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AD9546/PCBZ Datasheet(PDF) 173 Page - Analog Devices

No. de pieza AD9546/PCBZ
Descripción Electrónicos  Dual DPLL Digitized Clock Synchronizer
PDF  205 Pages
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Fabricante Electrónico  AD [Analog Devices]
Página de inicio  http://www.analog.com
Logo AD - Analog Devices

AD9546/PCBZ Datasheet(HTML) 173 Page - Analog Devices

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Data Sheet
AD9546
Rev. 0 | Page 173 of 205
Using C1 as an example, convert C1 = −2.8927765 × 10−6 to its
corresponding register values. First, check that C1 is greater
than the quantization limit.
1
log |
|
1
log 2
C
E

=
+


=
1
2
log
|
10
8927765
.
2
|
log
6
+
×
−
−
= −18
This value is greater than the quantization limit of −127.
Therefore,
C1_E = E = −18 = 0xEE (hexadecimal)
C1_S = round(C1 × 215 − E)
= round(−2.8927765 × 10−6 × 215 − (−18))
= −24,849
= 0x9EEF (hexadecimal)
Using C2 as an example, convert C2 = –1.9110192 × 10−7 to its
corresponding register values. First, check that C2 is greater
than the quantization limit.
2
log |
|
1
log 2
C
E

=
+


=
1
2
log
|
10
9110192
.
1
|
log
7
+
×
−
−
= −22
This value is greater than the quantization limit of −127.
Therefore,
C2_E = E = −22 = 0xEA (hexadecimal)
C2_S = round(C2 × 215 − E)
= round(−1.9110192 × 10−7 × 215 − (−22))
= −26,265
= 0x9967 (hexadecimal)
Using C3 as an example, convert C3 = 2.2493907 × 10−9 to its
corresponding register values. First, check that C3 is greater
than the quantization limit.
E =
1
2
log
|
|
log
3
+
C
=
1
2
log
|
10
2493907
.
2
|
log
9
+
× −
= −28
This value is greater than the quantization limit of −127.
Therefore,
C3_E = E = −28 = 0xE4 (hexadecimal)
C3_S = round(C3 × 215 – E)
= round(2.2493907 × 10−9 × 215− (−28))
= 19,786
= 0x4D4A (hexadecimal)
Using C4 as an example, convert C4 = 2.7943570 × 10−11 to its
corresponding register values. First, check that C4 is greater
than the quantization limit.
4
log |
|
1
log 2
C
E

=
+


=
1
2
log
|
10
7943570
.
2
|
log
11
+
×
−
= −35
This value is greater than the quantization limit of −127.
Therefore,
C4_E = E = −35 = 0xDD (hexadecimal)
C4_S = round(C4 × 215 − E)
= round(2.7943570 × 10−11 × 215 − (−35))
= 31,462
= 0x7AE6 (hexadecimal)
Using C5 as an example, convert C5 = −2.4817310 × 10−13 to its
corresponding register values. First, check that C5 is greater
than the quantization limit.
5
log |
|
1
log 2
C
E

=
+


13
log | 2.4817310 10
|
1
log 2
−

−×
=
+


= −41
This value is greater than the quantization limit of −127.
Therefore,
C5_E = E = −41 = 0xD7 (hexadecimal)
C5_S = round(C5 × 215 − E)
= round(−2.4817310 × 10−13 × 215 − (−41))
= −17,883
= 0xBA25 (hexadecimal)



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