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Minimize instantaneous squared error

e k w 2 y k x k w 2

Lms algorithm

w k w k 1 μ x k e k
Where w k is the new weight vector, w k 1 is the old weight vector, and μ x k e k is a small step in the instantaneous error gradient direction.

Interpretation in terms of weight error vector


v k w k w opt
Where w opt is the optimal weight vector and
ε k y k x k w opt
where ε k is the minimum error. The stochastic difference equation is:
v k I μ x k x k v k 1 μ x k ε k

Convergence/stability analysis

Show that (tightness)

B v k B 0
With probability 1, the weight error vector is bounded for all k .

Chebyshev's inequality is

v k B v k 2 B 2
v k B 1 B 2 v k 2 v k
where v k 2 is the squared bias. If v k 2 v k is finite for all k , then B v k B 0 for all k .


v k tr v k v k
Therefore v k is finite if the diagonal elements of Γ k v k v k are bounded.

Convergence in mean

v k 0 as k . Take expectation of using smoothing property to simplify the calculation. We haveconvergence in mean if

  • R xx is positive definite (invertible).
  • μ 2 λ max R xx .

Bounded variance

Show that Γ k v k v k , the weight vector error covariance is bounded for all k .

We could have v k 0 , but v k ; in which case the algorithm would not be stable.
Recall that it is fairly straightforward to show that the diagonal elements of the transformed covariance C k U Γ k U tend to zero if μ 1 λ max R xx ( U is the eigenvector matrix of R xx ; R xx U D U ). The diagonal elements of C k were denoted by γ k , i i i 1 p .
v k tr Γ k tr U C k U tr C k i 1 p γ k , i
Thus, to guarantee boundedness of v k we need to show that the "steady-state" values γ k , i γ i .

We showed that

γ i μ α σ ε 2 2 1 μ λ i
where σ ε 2 ε k 2 , λ i is the i th eigenvalue of R xx ( R xx U λ 1 0 0 λ p U ), and α c σ ε 2 1 c .
0 c 1 2 i 1 p μ λ i 1 μ λ i 1
We found a sufficient condition for μ that guaranteed that the steady-state γ i 's (and hence v k ) are bounded: μ 2 3 i 1 p λ i Where i 1 p λ i tr R xx is the input vector energy.

With this choice of μ we have:

  • convergence in mean
  • bounded steady-state variance
This implies
B v k B 0
In other words, the LMS algorithm is stable about the optimumweight vector w opt .

Learning curve

Recall that

e k y k x k w k 1
and . These imply
e k ε k x k v k 1
where v k w k w opt . So the MSE
e k 2 σ ε 2 v k 1 x k x k v k 1 σ ε 2 x n ε n n n k v k 1 x k x k v k 1 σ ε 2 v k 1 R xx v k 1 σ ε 2 tr R xx v k 1 v k 1 σ ε 2 tr R xx Γ k 1
Where tr R xx Γ k 1 α k 1 α c σ ε 2 1 c . So the limiting MSE is
ε k e k 2 σ ε 2 c σ ε 2 1 c σ ε 2 1 c
Since 0 c 1 was required for convergence, ε σ ε 2 so that we see noisy adaptation leads to an MSE larger than the optimal
ε k 2 y k x k w opt 2 σ ε 2
To quantify the increase in the MSE, define the so-called misadjustment :
M ε σ ε 2 σ ε 2 ε σ ε 2 1 α σ ε 2 c 1 c
We would of course like to keep M as small as possible.

Learning speed and misadjustment trade-off

Fast adaptation and quick convergence require that we take steps as large as possible. In other words,learning speed is proportional to μ ; larger μ means faster convergence. How does μ affect the misadjustment?

To guarantee convergence/stability we require μ 2 3 i 1 p λ i R xx Let's assume that in fact μ 1 i 1 p λ i so that there is no problem with convergence. This condition implies μ 1 λ i or μ λ i 1 i i 1 p . From here we see that

c 1 2 i 1 p μ λ i 1 μ λ i 1 2 μ i 1 p λ i 1
This misadjustment
M c 1 c c 1 2 μ i 1 p λ i
This shows that larger step size μ leads to larger misadjustment.

Since we still have convergence in mean, this essentially means that with a larger step size we "converge"faster but have a larger variance (rattling) about w opt .


small μ implies

  • small misadjustment in steady-state
  • slow adaptation/tracking
large μ implies
  • large misadjustment in steady-state
  • fast adaptation/tracking

w opt 1 1 x k 0 1 0 0 1 y k x k w opt ε k ε k 0 0.01

Lms algorithm

initialization w 0 0 0 and w k w k 1 μ x k e k k k 1 , where e k y k x k w k 1

Learning curve

μ 0.05

Lms learning curve

μ 0.3

Comparison of learning curves

Questions & Answers

can someone help me with some logarithmic and exponential equations.
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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
I'm not sure why it wrote it the other way
I got X =-6
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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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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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
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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.
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Source:  OpenStax, Statistical signal processing. OpenStax CNX. Jun 14, 2004 Download for free at http://cnx.org/content/col10232/1.1
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