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Examing Reconstruction Relations

Module by: Justin Romberg. E-mail the author

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Summary: This module focus on examing reconstruction relations.

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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 FⅈΩ f t CTFT F Ω fnT= f s n CTFT F s ⅇⅈω f n T f s n CTFT F s ω ft=12π∫-∞∞FⅈΩⅇⅈωndn f t 1 2 n F Ω ω n

F s ⅇⅈω=∑n=-∞∞ f s nⅇ-ⅈωn=∑n=-∞∞fnTⅇ-ⅈωn=∑n=-∞∞12π∫-∞∞FⅈΩⅇⅈΩnTdΩⅇ-ⅈωn=12π∫-∞∞FⅈΩ∑n=-∞∞ⅇⅈnΩT−ωdΩ=12π∫-∞∞FⅈΩ2π∑r=-∞∞δω−(ΩT−2πr)dΩ=∫-∞∞FⅈΩ∑r=-∞∞δω−(ΩT−2πr)dΩ=∫-∞∞FⅈΩ∑r=-∞∞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 ⅇⅈω=∫-∞∞FⅈΩ∑r=-∞∞1TδΩ−ω−2πrTdΩ=1T∑r=-∞∞∫-∞∞FⅈΩδΩ−ω−2πrTdΩ=1T∑r=-∞∞Fⅈω−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 ⅇⅈω F s ω is a sum of shifted (and scaled) Fⅈω 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)

Fsⅇⅈω=1T∑r=∞∞Fⅈω−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

FⅈΩ→FⅈωT → F Ω F ω T

Figure 4
Figure 4 (exrec_f2.png)

note:

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

Periodic repetition

Fsⅇⅈω=1T∑r=-∞∞Fⅈω−2πrT Fs ω 1 T r F ω 2 r T What this equation says is that we "plop down" a copy of 1TFⅈωT 1 T F ω T at every multiple of 2π 2 (and sum up)..

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

note:

Fsⅇⅈω 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 T≤πΩB 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 FⅈωT F ω T , T≤πΩB T ΩB is zero outside of -ππ

Figure 6
(a) The CTFT of signal ft ft
Figure 6(a) (mrec_f1.png)
(b) Signal is zero outside of -ππ
Figure 6(b) (mrec_f2.png)

  • then 1T∑r=∞∞Fⅈω−2πrT 1 T r F ω 2 r T = DTFT of fsn=fnT fs n f n T is just a periodic version of 1TFⅈωT 1 T F ω T Fsⅇⅈω=1T∑r=-∞∞Fⅈω−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↔∫-∞∞fimptⅇ-ⅈωtdt=∫-∞∞∑n=-∞∞fsnδt−nTⅇ-ⅈωtdt ↔ fimp t t fimp t ω t t n fs n δ t n T ω t so, fimpt=∑n=-∞∞fsn∫-∞∞δt−nTⅇ-ⅈωtdt=∑n=-∞∞fsnⅇ-ⅈωTn=FsⅇⅈωT 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 fimpⅈω 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
(a) CTFT of fimpⅈω fimp ω
Figure 11(a) (mrec_f5.png)
(b) Lowpass filter
Figure 11(b) (mrec_f6.png)
(c) Reconstructed signal, which is identical to the original
Figure 11(c) (mrec_f7.png)

If Gⅈω G ω is "ideal", then = F ~ ⅈω 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, Fⅈω 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)

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