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Unformatted text preview: EE263s Summer 200910 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 y = k T x , 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 + x a bardbl bardbl x a bardbl k T x. The relative error of the approximation is given by = e/ bardbl x a bardbl . We know, of course, that the absolute value of the relative error is very small provided x 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 2 2 where = bardbl x bardbl / bardbl x a bardbl is the relative size of x . For example, for a relative displacement of = 1%, we have . 00005, i.e. , the linearized model is accurate to about 0 . 005%. To prove this bound you can proceed as follows: Show that = 1 + radicalbig 1 + 2 + 2 where = k T x/ bardbl x a bardbl . Verify that   . Consider the function g ( ) = 1 + radicalbig 1 + 2 + 2 with   . By maximizing and minimizing g over the interval show that 2 2 . Solution. (a) For the linearized model we have y = parenleftbigg y x parenrightbigg x 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 y = k T x 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 x k T x x (b) First we show that = 1 + radicalbig 1 + 2 + 2 where = k T x/ bardbl x a bardbl . Note that e = bardbl x + x a bardbl bardbl x a bardbl k T x = bardbl x a bardbl parenleftbigg bardbl x a bardbl x a bardbl + x bardbl x a bardbl bardbl 1 k T x bardbl x a bardbl parenrightbigg , and after dividing both sides by bardbl x a bardbl and using k = ( x a ) / bardbl x a bardbl , = k T x/ bardbl x a bardbl and = e/ bardbl x a bardbl we get = bardbl k + x bardbl x a bardbl bardbl 1 . (1) But bardbl k + x bardbl x a bardbl bardbl = radicalBigg parenleftbigg k + x bardbl x a bardbl parenrightbigg...
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 Summer '08
 BOYD,S

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