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Fourier transform of sinx

The fourier transform (FT) of a sin function (from time domain to frequency domain) would be characterised by two vertical lines of equal magnitude at +f and -f (where f = frequency). I am not sure, though, why the y value at +f and -f are +ve A and -ve A respectively (where A = ampltidue). Why would ampltidue be in the negative direction for -f and not +f

(And why for the FT of a cos(x) graph would both +f and -f be +ve A)
Thanks
Hi, I've moved your thread to the maths section. :smile:
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Arbolus
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What is the relationship between Asin(fx) and -Asin(-fx)? And between Acos(fx) and Acos(-fx)?

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Original post by jsmith6131
The fourier transform (FT) of a sin function (from time domain to frequency domain) would be characterised by two vertical lines of equal magnitude at +f and -f (where f = frequency). I am not sure, though, why the y value at +f and -f are +ve A and -ve A respectively (where A = ampltidue). Why would ampltidue be in the negative direction for -f and not +f

(And why for the FT of a cos(x) graph would both +f and -f be +ve A)
Thanks


I think that this follows easily from the fact that sin⁡ωx=12i(eiωx−e−iωx)\sin \omega x = \frac{1}{2i}(e^{i\omega x}-e^{-i\omega x})

You end up with a sum of translated delta functions of opposite amplitude due to the subtraction. And for the cosine, you start with cos⁡ωx=12(eiωx+e−iωx)\cos \omega x = \frac{1}{2}(e^{i\omega x}+e^{-i\omega x})
Original post by atsruser
I think that this follows easily from the fact that sin⁡ωx=12i(eiωx−e−iωx)\sin \omega x = \frac{1}{2i}(e^{i\omega x}-e^{-i\omega x})

You end up with a sum of translated delta functions of opposite amplitude due to the subtraction. And for the cosine, you start with cos⁡ωx=12(eiωx+e−iωx)\cos \omega x = \frac{1}{2}(e^{i\omega x}+e^{-i\omega x})


thanks

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