ak4373 ETC-unknow, ak4373 Datasheet - Page 49

• Manufacturer Part Number: ak4373
• Description: Low Power Stereo Dac With Hp/spk-amp
• Manufacturer: ETC-unknow

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This block can be used as Equalizer or Notch Filter. 5-band Equalizer (EQ1, EQ2, EQ3, EQ4 and EQ5) is ON/OFF
independently by EQ1, EQ2, EQ3, EQ4 and EQ5 bits. When the Equalizer is OFF, the audio data passes this block by
0dB gain. E1A15-0, E1B15-0 and E1C15-0 bits set the coefficient of EQ1. E2A15-0, E2B15-0 and E2C15-0 bits set the
coefficient of EQ2. E3A15-0, E3B15-0 and E3C15-0 bits set the coefficient of EQ3. E4A15-0, E4B15-0 and E4C15-0
bits set the coefficient of EQ4. E5A15-0, E5B15-0 and E5C15-0 bits set the coefficient of EQ5. The EQx (x=1∼5)
coefficient should be set when EQx bit = “0” or PMDAC bit = “0”.
Note 42. [Translation the filter coefficient calculated by the equations above from real number to binary code (2’s
MS0991-E-00
Five Programmable Biquads
A
Transfer function
H(z) = 1 + h
h
The center frequency should be set as below.
fs: Sampling frequency
fo
fb
K
Register setting
X = (Real number of filter coefficient calculated by the equations above) x 2
X should be rounded to integer, and then should be translated to binary code (2’s complement).
MSB of each filter coefficient setting register is sign bit.
(n = 1, 2, 3, 4, 5)
n
n
(z) = A
1
= K
1
1
(n = 1, 2, 3, 4, 5)
EQ1: E1A[15:0] bits =A
EQ2: E2A[15:0] bits =A
EQ3: E3A[15:0] bits =A
EQ4: E4A[15:0] bits =A
EQ5: E5A[15:0] bits =A
(MSB=E1A15, E1B15, E1C15, E2A15, E2B15, E2C15, E3A15, E3B15, E3C15, E4A15, E4B15, E4C15,
fo
E5A15, E5B15, E5C15; LSB= E1A0, E1B0, E1C0, E2A0, E2B0, E2C0, E3A0, E3B0, E3C0, E4A0, E4B0,
E4C0, E5A0, E5B0, E5C0)
~ K
~ fo
~ fb
complement)]
n
n
/ fs < 0.497
5
x
5
5
n
: Center frequency
: Band width where the gain is 3dB different from center frequency
: Gain (−1 ≤ K
1
(z) + h
1− B
1 + tan (πfb
tan (πfb
1 − z
(Note
n
z
2
(z) + h
−1
− C
−2
n
42)
n
/fs)
n
3
n
z
≤ 3)
(z) + h
/fs)
−2
1
2
3
4
5
, E1B[15:0] bits =B
, E2B[15:0] bits =B
, E3B[15:0] bits =B
, E4B[15:0] bits =B
, E5B[15:0] bits =B
4
(z) + h
,
B
n
5
= cos(2π fo
(z)
1
2
3
4
5
, E1C[15:0] bits =C
, E2C[15:0] bits =C
, E3C[15:0] bits =C
, E4C[15:0] bits =C
, E5C[15:0] bits =C
n
/fs) x
- 49 -
1 + tan (πfb
2
1
2
3
4
5
n
/fs)
13
,
C
n
=
1 + tan (πfb
1 − tan (πfb
n
n
[AK4373]
/fs)
/fs)
2008/09