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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

Exercise 1

What CT signal ft f t has the CTFT shown below? ft=12π-Fwwtdw f t 1 2 w F w w t

Figure 1: The CTFT of ft f t .
Figure 1 (samp1.png)

Hint: Fw= F 1 w* F 2 w F w F 1 w F 2 w where the two parts of Fw F w are:

Figure 2
Subfigure 2.1Subfigure 2.2
Subfigure 2.1 (samp2.png)Subfigure 2.2 (samp3.png)

Solution 1

ft=12π-Fwwtdw f t 1 2 w F w w t

Exercise 2

What DT signal f s n f s n has the DTFT shown below? f s n=12π-ππ f s wwndw f s n 1 2 w f s w w n

Figure 3: DTFT that is a periodic (with period=2π period 2 ) version of Fw F w in Figure 1.
Figure 3 (samp4.png)

Solution 2

Since Fw=0 F w 0 outside of -22 -2 2 ft=12π-22Fwwtdw 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 wwndw f s n 1 2 w -2 2 f s w w n Since Fw= F s w 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 .

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
Figure 4 (samp_big.png)

Generalization

Of course, the results from the above examples can be generalized to any ft f t with Fw=0 F w 0 , |w|>π w , where ft f t is bandlimited to -ππ .

Figure 5: Fw F w is the CTFT of ft f t .
Subfigure 5.1Subfigure 5.2
Subfigure 5.1 (samp_g1.png)Subfigure 5.2 (samp_g2.png)
Figure 6: F s w F s w is the DTFT of f s n f s n .
Subfigure 6.1Subfigure 6.2
Subfigure 6.1 (samp_g3.png)Subfigure 6.2 (samp_g4.png)

F s w F s w is a periodic (with period 2π 2 ) version of Fw F w . F s w F s w is the DTFT of signal sampled at the integers. Fw 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 Fw 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 w F s w

Figure 7: F s w F s w is the DTFT of f s n f s n .
Subfigure 7.1Subfigure 7.2
Subfigure 7.1 (samp_e1.png)Subfigure 7.2 (samp_e2.png)

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 .

Figure 8
Figure 8 (samp_e3.png)

Exercise 3

What is the CTFT of f imp t f imp t ?

Solution 3

f imp t=n=- f s nδt-n f imp t n f s n δ t n

F imp w=- f imp t-wtdt=-n=- f s nδt-n-wtdt=n=- f s n-δt-n-wtdt=n=- f s n-wn= F s w 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-n-wtdt=-wn t δ t n w t w n

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 .

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?
Figure 9 (samp_blk.png)

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