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and so E [ x - x ˆ 2 2 ] , the mean square error, is double the MMSE  [link] .

Let us pause to reflect about this result. When the SNR is high, i.e., x 2 2 z 2 2 , the MMSE should be rather low, and double the MMSE seems pretty good. On the other hand, when the SNR is low, the MMSE could be almost as large as E [ x 2 2 ] , and double the MMSE could be larger – as much as twice larger – than E [ x 2 2 ] . That is, gussing x ˆ = E [ x ] could give better signal estimation performance than using the Kolmogorov sampler. This pessimistic result encourages us to search for better signal reconstruction methods.

Arbitrary channels: So far we considered the Kolmogorov sampler for the white scalar channel, y = x + z . Suppose instead that x is processed or measured by a more complicated system,

y = J ( x ) + z .

Note that J is known, e.g., in a compressed sensing application  [link] , [link] J would be a known matrix. An even more involved system would be y = ( J x ) z , where J x is application of a mapping to x, J x R L , and z denotes application of a random noise operator to J x . To keep the presentation simple, we use the additive noise setting [link] .

How can the Kolmogorov sampler [link] be applied to the additive noise setting? Recall that for the scalar channel, the Kolmogorov sampler minimizes K ( w ) subject to y - w 2 2 n . For the arbitrary mapping J with additive noise [link] , this implies y - J ( w ) 2 2 n . Therefore, we get

x ˆ = arg min s . t . Y - J ( w ) 2 2 n K ( w ) .

Another similar approach relies on optimization via Lagrange multipliers,

x ˆ = arg min K ( w ) - log 2 ( f z ( y - J ( w ) ) ) ,

where the lagrange multiplier is 1, because both K ( w ) and log 2 ( f z ( y - J ( w ) ) ) are quantified in bits.

What is the performance of the Kolmogorov sampler for an arbitrary J ? We speculate  [link] , [link] that x ˆ is generated by the posterior, and so E [ x - ˆ 2 2 ] is double the MMSE, where expectation is taken over the source X and noise z . These results remain to be shown rigorously.

Convergence of mcmc algorithm

We will now prove a substantial result – that the MCMC algorithm  [link] , [link] , [link] converges to the globally minimal energy solution for the specific case of compressed sensing  [link] , [link] . An extension of this proof to arbitrary channels J is in progress.

If the operator J in [link] is a matrix, and we denote it by Φ R m × n where m n , then the setup is known as compressed sensing (CS)  [link] , [link] and the estimation problem is commonly referred to as recovery or reconstruction.By posing a sparsity or compressibility requirement on the signal and using it as a prior during recovery, it is indeed possible to accurately estimate x from y in the CS setting.

With the quantization alphabet definition in [link] , α ˆ will quantize x to a greater resolution as N increases. We will show that under suitable conditions on f X , performing maximum a posteriori (MAP) estimation over the discrete alphabet α ˆ asymptotically converges to the MAP estimate over the continuous distribution f X . This reduces the complexity of the estimation problem from continuous to discrete.

We assume for exposition that we know the input statistics f X . Given the measurements y , the MAP estimator for x has the form

x M A P arg max w f X ( w ) f Y | X ( y | w ) .

Because z is i.i.d. Gaussian with mean zero and known variance σ Z 2 ,

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Universal algorithms in signal processing and communications. OpenStax CNX. May 16, 2013 Download for free at http://cnx.org/content/col11524/1.1
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