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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