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 Exercise [07.06] 
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Joined: 09 Jul 2008, 15:25
Posts: 2
Post Exercise [07.06]
For solution see wordfile


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File comment: This solution also explains why the integral = 0
Solution of problem 7 6 more.doc [128 KiB]
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09 Jul 2008, 15:28
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Joined: 30 Jun 2008, 22:14
Posts: 25
Post Re: Exercise [07.06]
This is brilliant, hans! I didn't know that poles and their orders can be found this way.

One question. In this solution, poles and their orders were first guessed, then proved. Does anybody know of a general procedure to arrive at these numbers without the guess work in the first place? Or is there any need for a general procedure at all?


13 Jul 2008, 11:03
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Joined: 09 Jul 2008, 15:25
Posts: 2
Post Re: Exercise [07.06]
To calculate the poles one has to calculate the zero’s of the denominator:

z^2.sin(az)=0
z=0 (2x) or sin(az)=0
z=0 2x) or az=kpi with k= ……,-3,-2,-1,0,1,2,3,….
z=0 (2x) or z= kpi/a
z=0 (3x) or z= kpi/a with k=….,-3,-2,-1,1,2,3,….


15 Jul 2008, 17:13
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