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Forums
Mathematics
Calculus
Magnitude and phase of the Fourier transform
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[QUOTE="PainterGuy, post: 6335444, member: 311618"] [B]TL;DR Summary:[/B] Trying to understand how a function f(t) could be generated using magnitude and phase information from its FT at basic level. Hi, A rectangular pulse having unit height and lasts from -T/2 to T/2. "T" is pulse width. Let's assume T=2π. The following is Fourier transform of the above mentioned pulse. F(ω)=2sin{(ωT)/2}/ω ; since T=2π ; therefore F(ω)=2sin(ωπ)/ω Magnitude of F(ω)=|F(ω)|=√[{2sin(ωπ)/ω}^2]=|2sin(ωπ)/ω| Phase of F(ω), ∠F(ω): phase of complex number x+iy is defined as: θ=tan⁻¹(y/x). In case of F(ω) "y" is zero. The expression "2sin(ωπ)/ω" would alternate between "+" and "-" values. θ=lim_(y→0){tan⁻¹(y/x)}=0, π. The phase, ∠F(ω), switches between "0" and "π" depending upon the sign of "x". I have always thought that summation or integral of: |F(ω)|cos{ωt+∠F(ω)} for ω ≥ 0 would produce the original function. Am I thinking correctly? Thank you! [/QUOTE]
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Forums
Mathematics
Calculus
Magnitude and phase of the Fourier transform
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