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# ps1sol - EE 261 The Fourier Transform and its Applications...

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EE 261 The Fourier Transform and its Applications Fall 2011 Solutions to Problem Set One 1. Some practice combining simple signals. (5 points each) The (scaled) triangle function with a parameter a > 0 is Λ a ( t ) = Λ( t/a ) = ( 1 - 1 a | t | , | t | ≤ a 0 , | t | > a The graph is ! a a 1 The parameter a speciﬁes the width, namely 2 a . Alternately, a determines the slopes of the sides: the left side has slope 1 /a and the right side has slope - 1 /a . We can further modify Λ a by scaling the height and shifting horizontally, forming b Λ a ( t - c ). The slopes of the sides of the scaled function are then ± b/a . The graph is: b Λ a ( t c ) b c acc + a 1

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Express each of the following as a sum of two shifted, scaled triangle functions b 1 Λ a 1 ( t - c 1 )+ b 2 Λ a 2 ( t - c 2 ). Think of the sum as ‘left-triangle’ plus a ‘right-triangle’ (‘right’ meaning to the right, not having an angle of 90 ). For part (d), the values x 1 , x 2 and x 3 cannot be arbitrary. Rather, to be able to express the plot as the sum of two Λ’s they must satisfy a relationship that you should determine. (Why do this? It’s superposition waiting to happen – the possibility of operating on a complicated signal by operating on simpler constituents, and adding up the results. Here you are called upon to ﬁnd the constituents. ) ! 2 0 1 2 0 ! 2 5 4 (a) (b) 1 1 3 7 6 ( c ) ( x 1 , 0) ( x 2 ,y 2 ) ( x 3 , 0) ( d ) Solutions Parts (a) and (b): To get a ﬂat piece in the sum of two Λ’s over a certain interval, as in both parts (a) and (b), we want the slopes to cancel over that interval – where one side is going down the other is going up, and vice versa, so the slopes add to 0. So both Λ a ’s should have the same a . For part (a), the max is 1, so it looks like we don’t need to scale the heights and can take b = 1 for both. The question is how much the Λ’s overlap, which is governed by the shift parameter c , more precisely how much one Λ is shifted relative to the other. We see that if half the triangles overlap we get just what we want; bring the left zero-point of the right-hand triangle over to hit at the middle of the base of the left-hand triangle (where the left-hand triangle is at its max). Then as the side of the left-hand triangle decreases from its max, the side of the right-hand triangle increases by the same amount, adding to 1. Here’s the picture. 2
The graph is given by Λ 2 ( t ) + Λ 2 ( t - 2). For part(b), again we want overlapping slopes to cancel and we take a = 2 for both Λ’s. We also scale up both Λ’s by multiplying each by 2. To get the ﬂat piece, this time the relative shift of the two Λ’s should be such that less than half of the triangles are overlapping. How much less? That’s determined by knowing that the height of the ﬂat piece is 1. Bring the left zero-point of the right-hand triangle over to the point where the left-hand triangle has height 1. Here’s the picture.

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## This note was uploaded on 12/04/2011 for the course EE 263 at Stanford.

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ps1sol - EE 261 The Fourier Transform and its Applications...

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