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(→3) Recovering the reconstructed signal from its resampling) |
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== Calculation == | == Calculation == | ||
The are 3 steps to recover the original function from its resampling. | The are 3 steps to recover the original function from its resampling. | ||
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=== 1) Multiply the function by an impulse train === | === 1) Multiply the function by an impulse train === | ||
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<math>x_p(t)=x(t)\sum_{n=-\infty}^\infty \delta(t-nT)</math> where T is the period of the function | <math>x_p(t)=x(t)\sum_{n=-\infty}^\infty \delta(t-nT)</math> where T is the period of the function | ||
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=== 2) Convolve Xp(t) with h1(t), the step function === | === 2) Convolve Xp(t) with h1(t), the step function === | ||
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<math>x_0(t)=x(nT)h_1(t-nT)</math> , which is our result, the shifted piecewise step functions. | <math>x_0(t)=x(nT)h_1(t-nT)</math> , which is our result, the shifted piecewise step functions. | ||
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=== 3) Recovering the reconstructed signal from its resampling === | === 3) Recovering the reconstructed signal from its resampling === | ||
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<math>x_r(t)=x_0(t)*h_1(t)*h_2(t)</math> | <math>x_r(t)=x_0(t)*h_1(t)*h_2(t)</math> | ||
| − | This is most easiest performed in the frequency domain with Fourier transforms. To most easily find h_2(t), make sure that : | + | This is most easiest performed in the frequency domain with Fourier transforms. To most easily find <math>h_2(t)</math>, make sure that : |
<math>H(\omega)=H_1(\omega)H_2(\omega)=\mathcal{F}(h_1(t)*h_2(t))</math> | <math>H(\omega)=H_1(\omega)H_2(\omega)=\mathcal{F}(h_1(t)*h_2(t))</math> | ||
and make sure that <math>H(\omega)</math> has the correct low pass filter properties listed above. | and make sure that <math>H(\omega)</math> has the correct low pass filter properties listed above. | ||
Latest revision as of 17:16, 17 November 2008
Contents
Sampling with Zero-Order Hold
English Definition
A signal can be sampled using piecewise step functions; The sample is extended directly across until the new sample is reached. This sampling of the function is called $ x_0(t) $. The reconstructed function $ x_r(t) $ can be recovered using a low pass filter created from the multiplication of the original step function generator $ \mathcal{H}_1(\omega) $ and another function $ \mathcal{H}_2(\omega) $
Calculation
The are 3 steps to recover the original function from its resampling.
1) Multiply the function by an impulse train
First, multiply this function x(t) by an impulse train p(t) This should yield:
$ x_p(t)=x(t)*p(t) $
$ x_p(t)=x(t)\sum_{n=-\infty}^\infty \delta(t-nT) $ where T is the period of the function
2) Convolve Xp(t) with h1(t), the step function
We have a series of impulses in the time domain, but we want them to extend across, so we must convolve them to get $ x_0(t) $ , or the actual sampling.
Beforehand, our $ h_1(t) $ is given as
$ h_1(t) = \begin{cases} 1 & 0 \le t < T \\ 0 & \mbox{else} \end{cases} $
$ x_0(t)=x_p(t)*h_1(t) $
$ x_0(t)=h_1(t)*x(t)\sum_{n=-\infty}^\infty \delta(t-nT) $
Now, convert $ x(t) $ to $ x(nT) $ for period T
$ x_0(t)=h_1(t)*x(nT)\sum_{n=-\infty}^\infty \delta(t-nT) $
$ x_0(t)=x(nT)h_1(t-nT) $ , which is our result, the shifted piecewise step functions.
3) Recovering the reconstructed signal from its resampling
To recover the signal $ x(t) $, we must find a signal $ h_2(t) $ so that the convolution of $ h_1(t) $ and $ h_2(t) $ represent a low pass filter with a gain of T and a cutoff frequency $ \omega_c $ between $ [\omega_m,\omega_s-\omega_m] $ to avoid aliasing.
In other words,
$ x_r(t)=x_0(t)*h_1(t)*h_2(t) $
This is most easiest performed in the frequency domain with Fourier transforms. To most easily find $ h_2(t) $, make sure that :
$ H(\omega)=H_1(\omega)H_2(\omega)=\mathcal{F}(h_1(t)*h_2(t)) $
and make sure that $ H(\omega) $ has the correct low pass filter properties listed above.
