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hw3_solutions - EE263s Summer 2009-10 Laurent Lessard...

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EE263s Summer 2009-10 Laurent Lessard EE263s homework 3 1. Linearizing range measurements. Consider a single (scalar) measurement y of the distance or range of x R n to a fixed point or beacon at a , i.e. , y = bardbl x a bardbl . (a) Show that the linearized model near x 0 can be expressed as δy = k T δx , where k is the unit vector ( i.e. , with length one) pointing from a to x 0 . Derive this analytically, and also draw a picture (for n = 2) to demonstrate it. (b) Consider the error e of the linearized approximation, i.e. , e = bardbl x 0 + δx a bardbl − bardbl x 0 a bardbl − k T δx. The relative error of the approximation is given by η = e/ bardbl x 0 a bardbl . We know, of course, that the absolute value of the relative error is very small provided δx is small. In many specific applications, it is possible and useful to make a stronger statement, for example, to derive a bound on how large the error can be. You will do that here. In fact you will prove that 0 η α 2 2 where α = bardbl δx bardbl / bardbl x 0 a bardbl is the relative size of δx . For example, for a relative displacement of α = 1%, we have η 0 . 00005, i.e. , the linearized model is accurate to about 0 . 005%. To prove this bound you can proceed as follows: Show that η = 1 + radicalbig 1 + α 2 + 2 β β where β = k T δx/ bardbl x 0 a bardbl . Verify that | β | ≤ α . Consider the function g ( β ) = 1 + radicalbig 1 + α 2 + 2 β β with | β | ≤ α . By maximizing and minimizing g over the interval α β α show that 0 η α 2 2 . Solution. (a) For the linearized model we have δy = parenleftbigg ∂y ∂x parenrightbigg δx so all we have to do is to compute the matrix ∂y/∂x . Since y = bardbl x a bardbl we have y 2 = ( x a ) T ( x a ) and differentiating both sides with respect to x gives 2 ∂y ∂x y = 2( x a ) T and therefore ∂y ∂x = ( x a ) T y = ( x a ) T bardbl x a bardbl , so δy = k T δx with k = ( x a ) / bardbl x a bardbl . Clearly, k points from a to x and is of length one since k T k = ( x a ) T ( x a ) bardbl x a bardbl 2 = 1 . 1
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a k bardbl x 0 a bardbl bardbl x a bardbl x 0 δx k T δx x (b) First we show that η = 1 + radicalbig 1 + α 2 + 2 β β where β = k T δx/ bardbl x 0 a bardbl . Note that e = bardbl x 0 + δx a bardbl − bardbl x 0 a bardbl − k T δx = bardbl x 0 a bardbl parenleftbigg bardbl x 0 a bardbl x 0 a bardbl + δx bardbl x 0 a bardbl bardbl − 1 k T δx bardbl x 0 a bardbl parenrightbigg , and after dividing both sides by bardbl x 0 a bardbl and using k = ( x 0 a ) / bardbl x 0 a bardbl , β = k T δx/ bardbl x 0 a bardbl and η = e/ bardbl x 0 a bardbl we get η = bardbl k + δx bardbl x 0 a bardbl bardbl − 1 β. (1) But bardbl k + δx bardbl x 0 a bardbl bardbl = radicalBigg parenleftbigg k + δx bardbl x 0 a bardbl parenrightbigg T parenleftbigg k + δx bardbl x 0 a bardbl parenrightbigg = radicalBigg bardbl k bardbl 2 + 2 k T δx bardbl x 0 a bardbl + bardbl δx bardbl 2 bardbl x 0 a bardbl 2 .
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