# 0.7 Compressed sensing

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This collection reviews fundamental concepts underlying the use of concise models for signal processing. Topics are presented from a geometric perspective and include low-dimensional linear, sparse, and manifold-based signal models, approximation, compression, dimensionality reduction, and Compressed Sensing.

A new theory known as Compressed Sensing (CS) has recently emerged that can also be categorized as a type of dimensionalityreduction. Like manifold learning, CS is strongly model-based (relying on sparsity in particular).However, unlike many of the standard techniques in dimensionality reduction (such as manifold learning or the JL lemma), the goal ofCS is to maintain a low-dimensional representation of a signal $x$ from which a faithful approximation to $x$ can be recovered. In a sense, this more closely resembles the traditional problem ofdata compression (see Compression ). In CS, however, the encoder requires no a priori knowledge of thesignal structure. Only the decoder uses the model (sparsity) to recover the signal. Wejustify such an approach again using geometric arguments.

## Motivation

Consider a signal $x\in {\mathbb{R}}^{N}$ , and suppose that the basis $\Psi$ provides a $K$ -sparse representation of $x$

$x=\Psi \alpha ,$
with ${\parallel \alpha \parallel }_{0}=K$ . (In this section, we focus on exactly $K$ -sparse signals, though many of the key ideas translate to compressible signals  [link] , [link] . In addition, we note that the CS concepts are also extendable totight frames.)

As we discussed in Compression , the standard procedure for compressing sparse signals, known as transformcoding, is to (i) acquire the full $N$ -sample signal $x$ ; (ii) compute the complete set of transform coefficients $\alpha$ ; (iii) locate the $K$ largest, significant coefficients and discard the (many) small coefficients; (iv) encode the values and locations of the largest coefficients.

This procedure has three inherent inefficiencies: First, for a high-dimensional signal, we must start with a large number ofsamples $N$ . Second, the encoder must compute all $N$ of the transform coefficients $\alpha$ , even though it will discard all but $K$ of them. Third, the encoder must encode the locations of the large coefficients, which requiresincreasing the coding rate since the locations change with each signal.

## Incoherent projections

This raises a simple question: For a given signal, is it possible to directly estimate the set of large $\alpha \left(n\right)$ 's that will not be discarded? While this seems improbable, Candès, Romberg,and Tao  [link] , [link] and Donoho [link] have shown that a reduced set of projections can contain enoughinformation to reconstruct sparse signals. An offshoot of this work, often referred to as Compressed Sensing (CS) [link] , [link] , [link] , [link] , [link] , [link] , [link] , has emerged that builds on this principle.

In CS, we do not measure or encode the $K$ significant $\alpha \left(n\right)$ directly. Rather, we measure and encode $M projections $y\left(m\right)=$ of the signal onto a second set of functions $\left\{{\phi }_{m}\right\},m=1,2,...,M$ . In matrix notation, we measure

$y=\Phi x,$
where $y$ is an $M×1$ column vector and the measurement basis matrix $\Phi$ is $M×N$ with each row a basis vector ${\phi }_{m}$ . Since $M , recovery of the signal $x$ from the measurements $y$ is ill-posed in general; however the additional assumption of signal sparsity makes recovery possible and practical.

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?
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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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
is it 3×y ?
J, combine like terms 7x-4y
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
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Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
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preparation of nanomaterial
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
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Damian
silver nanoparticles could handle the job?
Damian
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Azam
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Uday
I'm interested in Nanotube
Uday
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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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