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Fast Fourier Transform

JNichols
New Contributor I
1,113 Views

I am using the 

call basic_dp_real_dft_1d(XF, num, numR, RC)

routine.  Your sample code in Fortran has the input matrix with a size 7, ie 1 to 7 in the array, but I thought I picked up that the C sample is 0 to 8 as expected.  

We normally stick to arrays with a size of 16384 -  2000 records per second for about 8 seconds.  This gives us 1000 Hz at 1000/8192 steps.  I normally use the Numerical Recipes FFT Solver, but I thought I would give this a go.  

I am looking at a data length of 960001 records, so it is not even close to a power or 2.  The FFT routine returns me a complex array of half the length as expected.   I just threw in the whole array to see what happened, I expected it to throw an error. But it did not. 

I spent an hour and read the MKL FFT manual chapter, I am none the wiser as to how you get away from the power of 2 requirement.  I checked the output  - the first column is the unit step, so the 9000 is about 10 Hz, for a concrete floor - reasonable, the second of 19000 is about 20 Hz, not unexpected and the 28000 is the electrical cables shaking the building.  We see this a lot. I have not looked for the higher order frequencies 120, 180 etc. yet. 

Am I correct, you do not need a power of 2 input array, in this case I can supply 524288 or 2 to the 19th?

Can I assume that the step is 1000/480000?

 TEST PASSED
     1    ********   3.142  ******
     2      77.753   1.777  ******
     3      43.402   1.754  ******
     4      37.835   1.710  ******
     5      19.610   1.479  ******
     6      18.179   1.167  ******
     7      23.374   1.554  ******
     8      12.224   2.323  ******
     9      17.760   0.986  ******
    18      11.204   1.820  ******
  7399      11.094   4.902  ******
  9070      13.475   3.112  ******
  9248      11.965   3.648  ******
 19147      10.214   2.266  ******
 28196      13.586   1.374  ******
148194      10.736   3.721  ******

 

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Ruqiu_C_Intel
Employee
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This issue is closing and we will no longer respond to this thread. If you require additional assistance from Intel, please start a new thread. Any further interaction in this thread will be considered community only. 


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10 Replies
JNichols
New Contributor I
1,103 Views

If I graph the frequency output, I get the following for the 960000 elements.  The 15 to 20 Hz is about what one could expect, the power at about 58 Hz means the Houston Supply is less than perfect, which does not surprise me as my old electrical clock changes time at a fairly annoying constant rate.  The 120 is double the 60 and quite good, the 180 is high, which is unusual.  Usually they drop below the 60 multiples. 

image_2021-09-13_153558.png

ShanmukhS_Intel
Moderator
1,066 Views

Hi,


Thanks for reaching out to us.


We are working on your issue. Could you please share us your environment details(OS details, MKL version etc..)


Best Regards,

Shanmukh.SS


JNichols
New Contributor I
1,061 Views

The dimension of the transform is a positive integer value represented in an integer scalar of Integer data
type. For a one-dimensional transform, the transform length is specified by a positive integer value
represented in an integer scalar of Integer data type.

I found this in your latest stuff on the FFT, when you are discussing lengths.  

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Screenshot 2021-09-14 104915.png

I think this is all you want.  

The status error messages from the FFT that are predefined do not include an error message for incorrect length. 

JNichols
New Contributor I
1,058 Views

This is the real sample from your files.  It has been around for a long time as I have older versions in my backup files.  The length of the real array is 7, I thought at one stage I went and had a look at the C version and there was a 0 to 7 array as I expected, I have not rechecked any my memory may be flawed.  

The first time I used this method, I stuck to 1048 as the input array, but I was lazy and just threw in the whole array, as your 7 appears to make it ok?

Your references include the standard such as Brigham, which is definitely 2 to the nth power.  

    !===============================================================================
    ! Copyright 2011-2019 Intel Corporation.
    !
    ! This software and the related documents are Intel copyrighted  materials,  and
    ! your use of  them is  governed by the  express license  under which  they were
    ! provided to you (License).  Unless the License provides otherwise, you may not
    ! use, modify, copy, publish, distribute,  disclose or transmit this software or
    ! the related documents without Intel's prior written permission.
    !
    ! This software and the related documents  are provided as  is,  with no express
    ! or implied  warranties,  other  than those  that are  expressly stated  in the
    ! License.
    !===============================================================================

    ! Content:
    ! A simple example of double-precision real-to-complex out-of-place 1D
    ! FFT using Intel(R) MKL DFTI
    !
    !*****************************************************************************
    program basic_dp_real_dft_1d

    use MKL_DFTI, forget => DFTI_DOUBLE, DFTI_DOUBLE => DFTI_DOUBLE_R

    ! Size of 1D transform
    integer, parameter :: N = 7

    ! Arbitrary harmonic used to test FFT
    integer, parameter :: H = 1

    ! Need double precision
    integer, parameter :: WP = selected_real_kind(15,307)

    ! Execution status
    integer :: status = 0, ignored_status
    integer i

    ! Data arrays
    real(WP), allocatable :: x_real (:)
    complex(WP), allocatable :: x_cmplx (:)

    type(DFTI_DESCRIPTOR), POINTER :: hand

    hand => null()

    print *,"Example basic_dp_real_dft_1d"
    print *,"Forward-Backward double-precision real-to-complex",               &
        &      " out-of-place 1D transform"
    print *,"Configuration parameters:"
    print *,"DFTI_PRECISION              = DFTI_DOUBLE"
    print *,"DFTI_FORWARD_DOMAIN         = DFTI_REAL"
    print *,"DFTI_DIMENSION              = 1"
    print '(" DFTI_LENGTHS                = /"I0"/" )', N
    print *,"DFTI_PLACEMENT              =  DFTI_NOT_INPLACE"
    print *,"DFTI_CONJUGATE_EVEN_STORAGE = DFTI_COMPLEX_COMPLEX"

    print *,"Create DFTI descriptor"
    status = DftiCreateDescriptor(hand, DFTI_DOUBLE, DFTI_REAL, 1, N)
    if (0 /= status) goto 999

    print *,"Set out-of-place"
    status = DftiSetValue(hand, DFTI_PLACEMENT, DFTI_NOT_INPLACE)
    if (0 /= status) goto 999

    print *,"Set CCE storage"
    status = DftiSetValue(hand, DFTI_CONJUGATE_EVEN_STORAGE,                   &
        &                    DFTI_COMPLEX_COMPLEX)
    if (0 /= status) goto 999

    print *,"Commit DFTI descriptor"
    status = DftiCommitDescriptor(hand)
    if (0 /= status) goto 999

    print *,"Allocate data arrays"
    allocate ( x_real(N), STAT = status)
    if (0 /= status) goto 999
    allocate ( x_cmplx(INT(N/2.0) + 1), STAT = status)
    if (0 /= status) goto 999

    print *,"Initialize data for real-to-complex FFT"
    call init_r(x_real, N, H)


    print *,"Compute forward transform"
    status = DftiComputeForward(hand, x_real, x_cmplx)
    if (0 /= status) goto 999

    print *,"Verify the result"
    status = verify_c(x_cmplx, N, H)
    if (0 /= status) goto 999

    print *,"Initialize data for complex-to-real FFT"
    call init_c(x_cmplx, N, H)

    
    do 101 i = 1, 4
        write(*,*)i,x_cmplx(i)
101 end do
    print *,"Compute backward transform"
    status = DftiComputeBackward(hand, x_cmplx, x_real)
    if (0 /= status) goto 999

    print *,"Verify the result"
    status = verify_r(x_real, N, H)
    if (0 /= status) goto 999

100 continue

    print *,"Release the DFTI descriptor"
    ignored_status = DftiFreeDescriptor(hand)

    if (allocated(x_real) .or. allocated(x_cmplx)) then
        print *,"Deallocate data arrays"
    endif
    if (allocated(x_real))     deallocate(x_real)
    if (allocated(x_cmplx))    deallocate(x_cmplx)

    if (status == 0) then
        print *, "TEST PASSED"
        call exit(0)
    else
        print *, "TEST FAILED"
        call exit(1)
    end if

999 print '("  Error, status = ",I0)', status
    goto 100


    contains

    ! Compute mod(K*L,M) accurately
    pure real(WP) function moda(k,l,m)
    integer, intent(in) :: k,l,m
    integer*8 :: k8
    k8 = k
    moda = real(mod(k8*l,m),WP)
    end function moda

    ! Initialize x to be inverse of unit peak at H
    subroutine init_r(x, N, H)
    integer N, H
    real(WP) :: x(:)

    real(WP), parameter:: TWOPI = 6.2831853071795864769_WP
    integer k

    forall (k=1:N)
        x(k) = 2 * cos( TWOPI * moda(H,k-1,N) / N) / N
    end forall
    if (mod(2*(N-H),N)==0) x(1:N) =  x(1:N) / 2
    end subroutine init_r

    ! Initialize x to be inverse of unit peak at H
    subroutine init_c(x, N, H)
    integer N, H
    complex(WP) :: x(:)

    complex(WP), parameter :: ITWOPI = (0.0_WP,6.2831853071795864769_WP)
    integer k

    forall (k=1:N/2+1)
        x(k) = exp( -ITWOPI * moda(H,k-1,N) / N) / N
    end forall
    end subroutine init_c

    ! Verify that x(N) is unit peak at H and N-H
    integer function verify_c(x, N, H)
    integer N, H
    complex(WP) :: x(:)

    integer k
    real(WP) err, errthr, maxerr
    complex(WP) res_exp, res_got

    ! Note, this simple error bound doesn't take into account error of
    ! input data
    errthr = 2.5 * log(real(N,WP)) / log(2.0_WP) * EPSILON(1.0_WP)
    print '("  Check if err is below errthr " G10.3)', errthr

    maxerr = 0.0_WP
    do k = 1, N/2+1
        if (mod(k-1-H,N)==0 .or. mod(-k+1-H,N)==0) then
            res_exp = 1.0_WP
        else
            res_exp = 0.0_WP
        end if
        res_got = x(k)
        err = abs(res_got - res_exp)
        maxerr = max(err,maxerr)
        if (.not.(err < errthr)) then
            print '("  x("I0"):"$)', k
            print '(" expected ("G24.17","G24.17"),"$)', res_exp
            print '(" got ("G24.17","G24.17"),"$)', res_got
            print '(" err "G10.3)', err
            print *," Verification FAILED"
            verify_c = 1
            return
        end if
    end do
    print '("  Verified,  maximum error was " G10.3)', maxerr
    verify_c = 0
    end function verify_c

    ! Verify that x is unit peak at H
    integer function verify_r(x, N, H)
    integer N, H
    real(WP) :: x(:)

    integer k
    real(WP) err, errthr, maxerr, res_exp

    ! Note, this simple error bound doesn't take into account error of
    ! input data
    errthr = 2.5 * log(real(N,WP)) / log(2.0_WP) * EPSILON(1.0_WP)
    print '("  Check if err is below errthr " G10.3)', errthr

    maxerr = 0.0_WP
    do k = 1, N
        if (mod(k-1-H, N)==0) then
            res_exp = 1.0_WP
        else
            res_exp = 0.0_WP
        end if
        err = abs(x(k) - res_exp)
        maxerr = max(err,maxerr)
        if (.not.(err < errthr)) then
            print '("  x("I0"):"$)', k
            print '(" expected "G24.17","$)', res_exp
            print '(" got "G24.17","$)', x(k)
            print '(" err "G10.3)', err
            print *," Verification FAILED"
            verify_r = 100
            return
        end if
    end do
    print '("  Verified,  maximum error was " G10.3)', maxerr
    verify_r = 0
    end function verify_r

    end program basic_dp_real_dft_1d

 

JNichols
New Contributor I
1,063 Views

The other thing that is an interesting annoyance is the KIND=WP, on the Fortran Forum, the usual is KIND=DP.  when you mix modules you need to fix the problem. 

I always wondered what WP stood for, I know what it means.  

JohnNichols
Valued Contributor II
1,031 Views

I reran the problem with a 2**19 data set, about 70% of the original and aside from the scaling factor it gives a similar answer. 

Ruqiu_C_Intel
Employee
894 Views

Hi,


MKL DFT supports both the input length is a power of 2, for example 2^19 = 524288, or not a power of 2, for example 1048.


I tested the sample code basic_dp_real_dft_1d for both input length 1048 and 524288. Both of them are passed as below:


 Example basic_dp_real_dft_1d

 Forward-Backward double-precision real-to-complex out-of-place 1D transform

 Configuration parameters:

 DFTI_PRECISION       = DFTI_DOUBLE

 DFTI_FORWARD_DOMAIN     = DFTI_REAL

 DFTI_DIMENSION       = 1

 DFTI_LENGTHS        = /1048/

 DFTI_PLACEMENT       = DFTI_NOT_INPLACE

 DFTI_CONJUGATE_EVEN_STORAGE = DFTI_COMPLEX_COMPLEX

 Create DFTI descriptor

 Set out-of-place

 Set CCE storage

 Commit DFTI descriptor

 Allocate data arrays

 Initialize data for real-to-complex FFT

 Compute forward transform

 Verify the result

 Check if err is below errthr 0.557E-14

 Verified, maximum error was 0.722E-15

 Initialize data for complex-to-real FFT

 Compute backward transform

 Verify the result

 Check if err is below errthr 0.557E-14

 Verified, maximum error was 0.111E-14


and


 Example basic_dp_real_dft_1d

 Forward-Backward double-precision real-to-complex out-of-place 1D transform

 Configuration parameters:

 DFTI_PRECISION       = DFTI_DOUBLE

 DFTI_FORWARD_DOMAIN     = DFTI_REAL

 DFTI_DIMENSION       = 1

 DFTI_LENGTHS        = /524288/

 DFTI_PLACEMENT       = DFTI_NOT_INPLACE

 DFTI_CONJUGATE_EVEN_STORAGE = DFTI_COMPLEX_COMPLEX

 Create DFTI descriptor

 Set out-of-place

 Set CCE storage

 Commit DFTI descriptor

 Allocate data arrays

 Initialize data for real-to-complex FFT

 Compute forward transform

 Verify the result

 Check if err is below errthr 0.105E-13

 Verified, maximum error was 0.884E-16

 Initialize data for complex-to-real FFT

 Compute backward transform

 Verify the result

 Check if err is below errthr 0.105E-13

 Verified, maximum error was 0.390E-16


​Not sure whether you have read this article below about the CCS output for MKL sample code before

https://software.intel.com/content/www/us/en/develop/articles/unpack-result-of-intel-mkl-fft-to-alig...



JohnNichols
Valued Contributor II
868 Views

By default, the FFT size is the first equal or superior power of 2 of the window size.

So I am assuming and this is a question, that the MKL FFT packs the 1048 array to 2048 with 1000 extra zeros. 

This should be true, is it?

I would have thought this should trigger a warning to say you are packing the data with zeros?

Ruqiu_C_Intel
Employee
852 Views

No, for 1048 size length which is not power of 2, MKL doesn't packet it to 2048 with extra 1000 zeros. Actually MKL will perform DFT 1048 API.


Ruqiu_C_Intel
Employee
672 Views

This issue is closing and we will no longer respond to this thread. If you require additional assistance from Intel, please start a new thread. Any further interaction in this thread will be considered community only. 


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