Difference between revisions of "Aufgaben:Exercise 5.3Z: Zero-Padding"
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'''(1)''' <u>Proposed solutions 1 and 3</u> are correct: | '''(1)''' <u>Proposed solutions 1 and 3</u> are correct: | ||
− | *Already with N=128 | + | *Already with N=128, TP=1.28⋅T, i.e. larger than the width of the rectangle. |
− | *Thus the | + | *Thus the truncation error plays no role at all here. |
*The MQF value is determined solely by the aliasing error. | *The MQF value is determined solely by the aliasing error. | ||
− | *The numerical values clearly confirm that MQF is (almost) independent of N | + | *The numerical values clearly confirm that MQF is (almost) independent of N. |
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'''(3)''' The <u>first statement</u> is correct: | '''(3)''' The <u>first statement</u> is correct: | ||
− | *For N=128 , the product is MQF⋅fA≈4.7⋅10−6/T. For N=512 , the product is smaller by a factor of about 4 . | + | *For N=128 , the product is MQF⋅fA≈4.7⋅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 | + | *This means that „zero padding” does not achieve greater DFT accuracy, but a finer "resolution" of the frequency range. |
*The product MQF⋅fA takes this fact into account; it should always be as small as possible. | *The product MQF⋅fA takes this fact into account; it should always be as small as possible. | ||
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'''(4)''' <u>Proposed solutions 1 and 4</u> are correct: | '''(4)''' <u>Proposed solutions 1 and 4</u> are correct: | ||
− | *Because of TA⋅fA⋅N=1 , a constant N always results in a smaller fA value when TA is increased. | + | *Because of TA⋅fA⋅N=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) | + | *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 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 | + | *The truncation error, on the other hand, continues to play no role with the rectangular pulse as long as TP=N⋅TA is greater than the pulse duration T. |
'''(5)''' <u>All statements are true</u>: | '''(5)''' <u>All statements are true</u>: | ||
− | *With the parameter values N=64 and TA/T=0.01 , an extremely large | + | *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 | + | *All time coefficients are 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
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=1N⋅N−1∑μ=0|X(μ⋅fA)−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:
- This task belongs to the chapter Possible errors when using DFT.
- The theory for this chapter is summarised in the (German language) learning video
Fehlermöglichkeiten bei Anwendung der DFT ⇒ "Possible errors when using DFT".
Questions
Solution
- Already with N=128, TP=1.28⋅T, 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 fP⋅T=100.
- The supporting values of X(f) thus lie in the range –50≤f⋅T<+50.
- For the distance between two samples in the frequency range, fA=fP/N applies. This gives the following results:
- N=128: fA⋅T≈0.780_,
- N=512: fA⋅T≈0.195_.
(3) The first statement is correct:
- For N=128 , the product is MQF⋅fA≈4.7⋅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 MQF⋅fA takes this fact into account; it should always be as small as possible.
(4) Proposed solutions 1 and 4 are correct:
- Because of TA⋅fA⋅N=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=N⋅TA 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.