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#37
Aug2707, 02:36 AM

P: 87




#39
Aug2707, 12:01 PM

P: 87

yes, why?
I dont think that they have it. 


#41
Aug2707, 02:25 PM

P: 87

As I said to you:
let assume that our function g(t) instead of H(t) So that, g(t)= (1)^n * e^(t^2) * d^n/dt^n * e^(t^2)* cos (2 pi fc t) we can say that g(t) can satisfied equations 15 and 17 because of its orthogonality properties. Is that true? 


#42
Aug2707, 02:32 PM

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P: 2,482

Is cos (2 pi fc t) positive over the relevant range?
You may want to post it in the homework section as a separate question. 


#43
Aug2707, 02:36 PM

P: 87

OK, I will.
Thank you very much! 


#44
Aug2707, 04:07 PM

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P: 2,482

If the value of fc will be chosen in such a way that still keep the orthogonality property of Hermite polynomial, then orthogonality is preserved. (15) will apply, although with a different normalization constant and different psi functions.
New psi = old psi * cos (2 pi fc t). 


#45
Aug2907, 02:55 AM

P: 87




#46
Aug2907, 09:10 AM

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P: 2,482

Correct; that's because the integral in (15) will evaluate to a different output.
Another possibility is N_n will remain the same, but the delta will be different. Or both might change. But the "qualitative" result will not change, as long as fc is chosen to preserve orthogonality. That is, you will get to (17) with the new psi functions. 


#47
Aug2907, 11:53 AM

P: 87

1. please, can tell me how to find the normalization coeffecient N_n?
2. you said different δ _n,m. I know thet δ _n,m is Kronecker delta function, how it can be changed? Thanks a lot! 


#48
Aug2907, 12:27 PM

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1. N is determined by the output of the integral in (15). If the integral evaluated to δ*K for arbitrary K, then the norm. constant would have been N = 1/sqrt(K).
2. The δ itself won't change; but you may have something like Integral = z(δ 2^{n}n![itex]\sqrt{\pi}[/itex]) for some function z. 


#49
Aug2907, 12:39 PM

P: 87

1. you mean for arbitrary n,m.
2. what do you mean by z. 3. can you tell me how to evalute eq(15) to get this result: δ_n,m 2^n n! sqr(pi). if I you will know how they get this result for Hn, Hm, so I can also evaluted for my equation with Hn * cos (...) but this is my problem I dont know how they get this general formula. 


#50
Aug2907, 12:50 PM

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P: 2,482

1. yes
2. arbitrary function that results from including the cos term in the integrand (I haven't tried to integrate (15) with or without the cos term, so I don't know what z actually "looks like," even if we assume that a closedform solution exists with the cos term) 3. I don't know; I think [9] might have the answer. Someone has suggested to look it up from an integration table (under another thread in the homework section). 


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