EE 461 - Section 4a
Numerically Controlled Oscillators
EE 461
Section 4a - Numerically Controlled Oscillators
1
Numerically Controlled Oscillators
• Principle of Operation:
݂௢
Frequency
Control
Phase
Generator
clk
߮݊
L.U.T.
cos 2ߨ߮ ݊
݂௢ ݊
‫݊ݕ‬
Output
Phase in Cycles
L.U.T. = Look Up Table
= cos 2 EE 461
= cos 2௢ Section 4a - Numerically Controlled Oscillators
2
1
Phase Generator
߮݊
Phase
Accumulator
Units cycles
߮ ݊−1
߮ ݊ = ߮ 0 + ݂݊௢
∑
sample
݂௢
cycles/sample
(Frequency control,
units cycles/sample)
clk
= − 1 + ௢ = 0 + ௢
EE 461
Section 4a - Numerically Controlled Oscillators
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Frequency Control
• set the frequency of the output sinusoid
• It has units cycles/sample
• Can be at most cycles/sample
• It makes no sense for to be negative
1 cycles
0 ≤ ≤ 2 sample
EE 461
Section 4a - Numerically Controlled Oscillators
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2
Frequency Control (cont.)
Suppose 0 = 0 and =
Then = =
Could wrap to 0
when ߮ ݊ = 1 cycle
1
߮݊
cycles
1
2
0
4
8
12
݊→
Note: cos 2 = cos 2 mod1
Therefore could wrap to 0 when it reaches 1
EE 461
Section 4a - Numerically Controlled Oscillators
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Frequency Control (cont.)
For the general case:
Slope = ݂௢
1
߮ ݊ mod 1
in cycles
1
2
0
4
8
݊→
12
To find the slope use 0 and 11 since they are round numbers,
0 = 0 and 11 = 2cycles
= =
EE 461
Section 4a - Numerically Controlled Oscillators
= 6
3
Example 4.1
A phase accumulator has input = cycles per sample
What is the average time (measured in samples) between rollovers?
The question is how many sample times does it take for the
accumulator to get from 0 to 1 cycle?
Answer:
೚
=
య ౙ౯ౙౢ౛౩
భల ౩౗ౣ౦ౢ౛
EE 461
=
Section 4a - Numerically Controlled Oscillators
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Frequency Resolution
• The phase, , starts at zero and increases with so will
always be positive
• This means it can be represented by an unsigned binary
number
• Since the phase is an input to a cosine function, when is
in units cycles, only the fraction portion is used
• For example:
= 1.25cycles
cos 2
radians
× 1.25cycles = cos 2radians + 2 0.25 radians
cycle
cos 2 0.25 radians
EE 461
Section 4a - Numerically Controlled Oscillators
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4
Frequency Resolution (cont.)
• This means the phase accumulator holds a binary number
that is greater than or equal to zero and less than one:
• The contents of the phase accumulator should be treated as an
unsigned fraction
• Suppose the phase accumulator is bits in length and its
contents are treated as an unsigned fraction:
ேೝ ିଵ
− 1 = ௞
௞ୀ଴
Phase
Accumulator
ܾ଴
1
2
binary
point
EE 461
ܾଵ
1
2
௞ାଵ
ேೝ ିଵ
= ௞ 2ିଵ 2ି௞
௞ୀ଴
⋯
ܾଶ
ܾேೝ ିଶ ܾேೝ ିଵ
1
2
1
4
ேೝ
Section 4a - Numerically Controlled Oscillators
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Frequency Resolution (cont.)
߮݊
Phase
Accumulator
߮ ݊−1
ܰ௥
+
ܰ௥
߮݊
∑
+
ܰ௥
݂௢
The carry-out for this
added is ignored
clk
• − 1 and are bit unsigned fractions
• The carry-out of the adder is the integer bit which has value 1
cycle and so can be dropped
EE 461
Section 4a - Numerically Controlled Oscillators
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Frequency Resolution (cont.)
• The allowable frequencies are those that can be represented
with an bit unsigned fraction
• The frequency resolution is
EE 461
૚ ࡺ࢘
૛
cycles / sample
Section 4a - Numerically Controlled Oscillators
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Example 4.2
Consider a phase generator with = 4 bits and =
.0011binary = decimal:
EE 461
Sample
Number
Accumulator
−
Adder Output
1
.0000
.0011 or 3⁄16
2
.0011
.0110 or 6⁄16
3
.0110
.1001 or 9⁄16
4
.1001
.1100 or 12⁄16
5
.1100
.1111 or 15⁄16
6
.1111
.0010 or 2⁄16
7
.0010
.0101 or 5⁄16
Section 4a - Numerically Controlled Oscillators
carry out
dropped
12
6
Sizing the Phase Accumulator
• If the phase accumulator has bits, then the available
frequencies are:
•
ಿೝ
, = 0,1, ⋯ ,
ಿೝ
• The steps in frequency are:
•
ಿೝ • The step size is referred to as the frequency resolution
• There are applications where the NCO must be frequency
agile (frequency of operation is not known in advance)
• The are applications where the frequency of operation is fixed
and known in advance
EE 461
Section 4a - Numerically Controlled Oscillators
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Sizing the Phase Accumulator (cont.)
• The tolerance of a frequency agile NCO is specified by the
maximum error, i.e. desired frequency ± the error:
• ݂௔ ± ∆݂
• Normally the application specification will indicate that an
NCO is required that can be set to any frequency within a
tolerance of ∆
• From this, the engineer needs to calculate the minimum
length for the phase accumulator
• If the frequency step size is less than 2∆, then can be set
to a value that is at most ∆ away from EE 461
Section 4a - Numerically Controlled Oscillators
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7
Sizing the Phase Accumulator (cont.)
Step Size
݇
2ேೝ
ଵ
ଶಿೝ
݂௔
ଵ
ଶ
Less than the step size
݇+1
2ேೝ
݇+2
2ேೝ
Frequencies available from NCO
• If is as shown above, then should be chosen ಿೝ
• Then will be slightly higher than • The difference between and is at most ಿೝ :
௢ − ௔ ≤
EE 461
1 1
2 2ேೝ
Section 4a - Numerically Controlled Oscillators
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Sizing the Phase Accumulator (cont.)
• The specification says that:
௢ − ௔ < ∆
• We know that:
௢ − ௔ ≤
1 1
2 2ேೝ
• Therefore:
1 1
< ∆
2 2ேೝ
1
< 2ேೝ
2∆
1
log
< ௥ log 2
2∆
−log 2∆
௥ >
log 2
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Section 4a - Numerically Controlled Oscillators
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8
Example 4.3
• An analog sinusoid is synthesized by sending the output of an
NCO to a DAC
• The application requires a frequency agile NCO whose
frequency can be set to an accurate of ±0.05Hz
• The sampling rate of the DAC is = 100MHz
• Find the minimum length for the phase accumulator
EE 461
Section 4a - Numerically Controlled Oscillators
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Example 4.3 Solution
∆ =
=
∆
0.05cycles/second
100 × 10 samples/second
= 5 × 10
cycles
sample
−log 2∆
−log(2 × 5 × 10 )
>
=
log 2
log(2)
> 29.9
Therefore the minimum value for is 30
EE 461
Section 4a - Numerically Controlled Oscillators
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9
ܰ௥ for a Fixed Frequency Application
• The frequency for the NCO is known in advance
• Finding the minimum is a two step process:
• First step identical to frequency agile application:
• ௥ >
ି୪୭୥ ଶ∆௙
୪୭୥ ଶ
• Second step:
• Find the ௥ bit unsigned fraction for ௢ is within ±∆ of ௢
• Then look at the least significant bit of ௢
• If it is zero, we do not need it, so we can shorten the accumulator
• We can remove as many of the least significant bits (that are zero)
as possible
EE 461
Section 4a - Numerically Controlled Oscillators
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ܰ௥ for a Fixed Frequency Application (cont.)
• Start by expressing the unsigned fraction by:
• ݂௢ =
௞
,
ଶಿ ೝ
where ݇ is a positive integer
• We want to find such that:
≈ ௔
2ேೝ
≈ ௔ 2ேೝ
• Since is an integer, the best to use is:
݇ = round ݂௔ 2ேೝ
• can be converted to binary and lowest bits checked for zero
• Alternately, if is evenly divisible by 2, it is a binary number
with least significant bit of zero:
• The number of times ݇ can be evenly divided by 2 determines
the number of zeros (bits) that can be removed
EE 461
Section 4a - Numerically Controlled Oscillators
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10
Example 4.4
• An application requires a fixed frequency sinusoid =
13,969,838.57Hz ± 0.01Hz
• The sinusoid is generated by an NCO and a DAC that is driven
at a sampling rate of 100 MHz
• Find the minimum length for the phase accumulator
EE 461
Section 4a - Numerically Controlled Oscillators
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Example 4.4 Solution
First Step:
−log 2∆
=
௥ >
log 2
௥ > 32.2
−log 2
0.01
10଼
log 2
Therefore use ௥ = 33
Second Step:
= round ௔ 2ேೝ = round
= 1,199,999,996
௔ ே
13,969,838.57
2 ೝ = round
× 2ଷଷ
௦
10଼
is evenly divisible by 2, = 599,999,998
௞
ଶ
௞
ସ
is evenly divisible by 4, = 299,999,999
is not evenly divisible by 8
EE 461
Section 4a - Numerically Controlled Oscillators
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Example 4.4 Solution (cont.)
Since is evenly divisible by 4, the binary unsigned fraction will end in two zeroes which are not needed and can be
removed
The minimum = 33 − 2 = 31
Check:
௢ =
=
1,199,999,996
2ଷଷ
1000111100001101000101111111100
2ଷଷ
= .001000111100001101000101111111100
Can use ௢ = .0010001111000011010001011111111 (31 bits)
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Section 4a - Numerically Controlled Oscillators
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