Why does Mathematica's Fourier command omit phase information?

[Niles]

Hi all

I have a function F, which depends on a discrete variable x, and I need to Fourier Transform it. I have put all the values of F in a table.

Then I have used the command "Fourier" on the table, which - according to http://reference.wolfram.com/mathematica/ref/Fourier.html - results in the discrete Fourier Transform. But the new table only contains pure numbers, no phase factor. Why is that, and is there a way to include the phase?


Niles.

 

[Niles]

Does anybody know this? I feel bad "bumping" the thread, but I am in real trouble if I can't work this out.

 

[shoehorn]

Post the relevant code.

 

[Niles]

Here you go (the values of F are in the table)

four = Fourier[{0.5,1.0,3.0,5.0}]

 

[Xitami]

Mathematica {4.75+0 i,-1.25-2. i,-1.25+0 i,-1.25+2. i}
Why not {9.50 + 0 i, -2.50 + 4. i, -2.50 + 0.i, -2.50 - 4.i} ?

 

[Niles]

But I really don't know what it is that Mathematica is calculating? The amplitudes or what?

 

[jackmell]

Hi Niles. It's easy to see what Mathematica is computing. Just do a help on Fourier and hit the more info button. For a list {u_n}, Mathematica creates a list {v_n} such that:

$v_s=\frac{1}{\sqrt{n}}\sum_{r=1}^n u_r e^{2\pi i(r-1)(s-1)/n}$

so for your list {0.5, 1.0, 3.0, 5.0},

the first value computed by Mathematica is:

$v_1=\frac{1}{\sqrt{n}}\sum_{r=1}^n u_r e^{2\pi i(r-1)(1-1)/n}=4.75$

Don't know what you mean by phase factor or amplitude though. Try reading through all the various options in the More info. Maybe you'll spot something that can help you.

 

[Niles]

I will have to think about this. Thanks!

 

[shoehorn]

If x(n) represents the time-domain data, the formula for an N-point discrete Fourier transform is given by

$$X(m) = \sum_{n=0}^{N-1} x(n) e^{-2\pi imn/N}, \qquad m=0\ldots N-1.$$

You'll often be interested in the magnitude and power contained in each X(m). If you represent an arbitrary DFT output value, X(m), by its real and imaginary parts:

$$X(m) = X_\textrm{real}(m) + iX_\textrm{imag}(m) = X_\textrm{mag}(m) \textrm{ at an angle of }X_\theta(m)$$

the magnitude of X(m) is simply

$$X_\textrm{mag}(m) = |X(m)| = \sqrt{X_\textrm{real}(m)^2 + X_\textrm{imag}(m)^2}.$$

The phase angle of X(m) is then defined by

$$X_\theta(m) = \tan^{-1} \left(\frac{X_\textrm{imag}(m)}{X_\textrm{real}(m)}\right).$$

Therefore, if you calculate the DFT of your data as

 In[1]:= four = Fourier[{0.5,1.0,3.0,5.0}]
Out[1]:= {4.75 + 0. I, -1.25 - 2. I, -1.25 + 0. I, -1.25 + 2. I}

the phase is obtained simply as

 In[2]:= phases = ArcTan[Im[four]/Re[four]]
Out[2]:= {0., 1.0122, 0., -1.0122}

Is that what you're looking for?

 

[Niles]

I will try and experiment with it; I'll keep you posted. Thanks so far.