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Sampling

Module by: Justin Romberg

Summary: This module deals with translating continuous time problem into discrete time

Introduction

The digital computer can process discrete time signals using extremely flexible and powerful algorithms. However, most signals of interest are continuous time, which is how the almost always appear in nature.
This module introduces the idea of translating continuous time problems into discrete time, and you can read on to learn more of the details and importance of sampling.
    Key Questions
  • How do we turn a continuous time signal into a discrete time signal (sampling, A/D)?
  • When can we reconstruct a CT signal exactly from its samples (reconstruction, D/A)?
  • Manipulating the DT signal does what to the reconstructed signal?

Sampling

Sampling (and reconstruction) are best understood in the frequency domain. We'll start by looking at some examples
Problem 1
What CT signal ft f t has the CTFT shown below? ft=12π∫-∞∞Fⅈwⅇⅈwtdw f t 1 2 w F w w t
samp1.png
Figure 1: The CTFT of ft f t .
Hint: Fⅈw= F 1 ⅈw* F 2 ⅈw F w F 1 w F 2 w where the two parts of Fⅈw F w are:
samp2.pngsamp3.png
Subfigure 2.1
Subfigure 2.2
Figure 2
[ Click for Solution 1 ]
Solution 1
ft=12π∫-∞∞Fⅈwⅇⅈwtdw f t 1 2 w F w w t
[ Hide Solution 1 ]
Problem 2
What DT signal f s n f s n has the DTFT shown below? f s n=12π∫-ππ f s ⅇⅈwⅇⅈwndw f s n 1 2 w f s w w n
samp4.png
Figure 3: DTFT that is a periodic (with period=2π period 2 ) version of Fⅈw F w in Figure 1.
[ Click for Solution 2 ]
Solution 2
Since Fⅈw=0 F w 0 outside of -22 -2 2 ft=12π∫-22Fⅈwⅇⅈwtdw f t 1 2 w -2 2 F w w t Also, since we only use one interval to reconstruct f s n f s n from its DTFT, we have f s n=12π∫-22 f s ⅇⅈwⅇⅈwndw f s n 1 2 w -2 2 f s w w n Since Fⅈw= F s ⅇⅈw F w F s w on -22 -2 2 f s n=ft|t=n f s n t n f t i.e. f s n f s n is a sampled version of ft f t .
[ Hide Solution 2 ]
samp_big.png
Figure 4: ft f t is the continuous-time signal above and f s n f s n is the discrete-time, sampled version of ft f t

Generalization

Of course, the results from the above examples can be generalized to any ft f t with Fⅈw=0 F w 0 , |w|>π w , where ft f t is bandlimited to -ππ .
samp_g1.pngsamp_g2.png
Subfigure 5.1
Subfigure 5.2
Figure 5: Fⅈw F w is the CTFT of ft f t .
samp_g3.pngsamp_g4.png
Subfigure 6.1
Subfigure 6.2
Figure 6: F s ⅇⅈw F s w is the DTFT of f s n f s n .
F s ⅇⅈw F s w is a periodic (with period 2π 2 ) version of Fⅈw F w . F s ⅇⅈw F s w is the DTFT of signal sampled at the integers. Fⅈw F w is the CTFT of signal.
conclusion: If ft f t is bandlimited to -ππ then the DTFT of the sampled version f s n=fn f s n f n is just a periodic (with period 2π 2 ) version of Fⅈw F w .

Turning a Discrete Signal into a Continuous Signal

Now, let's look at turning a DT signal into a continuous time signal. Let f s n f s n be a DT signal with DTFT F s ⅇⅈw F s w
samp_e1.pngsamp_e2.png
Subfigure 7.1
Subfigure 7.2
Figure 7: F s ⅇⅈw F s w is the DTFT of f s n f s n .
Now, set f imp t=∑n=-∞∞ f s nδt-n f imp t n f s n δ t n The CT signal, f imp t f imp t , is non-zero only on the integers where there are impulses of height f s n f s n .
samp_e3.png
Figure 8
Problem 3
What is the CTFT of f imp t f imp t ?
[ Click for Solution 3 ]
Solution 3
f imp t=∑n=-∞∞ f s nδt-n f imp t n f s n δ t n
F ∼ imp ⅈw=∫-∞∞ f imp tⅇ-ⅈwtdt=∫-∞∞∑n=-∞∞ f s nδt-nⅇ-ⅈwtdt=∑n=-∞∞ f s n∫-∞∞δt-nⅇ-ⅈwtdt=∑n=-∞∞ f s nⅇ-ⅈwn= F s ⅇⅈw F ∼ imp w t f imp t w t t n f s n δ t n w t n f s n t δ t n w t n f s n w n F s w (1)
So, the CTFT of f imp t f imp t is equal to the DTFT of f s n f s n
note: We used the sifting property to show ∫-∞∞δt-nⅇ-ⅈwtdt=ⅇ-ⅈwn t δ t n w t w n
[ Hide Solution 3 ]
Now, given the samples f s n f s n of a bandlimited to -ππ signal, our next step will be to see how we can reconstruct ft f t .
samp_blk.png
Figure 9: Block diagram showing the very basic steps used to reconstruct ft f t . Can we make our results equal ft f t exactly?

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