Skip to content Skip to navigation

Connexions

You are here: Home » Content » Examing Reconstruction Relations

Navigation

Content Actions

  • Download module PDF
  • Add to ...
    Add the module to:
    • My Favorites
    • A lens
    • An external social bookmarking service
    • My Favorites (What is 'My Favorites'?)
      'My Favorites' is a special kind of lens which you can use to bookmark modules and collections directly in Connexions. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need a Connexions account to use 'My Favorites'.
    • A lens (What is a lens?)

      Definition of a lens

      Lenses

      A lens is a custom view of Connexions content. You can think of it as a fancy kind of list that will let you see Connexions through the eyes of organizations and people you trust.

      What is in a lens?

      Lens makers point to Connexions materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

      Who can create a lens?

      Any individual Connexions member, a community, or a respected organization.

    • External bookmarks
  • E-mail the author

Recently Viewed

Examing Reconstruction Relations

Module by: Justin Romberg

Summary: This module focus on examing reconstruction relations.

Introduction

We've seen that if ft f t is bandlimited to -ππ , we can reconstruct it from its samples at the integers exactly. The key to this was the fact that the DTFT of f s n f s n is just a periodic version of the CTFT of ft f t

Let's examine the situation in general

Figure 1
Figure 1 (fig2.png)

Relationship of CTFT and DTFT

What is the relationship of the CTFT of ft f t to the DTFT of f s n=ft f s n f t sampled at nT n T ?

Figure 2
Figure 2 (fig3.png)

We will use the following notation. Note that we will use different variables, ΩΩ and ωω, so that we do not get confused. ft CTFT FΩ f t CTFT F Ω fnT= f s n CTFT F s ω f n T f s n CTFT F s ω ft=12π-FΩωndn f t 1 2 n F Ω ω n

F s ω=n=- f s n-ωn=n=-fnT-ωn=n=-12π-FΩΩnTdΩ-ωn=12π-FΩn=-nΩT-ωdΩ=12π-FΩ2πr=-δω-ΩT-2πrdΩ=-FΩr=-δω-ΩT-2πrdΩ=-FΩr=-δ-TΩ-ω-2πrTdΩ F s ω n f s n ω n n f n T ω n n 1 2 Ω F Ω Ω n T ω n 1 2 Ω F Ω n n Ω T ω 1 2 Ω F Ω 2 r δ ω Ω T 2 r Ω F Ω r δ ω Ω T 2 r Ω F Ω r δ T Ω ω 2 r T (1)
Recall, property of δt δ t
δaΩ=1|a|δΩ δ a Ω 1 a δ Ω (2)
F s ω=-FΩr=-1TδΩ-ω-2πrTdΩ=1Tr=--FΩδΩ-ω-2πrTdΩ=1Tr=-Fω-2πrT F s ω Ω F Ω r 1 T δ Ω ω 2 r T 1 T r Ω F Ω δ Ω ω 2 r T 1 T r F ω 2 r T (3)
F s ω F s ω is a sum of shifted (and scaled) Fω F ω 's. In order to get to our final results in Equation 3, we used the sifting property.

Figure 3: Illustration of sampling ft f t with a period of TT.
Figure 3 (exrec_f1.png)

Fsω=1Tr=Fω-2πrT Fs ω 1 T r F ω 2 r T

Relationship

Let us take a minute to examine this relation in 2 steps:

Axis Scaling

FΩFωT F Ω F ω T

Figure 4
Figure 4 (exrec_f2.png)

note:

Ω=πTω=π Ω T ω , this covers the r=0 r 0 case in the sum above.

Periodic repetition

Fsω=1Tr=-Fω-2πrT Fs ω 1 T r F ω 2 r T What this equation says is that we "plop down" a copy of 1TFωT 1 T F ω T at every multiple of 2π 2 (and sum up)..

Figure 5: You can see the periodic nature of Fsω Fs ω . Note that FωT F ω T is zero for everything beyond the interval of -ππ .
 (exrec_f3.png)

note:

Fsω Fs ω is 2π 2 -periodic, just like we know it should be.
Recall, T=sample period T sample period (i.e. each sample in the discrete-time has a distance of TT between them). Before, we saw that if T=1 T 1 (sample on the integers) and ft f t is bandlimited to -ππ we can reconstruct ft f t exactly from its samples fsn fs n .

General Case

Say ft f t is bandlimited to -ΩBΩB ΩB ΩB . Then if we sample with period TπΩB T ΩB , we can reconstruct ft f t exactly from its samples fsn fs n . Show this following the exact same steps as in the -ππ , T=1 T 1 case.

  • If ft f t is bandlimited to -ΩBΩB ΩB ΩB , then FωT F ω T , TπΩB T ΩB is zero outside of -ππ

Figure 6
Subfigure 6.1: The CTFT of signal ft ft
Subfigure 6.1 (mrec_f1.png)
Subfigure 6.2: Signal is zero outside of -ππ
Subfigure 6.2 (mrec_f2.png)

  • then 1Tr=Fω-2πrT 1 T r F ω 2 r T = DTFT of fsn=fnT fs n f n T is just a periodic version of 1TFωT 1 T F ω T Fsω=1Tr=-Fω-2πrT Fs ω 1 T r F ω 2 r T

Figure 7: DTFT of Sampled Signal
Figure 7 (mrec_f3.png)

  • turning fsn fs n into an impulse sequence fimpt=n=-fsnδt-nT fimp t n fs n δ t n T

Figure 8
Figure 8 (mrec_f4.png)

  • From here we make some calculations and we get fimpt-fimpt-ωtdt=-n=-fsnδt-nT-ωtdt fimp t t fimp t ω t t n fs n δ t n T ω t so, fimpt=n=-fsn-δt-nT-ωtdt=n=-fsn-ωTn=FsωT fimp t n fs n t δ t n T ω t n fs n ω T n Fs ω T

Figure 9
Figure 9 (mrec_f3.png)
Figure 10: CTFT of fimpω fimp ω is just a stretched out version of DTFT of fsn fs n
Figure 10 (mrec_f5.png)

  • to recover f ~ t=ft f ~ t f t from fimpt fimp t , we lowpass filter with cutoff frequency πT T (also the filter has a gain of 1T 1 T ).

Figure 11
Subfigure 11.1: CTFT of fimpω fimp ω
Subfigure 11.1 (mrec_f5.png)
Subfigure 11.2: Lowpass filter
Subfigure 11.2 (mrec_f6.png)
Subfigure 11.3: Reconstructed signal, which is identical to the original
Subfigure 11.3 (mrec_f7.png)

If Gω G ω is "ideal", then = F ~ ω F ~ ω and the signal is reconstructed perfectly.

Figure 12: Block diagram briefly illustrating the reconstruction process.
Figure 12 (recon_blk3.png)

Example 1: You try one ...

Starting with the following signal, ft ft, and its CTFT, Fω F ω , use the same steps above to sample the signal, convert it into an impulse train, and then filter it to recreate the original signal.

Figure 13: Sample and reconstruct the above signal, ft ft.
Figure 13 (mrec_eg1.png)

Comments, questions, feedback, criticisms?

Send feedback