TMC2249A

PRODUCT SPECIFICATION

REV. 1.0.2 7/6/00

11

Once all of the products of the desired taps have been

summed, the result is available at the output. The user then

"pushes" a new time-data sample on to the appropriate even

or odd data register "stack" and reiterates the summation.

Note that the coefficient bank "pointers", the BDEL and

DDEL delay words, are alternately incremented and decre-

mented on successive filter passes to maintain alignment

between the incoming data samples and their respective

coefficients.

The effective filter speed is calculated by dividing the clock

rate by one-half the number of taps implemented.

Alternatively, non-symmetric FIR filters can be implemented

using the TMC2249A in a similar fashion. Here, a shift reg-

ister is used to delay the incoming data fed to the A input by

an amount equal to one-half the length of the filter (the

length of the A delay register).

As shown in Figure 5, the data is then sent to the C input,

thus "stacking" the A and C delay registers to create a single

N-tap FIR filter. The incremented delay words (ADEL-

DDEL) for all four inputs are identical. Again, the filter

throughput is equal to the clock speed divided by one-half

the number of taps implemented.

Figure 5. Non-Symmetric 32-Tap FIR Filtering Using the

TMC2249A

TMC2011A

16-Stage Shift Register

Filter Output

AB

C

D

x(m+0)

x(m)

h(0)

x(m+16)

h(16)

x(m+15)

h(15)

x(m+31)

TMC2249A

h(31)

Table 4. FIR Filtering – Operation Sequence

Notes:

1. If only the 16 MSBs of the result are used, the user may leave RND HIGH and SWAP low. If the 16 LSBs or all 24 bits of the

result are used, then RND should be set low.

2.

Cycle

Push

AB

Push

C

D

ADEL

CDEL

BDEL

DDEL

ACC

ENA

ENB

ENC

END

Convolutional Sum

Resultant

Output

1

–

–

–

–

0

0

0

0

L

H

H

H

H

x(31)

•h(0)+x(30)•h(1)

See Note 2

2

–

–

–

–

1

1

1

1

H

H

H

H

H

+x(29)

•h(2)+x(28)•h(3)

3

–

–

–

–

2

2

2

2

H

H

H

H

H

+x(27)

•h(4)+x(26)•h(5)

4

–

–

–

–

3

3

3

3

H

H

H

H

H

+x(25)

•h(6)+x(24)•h(7)

5

–

–

–

–

4

4

4

4

H

H

H

H

H

+x(23)

•h(8)+x(22)•h(9)

6

–

–

–

–

5

5

5

5

H

H

H

H

H

+x(21)

•h(10)+x(20)•h(11)

7

–

–

–

–

6

6

6

6

H

H

H

H

H

+x(19)

•h(12)+x(18)•h(13)

8

–

–

–

–

7

7

7

7

H

H

H

H

H

+x(17)

•h(14)+x(16)•h(15)

9

–

–

–

–

8

8

8

8

H

H

H

H

H

+x(15)

•h(15)+x(14)•h(14)

10

–

–

–

–

9

9

9

9

H

H

H

H

H

+x(13)

•h(13)+x(12)•h(12)

11

–

–

–

–

A

A

A

A

H

H

H

H

H

+x(11)

•h(11)+x(10)•h(10)

12

–

–

–

–

B

B

B

B

H

H

H

H

H

+x(9)

•h(9)+x(8)•h(8)

13

–

–

–

–

C

C

C

C

H

H

H

H

H

+x(7)

•h(7)+x(6)•h(6)

14

–

–

–

–

D

D

D

D

H

H

H

H

H

+x(5)

•h(5)+x(4)•h(4)

15

–

–

–

–

E

E

E

E

H

H

H

H

H

+x(3)

•h(3)+x(2)•h(2)

16

–

–

x(32)

–

F

F

F

F

H

H

H

L

H

+x(1)

•h(1)+x(0)•h(0)

17

–

–

–

–

0

0

F

F

H

H

H

H

H

+x(31)

•h(1)+x(32)•h(0)

18

–

–

–

–

1

1

E

E

H

H

H

H

H

+x(29)

•h(3)+x(30)•h(2)

19

–

–

–

–

2

2

D

D

H

H

H

H

H

+x(27)

•h(5)+x(28)•h(4)

20

–

–

–

–

3

3

C

C

H

H

H

H

H

+x(25)

•h(7)+x(26)•h(6)

21

–

–

–

–

4

4

B

B

H

H

H

H

H

+x(23)

•h(9)+x(24)•h(8)

•

•

sx k

()hk

() xk 16

+

()hk

()

+

()

K0

=

15

∑

=