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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 ft 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 ft f t

Let's examine the situation in general

### 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 ?

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=12πFiΩeiωnd n f t 1 2 n F Ω ω n

F s eiω=n= f s ne(iωn)=n=fnTe(iωn)=n=12πFiΩeiΩnTdΩe(iωn)=12πFiΩn=ein(ΩTω)dΩ=12πFiΩ2πr=δω(ΩT2πr)dΩ=FiΩr=δω(ΩT2πr)dΩ=FiΩr=δ(T)(Ωω2πrT)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 δt δ t
δaΩ=1|a|δΩ δ a Ω 1 a δ Ω
(2)
F s eiω=FiΩr=1TδΩω2πrTdΩ=1Tr=FiΩδΩω2πrTdΩ=1Tr=Fiω2πrT 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.

Fs eiω=1Tr=,Fiω2πrT 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

FiΩFiωT F Ω F ω T

#### note:

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

### Periodic repetition

Fs eiω=1Tr=Fiω2πrT 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)..

#### note:

Fs eiω 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 π π we can reconstruct ft f t exactly from its samples fs n fs n .

## General Case

Say ft f t is bandlimited to ΩB ΩB ΩB ΩB . Then if we sample with period Tπ ΩB T ΩB , we can reconstruct ft 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 ft f t is bandlimited to ΩB ΩB ΩB ΩB , then FiωT F ω T , Tπ ΩB T ΩB is zero outside of π π

• then 1Tr=,Fiω2πrT 1 T r F ω 2 r T = DTFT of fs n=fnT fs n f n T is just a periodic version of 1TFiωT 1 T F ω T Fs eiω=1Tr=Fiω2πrT Fs ω 1 T r F ω 2 r T

• turning fs n fs n into an impulse sequence fimp t=n= fs nδtnT fimp t n fs n δ t n T

• From here we make some calculations and we get fimp t fimp te(iωt)dt=n= fs nδtnTe(iωt)dt fimp t t fimp t ω t t n fs n δ t n T ω t so, fimp t=n= fs nδtnTe(iωt)dt=n= fs ne(iωTn)= Fs ei(ωT) fimp t n fs n t δ t n T ω t n fs n ω T n Fs ω T

• to recover f ~ t=ft 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 ).

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

### 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.

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