Discrete Cosine Transform (DCT)

Discrete Cosine Transform (DCT)
1

DCT is a Fourier-related transform similar to
DFT, but using only real numbers. It is
equivalent to a DFT roughly twice the
transform length, operating on real data with
“Even symmetry” (since Fourier transform of
a real and even function is real and even).
The most common variant of DCT is the type-
II DCT and its inverse is the type-III DCT.
2

Two related transform are the discrete sine
transform (DST), which is equivalent to a DFT
of “Real and odd” functions, and the modified
DCT (MDCT), which is based on a DCT of
overlapping data.
The other interesting transform is the discrete
Hartley transform (DHT) in which
Even part of DHT = Real part of DFT
Odd part of DHT = Imaginary part of DFT
3

DCT is often used in signal and image processing,
especially for lossy data compression, because it has
a strong “Energy Compaction” property: most of the
signal information tends to be concentrated in a few
low-frequency components of the DCT, approaching
the KLT for signals based on certain limits of Markov
processes.
DCT is used in JPEG image compression, MJPEG and
MPEG video compression.
MDCT is used in MP3, AAC, etc. audio compression.
4

Formulation:
DCT is a linear and invertible function
F: R n →R n (where R denotes the set of real
numbers.), or equivalently on nxn matrix.
DCT-I:
1
f j = ( x
2
0
+ ( −1)
j
x
n−1
)
+
∑ − n 2
k = 1
x
k
π
cos[
n −1
jk]
5

A DCT-I of n=5 real numbers abcde is exactly
equivalent to a DFT of 8 real numbers
abcdedcb (even symmetry), here divided by 2.
(In contrast, DCT-II ~ IV involve a halfsample
shift in the equivalent DFT.)
Note that DCT-I is not defined for n less than
2. (All other DCT types are defined for any
positive n.)
6

DCT-II:
f
j
=
∑ − n 1
k = 0
x
k
π
cos[ j(
k +
n
1
)]
2
Some authors further multiply the f o term by
1/
2
(see below for the corresponding
change in DCT-III) . This makes the DCT-II
matrix orthogonal (up to a scale factor), but
breaks the direct correspondence with a realeven
DFT of half-shifted input.
7

DCT-III:
f
j
=
1
2
x
0
+
∑ − n 1
k = 1
x
k
π
cos[ ( j
n
Some authors further multiply the x 0 term
by 1/
2 , this makes the DCT-III matrix
orthogonal (up to a scale factor), but breaks
the direct correspondence with a real even
DFT of half-shifted output.
+
1
) k]
2
8

DCT-IV:
f
j
=
∑ − n 1
k = 0
x
k
π
cos[ ( j
n
DCT-IV matrix is orthogonal (up to a
scale factor).
MDCT is based on DCT-IV with
overlapped data.
+
1
)(
2
k
+
1
)]
2
9

DCT V-VIII:
DCT types I-IV are equivalent to real-even
DFTs of even order; therefore, there are 4
additional types of DCT corresponding to
real-even DFTs of logically odd order,
which have factors of (n + 1/2) in the
denominators of the cosine arguments.
These variants seem to be rarely used in
practice.
10

Inverse Transforms:
IDCT-I is DCT-I multiplied by 2/(n-1).
IDCT-IV is DCT-IV multiplied by 2/n.
IDCT-II is DCT-III multiplied by 2/n (and
versa).
11

Computation
Direct application of the above formulas
would require O(n 2 ) operations, as in the
FFT it is possible to compute the same
thing with only O(nlogn) complexity by
factorizing the computation. (One can also
compute DCTs via FFTs combined with
O(n)) pre- and post-processing steps.)
12

References:
1. Rao and Yip, Discrete Cosine Transform:
Algorithms, Advantages, Applications; Academic
Press, Boston, 1990.
2. Arai, Agui, Nakajima, A Fast DCT-SQ scheme
for Images, Trans. On IEICE-E, 71(11), 1095, Nov.
1998.
3. Tseng and Millen, On Computing the DCT, IEEE
Trans. On Computers, pp. 966-968, Oct. 1978.
4. Frigo and Johnson, The Design and
Implementation of FFTW3, IEEE Proceedings, vol.
93, no. 2, pp. 216-231, 2005.
13

The implementation of the 2D-IDCT
Let the 8-point 1-D DCT of input data
f(x) be:
S
8
( u)
=
C
2
u
7
∑
x=
0
( 2x
+ 1)
πu
f ( x)
cos
16
First the 1-D DCT is applied to all the
rows of the 2-D input f(y,x):
S
8r
( y,
u)
=
C
2
u
7
∑
x=
0
( 2x
+ 1)
πu
f ( y,
x)
cos
16
(1)
(2)
14

Then the 1-D DCT is applied to the
columns of the results of (2):
S(
v,
u)
=
C
2
u
7
∑
y=
0
S
8r
( y,
u)
cos
y + 1)
πv
16
By substitution we get the formulation
of the 2-D DCT:
S(
v,
u)
=
C
2
v
C
2
u
7
7
∑∑
y= 0 x=
0
( 2
( 2x
+ 1)
πu
( 2y
+ 1)
πv
f ( y,
x)
cos cos
16 16
(3)
(4)
15

Fast 1-D DCT Algorithms
Define
Eqn. (1) can be written as:
Since
2πux
πu
α = , β = , and H = α + β.
16 16
7
7
Cu
( 2x
+ 1)
πu
Cu
S8 ( u)
= ∑ f ( x)
cos = ∑ f ( x)
cos( α + β )
2
16 2
2cos
x=
0
H cos β = 2cos(
α + β ) cos β =
↑
( HW )
x=
0
2xπu
2(
15 − x)
πu
cos + cos
16 16
16

4
C
u
πu
cos S
16
8
( u)
=
7
∑
x=
0
2xπu
2(
15 − x)
πu
f ( x)[cos
+ cos ]
16 16
If we constitute a sequence of elements
f(k), k = 0, 1, …, 15 with
f
( k)
=
⎧
⎨
⎩
f ( k),
∀k
< 8
f ( 15 − k),
∀k
≥ 8
(5)
17

We can re-write (5) as
4
C
u
πu
cos S
16
When
j
8
=
( u)
=
−1
15
2kπu
16
∑ f ( k)
cos = Re{ ∑
k= 0
k=
0
2kπu
− j
16
Decaese the 16-point DFT is defined by
F
16
( u)
=
15
∑
k = 0
f
( k)
e
2kπu
− j
16
15
f ( k)
e
}
18

We have the following DCT v.s. DFT
relationship
4
C u
u
cos
16
π : only the first 8 values
S ( u)
= Re{ F ( u)}
are needed.
8
Instead of performing an IDCT an IDFT of
twice the length is performed. IDFT can be
implemented by IFFT with complexity
O(NlogN)!
16
19

4 πu
cos
The multiplication by Cu 16 seems not so
efficient. However, bear in mind that the last
operation before the IDCT is the
“Quantization”. That means every value is to
be multiplied with a certain constant
(1/Quantization factor) depending on its
position in DCT matrix. So we can merge the
4 πu
multiplications by cos
C and the multiplication
u 16
by Quantizer dependent constant together.
As a result, the Quantization and DCT-DFT
transform is performed in one step.
20

Let
X M = (f(0), f(1), f(2), f(3), f(4), f(5), f(6),
f(7)): original data
1 1 1 1 1
FM = ( 16 F(0), F(1), F(2), F(3), F(4),
8
1
8 8
1
8
1
8 8 8
F(5), F(6), F(7)): scaled transformed
data
Define P(a,b) =
2π
a 2πb
cos +
cos
16 16
We can establish a matrix T M:
21

22
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
=
)
56
,
49
(
)
63
,
42
(
)
70
,
35
(
)
77
,
28
(
)
84
,
21
(
)
91
,
14
(
)
98
,
7
(
)
105
,
0
(
)
48
,
42
(
)
54
,
36
(
)
60
,
30
(
)
66
,
24
(
)
72
,
18
(
)
78
,
12
(
)
84
,
6
(
)
90
,
0
(
)
40
,
35
(
)
45
,
30
(
)
50
,
25
(
)
55
,
20
(
)
60
,
15
(
)
65
,
10
(
)
70
,
5
(
)
75
,
0
(
)
32
,
28
(
)
36
,
24
(
)
40
,
20
(
)
44
,
16
(
)
48
,
12
(
)
52
,
8
(
)
56
,
4
(
)
60
,
0
(
)
24
,
21
(
)
27
,
18
(
)
30
,
15
(
)
33
,
12
(
)
36
,
9
(
)
39
,
6
(
)
42
,
3
(
)
045
(
)
16
,
14
(
)
18
,
12
(
)
20
,
10
(
)
22
,
8
(
)
24
,
6
(
)
26
,
4
(
)
28
,
2
(
)
30
,
0
(
)
8
,
7
(
)
9
,
6
(
)
10
,
5
(
)
11
,
4
(
)
12
,
3
(
)
13
,
2
(
)
14
,
1
(
)
15
,
0
(
2
)
0
,
0
(
2
)
0
,
0
(
2
)
0
,
0
(
2
)
0
,
0
(
2
)
0
,
0
(
2
)
0
,
0
(
2
)
0
,
0
(
2
)
0
,
0
(
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
T M

F M = X M x (1/8)T M
Because
cosα
= cos( −α
);
cos( 2πn
+ α)
= cosα;
Define
k
2
= −cosα
= cos( π −α
) = cos( π + α)
π 2π
π 2
= cos ; k4
= cos = cos = ; k6
8 8 4 2
3π
π
= cos = sin
8 8
23

24
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
+
−
−
−
−
+
−
−
+
−
−
+
−
−
−
−
−
+
+
−
+
−
−
−
−
−
−
−
−
−
−
−
+
−
−
−
−
+
+
−
−
−
−
−
−
+
−
−
−
−
−
−
−
+
+
+
=
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
2
4
4
6
6
6
6
4
4
2
2
4
4
4
4
4
4
4
4
6
6
4
4
2
2
2
2
4
4
6
6
6
6
4
4
2
2
2
2
4
4
6
6
4
4
4
4
4
4
4
4
2
2
4
4
6
6
6
6
4
4
2
2
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
T M

25
X M = F M x (8T M -1 ) = FM x L; (L = 8T M -1 )
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
+
−
−
−
+
+
+
−
−
−
−
+
+
−
−
−
−
+
−
−
+
−
−
−
+
−
+
−
+
−
−
−
−
+
−
+
−
−
−
−
−
−
−
−
+
+
−
−
−
+
−
+
+
−
−
−
−
+
−
+
−
−
−
+
−
+
−
−
−
+
−
−
−
+
−
−
+
−
+
−
−
=
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
2
4
2
4
6
2
4
6
4
2
2
4
4
4
4
6
6
4
6
4
2
6
4
2
4
6
6
6
6
4
6
4
2
6
4
2
4
6
6
4
4
4
4
2
2
4
2
4
6
2
4
6
4
2
2
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
L

Where
C
C
2
6
π 2π
= 2cos
; C4
= 2cos
=
8
8
3π
π
= 2cos
= 2sin
8 8
The matrix L can be factored as:
L = B1 x M x A1 x A2 x A3
with:
2
26

27
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
=
1
1
1
0
0
0
0
0
0
0
0
0
1
1
0
0
1
0
1
1
0
0
0
0
0
0
0
0
0
0
1
0
1
0
1
1
0
0
0
0
0
0
0
0
1
1
0
0
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
1
B
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
=
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
2
6
4
6
2
4
C
C
C
C
C
C
M

28
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
−
=
1
0
0
0
0
0
0
0
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
1
0
0
0
0
1
1
0
1
0
1
A
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
=
0
0
0
0
0
0
0
1
0
1
0
1
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
1
0
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
1
0
0
0
0
0
1
0
2
A

29
So the IDFT can be
performed by 5
steps:
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
−
−
=
0
0
0
1
1
0
0
0
0
0
0
1
1
0
0
0
0
0
1
0
0
1
0
0
1
0
0
0
0
0
0
1
0
1
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
0
1
0
0
1
0
0
1
0
0
0
0
0
0
1
3
A

30
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎧
+
=
−
=
−
=
+
=
+
=
−
=
+
=
−
=
=
=
2
1
)
7
(
8
1
)
1
(
8
1
2
1
)
5
(
8
1
)
3
(
8
1
2
)
7
(
8
1
)
1
(
8
1
1
)
3
(
8
1
)
5
(
8
1
)
6
(
8
1
)
2
(
8
1
)
6
(
8
1
)
2
(
8
1
)
4
(
8
1
)
0
(
16
1
:
1
7
6
5
4
3
2
1
0
temp
temp
a
F
F
a
temp
temp
a
F
F
temp
F
F
temp
F
F
a
F
F
a
F
F
a
F
a
F
a
B
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
=
+
−
=
=
+
−
=
=
=
=
=
7
7
2
6
6
4
6
4
5
5
6
6
2
4
4
3
3
4
2
2
1
1
0
0
)
(
)
(
:
a
b
C
a
C
a
b
C
a
b
C
a
C
a
b
a
b
c
a
b
a
b
a
b
M

31
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎧
=
=
=
=
+
=
−
=
−
=
−
=
−
=
7
7
3
6
4
5
4
1
0
3
3
2
2
1
0
1
5
0
7
6
3
3
3
:
1
b
n
b
n
b
n
temp
n
b
b
n
b
b
n
b
b
n
b
temp
n
b
b
temp
A
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
−
=
−
=
−
=
+
=
+
=
=
=
=
0
5
7
6
3
6
2
1
5
6
3
4
2
1
3
4
2
0
1
7
0
:
2
n
n
m
n
n
m
n
n
m
n
n
m
n
n
m
n
m
n
m
n
m
A
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
−
=
−
=
+
=
+
=
−
=
−
=
+
=
+
=
0
4
2
3
1
5
7
6
7
6
1
5
2
3
0
4
)
7
(
)
6
(
)
5
(
)
4
(
)
3
(
)
2
(
)
1
(
)
0
(
:
3
m
m
f
m
m
f
m
m
f
m
m
f
m
m
f
m
m
f
m
m
f
m
m
f
A

32
In part M, a simplification can be made:
So define
)
(
)
(
)
(
)
(
)
(
6
2
6
6
4
6
6
6
6
6
2
6
6
4
2
6
6
4
2
6
2
4
6
4
6
6
4
6
4
6
6
2
4
6
6
2
4
4
C
C
a
a
a
C
C
a
C
a
C
a
C
a
C
a
C
a
b
C
C
a
a
a
C
C
a
C
a
C
a
C
a
C
a
C
a
b
+
+
+
−
=
+
−
+
−
=
+
−
=
−
−
+
−
=
+
−
−
−
=
+
−
=
4
4
)
(
4
and
6
6
4
4
6
4
6
6
2
6
2
temp
Ra
b
temp
Qa
b
a
a
C
temp
C
C
R
C
C
Q
−
=
−
−
=
+
=
+
=
−
=

If you count the operations you will get 29
additions and 5 multiplications.
Because an 8x8 2-D DCT can be computed by
applying 8-point 1-D DCT to each row and
each column of the 2D data. Therefore,
2x8x29 = 464 additions and 2x8x5 = 80
multiplications are required in total.
A more efficient DCT algorithm has been
proposed by Feig and Winograd in IEEE Trans.
on ASSP, pp. 2174-2193, 1992.
33