Frequency Domain View of Upsampling

Why Interpolator needs a LPF after Upsampling


A slecture by ECE student Chloe Kauffman

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.


Outline

  1. Background
  2. Introduction
  3. Derivation
  4. Example
  5. Conclusion


Background

$ {f}_{s} $ = sampling frequency (number of samples/second) Hz
$ {T}_{s} $ = sampling period (number of seconds/sample) seconds
$ {f}_{s} = {\frac{1}{{T}_{s}}} $

Sampling above Nyquist frequency guarantees a bandlimited sampled CT signal's reconstruction. **add source**
Define Nyquist Sampling rate as $ {f}_{Nyquist} = 2{f}_{M} $
$ {f}_{M} $ is max frequency of CT signal




Introduction

Sampling at frequencies much larger than Nyquist requires a filter for reconstruction with a less sharp cutoff. A digital LPF can be used to then obtain the reconstructed signal. *add source*

Assume $ {x}_{c}(t) $ is a bandlimited CT signal, $ {x}_{1}[n] $ is a DT sampled signal of $ {x}_{c}(t) $ with sampling period $ {T}_{1} $

This leads to the question, can you use

$ {x}_{1}[n] = x_{c}(n{T}_{1}) $

to obtain

$ {x}_{u}[n] = {x}_{c}(n{T}_{u}) $, a signal sampled at a HIGHER sampling frequency than $ {x}_{1}[n] $, without having to fully reconstruct $ {x}_{c}(t) $




Derivation

We want $ {f}_{u} > {f}_{Nyquist} $. In this situation, this means $ {f}_{u} > {f}_{1} $.
Therefore, we want $ {T}_{u} < {T}_{1} $. (i.e. $ {x}_{u}[n] $ is sampled at a higher frequency than $ {x}_{1}[n] $)
In other words,
$ {T}_{u} = {\frac{{T}_{1}}{D}} $ for some integer D.

$ {x}_{1}[n] = x_{c}(n{T}_{1}) $
$ {x}_{u}[n] = {x}_{c}(n{T}_{u}) $

$ {x}_{u}[n] ={x}_{1}[n/D] = {x}_{c}(n{T}_{1}({T}_{u}/{T}_{1})) = {x}_{c}(n{T}_{u}) $if n/D is an integer
$ {x}_{u}[n] = 0 $ else.

In frequency domain:
$ {X}_{u}({\omega}) = {\sum_{n = -{\infty}}^{\infty} {x}_{u}[n]e^{-j{\omega}n}} $
$ {X}_{u}({\omega}) = {\sum_{n = -{\infty}}^{\infty} {x}_{1}[n/D]e^{-j{\omega}n}} $
let n=mD
$ {X}_{u}({\omega}) = {\sum_{m = -{\infty}}^{\infty} {x}_{1}[m]e^{-j{\omega}mD}} $
$ {X}_{u}({\omega}) ={X}_{1}(D{\omega}) $ notice this is a rescaled version of $ {X}_{1} $



In order to get $ {x}_{int}(n) $, the reconstructed signal, we need to LPF $ {X}_{u}({\omega}) $.
$ {x}_{int}(n) = {x}_{u} * h(n) $
$ h(n) = sinc(n/D) $




Example

X1w.jpg
Xuw.jpg
Hlpf.jpg
Xint.jpg




Source: Prof. Mireille Boutin



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