Image Processing Reference
In-Depth Information
Routines E and F from Fig. 5 have been merged into single routine E (Fig. 6) to short an
amount of pipeline stages and remove unnecessary shift registers.
D
45p24
=C
45
+C
24
A
4
=x
3
-x
4
D
45p64
=C
45
+C
64
B
4
=A
4
+A
5
C
64
=S
6
B
4
C
24
=S
2
B
4
S
3
X
3
= D
7p66
-D
45p24
D
45m64
=C
45
-C
24
D
45m64
=C
45
-C
64
D
7p66
=C
7
+C
66
S
7
X
7
= D
7m26
+D
45m64
A
5
=x
2
-x
5
B
5
=A
5
+A
6
C
45
=S
4
B
5
C
66
=S
6
B
6
C
26
=S
2
B
6
B
6
=A
6
+A
7
D
7p26
=C
7
+C
26
S
1
X
1
= D
7p26
+ D
45p64
A
6
=x
1
-x
6
D
7m26
=C
7
-C
26
A
7
=x
0
-x
7
B
7
= A
7
C
7
= B
7
D
7m66
=C
7
-C
66
S
5
X
5
= D
7m66
-D
45m64
Fig. 7. Optimized, shorter pipeline chain based on the classical approach. The reduction of
the length of the chain at the cost of an additional multiplier.
A classical approach reduces a length of the chain from 6 to 5 stages only, at the cost of one
additional multipliers. An abridgement of the pipeline chain and in a consequence a reduction
of the shift registers needed for synchronization allows saving significant amount of logic
blocks, especially for wide data bus. In order to reduce an approximation errors, the data bus
in the intermediate stages is enlarged.
6. 16-point DCT algorithm
The 16-point DCT algorithm will be implemented according to the classical approach with
an optimization of the number of pipeline stages at the cost of an utilization of embedded
multipliers (Szadkowski, 2009). The 1st and the 2nd pipeline stages utilize the set of variables
(12) and (17) respectively. For N = 16 the fractional angle of the twiddle factor in the 1st step
of minimization equals to
. The same fractional angle corresponds to the 2nd step of
minimization for even indices corresponded to
A
n
.
β
=
π
B
0,1,2,3
=
A
0,1,2,3
+
A
7,6,5,4
B
4,5,6,7
=
A
3,2,1,0
−
A
4,5,6,7
(24)
The scaling procedure used for odd indices of
X
k
with the fractional angles
=
k
32
gives:
β
B
15
=
A
15
B
14,...,8
=
A
15,...,9
+
A
14,...,8
(25)
Coefficients
X
k
for even indices can be expressed by variables (24) and scaling factor (21)
⎡
⎣
⎤
⎦
=
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
X
0
X
8
X
4
X
12
S
4
S
4
S
4
S
4
B
0
B
1
B
2
B
3
1
2
√
2
S
4
−
S
4
−
S
4
S
4
(26)
S
2
S
6
−
S
6
−
S
2
S
6
−
S
2
S
2
−
S
6
⎡
⎣
⎤
⎦
=
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
X
2
X
14
X
6
X
10
S
7
S
5
S
3
S
1
B
4
B
5
B
6
B
7
1
2
√
2
−
−
S
1
S
3
S
5
S
7
(27)
−
−
−
S
5
S
1
S
7
S
3
−
S
3
S
7
S
1
S
5
