Thread: TRANSFER freqlimits

Started: 2010-09-13 19:07:04
Last activity: 2010-09-14 17:54:46
Topics: SAC Help
Geo .
2010-09-13 19:07:04
Greetings,
Could someone please clarify the physical meaning of freqlimits when using
TRANSFER function in SAC? I have read the documentation but for me it still
is a little fuzzy. Does it mean that when we use certain freqlimits, for
example freqlimits 0.01 0.015 100 200, the result we see is a narrow band
passed record with corner frequencies of, in this example 0.15Hz and 100Hz?
If that were the case and if I wanted to have the broadest bandpass
possible, what would be the limits to use? When I use lower freqlimits like
0.001 0.005 the resulting record seems to have a trend and sometimes
amplified long period signals. Therefore, I am forced to narrow the
freqlimits to 0.01 0.15 (lower end) 100 200 (higher end). I believe if
freqlimits is acting as a bandpass I am loosing some lower and higher
frequency information. Any feed back is much appreciated!

Thank you!

Januka - using SAC 101.3b

  • Arthur Snoke
    2010-09-13 21:28:43
    Januka,

    I suggest looking at the help file for transfer that was included with
    101.4 and is posted on the IRIS SAC Web site at
    http://www.iris.edu/software/sac/commands/transfer.html. I think/hope
    the description is clearer than it was earlier, although the functionality
    is probably the same.

    Dealing with your example: freqlimits 0.01 0.015 100 200. Assuming your
    Nyquist is above 200.0, what you have here is a filter that has unit
    weight between f2 and f3: 0.015 and 100.0. That is, f2 and f3 are not
    corners as in a "normal" bandpass. There is a cosine taper from unity to
    0 between 100.0 and 200.0 and between 0.015 and 0.01. The response is
    "zero" for frequencies below 0.01 and above 200.0. I put quotes around
    the zero, because unless your npts is a power of 2, because (apparently)
    of the nature of the FFT, the response will not be zero.

    I hope this helps. If not, let me know.

    Arthur

    On Mon, 13 Sep 2010, Januka Attanayake wrote:

    Greetings,
    Could someone please clarify the physical meaning of freqlimits when
    using TRANSFER function in SAC? I have read the documentation but for me
    it still is a little fuzzy. Does it mean that when we use certain
    freqlimits, for example freqlimits 0.01 0.015 100 200, the result we see
    is a narrow band passed record with corner frequencies of, in this
    example 0.15Hz and 100Hz? If that were the case and if I wanted to have
    the broadest bandpass possible, what would be the limits to use? When I
    use lower freqlimits like 0.001 0.005 the resulting record seems to have
    a trend and sometimes amplified long period signals. Therefore, I am
    forced to narrow the freqlimits to 0.01 0.15 (lower end) 100 200 (higher
    end). I believe if freqlimits is acting as a bandpass I am loosing some
    lower and higher frequency information. Any feed back is much
    appreciated!

    Thank you!

    Januka - using SAC 101.3b


    • Geo .
      2010-09-14 17:06:40
      Thanks Arthur, the process is clear now. However, one of the questions I
      raised still remains. That is how would you determine what low/high pass
      frequencies (f1 f2 f3 f4) would you use? Documentation says "for a Nyquist
      frequency of 0.5, set f3=10. and f4=20". When processing seismograms in bulk
      with different instruments
      (1) you wouldn't know what the Nyquist freq. is
      (2) If the answer to (1) is " yes you don't" how would you determine
      f1,f2,f3,f4 for a given Nyquist freq on general terms.

      Thanks again !

      Januka.

      On Mon, Sep 13, 2010 at 2:28 PM, Arthur Snoke <snoke<at>vt.edu> wrote:

      Januka,

      I suggest looking at the help file for transfer that was included with
      101.4 and is posted on the IRIS SAC Web site at <
      http://www.iris.edu/software/sac/commands/transfer.html. I think/hope
      the description is clearer than it was earlier, although the functionality
      is probably the same.

      Dealing with your example: freqlimits 0.01 0.015 100 200. Assuming your
      Nyquist is above 200.0, what you have here is a filter that has unit weight
      between f2 and f3: 0.015 and 100.0. That is, f2 and f3 are not corners as
      in a "normal" bandpass. There is a cosine taper from unity to 0 between
      100.0 and 200.0 and between 0.015 and 0.01. The response is "zero" for
      frequencies below 0.01 and above 200.0. I put quotes around the zero,
      because unless your npts is a power of 2, because (apparently) of the nature
      of the FFT, the response will not be zero.

      I hope this helps. If not, let me know.

      Arthur


      On Mon, 13 Sep 2010, Januka Attanayake wrote:

      Greetings,
      Could someone please clarify the physical meaning of freqlimits when using
      TRANSFER function in SAC? I have read the documentation but for me it still
      is a little fuzzy. Does it mean that when we use certain freqlimits, for
      example freqlimits 0.01 0.015 100 200, the result we see is a narrow band
      passed record with corner frequencies of, in this example 0.15Hz and 100Hz?
      If that were the case and if I wanted to have the broadest bandpass
      possible, what would be the limits to use? When I use lower freqlimits like
      0.001 0.005 the resulting record seems to have a trend and sometimes
      amplified long period signals. Therefore, I am forced to narrow the
      freqlimits to 0.01 0.15 (lower end) 100 200 (higher end). I believe if
      freqlimits is acting as a bandpass I am loosing some lower and higher
      frequency information. Any feed back is much appreciated!

      Thank you!

      Januka - using SAC 101.3b




      --
      "Nothing can start to exist nor can cease to exist, only transformation is
      possible"

      Januka Attanayake
      Solid Earth Geophysics & Economics
      University of Connecticut
      Beach Hall U-2045
      354,Mansfield Rd;
      Storrs.CT 06269

      Tel: +1 860 486 0475 (Office)
      +1 860 486 3914 (Lab - P405)
      URL: http://sites.google.com/site/janukaattanayake/

      • Arthur Snoke
        2010-09-14 17:54:46
        Comments interspersed.

        On Tue, 14 Sep 2010, Januka Attanayake wrote:

        Thanks Arthur, the process is clear now. However, one of the questions I
        raised still remains. That is how would you determine what low/high pass
        frequencies (f1 f2 f3 f4) would you use? Documentation says "for a
        Nyquist frequency of 0.5, set f3=10. and f4=20". When processing
        seismograms in bulk with different instruments (1) you wouldn't know
        what the Nyquist freq. is (2) If the answer to (1) is " yes you don't"
        how would you determine f1,f2,f3,f4 for a given Nyquist freq on general
        terms.

        In the help file that comes with 101.4 and which can bde accessed from the
        IRIS SAC manual pages, it says:

        If you want to do a low-pass filter but have no filtering at low
        frequencies, one way is to set f1=-2 and f2=-1. If you want to do a
        high-pass fiter but have no filtering at the high frequencies, for a
        Nyquist frequency of 0.5, set f3=10. and f4=20.

        The point of this passage is how do you use freqlimits if you want to have
        no filtering at one end of the spectrum but you do at the other. The
        program as written says that if f2 > f1, but f2 is less than df, there is
        no filtering at low frequencies (a negative frequency is always less than
        any df.). Similarly, for a Nyquist of 0.5 Hz, if f4 > f3 and f3 is
        greater than 0.5, there is no filtering at high frequencies (10 is much
        greater than 0.5). Note that it is important that f2 > f1, or the program
        wil think this is dealing with the high-frequency end.

        You ask how one knows the Nyquist if one has many instruments, etc. DF
        and the Nyquist are related to the time series itself. Here I read in a
        time series and do an lh:

        SAC> r N11A.lhz
        SAC> lh


        FILE: N11A.lhz - 1
        --------------

        NPTS = 3101 B = -9.980005e+01
        E = 3.000200e+03 IFTYPE = TIME SERIES FILE
        LEVEN = TRUE DELTA = 1.000000e+00
        DEPMIN = -1.928006e+03 DEPMAX = 1.540401e+03

        Now I do an fft and a lh

        SAC> fft
        DC level after DFT is -724.33
        SAC> lh


        FILE: N11A.lhz - 1
        --------------

        NPTS = 4096 B = 0.000000e+00
        E = 5.000000e-01 IFTYPE = SPECTRAL
        FILE-AMPL/PHASE
        LEVEN = TRUE DELTA = 2.441406e-04
        DEPMIN = -1.928006e+03 DEPMAX = 1.540401e+03

        In the first lh, we see that there are 3101 points with a sampling rate of
        1 sps for a total time, then, of (NPTS-1)*DELTA =3100 seconds. After the
        fft, there are
        4096 points because the fft adds zeroes to the time serie to the next
        power of 2. The lowest frequency is 0, and E is now the Nyquist of 0.5 Hz
        (0.5/DELTA). Now, DELTA = 2.441406e-04 is DF, the frequency interval.

        I hope this helps. If you still have questions ...

        • Januka Attanayake
          2010-09-14 15:31:44
          GREAT ! Thanks again Arthur, I haven't got a better explanation than the one that you just gave me! May be you should include this in SAC documentation.

          Serenity isn't freedom from the storm, but peace within the storm
          Januka Attanayake  Solid Earth Geophysics & Economics
          The University of Connecticut
          Beach Hall - U2045
          354, Mansfield Rd;
          Storrs, CT 06269
          Tel :  860 486 0475 (Office)
                    860 486 3914 (Lab - P405)
          URL: http://sites.google.com/site/janukaattanayake/






          --- On Tue, 9/14/10, Arthur Snoke <snoke<at>vt.edu> wrote:

          From: Arthur Snoke <snoke<at>vt.edu>
          Subject: Re: [SAC-HELP] TRANSFER freqlimits
          To: "Januka Attanayake" <jattanayake<at>gmail.com>
          Cc: sac-help<at>iris.washington.edu
          Date: Tuesday, September 14, 2010, 10:54 AM

          Comments interspersed.

          On Tue, 14 Sep 2010, Januka Attanayake wrote:

          Thanks Arthur, the process is clear now. However, one of the questions I raised still remains. That is how would you determine what low/high pass frequencies (f1 f2 f3 f4) would you use? Documentation says "for a Nyquist frequency of 0.5, set f3=10. and f4=20". When processing seismograms in bulk with different instruments (1) you wouldn't know what the Nyquist freq. is (2) If the answer to (1) is " yes you don't" how would you determine f1,f2,f3,f4 for a given Nyquist freq on general terms.

          In the help file that comes with 101.4 and which can bde accessed from the IRIS SAC manual pages, it says:

          If you want to do a low-pass filter but have no filtering at low frequencies, one way is to set f1=-2 and f2=-1. If you want to do a high-pass fiter but have no filtering at the high frequencies, for a Nyquist frequency of 0.5, set f3=10. and f4=20.

          The point of this passage is how do you use freqlimits if you want to have no filtering at one end of the spectrum but you do at the other.  The program as written says that if f2 > f1, but f2 is less than df, there is no filtering at low frequencies (a negative frequency is always less than any df.).  Similarly, for a Nyquist of 0.5 Hz, if f4 > f3 and f3 is greater than 0.5, there is no filtering at high frequencies (10 is much greater than 0.5).  Note that it is important that f2 > f1, or the program wil think this is dealing with the high-frequency end.

          You ask how one knows the Nyquist if one has many instruments, etc.  DF and the Nyquist are related to the time series itself.  Here I read in a time series and do an lh:

          SAC> r N11A.lhz
          SAC> lh


            FILE: N11A.lhz - 1
          --------------

                 NPTS = 3101                                B = -9.980005e+01
                    E = 3.000200e+03                   IFTYPE = TIME SERIES FILE
                LEVEN = TRUE                            DELTA = 1.000000e+00
               DEPMIN = -1.928006e+03                  DEPMAX = 1.540401e+03

          Now I do an fft and a lh

          SAC> fft
          DC level after DFT is -724.33
          SAC> lh


            FILE: N11A.lhz - 1
          --------------

                 NPTS = 4096                                B = 0.000000e+00
                    E = 5.000000e-01                   IFTYPE = SPECTRAL FILE-AMPL/PHASE
                LEVEN = TRUE                            DELTA = 2.441406e-04
               DEPMIN = -1.928006e+03                  DEPMAX = 1.540401e+03

          In the first lh, we see that there are 3101 points with a sampling rate of 1 sps for a total time, then, of (NPTS-1)*DELTA =3100 seconds.  After the fft, there are 4096 points because the fft adds zeroes to the time serie to the next power of 2.  The lowest frequency is 0, and E is now the Nyquist of 0.5 Hz (0.5/DELTA).  Now, DELTA = 2.441406e-04 is DF, the frequency interval.

          I hope this helps.  If you still have questions ...
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          sac-help mailing list
          sac-help<at>iris.washington.edu
          http://www.iris.washington.edu/mailman/listinfo/sac-help




05:08:19 v.01697673