# periodic function foureir transform in practice

Discussion in 'Homework Help' started by perchick, Jan 10, 2013.

1. ### perchick Thread Starter New Member

Oct 24, 2012
3
0
its seams like a very simple question but after 2 hours online i still don't understand this.
for function Asin(2∏*f1t) the Fourier transporm is A/2j[δ(f+f1)-δ(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|, δ(f-f1)=δ(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)

2. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
655
Did you see this Java applet? I don't pretend to fully understand it, but I believe the two counter-rotating blue vectors represent the positive and negative frequencies.
I believe the δ function corresponds to the frequency "impulse" (vertical line) that you see when a sine wave is plotted in the frequency domain.

3. ### perchick Thread Starter New Member

Oct 24, 2012
3
0
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/7eba64d...fa5c2aed20.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. A*delta(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 trouble understanding the Fourier transform when trying to transform theory into practice
perchick

4. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
The first thing to know is that the spectrum analyzer will only show you the positive frequency without the negative. It will usually superimpose the negative side onto the positive, giving you A, instead of A/2.

If you were to modulate the sine wave with another of a higher frequency, say, 1kHz modulated(using AM) with a 100kHz carrier, you'd see your signal at 99kHz and 101kHz, with the carrier at 100kHz. Basically, what this is doing is shifting your function, A/2δ(f-f1) + A/2δ(f+f1), out to be centered at 100kHz, so that you'd be able to see the +-f1 on a spectrum analyzer.

See here

Forgive me if I misunderstood your question

5. ### WBahn Moderator

Mar 31, 2012
17,446
4,698
First, I have no idea how the Fourier transform applies to realty, but I'm sure some realtor somewhere has applied it somehow. Now, as for how it applies to reality, consider the following:

sin(-x) = -sin(x)

6. ### MrChips Moderator

Oct 2, 2009
12,227
3,279
Just like imaginary numbers are real, so are negative frequencies. And so are delta functions.

Just think of negative frequencies as being to the left hand side of a central reference frequency on a graph and positive frequencies on the right. It's a bit like traveling back in time.

7. ### BillO Well-Known Member

Nov 24, 2008
985
136
A function such as:

$sin(\omega t)$

Does not carry any of the usual baggage you and I carry with respect to time. Neither does the transform care about which way the t variable is varying. A negative frequency is therefore just as valid a solution to the problem as is a positive frequency. After all, the Fourier transform of sin() is entirely imaginary. But we know better, don't we? We know that a sine wave at 1kHz certainly does not have an imaginary frequency and in fact sounds just like a cosine wave of 1kHz. The Fourier transform provides us with complex valued results. Complex as in the complex numbers. The correct interpetation, for the real world, would be to take the modulus of the resulting transform.