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ECE108HW%237sol_06

ECE108HW%237sol_06 - ECE 108 HW#7 Due Solution 1 Use the...

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ECE 108 HW#7 Due 3/27/06 — Solution 1. Use the differentiation property of the Fourier transform to determine the Fourier transform of the following signal. () ( ) ( ) 22 gt t t δδ =+ −− Then () ( ) 2s i n2 jj Gj e e j ωω ω =− = and () () 00 Gg t d t −∞ == Then ( ) 2sin 2 Xj j 2. Problem 4.7 To determine whether a signal is purely real or imaginary observe that * * 11 R I xt x t x t Using the conjugate property

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2 () * * 11 22 R I X jX j X j ω ωω =+ =− But a signal will be real iff ( ) ( ) ( ) ( ) ( ) ** 0 I xt Xj X j X j =⇒ = Similarly a signal will be imaginary iff ( ) ( ) ( ) ( ) ( ) r X j X j + = Next observe that 1 2 1 2 jt X j e d Xje d π −∞ −∞ = −= Now if into this second integral we substitute , we obtain the result that ′=− X j ed t X j t ππ ∞∞ −∞ −∞ = ∫∫ Using this expression and the expression for x ( t ) we get the following result. If x ( t ) is an even function, then ( ) ( ) ( ) ( ) xt x t X j X j =− ⇒ − = and if x ( t ) is an odd function, then ( ) ( ) ( ) ( ) x t X j =− − Now we can apply these to the signals given in (a) . Here ( ) ( ) ( ) 1 2 Xj u u ωωω ( ) ( ) ( ) ( ) 1 1 and X j X j ≠− −−
3 Therefore the signal is complex valued. Similarly, ( ) ( ) ( ) ( ) 11 1 1 and Xj X j X j ω ωω −≠ Therefore the signal has neither even nor odd symmetry. (b) . Here () 2 cos 2 sin 2  =   ( ) * 22 cos 2 sin 2 X j −= = Therefore the signal is purely imaginary. Also ( ) cos 2 sin 2 X j = Therefore the signal has odd symmetry (c) 2 2 2 3 sin sin j j e j e π + == Therefore ( ) ( ) * X j and so the signal is purely imaginary. Also ( ) ( ) ( ) ( ) and X j X j and so the signal has neither even nor odd symmetry. 3. Problem 4.19 1 1 Hj j = + () () () 34 tt yt e ut e ut −− =− Then

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4 () ( ) () () 34 00 3 4 0 11 4 3 1 3 4 3 4 jt t t y t Y j y t ed te t t t e e jj j j ωωω ωω ω ∞∞ −− −∞ −+ ℑ= = =
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ECE108HW%237sol_06 - ECE 108 HW#7 Due Solution 1 Use the...

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