hmk4_sol

# hmk4_sol - Math 267 Assignment 4 1 In each case below...

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Math 267 – Assignment 4 1. In each case below, sketch the graph of the function f ( t ) and ﬁnd the (complex) Fourier series of f ( t ). (a) f ( t ) is periodic with period 4 π , and f ( t ) = - 1 if - 2 π t < - π 1 if - π t < π - 1 if π t < 2 π (b) f ( t ) is periodic with period 2, and f ( t ) = t if - 1 t < 1. (c) f ( t ) is periodic with period 2, and f ( t ) = ± 1 if 0 t < 1 0 if 1 t < 2 Solution. (a) The graph of f ( t ) is shown in Figure 1. 1 0 1 ! 9 ! 5 ! 3 ! 7 ! Figure 1: Graph of f ( t ) in problem 1 (a). Next we calculate the complex Fourier coeﬃcients of f ( t ). Method 1: Direct calculation Because f ( t ) is periodic with period 2 = 4 π , we have = 2 π and b f ( k ) = 1 2 Z h 2 i f ( t ) e - ik π t dt = 1 4 π Z h 4 π i f ( t ) e - ik t 2 dt. Above R h 2 i means we are integrating over any interval [ 1 ,‘ 2 ] of length l 2 - l 1 = 2 , i.e., it does not matter what the precise values of integration limits are. The crucial thing is that the upper limit is exactly 2 more than the lower limit. (This is true because the integrand is 2 periodic.) In this problem, we choose the limits to be - π and 3 π (easier to calculate with this choice). We then have b f ( k ) = 1 4 π 3 π Z - π f ( t ) e - ik t 2 dt = 1 4 π Z π - π e - ik t 2 dt | {z } I 1 ( k ) - Z 3 π π e - ik t 2 dt | {z } I 2 ( k ) (1) 1

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Next, for k 6 = 0: I 1 ( k ) = " e - i k 2 t - i k 2 t # π - π = 2 i k ± e - i k 2 π - e i k 2 π ² = 4 k sin( k 2 π ) = 4 k if k = ..., - 3 , 1 , 5 ,... - 4 k if k = - 1 , 3 , 7 0 if k is even, k 6 = 0. (2) Similarly, I 2 ( k ) = " e - i k 2 t - i k 2 t # 3 π π = 2 i k ± e - i k 2 3 π - e - i k 2 π ² = 2 i k e - ikπ | {z } =( - 1) k ± e - i k 2 π - e i k 2 π ² = ( - 1) k I 1 ( k ) = - 4 k if k = - 3 , 1 , 5 4 k if k = - 1 , 3 , 7 0 if k is even, k 6 = 0. (3) Noting b f (0) = 1 4 π 3 π R - π f ( t ) dt = 0, we conclude that the complex Fourier coeﬃcients of f ( t ) are b f ( k ) = 2 πk if k = - 3 , 1 , 5 - 2 πk if k = - 1 , 3 , 7 0 if k is even . Finally, the complex Fourier series of f ( t ) is f ( t ) = X k = -∞ b f ( k ) e ik t 2 with b f ( k ) as above. Method 2: Using “time-scaling” property. In class, we showed that b f sqw ( k ) = ( 2 iπk if k odd 0 otherwise.
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hmk4_sol - Math 267 Assignment 4 1 In each case below...

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