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Difference between revisions of "Aufgaben:Exercise 5.3Z: Zero-Padding"

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[[File:P_ID1146__Sig_Z_5_3_neu.png|right|frame|MQF–Werte als Funktion von  TA/T  und  N]]
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[[File:P_ID1146__Sig_Z_5_3_neu.png|right|frame|MQF values as a function <br>of&nbsp; TA/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; TA=0.01T&nbsp; bzw.&nbsp; TA=0.05T&nbsp;.
+
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; TA=0.01T&nbsp; or&nbsp; TA=0.05T.
 +
 
 +
The resulting values for the "mean square error"&nbsp; (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; MSE&nbsp; the designation&nbsp; MQF &nbsp; &rArr; &nbsp; (German:&nbsp; "Mittlerer Quadratischer Fehler"):
  
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}
 
:$${\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}.$$
 
  \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.
 
:* 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.
 
 
:*This is then referred to as ''„zero padding”''.
 
 
 
 
 
  
 +
:*This is then referred to as&nbsp; zero padding.
  
  
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''Hints:''  
 
''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".
 +
 
  
  
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<quiz display=simple>
 
<quiz display=simple>
{Which statements can be derived from the given MQF values&nbsp; (valid for&nbsp; TA/T=0.01&nbsp; and&nbsp; N128)&nbsp; abgeleitet werden?
+
{Which statements can be derived from the given MQF values&nbsp; (valid for&nbsp; TA/T=0.01&nbsp; and&nbsp; N128)?
 
|type="[]"}
 
|type="[]"}
+ The&nbsp; MQF value here is almost independent o&nbsp; N.
+
+ The&nbsp; MQF value here is almost independent of&nbsp; N.
- The&nbsp; MQF value is determined by the termination error.
+
- The&nbsp; MQF value is determined by the truncation error.
 
+ The&nbsp; MQF value is determined by the aliasing error.
 
+ The&nbsp; MQF value is determined by the aliasing error.
  
  
{Let&nbsp; TA/T=0.01. What is the distance&nbsp; fA&nbsp; of adjacent samples in the frequency domain for&nbsp; N=128&nbsp; and&nbsp; N=512?
+
{Let&nbsp; TA/T=0.01.&nbsp; What is the distance&nbsp; fA&nbsp; of adjacent samples in the frequency domain for&nbsp; N=128&nbsp; and&nbsp; N=512?
 
|type="{}"}
 
|type="{}"}
 
N=128: &nbsp; &nbsp;  fAT =   { 0.781 3% }
 
N=128: &nbsp; &nbsp;  fAT =   { 0.781 3% }
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- The product&nbsp; MQFfA&nbsp; should be as large as possible.
 
- The product&nbsp; MQFfA&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; TA/T=0.01&nbsp; und&nbsp; TA/T=0.05 ?
+
{&nbsp; N=128&nbsp; is now fixed.&nbsp; Which statements apply to the comparison of the DFT results with&nbsp; TA/T=0.01&nbsp; und&nbsp; TA/T=0.05 ?
 
|type="[]"}
 
|type="[]"}
+ Mit&nbsp; TA/T=0.05&nbsp; you get a finer frequency resolution.
+
+ With&nbsp; TA/T=0.05&nbsp; you get a finer frequency resolution.
- Mit&nbsp; TA/T=0.05&nbsp; the&nbsp; MQF value is smaller.
+
- With&nbsp; TA/T=0.05&nbsp; the&nbsp; MQF value is smaller.
- Mit&nbsp; TA/T=0.05&nbsp; the influence of the termination error decreases.
+
- With&nbsp; TA/T=0.05&nbsp; the influence of the truncation error decreases.
+ Mit&nbsp; TA/T=0.05&nbsp; the influence of the aliasing error increases.
+
+ With&nbsp; TA/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; TA/T=0.01&nbsp; und&nbsp; TA/T=0.05&nbsp;?
+
{Now&nbsp; N=64.&nbsp; Which statements are true for the comparison of the DFT results with&nbsp; TA/T=0.01&nbsp; und&nbsp; TA/T=0.05&nbsp;?
 
|type="[]"}
 
|type="[]"}
 
+ With&nbsp; TA/T=0.05&nbsp; you get a finer frequency resolution.
 
+ With&nbsp; TA/T=0.05&nbsp; you get a finer frequency resolution.
 
+ With&nbsp; TA/T=0.05&nbsp; the&nbsp; MQF value is smaller.
 
+ With&nbsp; TA/T=0.05&nbsp; the&nbsp; MQF value is smaller.
+ With&nbsp; TA/T=0.05&nbsp; the influence of the termination error decreases.
+
+ With&nbsp; TA/T=0.05&nbsp; the influence of the truncation error decreases.
 
+ With&nbsp; TA/T=0.05&nbsp; the influence of the aliasing error increases.
 
+ With&nbsp; TA/T=0.05&nbsp; the influence of the aliasing error increases.
  
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===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{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; TP=1.28T, also größer als die Breite des Rechtecks.  
+
*Already with&nbsp; N=128,&nbsp; TP=1.28T, i.e. larger than the width of the rectangle.
*Somit spielt hier der Abbruchfehler überhaupt keine Rolle.  
+
*Thus the truncation error plays no role at all here.  
*Der&nbsp; MQF–Wert wird allein durch den Aliasingfehler bestimmt.  
+
*The&nbsp; MQF value is determined solely by the aliasing error.  
*Die Zahlenwerte bestätigen eindeutig, dass&nbsp; MQF&nbsp; (nahezu) unabhängig von&nbsp; N&nbsp; ist.  
+
*The numerical values clearly confirm that&nbsp; MQF&nbsp; is (almost) independent of&nbsp; N.  
  
  
  
'''(2)'''&nbsp;  Aus&nbsp; TA/T=0.01&nbsp; folgt&nbsp; fPT=100.  
+
'''(2)'''&nbsp;  From&nbsp; TA/T=0.01&nbsp; follows&nbsp; fPT=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 = 128: &nbsp; f_{\rm A} \cdot T \; \underline{\approx 0.780},
 
:*N = 512: &nbsp; f_{\rm A} \cdot T \; \underline{\approx 0.195}.
 
:*N = 512: &nbsp; f_{\rm A} \cdot T \; \underline{\approx 0.195}.
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'''(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.&nbsp; 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}&nbsp; value when&nbsp; T_{\rm A}&nbsp; 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).  
*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 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>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 truncation 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 rectangular function.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
 
__NOEDITSECTION__
 
__NOEDITSECTION__
[[Category:Exercises for Signal Representation|^5.3 Possible DFT Errors^]]
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[[Category:Signal Representation: Exercises|^5.3 Possible DFT Errors^]]

Latest revision as of 13:47, 22 September 2021

\rm MQF values as a function
of  T_{\rm A} /T  and  N

We consider the DFT of a rectangular pulse  x(t)  of height  A =1  and duration  T.  Thus the spectral function  X(f)  has a  \sin(f)/f–shaped course.

For this special case the influence of the DFT parameter  N  is to be analyzed, whereby the interpolation point distance in the time domain should always be  T_{\rm A} = 0.01T  or  T_{\rm A} = 0.05T.

The resulting values for the "mean square error"  \rm (MSE)  of the grid values in the frequency domain are given opposite for different values of   N.  Here, we use instead of  \rm MSE  the designation  \rm MQF   ⇒   (German:  "Mittlerer Quadratischer Fehler"):

{\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  T_{\rm A}/T = 0.01 ,  101  of the DFT coefficients  d(ν)  are always different from zero.

  • Of these,   99  have the value  1  and the two marginal coefficients are each equal to  0.5.
  • If  N  is increased, the DFT coefficient field is filled with zeros.
  • This is then referred to as  \text{zero padding}.




Hints:




Questions

1

Which statements can be derived from the given MQF values  (valid for  T_{\rm A}/T = 0.01  and  N ≥ 128)?

The  \rm MQF value here is almost independent of  N.
The  \rm MQF value is determined by the truncation error.
The  \rm MQF value is determined by the aliasing error.

2

Let  T_{\rm A}/T = 0.01.  What is the distance  f_{\rm A}  of adjacent samples in the frequency domain for  N = 128  and  N = 512?

N = 128:     f_{\rm A} \cdot T \ = \

N = 512:     f_{\rm A} \cdot T \ = \

3

What does the product  \text{MQF} \cdot f_{\rm A}  indicate in terms of DFT quality?

The product  \text{MQF} \cdot f_{\rm A}  considers the accuracy and density of the DFT values.
The product  \text{MQF} \cdot f_{\rm A}  should be as large as possible.

4

  N = 128  is now fixed.  Which statements apply to the comparison of the DFT results with  T_{\rm A}/T = 0.01  und  T_{\rm A}/T = 0.05 ?

With  T_{\rm A}/T = 0.05  you get a finer frequency resolution.
With  T_{\rm A}/T = 0.05  the  \rm MQF value is smaller.
With  T_{\rm A}/T = 0.05  the influence of the truncation error decreases.
With  T_{\rm A}/T = 0.05  the influence of the aliasing error increases.

5

Now  N = 64.  Which statements are true for the comparison of the DFT results with  T_{\rm A}/T = 0.01  und  T_{\rm A}/T = 0.05 ?

With  T_{\rm A}/T = 0.05  you get a finer frequency resolution.
With  T_{\rm A}/T = 0.05  the  \rm MQF value is smaller.
With  T_{\rm A}/T = 0.05  the influence of the truncation error decreases.
With  T_{\rm A}/T = 0.05  the influence of the aliasing error increases.


Solution

(1)  Proposed solutions 1 and 3 are correct:

  • Already with  N = 128T_{\rm P} = 1.28 \cdot T, i.e. larger than the width of the rectangle.
  • Thus the truncation error plays no role at all here.
  • The  \rm MQF value is determined solely by the aliasing error.
  • The numerical values clearly confirm that  \rm MQF  is (almost) independent of  N.


(2)  From  T_{\rm A}/T = 0.01  follows  f_{\rm P} \cdot T = 100.

  • The supporting values of  X(f) thus lie in the range  –50 ≤ f \cdot T < +50.
  • For the distance between two samples in the frequency range,   f_{\rm A} = f_{\rm P}/N applies. This gives the following results:
  • N = 128:   f_{\rm A} \cdot T \; \underline{\approx 0.780},
  • N = 512:   f_{\rm A} \cdot T \; \underline{\approx 0.195}.


(3)  The first statement is correct:

  • For  N = 128 , the product is  \text{MQF} \cdot f_{\rm A} \approx 4.7 \cdot 10^{-6}/T.  For  N = 512 , the product is smaller by a factor of about  4 .
  • This means that „zero padding” does not achieve greater DFT accuracy, but a finer "resolution" of the frequency range.
  • The product  \text{MQF} \cdot f_{\rm A}  takes this fact into account; it should always be as small as possible.


(4)  Proposed solutions 1 and 4 are correct:

  • Because of  T_{\rm A} \cdot f_{\rm A} \cdot N = 1 , a constant  N  always results in a smaller  f_{\rm A}  value when  T_{\rm A}  is increased.
  • From the table on the information page, one can see that the mean square error  \rm (MQF)  is significantly increased  (by a factor of about  400).
  • The effect is due to the aliasing error, since the transition from  T_{\rm A}/T = 0.01  auf  T_{\rm A}/T = 0.05  reduces the frequency period by a factor of  5 .
  • The truncation error, on the other hand, continues to play no role with the rectangular pulse as long as  T_{\rm P} = N \cdot T_{\rm A}  is greater than the pulse duration  T.


(5)  All statements are true:

  • With the parameter values  N = 64  and  T_{\rm A}/T = 0.01 , an extremely large truncation error occurs.
  • All time coefficients are  1, so the DFT incorrectly interprets a DC signal instead of the rectangular function.