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Discussion Groups | Comp.DSP | Frequency Interpolation - Grandke's Method

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Frequency Interpolation - Grandke's Method - creekmor - 2011-04-20 15:28:00

I have a question about Grandke's method of frequency interpolation.  (I suspect this question has been asked before).  Using the Hanning window (and Schoukens, 1991 notation), he derives

f_interp = (i + delta) * fo, where 

delta = (2*alpha - 1) / (alpha + 1), and

alpha = abs(X(i+1) / X(i)), and

{X(i), X(i+1)} includes the peak.  By definition,

0 < delta < 1, which implies 1/2 < alpha < 2.  Why must

alpha satisfy this?



---


Re: Frequency Interpolation - Grandke's Method - Eric Jacobsen - 2011-04-20 16:32:00

On Wed, 20 Apr 2011 14:28:13 -0500, creekmor <u...@compgroups.net/>
wrote:

>I have a question about Grandke's method of frequency interpolation.  (I suspect this question has been asked before).  Using the Hanning window (and Schoukens, 1991 notation), he derives
>
>f_interp = (i + delta) * fo, where 
>
>delta = (2*alpha - 1) / (alpha + 1), and
>
>alpha = abs(X(i+1) / X(i)), and
>
>{X(i), X(i+1)} includes the peak.  By definition,
>
>0 < delta < 1, which implies 1/2 < alpha < 2.  Why must
>
>alpha satisfy this?

The units on delta are DFT bins, so if delta > 1 then you picked the
wrong bin for the maximum.   The correction is just to find a finer
estimate between bins, so the assumption is that the peak is already
determined within one bin.


Eric Jacobsen
http://www.ericjacobsen.org
http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php


Re: Frequency Interpolation - Grandke's Method - robert bristow-johnson - 2011-04-20 17:01:00



Re: Frequency Interpolation - Grandke's Method - dbd - 2011-04-21 13:42:00



Re: Frequency Interpolation - Grandke's Method - jcreek - 2011-04-29 09:02:00

>On Wed, 20 Apr 2011 14:28:13 -0500, creekmor <u...@compgroups.net/>
>wrote:
>
>>I have a question about Grandke's method of frequency interpolation.  (I
suspect this question has been asked before).  Using the Hanning window
(and Schoukens, 1991 notation), he derives
>>
>>f_interp = (i + delta) * fo, where 
>>
>>delta = (2*alpha - 1) / (alpha + 1), and
>>
>>alpha = abs(X(i+1) / X(i)), and
>>
>>{X(i), X(i+1)} includes the peak.  By definition,
>>
>>0 < delta < 1, which implies 1/2 < alpha < 2.  Why must
>>
>>alpha satisfy this?
>
>The units on delta are DFT bins, so if delta > 1 then you picked the
>wrong bin for the maximum.   The correction is just to find a finer
>estimate between bins, so the assumption is that the peak is already
>determined within one bin.
>
>
>Eric Jacobsen
>http://www.ericjacobsen.org
>http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
>
I agree with that; however, the quantity

alpha = abs(X(i+1) / X(i))

clearly could be between 0 and infty.  For example, if the peak is located
at bin # i, and the next largest bin # is i+1, then

0 < alpha < 1

while if the peak is located at bin # i+1 and the next largest bin # is i,
then

1 < alpha < infty

However, to ensure that the fractional bin spacing delta satisfies

0 < delta < 1,

alpha must then satisfy

1/2 < alpha < 2

which contradicts the above.  In comparison, the interpolation formula for
the rectangular window is

delta = alpha / (alpha+1)

which causes no similar problem.