Ph12a_s3_2009

Ph12a_s3_2009 - Ph12a Solution Set 3 October 21, 2009 2.30...

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Ph12a Solution Set 3 October 21, 2009 2.30 (a) Choosing the origin of time at the center of Δ t , the sine terms all vanish (see eqn (52) on p. 66, with F ( t ) an even function). (b) B 0 = 1 T 1 R T 1 / 2 - T 1 / 2 f ( t ) dt = Δ t T 1 B n = 2 T 1 R T 1 / 2 - T 1 / 2 f ( t )cos 2 πnt T 1 = 2 T 1 R Δ t/ 2 - Δ t/ 2 cos 2 πnt T 1 = 2 πn sin πn Δ t T 1 (c) For Δ t << T 1 , one gets approximately B n = 2 πn πn Δ t T 1 = t T 1 for small n , which is independent of n . (d) The formula for B n derived in (b) can be expressed in the form B n = 2 ν 1 Δ t sin( nπν 1 Δ t ) nπν 1 Δ t Pretending for a moment that n is a continuous variable, we can produce the plot shown in Figure 1. Precisely which points on this plot correspond to integer values of n , depends on the value of ν 1 Δ t , so we can’t mark those points on the plot. However, we can note that there must be many of those points within a unit interval of the x -axis of the plot, because ν 1 Δ t = Δ t/T 1 << 1. (e)
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Ph12a_s3_2009 - Ph12a Solution Set 3 October 21, 2009 2.30...

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