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| − | =HW3_Signal Reconstruction_Interpolation= | + | =HW3_Signal Reconstruction_Interpolation (Band-limited)= |
| + | After having creating a sampled version of the original function, '''<math>X_{s}</math>''', it needs to be reconstructed into the original function '''<math>x(t)</math>'''. To do this, the Whittaker-Shannon interpolation formula is utilized. | ||
| + | The sampling theorem says that given a function that meets two requirements: | ||
| + | :1) It is band-limited. This means that the Fourier transform of the original signal, also known as the spectrum, is 0 for |f| > B, where B is the bandwidth. | ||
| + | :2) It is sampled at the Nyquist frequency, <math>f_s > 2B</math> | ||
| − | + | it can be exactly reconstructed from its samples. If it does not meet these two requirements, aliasing occurs. | |
| + | To recover the signal, the sampled function is simply multiplied by a low-pass filter with height equal to the sampling period T, <math>{H_r}(f)</math>, to isolate the original signal. | ||
| + | <math>{X_r}(f) = {H_r}(f){X_s}(f)</math> | ||
| + | |||
| + | [[Image:ReconstructFilter.png|500px|thumb|center|Sampled function with lowpass filter]] | ||
| + | |||
| + | '''Mathematical Process for finding <math>{X_r}(f)</math>:''' | ||
| + | |||
| + | <math>{X_r}(f) = {H_r}(f) \cdot {X_s}(f)</math> where <math>{H_r}(f) = T {rect}(Tf)</math> | ||
| + | |||
| + | Convert to the time domain. <math>X_s(f)</math> in the time domain is the sum of a train of deltas at every period of sampling, T | ||
| + | |||
| + | :<math>x_r(t) = h_r(t) * x_s(t) = sinc(t/T) * \sum_{k=-\infty}^{\infty} x(kT)\cdot \delta(t - kT)</math> | ||
| + | |||
| + | Properties of convolution allow | ||
| + | |||
| + | :<math>x_r(t) = \sum_{k=-\infty}^{\infty} x(kT) \cdot sinc(t/T) * \delta(t - kT)</math> | ||
| + | |||
| + | When convolving a function with a shifted delta, the result is the function shifted | ||
| + | |||
| + | :<math>x_r(t) = \sum_{k=-\infty}^{\infty} x(kT) \cdot sinc(\frac{t-kT}{T})</math> | ||
| + | |||
| + | It should be noted that while this may work in the theoretical, it is impossible to make a low-pass filter with an infinite cut-off slope. It is for this reason oversampling is used when reconstructing the signal. Oversampling inserts zeros after every sample according to the oversampling factor D. As an example, if D=4, the system would insert three samples of value zero in between each of the old samples. The sampled output spectrum will still repeat periodically, just with 4 times the period. Additionally an analog filter will have to be used to filter the sampled output. | ||
| + | |||
| + | Reference: http://www.silcom.com/~aludwig/Signal_processing/Signal_processing.htm#Resolution_and_bandwidth | ||
[[ Hw3ECE438F09boutin|Back to Hw3ECE438F09boutin]] | [[ Hw3ECE438F09boutin|Back to Hw3ECE438F09boutin]] | ||
Latest revision as of 18:51, 22 September 2009
HW3_Signal Reconstruction_Interpolation (Band-limited)
After having creating a sampled version of the original function, $ X_{s} $, it needs to be reconstructed into the original function $ x(t) $. To do this, the Whittaker-Shannon interpolation formula is utilized.
The sampling theorem says that given a function that meets two requirements:
- 1) It is band-limited. This means that the Fourier transform of the original signal, also known as the spectrum, is 0 for |f| > B, where B is the bandwidth.
- 2) It is sampled at the Nyquist frequency, $ f_s > 2B $
it can be exactly reconstructed from its samples. If it does not meet these two requirements, aliasing occurs.
To recover the signal, the sampled function is simply multiplied by a low-pass filter with height equal to the sampling period T, $ {H_r}(f) $, to isolate the original signal.
$ {X_r}(f) = {H_r}(f){X_s}(f) $
Mathematical Process for finding $ {X_r}(f) $:
$ {X_r}(f) = {H_r}(f) \cdot {X_s}(f) $ where $ {H_r}(f) = T {rect}(Tf) $
Convert to the time domain. $ X_s(f) $ in the time domain is the sum of a train of deltas at every period of sampling, T
- $ x_r(t) = h_r(t) * x_s(t) = sinc(t/T) * \sum_{k=-\infty}^{\infty} x(kT)\cdot \delta(t - kT) $
Properties of convolution allow
- $ x_r(t) = \sum_{k=-\infty}^{\infty} x(kT) \cdot sinc(t/T) * \delta(t - kT) $
When convolving a function with a shifted delta, the result is the function shifted
- $ x_r(t) = \sum_{k=-\infty}^{\infty} x(kT) \cdot sinc(\frac{t-kT}{T}) $
It should be noted that while this may work in the theoretical, it is impossible to make a low-pass filter with an infinite cut-off slope. It is for this reason oversampling is used when reconstructing the signal. Oversampling inserts zeros after every sample according to the oversampling factor D. As an example, if D=4, the system would insert three samples of value zero in between each of the old samples. The sampled output spectrum will still repeat periodically, just with 4 times the period. Additionally an analog filter will have to be used to filter the sampled output.
Reference: http://www.silcom.com/~aludwig/Signal_processing/Signal_processing.htm#Resolution_and_bandwidth
