Calculating FFT for discrete values

[electronic engineer]

Homework Statement

We have a set of values: f(n)=f(0,1,2)=(1,3,2)

so f(0)=1, f(1)=3, f(2)=2, where n=0,1,2the number of values N=3

The question is to calculate the FFT of this signal, the Fourier spectrum the power spectrum and phase spectrum.

I'm not sure concerning FFT. And also about the Fourier Spectrum what should I use to conduct it? the DFT or FFT?

Homework Equations

[/B]
The calculation of DFT for these values.

The Attempt at a Solution

I personally calculated the Discrete Fourier Transform in this way:

$$ F[n]=\sum_{k=0}^{N-1}f(k)e^{-j \frac{2\pi}{N}nk} $$

so I had a set of values: F(0),F(1),F(2)

 

[RUber]

For only 3 points, it seems unnecessary to try to do the FFT. However, the point of the FFT is to take advantage of symmetry.
F(1) = f(0) +f(1) e^{-j 2pi/N * 1 }+f(2) e^{-j 2pi/N * 2 }
F(2) = f(0) +f(1) e^{-j 2pi/N * 2 }+f(2) e^{-j 2pi/N * 4 }= f(0) +f(1) e^{-j 2pi/N * 2 }+f(2) e^{-j 2pi/N * 1 }
So whay you really have is (for real-valued inputs):
$F(0) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\1\\1} ,$
$F(1) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\ e^{-j 2\pi/N }\\ e^{ j 2\pi/N }}$
$F(2) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\ e^{j 2\pi/N }\\ e^{- j 2\pi/N }} = F(1) ^*$
As I said, for 3 values, this is more a demonstration of the concept that F(N-k) = F(k)*, which should save you about 1 calculation by just replacing F(1) with its conjugate for F(2).

 

[electronic engineer]

For only 3 points, it seems unnecessary to try to do the FFT. However, the point of the FFT is to take advantage of symmetry.
F(1) = f(0) +f(1) e^{-j 2pi/N * 1 }+f(2) e^{-j 2pi/N * 2 }
F(2) = f(0) +f(1) e^{-j 2pi/N * 2 }+f(2) e^{-j 2pi/N * 4 }= f(0) +f(1) e^{-j 2pi/N * 2 }+f(2) e^{-j 2pi/N * 1 }
So whay you really have is (for real-valued inputs):
$F(0) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\1\\1} ,$
$F(1) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\ e^{-j 2\pi/N }\\ e^{ j 2\pi/N }}$
$F(2) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\ e^{j 2\pi/N }\\ e^{- j 2\pi/N }} = F(1) ^*$
As I said, for 3 values, this is more a demonstration of the concept that F(N-k) = F(k)*, which should save you about 1 calculation by just replacing F(1) with its conjugate for F(2).

That is a very good and summarized answer. Thank you RUber :)

 

[electronic engineer]

But I still don't understand how e^(-j*2*pi/N *2)= e^(j*2*pi/N). Are these two angles identical??

 

[RUber]

This is the symmetry of the function. $e^{ix}$ is 2pi periodic, so $e^{ix}=e^{ix+i 2\pi}$
$e^{-j2\pi/N 2+j 2\pi}= e^{-j4\pi/3 +j6\pi/3}=e^{j*2*pi/N}.$

 

[electronic engineer]

This is the symmetry of the function. $e^{ix}$ is 2pi periodic, so $e^{ix}=e^{ix+i 2\pi}$
$e^{-j2\pi/N 2+j 2\pi}= e^{-j4\pi/3 +j6\pi/3}=e^{j*2*pi/N}.$

OK, thank you again.