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periodic function foureir transform in practice

perchick

Nov 2, 2012
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its seams like a very simple question but after 2 hours online i still don't understand this.
for function Asin(2pi*f1t) the Fourier transporm is A/2j[dlta(f+f1)-dlta(f-f1)].
how do i look at this in realty? i mean, the delta function has no physical meaning and there is no negative frequency's. the meaning of the function is that my power is in the frequency f1 and the amplitude is A. how do i get this result? if i look at the negative frequency's as positive and taking the absolute value of the fourier transform (|F|, delta(f-f1)=delta(f+f1)) I still get just 1/2 of the amplitude not to mention i cant get rid of the delta function which has no meaning seen that its discontinuity at f1.
i need to explain results i got in the lab and i cant find the mathematical tools (i dont want to use fourier series, i want to explain everything with the transform)
 

Laplace

Apr 4, 2010
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If you are looking at power in the Fourier transform, Parseval's theorem relates energy in the time domain to energy in the frequency domain. But getting energy in the frequency domain requires integrating the square of the Fourier transform, which for this signal means squaring the Dirac delta function. However, the Dirac delta is not really a function - it is a distribution, so squaring it has no meaning. The best you might be able to do is find a function that approximates the delta distribution, square that, then take the limit as the width approaches zero. Good luck with that.
 

perchick

Nov 2, 2012
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Nov 2, 2012
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let me rephrase my question so it will be more obvious what i don't understand:
mathematically i get after the Fourier transform 2 delta function with an amplitude of 0.5A in -f1 and f1. in the lab i got 1 pulse with an amplitude of A in f1 ( http://www.israup.net/images/7eba64dab30ea8b8463a93fa5c2aed20.png) .
i want to explain this result using Fourier transform. meaning i want to get some mathematical expression which will give me what i got. Adelta(f-f1) will do the trick but i cant get to this expression.
i wonder if the Fourier transform gives me just the mathematical tool for calculating the power. is this correct? parseval theorem for periodic function states that i have to get the square of the coefficients of the Fourier series to get the power so it doesnt help me there, or does it?
i guess Im just having problems understanding the Fourier transform when trying to transform theory into practice
thank you Laplace for all your help
 

Laplace

Apr 4, 2010
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You asked, "how do i look at this in reality?" What if the instrument you were using in the lab was designed to take a sine wave of amplitude 'A' and display its Fourier spectrum with an amplitude of 'A'? That would be my reality.
 
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