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Exercise 5.3Z: Zero-Padding

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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 analyzed, 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  \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.