# 4.2 Signal recovery in noise

 Page 1 / 2
This module establishes a number of results concerning various L1 minimization algorithms designed for sparse signal recovery from noisy measurements. The results in this module apply to both bounded noise as well as Gaussian (or more generally, sub-Gaussian) noise.

The ability to perfectly reconstruct a sparse signal from noise-free measurements represents a promising result. However, in most real-world systems the measurements are likely to be contaminated by some form of noise. For instance, in order to process data in a computer we must be able to represent it using a finite number of bits, and hence the measurements will typically be subject to quantization error. Moreover, systems which are implemented in physical hardware will be subject to a variety of different types of noise depending on the setting.

Perhaps somewhat surprisingly, one can show that it is possible to modify

$\stackrel{^}{x}=arg\underset{z}{min}\phantom{\rule{4pt}{0ex}}{∥z∥}_{1}\phantom{\rule{1.em}{0ex}}\mathrm{subject to}\phantom{\rule{1.em}{0ex}}z\in \mathcal{B}\left(y\right).$

to stably recover sparse signals under a variety of common noise models  [link] , [link] , [link] . As might be expected, the restricted isometry property (RIP) is extremely useful in establishing performance guarantees in noise.

In our analysis we will make repeated use of Lemma 1 from "Noise-free signal recovery" , so we repeat it here for convenience.

Suppose that $\Phi$ satisfies the RIP of order $2K$ with ${\delta }_{2K}<\sqrt{2}-1$ . Let $x,\stackrel{^}{x}\in {\mathbb{R}}^{N}$ be given, and define $h=\stackrel{^}{x}-x$ . Let ${\Lambda }_{0}$ denote the index set corresponding to the $K$ entries of $x$ with largest magnitude and ${\Lambda }_{1}$ the index set corresponding to the $K$ entries of ${h}_{{\Lambda }_{0}^{c}}$ with largest magnitude. Set $\Lambda ={\Lambda }_{0}\cup {\Lambda }_{1}$ . If ${∥\stackrel{^}{x}∥}_{1}\le {∥x∥}_{1}$ , then

${∥h∥}_{2}\le {C}_{0}\frac{{\sigma }_{K}{\left(x\right)}_{1}}{\sqrt{K}}+{C}_{1}\frac{\left|〈\Phi ,{h}_{\Lambda },,,\Phi ,h〉\right|}{{∥{h}_{\Lambda }∥}_{2}}.$

where

${C}_{0}=2\frac{1-\left(1-\sqrt{2}\right){\delta }_{2K}}{1-\left(1+\sqrt{2}\right){\delta }_{2K}},\phantom{\rule{1.em}{0ex}}{C}_{1}=\frac{2}{1-\left(1+\sqrt{2}\right){\delta }_{2K}}.$

## Bounded noise

We first provide a bound on the worst-case performance for uniformly bounded noise, as first investigated in  [link] .

## (theorem 1.2 of [link] )

Suppose that $\Phi$ satisfies the RIP of order $2K$ with ${\delta }_{2K}<\sqrt{2}-1$ and let $y=\Phi x+e$ where ${∥e∥}_{2}\le ϵ$ . Then when $\mathcal{B}\left(y\right)=\left\{z:{∥\Phi ,z,-,y∥}_{2}\le ϵ\right\}$ , the solution $\stackrel{^}{x}$ to [link] obeys

${∥\stackrel{^}{x},-,x∥}_{2}\le {C}_{0}\frac{{\sigma }_{K}{\left(x\right)}_{1}}{\sqrt{K}}+{C}_{2}ϵ,$

where

${C}_{0}=2\frac{1-\left(1-\sqrt{2}\right){\delta }_{2K}}{1-\left(1+\sqrt{2}\right){\delta }_{2K}},\phantom{\rule{1.em}{0ex}}{C}_{2}=4\frac{\sqrt{1+{\delta }_{2K}}}{1-\left(1+\sqrt{2}\right){\delta }_{2K}}.$

We are interested in bounding ${∥h∥}_{2}={∥\stackrel{^}{x},-,x∥}_{2}$ . Since ${∥e∥}_{2}\le ϵ$ , $x\in \mathcal{B}\left(y\right)$ , and therefore we know that ${∥\stackrel{^}{x}∥}_{1}\le {∥x∥}_{1}$ . Thus we may apply [link] , and it remains to bound $\left|〈\Phi ,{h}_{\Lambda },,,\Phi ,h〉\right|$ . To do this, we observe that

${∥\Phi ,h∥}_{2}={∥\Phi ,\left(,\stackrel{^}{x},-,x,\right)∥}_{2}={∥\Phi ,\stackrel{^}{x},-,y,+,y,-,\Phi ,x∥}_{2}\le {∥\Phi ,\stackrel{^}{x},-,y∥}_{2}+{∥y,-,\Phi ,x∥}_{2}\le 2ϵ$

where the last inequality follows since $x,\stackrel{^}{x}\in \mathcal{B}\left(y\right)$ . Combining this with the RIP and the Cauchy-Schwarz inequality we obtain

$\left|〈\Phi ,{h}_{\Lambda },,,\Phi ,h〉\right|\le {∥\Phi ,{h}_{\Lambda }∥}_{2}{∥\Phi ,h∥}_{2}\le 2ϵ\sqrt{1+{\delta }_{2K}}{∥{h}_{\Lambda }∥}_{2}.$

Thus,

${∥h∥}_{2}\le {C}_{0}\frac{{\sigma }_{K}{\left(x\right)}_{1}}{\sqrt{K}}+{C}_{1}2ϵ\sqrt{1+{\delta }_{2K}}={C}_{0}\frac{{\sigma }_{K}{\left(x\right)}_{1}}{\sqrt{K}}+{C}_{2}ϵ,$

completing the proof.

In order to place this result in context, consider how we would recover a sparse vector $x$ if we happened to already know the $K$ locations of the nonzero coefficients, which we denote by ${\Lambda }_{0}$ . This is referred to as the oracle estimator . In this case a natural approach is to reconstruct the signal using a simple pseudoinverse:

$\begin{array}{cc}\hfill {\stackrel{^}{x}}_{{\Lambda }_{0}}& ={\Phi }_{{\Lambda }_{0}}^{†}y={\left({\Phi }_{{\Lambda }_{0}}^{T}{\Phi }_{{\Lambda }_{0}}\right)}^{-1}{\Phi }_{{\Lambda }_{0}}^{T}y\hfill \\ \hfill {\stackrel{^}{x}}_{{\Lambda }_{0}^{c}}& =0.\hfill \end{array}$

The implicit assumption in [link] is that ${\Phi }_{{\Lambda }_{0}}$ has full column-rank (and hence we are considering the case where ${\Phi }_{{\Lambda }_{0}}$ is the $M×K$ matrix with the columns indexed by ${\Lambda }_{0}^{c}$ removed) so that there is a unique solution to the equation $y={\Phi }_{{\Lambda }_{0}}{x}_{{\Lambda }_{0}}$ . With this choice, the recovery error is given by

${∥\stackrel{^}{x},-,x∥}_{2}={∥{\left({\Phi }_{{\Lambda }_{0}}^{T}{\Phi }_{{\Lambda }_{0}}\right)}^{-1},{\Phi }_{{\Lambda }_{0}}^{T},\left(\Phi x+e\right),-,x∥}_{2}={∥{\left({\Phi }_{{\Lambda }_{0}}^{T}{\Phi }_{{\Lambda }_{0}}\right)}^{-1},{\Phi }_{{\Lambda }_{0}}^{T},e∥}_{2}.$

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply 12, 17, 22.... 25th term Alexandra Reply 12, 17, 22.... 25th term Akash College algebra is really hard? Shirleen Reply Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table. Carole I'm 13 and I understand it great AJ I am 1 year old but I can do it! 1+1=2 proof very hard for me though. Atone hi Adu Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily. Vedant hi vedant can u help me with some assignments Solomon find the 15th term of the geometric sequince whose first is 18 and last term of 387 Jerwin Reply I know this work salma The given of f(x=x-2. then what is the value of this f(3) 5f(x+1) virgelyn Reply hmm well what is the answer Abhi If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10 Augustine how do they get the third part x = (32)5/4 kinnecy Reply make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be AJ how Sheref can someone help me with some logarithmic and exponential equations. Jeffrey Reply sure. what is your question? ninjadapaul 20/(×-6^2) Salomon okay, so you have 6 raised to the power of 2. what is that part of your answer ninjadapaul I don't understand what the A with approx sign and the boxed x mean ninjadapaul it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared Salomon I'm not sure why it wrote it the other way Salomon I got X =-6 Salomon ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6 ninjadapaul oops. ignore that. ninjadapaul so you not have an equal sign anywhere in the original equation? ninjadapaul hmm Abhi is it a question of log Abhi 🤔. Abhi I rally confuse this number And equations too I need exactly help salma But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends salma Commplementary angles Idrissa Reply hello Sherica im all ears I need to learn Sherica right! what he said ⤴⤴⤴ Tamia hii Uday hi salma hi Ayuba Hello opoku hi Ali greetings from Iran Ali salut. from Algeria Bach hi Nharnhar A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!