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Y ( e j ω ) = 1 T s X ω 2 π T s for ω [ - π , π ] .

Here, it is understood that Y ( e j ω ) is periodic with period 2 π . Note in this expression that Y ( e j ω ) and X ( f ) are related by a simple scaling of the frequency and magnitude axes.Also note that ω = π in Y ( e j ω ) corresponds to the Nyquist frequency, f = 1 / ( 2 T s ) in X ( f ) .

Sometimes after the sampled signal has been digitally processed, it must then converted back to an analog signal.Theoretically, this can be done by converting the discrete-time signal to a sequenceof continuous-time impulses that are weighted by the sample values. If this continuous-time “impulse train” is filtered with anideal low pass filter, with a cutoff frequency equal to the Nyquist frequency, a scaled version of the original low pass filtered signalwill result. The spectrum of the reconstructed signal S ( f ) is given by

S ( f ) = Y ( e j 2 π f T s ) for | f | < 1 2 T s 0 otherwise.

Sampling and reconstruction using sample-and-hold

Sampling and reconstruction using a sample-and-hold.

In practice, signals are reconstructed using digital-to-analog converters.These devices work by reading the current sample, and generating a corresponding output voltage for a period of T s seconds. The combined effect of sampling and D/A conversion may be thoughtof as a single sample-and-hold device. Unfortunately, the sample-and-hold process distortsthe frequency spectrum of the reconstructed signal. In this section, we will analyze the effectsof using a zero th - order sample-and-hold in a sampling and reconstruction system. Later in the laboratory, we will see how the distortionintroduced by a sample-and-hold process may be reduced through the use of discrete-time interpolation.

[link] illustrates a system with a low-pass input filter,a sample-and-hold device, and a low-pass output filter. If there were no sampling, this system would simply betwo analog filters in cascade. We know the frequency response for this simpler system. Any differences between thisand the frequency response for the entire system is a result of the samplingand reconstruction. Our goal is to compare the two frequency responses using Matlab.For this analysis, we will assume that the filters are N t h order Butterworth filters with a cutoff frequency of f c , and that the sample-and-hold runs at a sampling rate of f s = 1 / T s  .

We will start the analysis by first examining the ideal case. Consider replacing the sample-and-hold with an ideal impulsegenerator, and assume that instead of the Butterworth filters we use perfect low-pass filters with a cutoff of f c   . After analyzing this case we will modify the results to account for thesample-and-hold and Butterworth filter roll-off.

If an ideal impulse generator is used in place of the sample-and-hold, then the frequencyspectrum of the impulse train can be computed by combining the sampling equation in [link] with the reconstruction equation in [link] .

S ( f ) = Y ( e j 2 π f T s ) = 1 T s k = - X 2 π f T s - 2 π k 2 π T s = 1 T s k = - X f - k f s , for | f | 1 2 T s . S ( f ) = 0 for | f | > 1 2 T s .

If we assume that f s > 2 f c , then the infinite sum reduces to one term.In this case, the reconstructed signal is given by

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Source:  OpenStax, Purdue digital signal processing labs (ece 438). OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10593/1.4
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