hw3_solutions

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

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Unformatted text preview: 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 can be expressed as δy = k T δx , where k is the unit vector ( i.e. , with length one) pointing from a to x . 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 + δx − a bardbl − bardbl x − a bardbl − k T δx. The relative error of the approximation is given by η = e/ bardbl x − 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 ≤ η ≤ α 2 2 where α = bardbl δx bardbl / bardbl x − a bardbl is the relative size of δx . For example, for a relative displacement of α = 1%, we have η ≤ . 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 − 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 ≤ 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 a k bardbl x − a bardbl bardbl x − a bardbl x 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 − a bardbl . Note that e = bardbl x + δx − a bardbl − bardbl x − a bardbl − k T δx = bardbl x − a bardbl parenleftbigg bardbl x − a bardbl x − a bardbl + δx bardbl x − a bardbl bardbl − 1 − k T δx bardbl x − a bardbl parenrightbigg , and after dividing both sides by bardbl x − a bardbl and using k = ( x − a ) / bardbl x − a bardbl , β = k T δx/ bardbl x − a bardbl and η = e/ bardbl x − a bardbl we get η = bardbl k + δx bardbl x − a bardbl bardbl − 1 − β. (1) But bardbl k + δx bardbl x − a bardbl bardbl = radicalBigg parenleftbigg k + δx bardbl x − a bardbl parenrightbigg...
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This note was uploaded on 07/28/2011 for the course EE 263 taught by Professor Boyd,s during the Summer '08 term at Stanford.

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

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