Complex variables with applications by S Ponnusamy; Herb Silverman

By S Ponnusamy; Herb Silverman

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As a practical matter, in motion planning the mobile robot does not always know exactly where it is. Using sensors, such as sonars, that bounce signals off walls to determine the robot’s distance from the nearest walls, an internal representation of the local environment can be constructed. 5 Workspace generation by sweeping. 6 Configuration-space obstacles of mobile robots. calculated. Given a global map of the environment, a generalized convolution of the local free space estimate with the global map will generate a number of likely locations of the robot.

10) The fact that a function is recovered from its Fourier transform is found by first observing that it 2 is true for the special case of g(x) = e−ax for a > 0. One way to calculate ∞ g(ω) ˆ = e−ax e−iωx dx 2 −∞ is to differentiate both sides with respect to ω to yield d g(ω) ˆ = −i dω ∞ −∞ xe−ax e−iωx dx = 2 Integrating by parts, and observing that e−iωx g(x) i 2a ∞ −∞ dg −iωx dx. e dx vanishes at the limits of integration yields d ω gˆ = − g(ω). ˆ dω 2a The solution of this first-order ordinary differential equation is of the form ω2 − 4a g(ω) ˆ = g(0)e ˆ , ∞ g(0) ˆ = −∞ e−ax dx = 2 π .

Using this property and the definition of the DFT it is easy to show that fˆk+N = fˆk . 36) k=0 by observing that for a geometric sum with r = 1 and |r| ≤ 1, N−1 rk = k=0 1 − rN . 1−r Setting r = ei2π(n−m)/N for n = m yields the property that r N = 1, and so the numerator in the above equation is zero in this case. When n = m, all the exponentials in the sum reduce to the number 1, and so summing N times and dividing by N yields 1. Equipped with Eq. 36), one observes that N−1 fˆk ei2πj k/N = k=0 N−1 k=0 1 N N−1 = fn n=0 N−1 fn e−i2πnk/N ei2πj k/N n=0 1 N N−1 ei2π(j −n)k/N k=0 N−1 fn δj,n = n=0 and thus the DFT inversion formula N−1 fj = fˆk ei2πj k/N .

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