Fourier Transform
Now for some scary math ...
Recalling some basics...
Complex numbers have two parts:
rectangular coordinates
R + jII
what’s this?
what’s this?
Recalling some basics...
Complex numbers have two parts:
rectangular coordinates
R + jII
real
imaginary
Recalling some basics...
Complex numbers have two parts:
rectangular coordinates
R + jII
real
imaginary
what kind of
transform is
this?
Recalling some basics...
Complex numbers have two parts:
rectangular coordinates
R + jII
real
imaginary
Polar
Recalling some basics...
Complex numbers have two parts:
rectangular coordinates
R + jII
real
imaginary
Alternative re-parameterization:
polar coordinates
r(cos + j sin )
How do you compute r and theta?
Recalling some basics...
Complex numbers have two parts:
rectangular coordinates
R + jII
imaginary
real
Alternative re-parameterization:
polar coordinates
r(cos + j sin )
.
polar transform
= tan
1
I
( )
R
r=
p
R2 + I 2
Recalling some basics...
Complex numbers have two parts:
rectangular coordinates
R + jII
imaginary
real
Alternative re-parameterization:
polar coordinates
r(cos + j sin )
How do you write this in exponential form?
.
polar transform
= tan
1
I
( )
R
r=
p
R2 + I 2
Recalling some basics...
Complex numbers have two parts:
rectangular coordinates
R + jII
imaginary
real
Alternative re-parameterization:
exponential form
polar coordinates
r(cos + j sin )
.
polar transform
= tan
1
I
( )
R
r=
p
R2 + I 2
OR
re
j
‘Euler’s formula’
ej = cos + j sin
This will help us understanding of the Fourier transform equations ...
Inverse Fourier transform
Discrete
Continuous
Fourier transform
F (k) =
Z
1
f (x)e
j2 kx
dx
f (x) =
1
Z
1
F (k)ej2
1
F (k) =
N
f (x)e
x=0
k = 0, 1, 2, . . . , N
j2 kx/N
.
f (x) =
N
X1
F (k)ej2
k=0
11
dk
1
.
N
X1
kx
x = 0, 1, 2, . . . , N
Where is the connection to the ‘summation of sine waves’ idea?
11
kx/N
Inverse Fourier transform
Discrete
Continuous
Fourier transform
F (k) =
Z
1
f (x)e
j2 kx
dx
f (x) =
1
Z
1
F (k)ej2
1
F (k) =
N
f (x)e
x=0
k = 0, 1, 2, . . . , N
j2 kx/N
.
f (x) =
N
X1
F (k)ej2
k=0
11
dk
1
.
N
X1
kx
x = 0, 1, 2, . . . , N
Where is the connection to the ‘summation of sine waves’ idea?
11
kx/N
Where is the connection to the ‘summation of sine waves’ idea?
.
N
X1
f (x) =
F (k)ej2
kx/N
k=0
f (x) =
N
X1
‘Euler’s formula’
ej = cos + j sin
.
scaling
parameter
⇢
F (k) cos(2 kx) + j sin(2 kx)
k=0
sum over
frequencies
wave components
“So how do you actually compute the DFT?”
–A. Student
Computing the Discrete Fourier Transform…
.
1
F (k) =
N
N
X1
j2 kx/N
f (x)e
x=0
…is just a matrix multiplication.
F = Wf
2
6
6
6
6
6
6
6
4
3
F (0)
F (1)
F (2)
F (3)
..
.
F (N
1)
2
7 6
7 6
7 6
7 6
7=6
7 6
7 6
5 4
0
W
W0
W0
W0
..
.
W0
W =e
0
0
W
W1
W2
W3
WN
0
W
W2
W4
W6
1
j2⇡/N
WN
···
···
···
···
..
.
W
W3
W6
W9
2
WN
3
···
W =W
2N
W
WN
WN
WN
..
.
0
W1
1
2
3
32
76
76
76
76
76
76
76
54
3
f (0)
f (1)
f (2)
f (3)
..
.
f (N
1)
7
7
7
7
7
7
7
5
Example
2
input signal
3
DFT
2
3
f (0)
8
6 f (1) 7 6 4 7
6
7=6
7
4 f (2) 5 4 8 5
f (3)
0
F (k) =
3
2
F (0)
1
6 F (1) 7 6 1
6
7=6
4 F (2) 5 4 1
F (3)
1
1
j
1
j
=
j2⇡xk/4
3
X
f (x)( j)
xk
x=0
32
1
1
1
1
3
2
3
1
f (0)
20
6 f (1) 7 6 j4 7
j 7
76
7=6
7
1 5 4 f (2) 5 4 12 5
j
f (3)
j4
Magnitude of DFT coefficients
20
15
10
5
0
0
f (x)e
x=0
Frequency Domain representation
2
3
X
1
2
3
The Convolution Theorem
• The Fourier transform of the convolution of two functions is the product of
their Fourier transforms
F{g h} = F{g}F{h}
!
• The inverse Fourier transform of the product of two Fourier transforms is the
convolution of the two inverse Fourier transforms
.
!
F
1
{gh} = F
1
{g} F
1
{h}
• Convolution in spatial domain is equivalent to multiplication in frequency
domain!