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Examing Reconstruction Relations

Module by: Justin Romberg. E-mail the author

Summary: This module focus on examing reconstruction relations.

Introduction

We've seen that if f⁢t f t is bandlimited to −π π , 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 f⁢t 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 f⁢t f t to the DTFT of f s ⁢n=f⁢t f s n f t sampled at n⁢T 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. f⁢t CTFT F⁢i⁢Ω f t CTFT F Ω f⁢n⁢T= f s ⁢n CTFT F s ⁢ei⁢ω f n T f s n CTFT F s ω f⁢t=12⁢π⁢∫−∞∞F⁢i⁢Ω⁢ei⁢ω⁢nd n f t 1 2 n F Ω ω n

F s ⁢ei⁢ω=∑n=−∞∞ f s ⁢n⁢e−(i⁢ω⁢n)=∑n=−∞∞f⁢n⁢T⁢e−(i⁢ω⁢n)=∑n=−∞∞12⁢π⁢∫−∞∞F⁢i⁢Ω⁢ei⁢Ω⁢n⁢TdΩ⁢e−(i⁢ω⁢n)=12⁢π⁢∫−∞∞F⁢i⁢Ω⁢∑n=−∞∞ei⁢n⁢(Ω⁢T−ω)dΩ=12⁢π⁢∫−∞∞F⁢i⁢Ω⁢2⁢π⁢∑r=−∞∞δ⁢ω−(Ω⁢T−2⁢π⁢r)dΩ=∫−∞∞F⁢i⁢Ω⁢∑r=−∞∞δ⁢ω−(Ω⁢T−2⁢π⁢r)dΩ=∫−∞∞F⁢i⁢Ω⁢∑r=−∞∞δ⁢(−T)⁢(Ω−ω−2⁢π⁢rT)dΩ 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 δ⁢t δ t
δ⁢a⁢Ω=1|a|⁢δ⁢Ω δ a Ω 1 a δ Ω
(2)
F s ⁢ei⁢ω=∫−∞∞F⁢i⁢Ω⁢∑r=−∞∞1T⁢δ⁢Ω−ω−2⁢π⁢rTdΩ=1T⁢∑r=−∞∞∫−∞∞F⁢i⁢Ω⁢δ⁢Ω−ω−2⁢π⁢rTdΩ=1T⁢∑r=−∞∞F⁢i⁢ω−2⁢π⁢rT 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 ⁢ei⁢ω F s ω is a sum of shifted (and scaled) F⁢i⁢ω F ω 's. In order to get to our final results in Equation 3, we used the sifting property.

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

Fs ⁢ei⁢ω=1T⁢∑r=,∞∞F⁢i⁢ω−2⁢π⁢rT 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

F⁢i⁢Ω→F⁢i⁢ωT → 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

Fs ⁢ei⁢ω=1T⁢∑r=−∞∞F⁢i⁢ω−2⁢π⁢rT Fs ω 1 T r F ω 2 r T What this equation says is that we "plop down" a copy of 1T⁢F⁢i⁢ωT 1 T F ω T at every multiple of 2⁢π 2 (and sum up)..

Figure 5: You can see the periodic nature of Fs ⁢ei⁢ω Fs ω . Note that F⁢i⁢ωT F ω T is zero for everything beyond the interval of −π π .
(exrec_f3.png)

note:

Fs ⁢ei⁢ω Fs ω is 2⁢π 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 f⁢t f t is bandlimited to −π π we can reconstruct f⁢t f t exactly from its samples fs ⁢n fs n .

General Case

Say f⁢t f t is bandlimited to −ΩB ΩB ΩB ΩB . Then if we sample with period T≤π ΩB T ΩB , we can reconstruct f⁢t f t exactly from its samples fs ⁢n fs n . Show this following the exact same steps as in the −π π , T=1 T 1 case.

  • If f⁢t f t is bandlimited to −ΩB ΩB ΩB ΩB , then F⁢i⁢ωT F ω T , T≤π ΩB T ΩB is zero outside of −π π

Figure 6
(a) The CTFT of signal f⁢t ft
Figure 6(a) (mrec_f1.png)
(b) Signal is zero outside of −π π
Figure 6(b) (mrec_f2.png)

  • then 1T⁢∑r=,∞∞F⁢i⁢ω−2⁢π⁢rT 1 T r F ω 2 r T = DTFT of fs ⁢n=f⁢n⁢T fs n f n T is just a periodic version of 1T⁢F⁢i⁢ωT 1 T F ω T Fs ⁢ei⁢ω=1T⁢∑r=−∞∞F⁢i⁢ω−2⁢π⁢rT Fs ω 1 T r F ω 2 r T

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

  • turning fs ⁢n fs n into an impulse sequence fimp ⁢t=∑n=−∞∞ fs ⁢n⁢δ⁢t−n⁢T 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 fimp ⁢t↔∫−∞∞ fimp ⁢t⁢e−(i⁢ω⁢t)dt=∫−∞∞∑n=−∞∞ fs ⁢n⁢δ⁢t−n⁢T⁢e−(i⁢ω⁢t)dt ↔ fimp t t fimp t ω t t n fs n δ t n T ω t so, fimp ⁢t=∑n=−∞∞ fs ⁢n⁢∫−∞∞δ⁢t−n⁢T⁢e−(i⁢ω⁢t)dt=∑n=−∞∞ fs ⁢n⁢e−(i⁢ω⁢T⁢n)= Fs ⁢ei⁢(ω⁢T) 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 fimp ⁢i⁢ω fimp ω is just a stretched out version of DTFT of fs ⁢n fs n
Figure 10 (mrec_f5.png)

  • to recover f ~ ⁢t=f⁢t f ~ t f t from fimp ⁢t 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 fimp ⁢i⁢ω 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 G⁢i⁢ω G ω is "ideal", then F ~ ⁢i⁢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, f⁢t ft, and its CTFT, F⁢i⁢ω 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, f⁢t ft.
Figure 13 (mrec_eg1.png)

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