Fourier Analysis. PSCI 702 November 2, 2005. Even and Odd Functions. Even Functions. Odd Functions. Even and Odd Functions. Kronecker’s Rule. Periodic Functions. Trigonometric System. Trigonometric System of period 2a. Fourier Series.
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source: f(x) =x, 8 <= x <= 8
n= 5
n= 1
n= 3
y
Successive approximations of f(x)
f(x)
x
1 + 2 + 3
1 + 2
1
1
0

2
3
1 + 2 + 3 + 4 + 5
1 + 2 + 3 + 4
Square wave: Y = 0 for  < x < 0 and Y=1 for 0 < x <
Y = 1/2 + 2/pi( sinx + sin3x/3 + sin5x/5 + sin7x/7 … + sin(2m+1)x/(2m+1) + …)
1 2 3 4 5
May do with sum of cosines too.
The inverse equations:
,
Using the formulas above and some properties of exponential function, the Fourier series can also be written as an expansion in terms of complex exponentials as:
,
Complex version of the Fourier expansionF(w)
w
6p
4p
2p
2p
4p
6p
t
t
t
t
t
t
0
Dualityf(t)
1
t
t/2
0
t/2
Find the Fourier transform of
Find the Fourier transform of
f(x) = Π(x /4) – Λ(x /2) + .5Λ(x)
Using the Fourier transforms of Π and Λ
and the linearity and scaling properties,
F(u) = 4sinc(4u)  2sinc2(2u) + .5sinc2(u)
Example problem: Alternative Answer.
*
–1 .5 0 .5 1
–2 1 0 1 2
Find the Fourier transform of
f(x) = Π(x /4) – 0.5((Π(x /3) * Π(x))

Using the Fourier transforms of Π and Λ
and the linearity and scaling and convolution properties ,
F(u) = 4sinc(4u) – 1.5sinc(3u)sinc(u)
The discrete Fourier transform
Motivation: computer applications of the Fourier transform require that all of the definitions and properties of Fourier transforms be translated into analogous statements appropriate to functions represented by a discrete set of sampling points rather than by continuous functions.
Let f(x) be a function.Let {fk = f(xk)} be a set of N function values, k = 0, 1, …, N1.Let be the separation of the equidistant sampling points.Assumption: N is even.
The discrete Fourier transform is:
The inverse discrete transform is:
The discrete Fourier transform(2)
Let’s examine more closely the formula of the discrete Fourier transform:
We know that (it’s called nth root of unity), so the formula above can be rewritten as:
Say we want to perform a 8 point DFT on a discretized version of a continuous input signal
having frequency components 1KHz and 2KHz
Calculation of Ts : Suppose
Period of x(t) = 1/1Khz = 1/1000
8 samples per period => sample time (Ts) = 1/8000 sec
Or sample rate = 8000 samples/s
t = nTs so
n = 0,1,…,7
Therefore :
DC Component
And so on...
Where X(k) = (k*8Khz)/8
Actually evaluating we get the values:
X(0) = 0 + i 0 (dc)
X(2) = 1.414 + i1.414 (2Khz)
X(4) = 0 + i0 (4Khz)
X(6) = 1.414 – i 1.414 (6Khz)
X(1) = 0 – i 4 (1KHz)
X(3) = 0 + i 0 (3Khz)
X(5) = 0 + i 0 (5Khz)
X(7) = 0 + i 4 (7KHz)