Thread: question on integration

Started: 2020-01-09 18:38:06
Last activity: 2020-01-10 18:21:17
Topics: SAC Help
Avinash
2020-01-09 18:38:06
Hi SAC users,
I have a somewhat stupid question about integration in SAC.
SAC provides two methods to do integration: time-domain (int) and
frequency-domain (divomega).
"int" uses the trapezoidal method and the first data point in the
integrated time-series is assumed to be zero.
"divomega" uses the Fourier transform properties integral=IFFT[ FFT[ x(t )]
/ i w ], in which w is angular frequency and x(t) is time series. Is this
method making any assumptions ? What happens at the first frequency point
(sum of x(t)) ?
Thanks for your help.
Regards,
Avinash Nayak

  • George Helffrich
    2020-01-10 18:21:17
    Dear All -

    A simple experiment with SAC:

    SAC> fg seismogram; rmean; int
    SAC> lh


    FILE: SEISMOGR
    --------------
    NPTS = 999 B = 9.4650
    E = 19.4450 IFTYPE = TIME SERIES FILE
    LEVEN = TRUE DELTA = 0.10E-01
    IDEP = UNKNOWN DEPMIN = -0.519559E-01
    DEPMAX = 0.524975E-01 DEPMEN = -0.192864E-02
    OMARKER = -43.2000 AMARKER = 10.4700
    FMARKER = 17.7800 KZDATE = MAR 29 (088), 1981
    KZTIME = 10:38:14.000 IZTYPE = BEGIN TIME
    KSTNM = CDV CMPAZ = 0.0
    CMPINC = 0.0 STLA = 48.0
    STLO = -120.0 KEVNM = K8108838
    EVLA = 48.0 EVLO = -125.0
    EVDP = 0.0 IEVTYP = AFTERSHOCK
    DIST = 373.063 AZ = 88.1471
    BAZ = 271.853 GCARC = 3.35746
    LOVROK = TRUE
    SAC> fg seismogram; rmean; fft; divomega; ifft
    (10.6d)FFT default change: not removing the mean
    DC level after DFT is 0.10729E-07
    SAC> lh


    FILE: SEISMOGR
    --------------
    NPTS = 1000 B = 9.460
    E = 19.450 IFTYPE = TIME SERIES FILE
    LEVEN = TRUE DELTA = 0.10E-01
    IDEP = VELOCITY (VOLTS) DEPMIN = -0.503674E-01
    DEPMAX = 0.547946E-01 DEPMEN = -0.450176E-04
    OMARKER = -43.2000 AMARKER = 10.4700
    FMARKER = 17.7800 KZDATE = MAR 29 (088), 1981
    KZTIME = 10:38:14.000 IZTYPE = BEGIN TIME
    KSTNM = CDV CMPAZ = 0.0
    CMPINC = 0.0 STLA = 48.0
    STLO = -120.0 KEVNM = K8108838
    EVLA = 48.0 EVLO = -125.0
    EVDP = 0.0 IEVTYP = AFTERSHOCK
    DIST = 373.063 AZ = 88.1471
    BAZ = 271.853 GCARC = 3.35746
    LOVROK = TRUE
    SAC>

    The difference between the “int” and “divomega” results is one data point less in the resulting trace. The two traces, though numerically different, are visually identical and may be overlaid for comparison.

    On 10 Jan 2020, at 02:39, Avinash <avinash07guddu<at>gmail.com> wrote:

    Hi SAC users,
    I have a somewhat stupid question about integration in SAC.
    SAC provides two methods to do integration: time-domain (int) and frequency-domain (divomega).
    "int" uses the trapezoidal method and the first data point in the integrated time-series is assumed to be zero.
    "divomega" uses the Fourier transform properties integral=IFFT[ FFT[ x(t )] / i w ], in which w is angular frequency and x(t) is time series. Is this method making any assumptions ? What happens at the first frequency point (sum of x(t)) ?
    Thanks for your help.
    Regards,
    Avinash Nayak


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    George Helffrich
    george<at>elsi.jp


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