Hi Group I have derived this estimator. The limited tests I have done shows high accuracy. Let k be the index of the DFT bin with the largest magnitude. Let j be the index of the DFT bin with the largest magnitude neigboring bin k. That is j=k-1 or j=k+1. Now let P=Tan(PI*k/N) and Q=Tan(PI*j/N) Then the normalized frequency is estimated by: Freq=(N/PI)*Arctan(Sqr(P*Q*(Im[k]*P-Im[j]*Q)/(Im[k]*Q-Im[j]*P))) where Sqr is Square root and Im[] is the imaginary part of X[]. I was wondering if anyone has a test suite, so it would be possible to compare with other estimators? Regards, Michael Plet
New frequency estimator from two DFT bins
Started by ●February 6, 2017
Reply by ●February 6, 20172017-02-06
>Hi Group > >I have derived this estimator. The limited tests I have done shows >high accuracy. > > >Let k be the index of the DFT bin with the largest magnitude. >Let j be the index of the DFT bin with the largest magnitude >neigboring bin k. >That is j=k-1 or j=k+1. > >Now let P=Tan(PI*k/N) and Q=Tan(PI*j/N) > >Then the normalized frequency is estimated by: > >Freq=(N/PI)*Arctan(Sqr(P*Q*(Im[k]*P-Im[j]*Q)/(Im[k]*Q-Im[j]*P))) > >where Sqr is Square root and Im[] is the imaginary part of X[]. > > >I was wondering if anyone has a test suite, so it would be possible to >compare with other estimators? > > >Regards, >Michael PletTo answer your question directly. You can find a set of SciLab (Matlab equivalent) routines by Julien Arzi here: http://www.tsdconseil.fr/log/scriptscilab/festim/index-en.html Here is the paper he wrote comparing several estimators with varying levels of noise: http://www.tsdconseil.fr/log/scriptscilab/festim/freqestim-comp.pdf I'm pretty sure all the FFT based frequency estimators are three bin equations. Here is a paper covering *exact* frequency equations using two bins: www.vicanek.de/articles/FreqFromTwoBins.pdf I, too, have an *exact* equation for the two bin case, but in my testing it did not fare as well as Martin's in the presence of noise so I have not published it. It is an overdetermined problem so multiple, different, exact solutions are possible. You can find the derivation for my exact three bin formula here: https://www.dsprelated.com/showarticle/773.php My blog articles leading up to this one cover the material the derivation is based on. When I have a little time, I will plug your equation into one of my test programs and see how it does. If you want me to send you some source code, send me an email to cedron at exede dot net. Ced --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●February 7, 20172017-02-07
On Mon, 06 Feb 2017 17:26:52 -0600, "Cedron" <103185@DSPRelated> wrote:>>Hi Group >> >>I have derived this estimator. The limited tests I have done shows >>high accuracy. >> >> >>Let k be the index of the DFT bin with the largest magnitude. >>Let j be the index of the DFT bin with the largest magnitude >>neigboring bin k. >>That is j=k-1 or j=k+1. >> >>Now let P=Tan(PI*k/N) and Q=Tan(PI*j/N) >> >>Then the normalized frequency is estimated by: >> >>Freq=(N/PI)*Arctan(Sqr(P*Q*(Im[k]*P-Im[j]*Q)/(Im[k]*Q-Im[j]*P))) >> >>where Sqr is Square root and Im[] is the imaginary part of X[]. >> >> >>I was wondering if anyone has a test suite, so it would be possible to >>compare with other estimators? >> >> >>Regards, >>Michael Plet > >To answer your question directly. You can find a set of SciLab (Matlab >equivalent) routines by Julien Arzi here: > >http://www.tsdconseil.fr/log/scriptscilab/festim/index-en.html > >Here is the paper he wrote comparing several estimators with varying >levels of noise: > >http://www.tsdconseil.fr/log/scriptscilab/festim/freqestim-comp.pdf > >I'm pretty sure all the FFT based frequency estimators are three bin >equations. > > >Here is a paper covering *exact* frequency equations using two bins: > >www.vicanek.de/articles/FreqFromTwoBins.pdf > >I, too, have an *exact* equation for the two bin case, but in my testing >it did not fare as well as Martin's in the presence of noise so I have not >published it. It is an overdetermined problem so multiple, different, >exact solutions are possible. > >You can find the derivation for my exact three bin formula here: > >https://www.dsprelated.com/showarticle/773.php > >My blog articles leading up to this one cover the material the derivation >is based on. > >When I have a little time, I will plug your equation into one of my test >programs and see how it does. > >If you want me to send you some source code, send me an email to cedron at >exede dot net. > >Ced >--------------------------------------- >Posted through http://www.DSPRelated.comCedron, I have read your articles (771,773 and 787) on dsprelated. They are very well written and informative! The math is easy to follow. I have tons of DSP books, but none of them has stuff like this. It would be a great help if you can find time to try my formula in your test program. Michael
Reply by ●February 7, 20172017-02-07
> >Cedron, I have read your articles (771,773 and 787) on dsprelated. >They are very well written and informative! The math is easy to >follow. >I have tons of DSP books, but none of them has stuff like this. > >It would be a great help if you can find time to try my formula in >your test program. > >MichaelHi Michael, Your formula performed better than I expected, I'm impressed. It appears to be exact in the noiseless case. Since your formula includes N, it is definitely a possibility. Any estimator that does not include N cannot be exact. Also, by inspection, your formula will give the correct answer for an integer value frequency. However, since you are only using the imaginary parts of the bins, your formula's susceptibility to noise is going to be very phase shift dependent. I ran the same test program with some different phase values and ended up with some worse results and even a few nans (Not a Number). When I coded your formula I always used the k = j+1 case. This should be correct by symmetry in the formula. They are still the two largest bins. As you can also see from the results, my two bin formula is also susceptible to certain frequency fraction/phase combinations. Thank you for the kind words. Ced ============================================= MP = Michael Plet CD = Cedron Dawg MV = Martin Vicanek The first column under each header is the average error (x100) and the second column is the standard deviation (x100). Here is one set of results: The sample count is 100 and the run size is 100 The phase value is 0.785398 Errors are shown at 100x actual value Target Noise Level = 0.000 Freq MP2Bin CD2Bin MV2BinComb ---- ------------- ------------- ------------- 3.0 -0.000 0.000 -0.000 0.000 -0.000 0.000 3.1 -0.000 0.000 0.000 0.000 -0.000 0.000 3.2 0.000 0.000 0.000 0.000 0.000 0.000 3.3 0.000 0.000 -0.000 0.000 -0.000 0.000 3.4 0.000 0.000 -0.000 0.000 0.000 0.000 3.5 0.000 0.000 -0.000 0.000 -0.000 0.000 3.6 -0.000 0.000 0.000 0.000 0.000 0.000 3.7 -0.000 0.000 0.000 0.000 0.000 0.000 3.8 0.000 0.000 -0.000 0.000 0.000 0.000 3.9 0.000 0.000 0.000 0.000 -0.000 0.000 Target Noise Level = 0.001 Freq MP2Bin CD2Bin MV2BinComb ---- ------------- ------------- ------------- 3.0 -0.002 0.019 -0.000 0.020 -0.001 0.014 3.1 -0.000 0.012 0.005 0.023 0.001 0.012 3.2 -0.002 0.009 -0.004 0.032 -0.002 0.009 3.3 -0.000 0.009 -0.021 0.138 0.000 0.009 3.4 0.000 0.009 0.001 0.037 0.001 0.008 3.5 -0.002 0.011 -0.001 0.015 -0.002 0.008 3.6 -0.004 0.019 -0.001 0.010 -0.002 0.007 3.7 0.011 0.067 -0.000 0.009 0.000 0.008 3.8 -0.006 0.072 0.001 0.011 0.001 0.011 3.9 -0.002 0.031 0.001 0.010 0.000 0.010 Target Noise Level = 0.010 Freq MP2Bin CD2Bin MV2BinComb ---- ------------- ------------- ------------- 3.0 0.000 0.169 0.018 0.210 0.011 0.154 3.1 -0.013 0.121 0.011 0.173 -0.006 0.106 3.2 0.001 0.105 -0.019 0.283 -0.001 0.107 3.3 -0.000 0.078 0.075 1.523 -0.001 0.077 3.4 0.002 0.089 0.010 0.311 0.003 0.082 3.5 -0.017 0.101 -0.017 0.164 -0.017 0.072 3.6 -0.014 0.171 0.010 0.098 0.002 0.082 3.7 -0.006 0.619 -0.006 0.083 -0.006 0.081 3.8 0.082 0.709 -0.006 0.092 -0.007 0.091 3.9 -0.016 0.317 -0.006 0.106 -0.008 0.110 Target Noise Level = 0.100 Freq MP2Bin CD2Bin MV2BinComb ---- ------------- ------------- ------------- 3.0 0.229 1.571 0.476 1.835 0.389 1.394 3.1 0.294 1.223 0.116 2.066 0.248 1.146 3.2 0.010 0.929 -0.519 3.283 -0.028 0.966 3.3 0.003 0.841 -6.080 37.449 0.003 0.818 3.4 0.078 0.773 -0.059 3.317 0.066 0.791 3.5 -0.021 1.090 0.148 1.581 0.049 0.836 3.6 0.027 1.765 0.026 0.965 0.027 0.699 3.7 0.902 5.282 -0.038 0.815 0.009 0.802 3.8 0.164 6.399 -0.017 0.907 -0.012 0.906 3.9 -0.686 2.638 -0.097 0.902 -0.178 0.906 Here's how I coded it: double theP = tan( argBaseBinNumber * M_PI / (double) myN ); double theQ = tan( ( argBaseBinNumber + 1 ) * M_PI / (double) myN ); double theNum = theP * argBaseOfBinValues[0].myImag - theQ * argBaseOfBinValues[1].myImag; double theDen = theQ * argBaseOfBinValues[0].myImag - theP * argBaseOfBinValues[1].myImag; double theTemp = sqrt( theP * theQ * theNum / theDen ); double theFrequency = ( (double) myN / M_PI ) * atan( theTemp ); --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●February 7, 20172017-02-07
On Tue, 07 Feb 2017 10:09:03 -0600, "Cedron" <103185@DSPRelated> wrote:>> >>Cedron, I have read your articles (771,773 and 787) on dsprelated. >>They are very well written and informative! The math is easy to >>follow. >>I have tons of DSP books, but none of them has stuff like this. >> >>It would be a great help if you can find time to try my formula in >>your test program. >> >>Michael > >Hi Michael, > >Your formula performed better than I expected, I'm impressed. It appears >to be exact in the noiseless case. Since your formula includes N, it is >definitely a possibility. Any estimator that does not include N cannot be >exact. Also, by inspection, your formula will give the correct answer for >an integer value frequency. > >However, since you are only using the imaginary parts of the bins, your >formula's susceptibility to noise is going to be very phase shift >dependent. I ran the same test program with some different phase values >and ended up with some worse results and even a few nans (Not a Number). > >When I coded your formula I always used the k = j+1 case. This should be >correct by symmetry in the formula. They are still the two largest bins. > >As you can also see from the results, my two bin formula is also >susceptible to certain frequency fraction/phase combinations. > >Thank you for the kind words. > >Ced > >============================================= > >MP = Michael Plet >CD = Cedron Dawg >MV = Martin Vicanek > >The first column under each header is the average error (x100) and the >second column is the standard deviation (x100). > >Here is one set of results: > >The sample count is 100 >and the run size is 100 > The phase value is 0.785398 > >Errors are shown at 100x actual value > >Target Noise Level = 0.000 > >Freq MP2Bin CD2Bin MV2BinComb >---- ------------- ------------- ------------- >3.0 -0.000 0.000 -0.000 0.000 -0.000 0.000 >3.1 -0.000 0.000 0.000 0.000 -0.000 0.000 >3.2 0.000 0.000 0.000 0.000 0.000 0.000 >3.3 0.000 0.000 -0.000 0.000 -0.000 0.000 >3.4 0.000 0.000 -0.000 0.000 0.000 0.000 >3.5 0.000 0.000 -0.000 0.000 -0.000 0.000 >3.6 -0.000 0.000 0.000 0.000 0.000 0.000 >3.7 -0.000 0.000 0.000 0.000 0.000 0.000 >3.8 0.000 0.000 -0.000 0.000 0.000 0.000 >3.9 0.000 0.000 0.000 0.000 -0.000 0.000 > > >Target Noise Level = 0.001 > >Freq MP2Bin CD2Bin MV2BinComb >---- ------------- ------------- ------------- >3.0 -0.002 0.019 -0.000 0.020 -0.001 0.014 >3.1 -0.000 0.012 0.005 0.023 0.001 0.012 >3.2 -0.002 0.009 -0.004 0.032 -0.002 0.009 >3.3 -0.000 0.009 -0.021 0.138 0.000 0.009 >3.4 0.000 0.009 0.001 0.037 0.001 0.008 >3.5 -0.002 0.011 -0.001 0.015 -0.002 0.008 >3.6 -0.004 0.019 -0.001 0.010 -0.002 0.007 >3.7 0.011 0.067 -0.000 0.009 0.000 0.008 >3.8 -0.006 0.072 0.001 0.011 0.001 0.011 >3.9 -0.002 0.031 0.001 0.010 0.000 0.010 > > >Target Noise Level = 0.010 > >Freq MP2Bin CD2Bin MV2BinComb >---- ------------- ------------- ------------- >3.0 0.000 0.169 0.018 0.210 0.011 0.154 >3.1 -0.013 0.121 0.011 0.173 -0.006 0.106 >3.2 0.001 0.105 -0.019 0.283 -0.001 0.107 >3.3 -0.000 0.078 0.075 1.523 -0.001 0.077 >3.4 0.002 0.089 0.010 0.311 0.003 0.082 >3.5 -0.017 0.101 -0.017 0.164 -0.017 0.072 >3.6 -0.014 0.171 0.010 0.098 0.002 0.082 >3.7 -0.006 0.619 -0.006 0.083 -0.006 0.081 >3.8 0.082 0.709 -0.006 0.092 -0.007 0.091 >3.9 -0.016 0.317 -0.006 0.106 -0.008 0.110 > > >Target Noise Level = 0.100 > >Freq MP2Bin CD2Bin MV2BinComb >---- ------------- ------------- ------------- >3.0 0.229 1.571 0.476 1.835 0.389 1.394 >3.1 0.294 1.223 0.116 2.066 0.248 1.146 >3.2 0.010 0.929 -0.519 3.283 -0.028 0.966 >3.3 0.003 0.841 -6.080 37.449 0.003 0.818 >3.4 0.078 0.773 -0.059 3.317 0.066 0.791 >3.5 -0.021 1.090 0.148 1.581 0.049 0.836 >3.6 0.027 1.765 0.026 0.965 0.027 0.699 >3.7 0.902 5.282 -0.038 0.815 0.009 0.802 >3.8 0.164 6.399 -0.017 0.907 -0.012 0.906 >3.9 -0.686 2.638 -0.097 0.902 -0.178 0.906 > > >Here's how I coded it: > > > double theP = tan( argBaseBinNumber * M_PI / (double) myN ); > double theQ = tan( ( argBaseBinNumber + 1 ) * M_PI / (double) myN >); > > double theNum = theP * argBaseOfBinValues[0].myImag > - theQ * argBaseOfBinValues[1].myImag; > > double theDen = theQ * argBaseOfBinValues[0].myImag > - theP * argBaseOfBinValues[1].myImag; > > double theTemp = sqrt( theP * theQ * theNum / theDen ); > > double theFrequency = ( (double) myN / M_PI ) * atan( theTemp ); > > >--------------------------------------- >Posted through http://www.DSPRelated.comHi Cedron, thank you so much for taking the time to test my formula! I'm not surprised that it is exact in the noiseless case because of the way I derived it. But I had no idea what would happen when noise was added. It seems that all three formulas are quite good (so don't be shy about publishing yours :-)) Sure k=j+1 is fine. Then k can be the largest magnitude and j the largest neighbor or j the largest magnitude and k the largest neighbor. I am working on another formula to use the real part. After that I will see if I can use both real and imaginary parts in one formula. I think a formula based on two complex values will be more robust. Thanks again for testing my formula against some others. Michael
Reply by ●February 7, 20172017-02-07
On Mon, 06 Feb 2017 19:17:27 +0100, Michael Plet <me@home.com> wrote:>Hi Group > >I have derived this estimator. The limited tests I have done shows >high accuracy. > > >Let k be the index of the DFT bin with the largest magnitude. >Let j be the index of the DFT bin with the largest magnitude >neigboring bin k. >That is j=k-1 or j=k+1. > >Now let P=Tan(PI*k/N) and Q=Tan(PI*j/N) > >Then the normalized frequency is estimated by: > >Freq=(N/PI)*Arctan(Sqr(P*Q*(Im[k]*P-Im[j]*Q)/(Im[k]*Q-Im[j]*P))) > >where Sqr is Square root and Im[] is the imaginary part of X[]. > > >I was wondering if anyone has a test suite, so it would be possible to >compare with other estimators? > > >Regards, >Michael PletThe link below is very old, but has some example estimators as well as some test code to run them. It would not be difficult to add your estimator and test it in the context of the other included estimators. http://www.ericjacobsen.org/fe2/fe2.htm Your estimator is quite complex on the face of it, and usually two-term estimators have difficulty at some point as the frequency offset varies, i.e., performance is highly offset dependent. Just something to look out for. --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus
Reply by ●February 7, 20172017-02-07
On Tue, 07 Feb 2017 21:29:48 GMT, eric.jacobsen@ieee.org wrote:>On Mon, 06 Feb 2017 19:17:27 +0100, Michael Plet <me@home.com> wrote: > >>Hi Group >> >>I have derived this estimator. The limited tests I have done shows >>high accuracy. >> >> >>Let k be the index of the DFT bin with the largest magnitude. >>Let j be the index of the DFT bin with the largest magnitude >>neigboring bin k. >>That is j=k-1 or j=k+1. >> >>Now let P=Tan(PI*k/N) and Q=Tan(PI*j/N) >> >>Then the normalized frequency is estimated by: >> >>Freq=(N/PI)*Arctan(Sqr(P*Q*(Im[k]*P-Im[j]*Q)/(Im[k]*Q-Im[j]*P))) >> >>where Sqr is Square root and Im[] is the imaginary part of X[]. >> >> >>I was wondering if anyone has a test suite, so it would be possible to >>compare with other estimators? >> >> >>Regards, >>Michael Plet > >The link below is very old, but has some example estimators as well as >some test code to run them. It would not be difficult to add your >estimator and test it in the context of the other included estimators. > >http://www.ericjacobsen.org/fe2/fe2.htm > >Your estimator is quite complex on the face of it, and usually >two-term estimators have difficulty at some point as the frequency >offset varies, i.e., performance is highly offset dependent. Just >something to look out for. > > > >--- >This email has been checked for viruses by Avast antivirus software. >https://www.avast.com/antivirusThank you very much for the link. I'll have a look at the other estimators and the code.
Reply by ●February 7, 20172017-02-07
> >The link below is very old, but has some example estimators as well as >some test code to run them. It would not be difficult to add your >estimator and test it in the context of the other included estimators. > >http://www.ericjacobsen.org/fe2/fe2.htm > >Your estimator is quite complex on the face of it, and usually >two-term estimators have difficulty at some point as the frequency >offset varies, i.e., performance is highly offset dependent. Just >something to look out for. > > > >--- >This email has been checked for viruses by Avast antivirus software. >https://www.avast.com/antivirusJacobsen's estimator is actually an approximation of my 3 bin formula. I've got a blog article in rough draft form that describes the differences and the conditions under which they matter. The rest of the estimators on the page you referenced are also approximations. The difference between having exact equations and approximations can be seen in the noiseless case. If I were to reduce the sample count, the differences would be even more pronounced. Although the exact formulas tend to be a little more computationally intensive, the extra work is swamped by the amount of calculations required for the DFT values so overall the extra computations are insignificant. I am looking forward to seeing the improved version of Michael Plet's formula which should also employs the real parts of the DFT bins. Just like Vicanek's 2 bin formula, I'll bet it won't suffer any frequency offset variations that you speak of. His current one, and my 2 bin one, certainly do in the presence of noise. Ced ========================================================================== The sample count is 100 and the run size is 100 The phase value is 0.785398 Errors are shown at 100x actual value Target Noise Level = 0.000 Freq MP2Bin CD3Bin Jacobsen3Bin ---- ------------- ------------- ------------- 3.0 -0.000 0.000 0.000 0.000 0.000 0.000 3.1 -0.000 0.000 -0.000 0.000 -0.169 0.000 3.2 0.000 0.000 0.000 0.000 -0.515 0.000 3.3 0.000 0.000 0.000 0.000 -0.720 0.000 3.4 0.000 0.000 -0.000 0.000 -0.543 0.000 3.5 0.000 0.000 0.000 0.000 -0.014 0.000 3.6 -0.000 0.000 -0.000 0.000 -0.325 0.000 3.7 -0.000 0.000 0.000 0.000 -0.428 0.000 3.8 0.000 0.000 0.000 0.000 -0.297 0.000 3.9 0.000 0.000 0.000 0.000 -0.092 0.000 Target Noise Level = 0.001 Freq MP2Bin CD3Bin Jacobsen3Bin ---- ------------- ------------- ------------- 3.0 0.000 0.017 0.000 0.011 0.000 0.011 3.1 -0.001 0.011 -0.001 0.010 -0.169 0.011 3.2 -0.000 0.009 0.001 0.009 -0.515 0.010 3.3 0.000 0.009 0.001 0.011 -0.719 0.012 3.4 -0.000 0.009 -0.001 0.011 -0.544 0.013 3.5 -0.002 0.012 -0.003 0.014 -0.017 0.016 3.6 -0.001 0.018 -0.001 0.014 -0.326 0.012 3.7 -0.006 0.060 -0.002 0.015 -0.429 0.013 3.8 0.001 0.069 0.001 0.011 -0.296 0.010 3.9 -0.001 0.032 -0.002 0.010 -0.094 0.010 Target Noise Level = 0.010 Freq MP2Bin CD3Bin Jacobsen3Bin ---- ------------- ------------- ------------- 3.0 0.019 0.154 0.003 0.094 0.002 0.092 3.1 0.001 0.125 -0.003 0.093 -0.173 0.094 3.2 -0.012 0.088 0.007 0.102 -0.503 0.110 3.3 0.005 0.081 0.004 0.103 -0.716 0.115 3.4 -0.002 0.084 -0.002 0.120 -0.542 0.139 3.5 -0.009 0.114 0.003 0.152 -0.009 0.179 3.6 0.006 0.168 0.033 0.138 -0.296 0.120 3.7 0.091 0.596 0.005 0.119 -0.423 0.108 3.8 0.140 0.690 -0.023 0.132 -0.316 0.121 3.9 0.035 0.285 0.018 0.107 -0.074 0.104 Target Noise Level = 0.100 Freq MP2Bin CD3Bin Jacobsen3Bin ---- ------------- ------------- ------------- 3.0 -0.176 1.766 0.035 1.132 0.053 1.111 3.1 0.238 1.089 0.173 1.014 -0.018 1.063 3.2 -0.081 0.968 -0.203 1.009 -0.737 1.073 3.3 0.077 0.824 0.053 1.094 -0.665 1.211 3.4 -0.042 0.856 -0.024 1.197 -0.551 1.381 3.5 -0.014 1.168 -0.043 1.355 -0.069 1.565 3.6 0.110 1.797 -0.150 1.594 -0.448 1.418 3.7 0.246 5.629 0.003 1.177 -0.445 1.087 3.8 1.384 8.476 -0.044 1.083 -0.348 1.020 3.9 -0.113 3.245 -0.040 0.985 -0.137 0.948 --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●February 8, 20172017-02-08
>Hi Group > >I have derived this estimator. The limited tests I have done shows >high accuracy. > > >Let k be the index of the DFT bin with the largest magnitude. >Let j be the index of the DFT bin with the largest magnitude >neigboring bin k. >That is j=k-1 or j=k+1. > >Now let P=Tan(PI*k/N) and Q=Tan(PI*j/N) > >Then the normalized frequency is estimated by: > >Freq=(N/PI)*Arctan(Sqr(P*Q*(Im[k]*P-Im[j]*Q)/(Im[k]*Q-Im[j]*P))) > >where Sqr is Square root and Im[] is the imaginary part of X[]. > > >I was wondering if anyone has a test suite, so it would be possible to >compare with other estimators? > > >Regards, >Michael PletHi Michael, your estimator seems to be equivalent to mine (http://vicanek.de/articles/FreqFromTwoBins.pdf) if you choose a = b = 1. I have uploaded an extended version of my writeup to facilitate translation between my centered DFT notation and the standard DFT. Perhaps you can get some inspiration from there for a formula based on Re[k] (and some combination of formulas to optimize SNR)? Best Martin --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●February 8, 20172017-02-08
On Tue, 07 Feb 2017 22:55:37 -0600, "martinvicanek" <99986@DSPRelated> wrote:>>Hi Group >> >>I have derived this estimator. The limited tests I have done shows >>high accuracy. >> >> >>Let k be the index of the DFT bin with the largest magnitude. >>Let j be the index of the DFT bin with the largest magnitude >>neigboring bin k. >>That is j=k-1 or j=k+1. >> >>Now let P=Tan(PI*k/N) and Q=Tan(PI*j/N) >> >>Then the normalized frequency is estimated by: >> >>Freq=(N/PI)*Arctan(Sqr(P*Q*(Im[k]*P-Im[j]*Q)/(Im[k]*Q-Im[j]*P))) >> >>where Sqr is Square root and Im[] is the imaginary part of X[]. >> >> >>I was wondering if anyone has a test suite, so it would be possible to >>compare with other estimators? >> >> >>Regards, >>Michael Plet > >Hi Michael, > >your estimator seems to be equivalent to mine >(http://vicanek.de/articles/FreqFromTwoBins.pdf) if you choose a = b = 1. >I have uploaded an extended version of my writeup to facilitate >translation between my centered DFT notation and the standard DFT. Perhaps >you can get some inspiration from there for a formula based on Re[k] (and >some combination of formulas to optimize SNR)? > >Best >Martin >--------------------------------------- >Posted through http://www.DSPRelated.comHi Martin If a = b = 1, wouldn't my formula include Re[k]? I have tried to insert some of my test bin values into your formula. The result was not what I expected. I think I need to rotate the DFT bins before applying your formula. I will look into that. Thank you for the link to the updated document. I am sure I will find inspiration in that. This subject is so basic and pure that I don't understand why it hasn't been treated in DSP litterature. Michael
