- #1

PainterGuy

- 940

- 69

- TL;DR Summary
- 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!

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!