How get amplitude and phase as a function of time?

[Jhenrique]

Given a function of the time like $f(t)$, its Fourier transform is $F(\nu)$ and by definition, $Abs(F(\nu))=A(\nu)$ and $Arg(F(\nu)) = \varphi(\nu)$, so I have amplitude and phase as function of the frequency, but make sense speak in amplitude and phase as function of the time? If yes? How get them?

 

[MisterX]

\text{abs}(f(t)) and \text{arg}(f(t)) of course

 

[Jhenrique]

\text{abs}(f(t)) and \text{arg}(f(t)) of course

But if f(t) = cos(t) thus the amplitude as function of the time is A(t) = 1 and the phase is φ(t) = 0 what is different of Abs(f(t)) and Arg(f(t)), respectively.

 

[bigfooted]

No, you only have a real part, so the modulus is abs(f(t)) = sqrt(cos^2(t) + 0) = cos(t) and the argument/phase is 0

http://en.wikipedia.org/wiki/Arg_(mathematics)

 

[Jhenrique]

No, you only have a real part, so the modulus is abs(f(t)) = sqrt(cos^2(t) + 0) = cos(t) and the argument/phase is 0

http://en.wikipedia.org/wiki/Arg_(mathematics)

argument is different of phase