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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

x ^ = arg min z z 1 subject to z B ( y ) .

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 Φ satisfies the RIP of order 2 K with δ 2 K < 2 - 1 . Let x , x ^ R N be given, and define h = x ^ - x . Let Λ 0 denote the index set corresponding to the K entries of x with largest magnitude and Λ 1 the index set corresponding to the K entries of h Λ 0 c with largest magnitude. Set Λ = Λ 0 Λ 1 . If x ^ 1 x 1 , then

h 2 C 0 σ K ( x ) 1 K + C 1 Φ h Λ , Φ h h Λ 2 .

where

C 0 = 2 1 - ( 1 - 2 ) δ 2 K 1 - ( 1 + 2 ) δ 2 K , C 1 = 2 1 - ( 1 + 2 ) δ 2 K .

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 Φ satisfies the RIP of order 2 K with δ 2 K < 2 - 1 and let y = Φ x + e where e 2 ϵ . Then when B ( y ) = { z : Φ z - y 2 ϵ } , the solution x ^ to [link] obeys

x ^ - x 2 C 0 σ K ( x ) 1 K + C 2 ϵ ,

where

C 0 = 2 1 - ( 1 - 2 ) δ 2 K 1 - ( 1 + 2 ) δ 2 K , C 2 = 4 1 + δ 2 K 1 - ( 1 + 2 ) δ 2 K .

We are interested in bounding h 2 = x ^ - x 2 . Since e 2 ϵ , x B ( y ) , and therefore we know that x ^ 1 x 1 . Thus we may apply [link] , and it remains to bound Φ h Λ , Φ h . To do this, we observe that

Φ h 2 = Φ ( x ^ - x ) 2 = Φ x ^ - y + y - Φ x 2 Φ x ^ - y 2 + y - Φ x 2 2 ϵ

where the last inequality follows since x , x ^ B ( y ) . Combining this with the RIP and the Cauchy-Schwarz inequality we obtain

Φ h Λ , Φ h Φ h Λ 2 Φ h 2 2 ϵ 1 + δ 2 K h Λ 2 .

Thus,

h 2 C 0 σ K ( x ) 1 K + C 1 2 ϵ 1 + δ 2 K = C 0 σ K ( x ) 1 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 Λ 0 . This is referred to as the oracle estimator . In this case a natural approach is to reconstruct the signal using a simple pseudoinverse:

x ^ Λ 0 = Φ Λ 0 y = ( Φ Λ 0 T Φ Λ 0 ) - 1 Φ Λ 0 T y x ^ Λ 0 c = 0 .

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

x ^ - x 2 = ( Φ Λ 0 T Φ Λ 0 ) - 1 Φ Λ 0 T ( Φ x + e ) - x 2 = ( Φ Λ 0 T Φ Λ 0 ) - 1 Φ Λ 0 T e 2 .

Questions & Answers

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Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
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To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
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A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
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Sharon
8m/s²
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What is Thermodynamics
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velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
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A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
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50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
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I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
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temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
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Source:  OpenStax, An introduction to compressive sensing. OpenStax CNX. Apr 02, 2011 Download for free at http://legacy.cnx.org/content/col11133/1.5
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