# 0.7 Compressed sensing  (Page 3/5)

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The second statement of the theorem differs from the first in the following respect: when $K , there will necessarily exist $K$ -sparse signals $x$ that cannot be uniquely recovered from the $M$ -dimensional measurement vector $y=\Phi x$ . However, these signals form a set of measure zero within the set of all $K$ -sparse signals and can safely be avoided if $\Phi$ is randomly generated independently of $x$ .

Unfortunately, as discussed in Nonlinear Approximation from Approximation , solving this ${\ell }_{0}$ optimization problem is prohibitively complex. Yet another challenge is robustness; in the setting ofTheorem "Recovery via ℓ 0 optimization" , the recovery may be very poorly conditioned. In fact, both of these considerations (computational complexity and robustness) can be addressed, but atthe expense of slightly more measurements.

## Recovery via convex optimization

The practical revelation that supports the new CS theory is that it is not necessary to solve the ${\ell }_{0}$ -minimization problem to recover $\alpha$ . In fact, a much easier problem yields an equivalent solution (thanks again to the incoherency of thebases); we need only solve for the ${\ell }_{1}$ -sparsest coefficients $\alpha$ that agree with the measurements $y$ [link] , [link] , [link] , [link] , [link] , [link] , [link] , [link]

$\stackrel{^}{\alpha }=argmin{\parallel \alpha \parallel }_{1}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\text{s.t.}\phantom{\rule{4.pt}{0ex}}y=\Phi \Psi \alpha .$
As discussed in Nonlinear Approximation from Approximation , this optimization problem, also known as Basis Pursuit [link] , is significantly more approachable and can be solved with traditionallinear programming techniques whose computational complexities are polynomial in $N$ .

There is no free lunch, however; according to the theory, more than $K+1$ measurements are required in order to recover sparse signals via Basis Pursuit. Instead, one typically requires $M\ge cK$ measurements, where $c>1$ is an oversampling factor . As an example, we quote a result asymptotic in $N$ . For simplicity, we assume that the sparsity scales linearly with $N$ ; that is, $K=SN$ , where we call $S$ the sparsity rate .

Theorem

[link] , [link] , [link] Set $K=SN$ with $0 . Then there exists an oversampling factor $c\left(S\right)=O\left(log\left(1/S\right)\right)$ , $c\left(S\right)>1$ , such that, for a $K$ -sparse signal $x$ in the basis $\Psi$ , the following statements hold:

1. The probability of recovering $x$ via Basis Pursuit from $\left(c\left(S\right)+ϵ\right)K$ random projections, $ϵ>0$ , converges to one as $N\to \infty$ .
2. The probability of recovering $x$ via Basis Pursuit from $\left(c\left(S\right)-ϵ\right)K$ random projections, $ϵ>0$ , converges to zero as $N\to \infty$ .

In an illuminating series of recent papers, Donoho and Tanner [link] , [link] , [link] have characterized the oversampling factor $c\left(S\right)$ precisely (see also "The geometry of Compressed Sensing" ). With appropriate oversampling, reconstruction via Basis Pursuit is also provably robust tomeasurement noise and quantization error [link] .

We often use the abbreviated notation $c$ to describe the oversampling factor required in various settings even though $c\left(S\right)$ depends on the sparsity $K$ and signal length $N$ .

A CS recovery example on the Cameraman test image is shown in [link] . In this case, with $M=4K$ we achieve near-perfect recovery of the sparse measured image.

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
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
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
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salma
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salma
Commplementary angles
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Sherica
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Sherica
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Tamia
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Uday
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salma
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
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?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
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Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
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Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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