Magnitude and phase of the Fourier transform

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Summary:

Trying to understand how a function f(t) could be generated using magnitude and phase information from its FT at basic level.

Main Question or Discussion Point

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!
 
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Answers and Replies

  • #2
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The Fourier transform requires all frequencies I believe.
Even for a real f(t) the transform F(ω) is not necessarily real. A sufficient condition is that that if f(t)=f(-t) then F(ω) is real. But any translation in time by t0 will give e-iωt0F(ω).
 
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I'm sorry but it looks like you're missing my question. Anyway, yes, FT requires all frequencies. Trigonometric form uses only positive frequencies while exponential form of FT used negative and positive frequencies to make things symmetric.
 
  • #4
jasonRF
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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!
No.

When you have such questions, the best thing to do is start with the definition and just do the algebra:
$$ \begin{eqnarray*}
f(t) & = & \frac{1}{2\pi}\int_{-\infty}^\infty F(\omega) \, e^{i\omega t} \, d\omega \\
& = & \frac{1}{2\pi}\int_{-\infty}^\infty \left| F(\omega)\right| \, e^{i \angle F(\omega)} \, e^{i\omega t} \, d\omega \\
& = & \frac{1}{2\pi}\int_{-\infty}^\infty \left| F(\omega)\right| \left[ \cos\left(\omega t + \angle F(\omega) \right) + i \sin \left(\omega t + \angle F(\omega) \right)\right]
\end{eqnarray*}
$$

Now, if your ##f(t)## has special properties, like being real, then there are symmetries that you can use to modify this expression and perhaps throw out terms, but usually that doesn't actually help you compute anything . As an electrical engineer I usually deal with signals that are complex valued, so I always use the full definition of the inverse transform.

Note: I edited this post significantly a few minutes after the initial posting.

jason
 
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As an electrical engineer I usually deal with signals that are complex valued, so I always use the full definition of the inverse transform.
Thank you for the help. Now I can see where I was going wrong. I should have used complex sinusoid.
 

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