(→Reconstructing a signal from its samples using Interpolation) |
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<math> h(t) = \frac{wcT sin(wct)}{\pi*wct} </math> | <math> h(t) = \frac{wcT sin(wct)}{\pi*wct} </math> | ||
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| + | such that | ||
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| + | <math> xr(t)= \sum_{n =-\infty}^{\infty} x(nT) \frac{wcT sin(wc(t-nT))}{\pi*wc(t-nT)} </math> | ||
Latest revision as of 10:38, 8 November 2008
Reconstructing a signal from its samples using Interpolation
We have learned in class that a signal can be reformed by obtaining multiple samples of its signal and using an important procedure we know as interpolation we can obtain the original signal of the function.
- it is noted that if the sampling instants are sufficiently close, then the signal can be reconstructed using a lowpass filter. the output is then considered to be:
$ xr(t)= xp(t) * h(t) $
or with xp(t):
$ xr(t)= \sum_{n =-\infty}^{\infty} x(nT)h(t-nT) $
-the following equation shows how to take a continuous curve and represent an interpolation formula for an ideal lowpass filter H(jw):
$ h(t) = \frac{wcT sin(wct)}{\pi*wct} $
such that
$ xr(t)= \sum_{n =-\infty}^{\infty} x(nT) \frac{wcT sin(wc(t-nT))}{\pi*wc(t-nT)} $
