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

Source: Prof. Mireille Boutin