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

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'''(1)'''&nbsp;  <u>Proposed solutions 1 and 3</u> are correct:
 
'''(1)'''&nbsp;  <u>Proposed solutions 1 and 3</u> are correct:
*Already with&nbsp; N=128&nbsp;,&nbsp; TP=1.28T, i.e. larger than the width of the rectangle.
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*Already with&nbsp; N=128,&nbsp; TP=1.28T, i.e. larger than the width of the rectangle.
*Thus the termination error plays no role at all here.  
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*Thus the truncation error plays no role at all here.  
 
*The&nbsp; MQF value is determined solely by the aliasing error.  
 
*The&nbsp; MQF value is determined solely by the aliasing error.  
*The numerical values clearly confirm that&nbsp; MQF&nbsp;  is (almost) independent of&nbsp; N&nbsp; ist.  
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*The numerical values clearly confirm that&nbsp; MQF&nbsp;  is (almost) independent of&nbsp; N.  
  
  
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'''(3)'''&nbsp;  The <u>first statement</u> is correct:
 
'''(3)'''&nbsp;  The <u>first statement</u> is correct:
*For&nbsp; N=128&nbsp;, the product is&nbsp; MQFfA4.7106/T. For&nbsp; N=512&nbsp;, the product is smaller by a factor of about&nbsp; 4&nbsp;.
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*For&nbsp; N=128&nbsp;, the product is&nbsp; MQFfA4.7106/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.  
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*This means that&nbsp;„zero padding” does not achieve greater DFT accuracy, but a finer "resolution" of the frequency range.  
 
*The product&nbsp; MQFfA&nbsp; takes this fact into account; it should always be as small as possible.  
 
*The product&nbsp; MQFfA&nbsp; takes this fact into account; it should always be as small as possible.  
  
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'''(4)'''&nbsp;  <u>Proposed solutions 1 and 4</u> are correct:
 
'''(4)'''&nbsp;  <u>Proposed solutions 1 and 4</u> are correct:
*Because of&nbsp; TAfAN=1&nbsp;, a constant&nbsp; N&nbsp; always results in a smaller&nbsp; fA value when TA is increased.
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*Because of&nbsp; TAfAN=1&nbsp;, a constant&nbsp; N&nbsp; always results in a smaller&nbsp; fA&nbsp; value when&nbsp; TA&nbsp; is increased.
*From the table on the information page, one can see that the mean square error&nbsp; (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; (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; TA/T=0.01&nbsp; auf&nbsp; TA/T=0.05&nbsp; reduces the frequency period by a factor of&nbsp; 5&nbsp;.  
 
*The effect is due to the aliasing error, since the transition from&nbsp; TA/T=0.01&nbsp; auf&nbsp; TA/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; TP=NTA&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; TP=NTA&nbsp; is greater than the pulse duration&nbsp; T.  
  
  
  
 
'''(5)'''&nbsp;  <u>All statements are true</u>:
 
'''(5)'''&nbsp;  <u>All statements are true</u>:
*With the parameter values&nbsp; N=64&nbsp; and&nbsp; TA/T=0.01&nbsp;, an extremely large termination error occurs.
+
*With the parameter values&nbsp; N=64&nbsp; and&nbsp; TA/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.
 
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Revision as of 17:55, 17 May 2021

MQF values as a function
of  TA/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 analysed, whereby the interpolation point distance in the time domain should always be  TA=0.01T  or  TA=0.05T.

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

MQF=1NN1μ=0|X(μfA)D(μ)fA|2.

Thus, for  TA/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  zero padding.




Hints:




Questions

1

Which statements can be derived from the given MQF values  (valid for  TA/T=0.01  and  N128)?

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

2

Let  TA/T=0.01.  What is the distance  fA  of adjacent samples in the frequency domain for  N=128  and  N=512?

N=128:     fAT = 

N=512:     fAT = 

3

What does the product  MQFfA  indicate in terms of DFT quality?

The product  MQFfA  considers the accuracy and density of the DFT values.
The product  MQFfA  should be as large as possible.

4

  N=128  is now fixed.  Which statements apply to the comparison of the DFT results with  TA/T=0.01  und  TA/T=0.05 ?

With  TA/T=0.05  you get a finer frequency resolution.
With  TA/T=0.05  the  MQF value is smaller.
With  TA/T=0.05  the influence of the truncation error decreases.
With  TA/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  TA/T=0.01  und  TA/T=0.05 ?

With  TA/T=0.05  you get a finer frequency resolution.
With  TA/T=0.05  the  MQF value is smaller.
With  TA/T=0.05  the influence of the truncation error decreases.
With  TA/T=0.05  the influence of the aliasing error increases.


Solution

(1)  Proposed solutions 1 and 3 are correct:

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


(2)  From  TA/T=0.01  follows  fPT=100.

  • The supporting values of  X(f) thus lie in the range  50fT<+50.
  • For the distance between two samples in the frequency range,   fA=fP/N applies. This gives the following results:
  • N=128:   fAT0.780_,
  • N=512:   fAT0.195_.


(3)  The first statement is correct:

  • For  N=128 , the product is  MQFfA4.7106/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  MQFfA  takes this fact into account; it should always be as small as possible.


(4)  Proposed solutions 1 and 4 are correct:

  • Because of  TAfAN=1 , a constant  N  always results in a smaller  fA  value when  TA  is increased.
  • From the table on the information page, one can see that the mean square error  (MQF)  is significantly increased  (by a factor of about  400).
  • The effect is due to the aliasing error, since the transition from  TA/T=0.01  auf  TA/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  TP=NTA  is greater than the pulse duration  T.


(5)  All statements are true:

  • With the parameter values  N=64  and  TA/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.