory of this chapter is summarised in the learning video&nbsp; [[Fehlermöglichkeiten_bei_Anwendung_der_DFT_(Lernvideo)|Possible Errors when Using DFT]]&nbsp;.
+*The theory for this chapter is summarised in the (German language) learning video <br> &nbsp; &nbsp; &nbsp;[[Fehlermöglichkeiten_bei_Anwendung_der_DFT_(Lernvideo)|Fehlermöglichkeiten bei Anwendung der DFT]] &nbsp; &rArr; &nbsp; "Possible errors when using DFT".

← Older revision		Revision as of 12:39, 23 March 2021
Line 116:		Line 116:

	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^5.3 Possible DFT Errors^]]	+	[[Category:Signal Representation: Exercises\|^5.3 Possible DFT Errors^]]

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1146__Sig_Z_5_3_neu.png|right|frame|$\rm MQF$&ndash;Werte als Funktion von&nbsp; $T_{\rm A} /T$&nbsp; und&nbsp; $N$]]
+[[File:P_ID1146__Sig_Z_5_3_neu.png|right|frame|$\rm MQF$&ndash; values as a fuction of&nbsp; $T_{\rm A} /T$&nbsp; and&nbsp; $N$]]
 We consider the DFT of a rectangular pulse&nbsp; $x(t)$&nbsp; of height&nbsp; $A =1$&nbsp; and duration&nbsp; $T$. Thus the spectral function&nbsp; $X(f)$&nbsp; a&nbsp; $\sin(f)/f$–shaped course.
@@ Line 78: / Line 78: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
+'''(1)'''&nbsp;  <u>Proposed solutions 1 and 3</u> are correct:
-*Bereits mit&nbsp; $N = 128$&nbsp; ist&nbsp; $T_{\rm P} = 1.28 \cdot T$, also größer als die Breite des Rechtecks.
+*Already with&nbsp; $N = 128$&nbsp;,&nbsp; $T_{\rm P} = 1.28 \cdot T$, i.e. larger than the width of the rectangle.
-*Somit spielt hier der Abbruchfehler überhaupt keine Rolle.
+*Thus the termination error plays no role at all here.
-*Der&nbsp; $\rm MQF$–Wert wird allein durch den Aliasingfehler bestimmt.
+*The&nbsp; $\rm MQF$ value is determined solely by the aliasing error.
-*Die Zahlenwerte bestätigen eindeutig, dass&nbsp; $\rm MQF$&nbsp; (nahezu) unabhängig von&nbsp; $N$&nbsp; ist.
+*The numerical values clearly confirm that&nbsp; $\rm MQF$&nbsp;  is (almost) independent of&nbsp; $N$&nbsp; ist.
-'''(2)'''&nbsp;  Aus&nbsp; $T_{\rm A}/T = 0.01$&nbsp; folgt&nbsp; $f_{\rm P} \cdot T = 100$.
+'''(2)'''&nbsp;  From&nbsp; $T_{\rm A}/T = 0.01$&nbsp; follows&nbsp; $f_{\rm P} \cdot T = 100$.
-*Die Stützwerte von&nbsp; $X(f)$ liegen also im Bereich&nbsp; $–50 ≤ f \cdot T < +50$.
+*The supporting values of&nbsp; $X(f)$ thus lie in the range&nbsp; $–50 ≤ f \cdot T < +50$.
-*Für den Abstand zweier Abtastwerte im Frequenzbereich gilt&nbsp; $f_{\rm A} = f_{\rm P}/N$. Daraus ergeben sich folgende Ergebnisse:
+*For the distance between two samples in the frequency range, &nbsp; $f_{\rm A} = f_{\rm P}/N$ applies. This gives the following results:
 :*$N = 128$: &nbsp; $f_{\rm A} \cdot T \; \underline{\approx 0.780}$,
 :*$N = 512$: &nbsp; $f_{\rm A} \cdot T \; \underline{\approx 0.195}$.
@@ Line 94: / Line 94: @@
-'''(3)'''&nbsp;  Richtig ist die <u>erste Aussage</u>:
+'''(3)'''&nbsp;  The <u>first statement</u> is correct:
-*Für&nbsp; $N = 128$&nbsp; ergibt sich für das Produkt&nbsp; $\text{MQF} \cdot f_{\rm A} \approx 4.7 \cdot 10^{-6}/T$. Für&nbsp; $N = 512$&nbsp; ist das Produkt etwa um den Faktor&nbsp; $4$&nbsp; kleiner.
+*For&nbsp; $N = 128$&nbsp;, the product is&nbsp; $\text{MQF} \cdot f_{\rm A} \approx 4.7 \cdot 10^{-6}/T$. For&nbsp; $N = 512$&nbsp;, the product is smaller by a factor of about&nbsp; $4$&nbsp;.
-*Das heißt: &nbsp; Durch „Zero–Padding” wird keine größere DFT-Genauigkeit erzielt, dafür aber eine feinere „Auflösung” des Frequenzbereichs.
+*This means that&nbsp;„zero padding” does not achieve greater DFT accuracy, but a finer „resolution” of the frequency range.
-*Das Produkt&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; berücksichtigt diese Tatsache; es sollte stets möglichst klein sein.
+*The product&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; takes this fact into account; it should always be as small as possible.
-'''(4)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 4</u>:
+'''(4)'''&nbsp;   <u>Proposed solutions 1 and 4</u> are correct:
-*Wegen&nbsp; $T_{\rm A} \cdot f_{\rm A} \cdot N = 1$&nbsp; ergibt sich bei konstantem&nbsp; $N$&nbsp; immer dann ein kleinerer&nbsp; $f_{\rm A}$–Wert, wenn man $T_{\rm A}$ vergrößert.
+*Because of&nbsp; $T_{\rm A} \cdot f_{\rm A} \cdot N = 1$&nbsp;, a constant&nbsp; $N$&nbsp; always results in a smaller&nbsp; $f_{\rm A}$ value when $T_{\rm A}$ is increased.
-*Aus der Tabelle auf der Angabenseite erkennt man, dass damit der mittlere quadratische Fehler&nbsp; $\rm (MQF)$&nbsp; signifikant&nbsp; $($etwa um den Faktor&nbsp; $400)$&nbsp; vergrößert wird.
+*From the table on the information page, one can see that the mean square error&nbsp; $\rm (MQF)$&nbsp; is significantly increased&nbsp; $($by a factor of about&nbsp; $400)$&nbsp;.
-*Der Effekt geht auf den Aliasingfehler zurück, da durch den Übergang von&nbsp; $T_{\rm A}/T = 0.01$&nbsp; auf&nbsp; $T_{\rm A}/T = 0.05$&nbsp; die Frequenzperiode um den Faktor&nbsp; $5$&nbsp; kleiner wird.
+*The effect is due to the aliasing error, since the transition from&nbsp; $T_{\rm A}/T = 0.01$&nbsp; auf&nbsp; $T_{\rm A}/T = 0.05$&nbsp; reduces the frequency period by a factor of&nbsp; $5$&nbsp;.
-*Der Abbruchfehler spielt dagegen  beim Rechteckimpuls weiterhin keine Rolle, solange&nbsp; $T_{\rm P} = N \cdot T_{\rm A}$&nbsp; größer ist als die Impulsdauer&nbsp; $T$.
+*The termination error, on the other hand, continues to play no role with the square pulse as long as&nbsp; $T_{\rm P} = N \cdot T_{\rm A}$&nbsp; is greater than the pulse duration&nbsp; $T$.
-'''(5)'''&nbsp;  <u>Alle Aussagen treffen zu</u>:
+'''(5)'''&nbsp;  <u>All statements are true</u>:
-* Mit den Parameterwerten&nbsp; $N = 64$&nbsp; und&nbsp; $T_{\rm A}/T = 0.01$&nbsp; tritt ein extrem großer Abbruchfehler auf.
+*With the parameter values&nbsp; $N = 64$&nbsp; and&nbsp; $T_{\rm A}/T = 0.01$&nbsp;, an extremely large termination error occurs.
-*Alle Zeitkoeffizienten sind hier&nbsp; $1$, so dass die DFT fälschlicherweise ein Gleichsignal anstelle der Rechteckfunktion interpretiert.
+*All time coefficients are&nbsp; $1$, so the DFT incorrectly interprets a DC signal instead of the square wave function.
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:P_ID1146__Sig_Z_5_3_neu.png|right|frame|$\rm MQF$&ndash;Werte als Funktion von&nbsp; $T_{\rm A} /T$&nbsp; und&nbsp; $N$]]
-Wir betrachten die DFT eines Rechteckimpulses&nbsp; $x(t)$&nbsp; der Höhe&nbsp; $A =1$&nbsp; und der Dauer&nbsp; $T$. Damit hat die Spektralfunktion&nbsp; $X(f)$&nbsp; einen&nbsp; $\sin(f)/f$–förmigen Verlauf.
+We consider the DFT of a rectangular pulse&nbsp; $x(t)$&nbsp; of height&nbsp; $A =1$&nbsp; and duration&nbsp; $T$. Thus the spectral function&nbsp; $X(f)$&nbsp; a&nbsp; $\sin(f)/f$–shaped course.
-Für diesen Sonderfall soll der Einfluss des DFT–Parameters&nbsp; $N$&nbsp; analysiert werden, wobei der Stützstellenabstand im Zeitbereich stets&nbsp; $T_{\rm A} = 0.01T$&nbsp; bzw.&nbsp; $T_{\rm A} = 0.05T$&nbsp; betragen soll.
+For this special case the influence of the DFT parameter&nbsp; $N$&nbsp; is to be analysed, whereby the interpolation point distance in the time domain should always be&nbsp; $T_{\rm A} = 0.01T$&nbsp; bzw.&nbsp; $T_{\rm A} = 0.05T$&nbsp;.
-Nebenstehend sind für unterschiedliche Werte von&nbsp; $N$&nbsp; die sich ergebenden Werte für den ''mittleren quadratischen Fehler''&nbsp; (MQF) der Stützwerte im Frequenzbereich angegeben:
+The resulting values for the ''mean square error'' &nbsp; (MSE, here MQF) of the grid values in the frequency domain are given opposite for different values of &nbsp; $N$&nbsp;:
 :$${\rm MQF} =  \frac{1}{N}\cdot \sum_{\mu = 0 }^{N-1}
   \left|X(\mu \cdot f_{\rm A})-\frac{D(\mu)}{f_{\rm A}}\right|^2 \hspace{0.05cm}.$$
-Für&nbsp; $T_A/T = 0.01$&nbsp; sind somit stets&nbsp; $101$&nbsp; der DFT–Koeffizienten&nbsp; $d(ν)$&nbsp; von Null verschieden.
+Thus, for&nbsp; $T_A/T = 0.01$&nbsp;,&nbsp; $101$&nbsp; of the DFT coefficients&nbsp; $d(ν)$&nbsp; are always different from zero.
-:* Davon besitzen&nbsp; $99$&nbsp; den Wert&nbsp; $1$&nbsp; und die beiden Randkoeffizienten sind jeweils gleich&nbsp; $0.5$.
+:* Of these, &nbsp; $99$&nbsp;  have the value&nbsp; $1$&nbsp; and the two marginal coefficients are each equal to&nbsp; $0.5$.
-:* Vergrößert man&nbsp; $N$, so wird das DFT–Koeffizientenfeld mit Nullen aufgefüllt.
+:* If&nbsp; $N$, is increased, the DFT coefficient field is filled with zeros.
-:*Man spricht dann von ''„Zero–Padding”''.
+:*This is then referred to as ''„zero padding”''.
@@ Line 27: / Line 27: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Signal_Representation/Possible_Errors_When_Using_DFT|Fehlermöglichkeiten bei Anwendung der DFT]].
+*This task belongs to the chapter&nbsp; [[Signal_Representation/Possible_Errors_When_Using_DFT|Possible Errors when Using DFT]].
-*Die Theorie zu diesem Kapitel ist im Lernvideo&nbsp; [[Fehlermöglichkeiten_bei_Anwendung_der_DFT_(Lernvideo)|Fehlermöglichkeiten bei Anwendung der DFT]]&nbsp; zusammengefasst.
+*The theory of this chapter is summarised in the learning video&nbsp; [[Fehlermöglichkeiten_bei_Anwendung_der_DFT_(Lernvideo)|Possible Errors when Using DFT]]&nbsp;.
@@ Line 36: / Line 36: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche Aussagen können aus den angegebenen MQF-Werten&nbsp; $($gültig für&nbsp; $T_{\rm A}/T = 0.01$&nbsp; und&nbsp; $N ≥ 128)$&nbsp; abgeleitet werden?
+{Which statements can be derived from the given MQF values&nbsp; $($valid for&nbsp; $T_{\rm A}/T = 0.01$&nbsp; and&nbsp; $N ≥ 128)$&nbsp; abgeleitet werden?
 |type="[]"}
-+ Der&nbsp; $\rm MQF$–Wert ist hier nahezu unabhängig von&nbsp; $N$.
++ The&nbsp; $\rm MQF$ value here is almost independent o&nbsp; $N$.
-- Der&nbsp; $\rm MQF$–Wert wird durch den Abbruchfehler bestimmt.
+- The&nbsp; $\rm MQF$ value is determined by the termination error.
-+ Der&nbsp; $\rm MQF$–Wert wird durch den Aliasingfehler bestimmt.
++ The&nbsp; $\rm MQF$ value is determined by the aliasing error.
-{Es gelte&nbsp; $T_{\rm A}/T = 0.01$. Wie groß ist der Abstand&nbsp; $f_{\rm A}$&nbsp; benachbarter Abtastwerte im Frequenzbereich für&nbsp; $N = 128$&nbsp; und&nbsp; $N = 512$?
+{Let&nbsp; $T_{\rm A}/T = 0.01$. What is the distance&nbsp; $f_{\rm A}$&nbsp; of adjacent samples in the frequency domain for&nbsp; $N = 128$&nbsp; and&nbsp; $N = 512$?
 |type="{}"}
 $N = 128$: &nbsp; &nbsp;  $f_{\rm A} \cdot T \ = \ $   { 0.781 3% }
@@ Line 52: / Line 52: @@
-{Was sagt das Produkt&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; hinsichtlich der DFT–Qualität aus?
+{What does the product&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; indicate in terms of DFT quality?
 |type="[]"}
-+ Das Produkt&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; berücksichtigt die Genauigkeit und die Dichte der DFT–Werte.
++ The product&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; considers the accuracy and density of the DFT values.
-- Das Produkt&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; sollte möglichst groß sein.
+- The product&nbsp; $\text{MQF} \cdot f_{\rm A}$&nbsp; should be as large as possible.
+{&nbsp; $N = 128$&nbsp; is now fixed. Which statements apply to the comparison of the DFT results with&nbsp; $T_{\rm A}/T = 0.01$&nbsp; und&nbsp; $T_{\rm A}/T = 0.05$ ?
 |type="[]"}
-+ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; erhält man eine feinere Frequenzauflösung.
++ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; you get a finer frequency resolution.
-- Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; ist der&nbsp; $\rm MQF$–Wert kleiner.
+- Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the&nbsp; $\rm MQF$ value is smaller.
-- Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; nimmt der Einfluss des Abbruchfehlers ab.
+- Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the termination error decreases.
-+ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; wächst der Einfluss des Aliasingfehlers.
++ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the aliasing error increases.
-{Nun gelte&nbsp; $N = 64$. Welche Aussagen treffen für den Vergleich der DFT–Ergebnisse mit&nbsp; $T_{\rm A}/T = 0.01$&nbsp; und&nbsp; $T_{\rm A}/T = 0.05$&nbsp; zu?
+{Now&nbsp; $N = 64$.Which statements are true for the comparison of the DFT results with&nbsp; $T_{\rm A}/T = 0.01$&nbsp; und&nbsp; $T_{\rm A}/T = 0.05$&nbsp;?
 |type="[]"}
-+ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; erhält man eine feinere Frequenzauflösung.
++ With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; you get a finer frequency resolution.
-+ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; ist der&nbsp; $\rm MQF$–Wert kleiner.
++ With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the&nbsp; $\rm MQF$ value is smaller.
-+ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; nimmt der Einfluss des Abbruchfehlers ab.
++ With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the termination error decreases.
-+ Mit&nbsp; $T_{\rm A}/T = 0.05$&nbsp; wächst der Einfluss des Aliasingfehlers.
++ With&nbsp; $T_{\rm A}/T = 0.05$&nbsp; the influence of the aliasing error increases.
@@ Line 77: / Line 76: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:

← Older revision	Revision as of 18:53, 29 November 2020
(No difference)

← Older revision		Revision as of 18:53, 29 November 2020
Line 117:		Line 117:

	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:Exercises for Signal Representation\|^5. ~~Zeit- und frequenzdiskrete Signaldarstellung~~^]]	+	[[Category:Exercises for Signal Representation\|^5.3 Possible DFT Errors^]]

← Older revision		Revision as of 13:32, 1 September 2020
Line 1:		Line 1:

−	{{quiz-Header\|Buchseite=~~Signaldarstellung~~/~~Fehlermöglichkeiten bei Anwendung der DFT~~	+	{{quiz-Header\|Buchseite=Signal_Representation/Possible_Errors_When_Using_DFT
	}}		}}