It is possible to sample a band limited (to W Hz) signal at a lower rate than 2W

Question

It is possible to sample a band limited (to W Hz) signal at a lower rate than 2W if a related signal is also sampled. One idea is to sample the signal x(t) and its derivative x.(t), and then somehow combine these sampled signals to reconstruct x(t). Show that both the signal and its derivative, taken separately, cannot be recovered from their under sampled versions. One idea is to use the derivative approximation and derive a digital filter from this expression to find the "missing" samples x (n - 1/2/W) and thus obtain a properly sampled signal What is the digital filter-: Do you think this scheme will work? Why or why not? A completely analog scheme involves passing the sampled signal and the sampled derivative through linear filters and then summing the results. By focusing on the frequency range 0 le f

Explanation / Answer

a) let X(f) is the fourier transform of x(t) then fourier transform of dx(t)/dt is j2fX(f) it means if X(f) is band limited then j2fX(f) is also band limited with same band width.there fore undersampling of X(f) and j2fX(f) leads distorted X(f) and j2fX(f)..there fore they cannot be reconstructed..

b) x is sampled at w rate ( but reqiured is 2w) x(n/w) are samples

distance between successive samples is 1/w i.e n/w - n-1/w

now y(t) = dx(t)/dt is also sampled at w rate

y(n/w) = (x(n/w) - x(n-0.5/w) )/ (1/2w)

it means we get the sample  x(n-0.5/w) = x(n/w) - y(n/w)/2w

now new distance between the samples is n/w - n-0.5/w = 0.5/w =1/2w indicates sampling rate 2w

it means effectively we are sampling by 2w

therefore signal is recovered

Digital filter converts the analog signal to digital signal..and signal can be recovered..since sampled values are now in digital form

C) x(n/w)H1(f) + y(n/w)H2(f) = x(n/w) + x(n- 0.5/w) for signal to be recovered

take H1(f) = 2 then H2(f) = x(n- 0.5/w) - x(n/w) / y(n/w)

H2(f) = -2w

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