(New page: x(t)*q(t) = T*SUM{x[k]*d(t-k*T)} .------. x(t)--->(*)------------------------------------>| H(f) |---> x(t) ^ '----...)
 
 
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Mohammed Almathami
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                 x(t)*q(t) = T*SUM{x[k]*d(t-k*T)}  .------.
 
                 x(t)*q(t) = T*SUM{x[k]*d(t-k*T)}  .------.
 
     x(t)--->(*)------------------------------------>| H(f) |---> x(t)
 
     x(t)--->(*)------------------------------------>| H(f) |---> x(t)

Latest revision as of 03:17, 30 July 2009

Mohammed Almathami



                x(t)*q(t) = T*SUM{x[k]*d(t-k*T)}   .------.
   x(t)--->(*)------------------------------------>| H(f) |---> x(t)
            ^                                      '------'
            |
            |                +inf
            '------- q(t) = T*SUM{ d(t - k*T) }
                            k=-inf


where d(t) = 'dirac' impulse function

    and T = 1/Fs = sampling period
       Fs = sampling frequency


              +inf
   q(t) = T * SUM{ d(t - k*T) }  is the "sampling function", is periodic
             k=-inf              with period T, and can be expressed as a
                                 Fourier series.  It turns out that ALL of
                                 the Fourier coefficients are equal to 1.
         +inf
   q(t) = SUM{ c[n]*exp(j*2*pi*n/T*t) }
        n=-inf

where c[n] is the Fourier series coefficient.

                   t0+T
   c[n] =  1/T * integral{q(t) * exp(-j*2*pi*n/T*t) dt}
                    t0

and t0 can be any time. We choose t0 to be -T/2.

                    T/2      +inf
   c[n] =  1/T * integral{T * SUM{ d(t - k*T) } * exp(-j*2*pi*n/T*t) dt}
                   -T/2     k=-inf
             T/2
        = integral{ d(t - 0*T) * exp(-j*2*pi*n/T*t) dt}
            -T/2
             T/2
        = integral{ d(t) * exp(-j*2*pi*n/T*t) dt}
            -T/2
        = exp(-j*2*pi*n/T*0) = 1

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva