EE 461 - Section 4b
Numerically Controlled Oscillators
EE 461
Section 4b - Numerically Controlled Oscillators
1
Building the Cosine Function
݂௢
Frequency
Control
Phase
Generator
clk
cos 2ߨ߮ ݊
Output
݂௢ ݊
Phase in Cycles
= cos 2 EE 461
‫݊ݕ‬
߮݊
= cos 2௢ Section 4b - Numerically Controlled Oscillators
2
1
Building the Cosine Function (cont.)
• There are several algorithms available for computing the
cosine of an angle
• These algorithms usually require several multipliers which
makes them slow to run on a DSP and expensive to
implement on an FPGA
• A fast yet economical circuit that implements a cosine
function is a ROM
• This is classified as a look-up-table (L.U.T.) algorithm
• In an FPGA implementation, the ௥ bit output of the phase
generator is connected to ௥ address lines of a ROM:
• The ROM is programmed so that its output is cos 2 each corresponding address EE 461
for
Section 4b - Numerically Controlled Oscillators
3
Building the Cosine Function (cont.)
• In a DSP chip implementation, the output of the phase
generator is used as an index to a buffer that has 2ேೝ words
• The buffer is initialized so that = cos 2 BUFF + 2ேೝ
FPGA
ROM
cos 2 index = BUFF
+ Computer
cos 2 Memory
(2ேೝ words
in length)
BUFF
is treated as an
௥ bit unsigned integer
EE 461
Section 4b - Numerically Controlled Oscillators
4
2
Word Length Quantization
• The size of the ROM measured in bits is the product of the
number or words and the number of bits in each word
• The number of words is 2ேೝ
• Let the number of bits in a word be ஽ bits/word:
ROM size = 2ேೝ words × ஽ bits/word = ஽ 2ேೝ bit
• ஽ should be as small as possible to reduce the size of the
ROM
• Making ஽ small causes severe quantization problems
• To make a decision on the word length, ஽ , the engineer
needs an equation that relates SNR to ஽
EE 461
Section 4b - Numerically Controlled Oscillators
5
Word Length Quantization (cont.)
• The model used for the quantization noise analysis is:
signed fraction
ROM
with infinite
word length
௥ bit number
connected to ௥
address lines
EE 461
∞
cos 2 Round
to ஽ bits
஽
cos 2 + ஽ Noise due to
rounding to ஽ bits
Section 4b - Numerically Controlled Oscillators
6
3
Word Length Quantization (cont.)
• Since −1 ≤ cos 2 ≤ 1, cos can not technically be
represented with an ஽ bit signed fraction as the largest
number that is possible is:
1
2
0.111 … 1 = 1 −
ேವ ିଵ
• There are two choices:
• Add an integer bit
• Store the value of 1 − 2
ଵ ேವ
ଶ
cos 2 • Makes the ROM output cos 2 in the ROM
where = 1 − 2
ଵ ேವ
ଶ
• The second choice is more economical so will be used here
EE 461
Section 4b - Numerically Controlled Oscillators
7
Word Length Quantization (cont.)
• The quantization noise due to rounding to ஽ − 1 fraction
bits is:
஽ ଶ =
ଶ
=
12
1
2
ேವ ିଵ
12
ଶ
=
2ିଶேವ
3
• The signal power for a sinusoid with amplitude is
஺మ
ଶ
• Therefore the SNR is:
஽ =
ଶ
2
2ିଶேವ
=
3 ଶ ଶே
2 ವ
2
3
EE 461
Section 4b - Numerically Controlled Oscillators
8
4
Word Length Quantization (cont.)
• For ஽ of any reasonable length, = 1 − 2
ଵ ேವ
ଶ
≈1
• Therefore the approximate SNR is:
஽ ≈
3 ଶே
2 ವ
2
EE 461
Section 4b - Numerically Controlled Oscillators
9
Further Reducing the Size of the ROM
• The size of the phase accumulator is determined by the
accuracy needed for the frequency
• This may be a number like 32 bits
• The L.U.T. would then need a ROM with 32 address bits (4
billion words), which is far too large to be practical
• To make the ROM a practical size, must be truncated
߮݊
Phase
Accumulator
߮ ݊−1
ܰ௥
+
ܰ௥
߮݊
∑
+
ܰ௥
݂௢
ܰ஺
்߮ ݊
ROM
with ஺
address
lines
cos 2 ் clk
EE 461
Section 4b - Numerically Controlled Oscillators
10
5
Further Reducing the Size of the ROM (cont.)
• is truncated to get ் • The most significant bits of ௥ are retained:
் = + ௘ Therefore:
Truncation noise
= cos 2 ் ≈ 1:
= cos 2 + 2௘ Phase noise
EE 461
Section 4b - Numerically Controlled Oscillators
11
Further Reducing the Size of the ROM (cont.)
• Using the identitycos + = cos cos − sin sin :
= cos 2 cos 2௘ − sin 2 sin 2௘ • The phase noise will certainly be small, so:
cos 2௘ ≈1
sin 2௘ ≈ 2௘ • Which leads to:
= cos 2 − 2௘ sin 2 • ௘ has a DC component and an AC component
• Let the DC component be and the AC component be ௘ :
௘ = + ௘ EE 461
Section 4b - Numerically Controlled Oscillators
12
6
Further Reducing the Size of the ROM (cont.)
= cos 2 •
•
− 2πsin 2 The DC component of ௘ causes a phase shift in the noise free
component of the output
Using the relation cos + sin = cos − , where =
ଶ + ଶ , and = tanିଵ
=
•
− 2௘ sin 2 ௕
௔
:
1ଶ + 2π ଶ cos 2 + tanିଵ
Since is sure to be small,
1ଶ + 2π
2π
1
ଶ
− 2௘ sin 2 = 1 and tanିଵ 2 = 2π:
= cos 2 + 2 − 2௘ sin 2 Noise free output
Additive noise
• The phase shift of 2 in the NCO output is not a problem
EE 461
Section 4b - Numerically Controlled Oscillators
13
Further Reducing the Size of the ROM (cont.)
= cos 2 + 2 + ஺ ஺ = 2௘ sin 2 Amplitude noise due to
truncating ஺ ଶ = 4 ଶ ௘ ଶ sinଶ 2 ஺ ଶ = 2 ଶ ௘ ଶ • ௘ is the AC component of the noise generated by truncating • If ௥ ≫ ௔ , then ௘ ଶ approaches
௘ ଶ =
2ିேೌ
12
஺ ଶ = 2 ଶ
EE 461
ଶ
=
௦మ
with = 2ିேಲ :
ଵଶ
2ିଶேೌ
12
2ିଶேೌ
ଶ ିଶே
ಲ
=
2
12
6
Section 4b - Numerically Controlled Oscillators
14
7
Further Reducing the Size of the ROM (cont.)
• The SNR due to quantizing is:
1
ୱ୧୥୬ୟ୪
2
=
஺ =
୬୭୧ୱୣ ଶ ିଶேಲ
2
6
஺ =
1
ଶ ିଶேಲ
2
3
=
3 × 2ଶேಲ
ଶ
• When all is considered:
= cos 2 + 2 + ஺ + ஽ where ஺ and ஽ are the noises resulting from quantizing and
the output of the ROM respectively
EE 461
Section 4b - Numerically Controlled Oscillators
15
SNR
• The total SNR is given by:
=
=
EE 461
ୱ୧୥୬ୟ୪
஺ ଶ + ஽ ଶ =
1
2
ଶ
6
2ିଶேಲ +
2ିଶேವ
3
1
ଶ ିଶேಲ 2 ିଶேವ
2
+ 2
3
3
Section 4b - Numerically Controlled Oscillators
16
8
SNR (cont.)
• Taking each of the terms (inequalities) separately:
஺ ≤
1
ଶ
3
஽ ≤
=
2ିଶேಲ
1
2
× 2ିଶேವ
3
3 × 2ଶேಲ
ଶ
=
3
× 2ଶேವ
2
• Taking 10log SNR yields:
ௗ஻ ≤ −5.17dB + 6.02
ௗ஻ ≤ 1.76dB + 6.02
EE 461
dB
bit ஺
dB
bit ஽
Section 4b - Numerically Controlled Oscillators
17
Example 4.5
• An NCO with an SNR greater than 70 dB is required
What are the dimensions of the ROM required for the lookup-table?
• Find minimum word size for ஽ :
ௗ஻ ≤ 1.76dB + 6.02
஽ ≥
dB
bit ஽
70 − 1.76
= 11.34bits
6.02
஽ ≥ 12bits
EE 461
Section 4b - Numerically Controlled Oscillators
18
9
Example 4.5 (cont.)
• Find minimum word size for ஺ :
ௗ஻ ≤ −5.17dB + 6.02
஺ ≥
dB
bit ஺
70 + 5.17
= 12.49bits
6.02
஺ ≥ 13bits
• Check overall SNR using ஽ = 12bits and ஺ = 13bits:
=
1
=
ଶ
1
ଶ
2
2
2ିଶேಲ + 2ିଶேವ
2ିଶ ଵଷ + 2ିଶ ଵଶ
3
3
3
1
=
= 11,266,336 → 70.5dB
4.902 × 10ି଼ + 3.974 × 10ି଼
3
EE 461
Section 4b - Numerically Controlled Oscillators
19
Example 4.6
• A system needs a frequency agile NCO:
• Operating at a sampling rate of ௦ = 10଼ samples/second
• The frequency must be settable within a tolerance of ±0.5Hz
• The SNR of the output sinusoid must be at least 55dB
Find the size of the smallest ROM and the minimum length for
the phase accumulator to achieve the required performance
• First find ௥ :
−log 2∆
௥ >
=
log 2
−log 2
0.5
10଼
log 2
= 26.6
• Therefore minimum ௥ is 27
• As a frequency agile NCO is specified, the length of ௥ cannot
be reduced
EE 461
Section 4b - Numerically Controlled Oscillators
20
10
Example 4.6 (cont.)
• Next find minimum values for ஽ and ஺ :
ௗ஻ ≤ 1.76dB + 6.02
஽ ≥
dB
bit ஽
55 − 1.76
= 8.84bits
6.02
஽ ≥ 9bits
ௗ஻ ≤ −5.17dB + 6.02
஺ ≥
dB
bit ஺
55 + 5.17
= 9.99bits
6.02
஺ ≥ 10bits
EE 461
Section 4b - Numerically Controlled Oscillators
21
Example 4.6 (cont.)
• Finally calculate SNR and check if it meets requirements:
1
1
=
ଶ ିଶேಲ 2 ିଶேವ ଶ ିଶ ଵ଴
2
2
+ 2
2
+ 2ିଶ ଽ
3
3
3
3
1
=
= 176,056 → 52.5dB
3.137 × 10ି଺ + 2.543 × 10ି଺
=
SNR is too low
• Next try ஽ = 9 + 1 = 10 and ஺ = 10:
=
1
ଶ
3
2ିଶ
ଵ଴
2
+ 2ିଶ
3
= 54.3dB
ଵ଴
SNR is still too low
EE 461
Section 4b - Numerically Controlled Oscillators
22
11
Example 4.6 (cont.)
• Next try ஽ = 9 and ஺ = 10 + 1 = 11:
=
1
ଶ
3
2ିଶ
ଵଵ
2
+ 2ିଶ
3
= 54.8dB
ଽ
SNR is also too low
• Next try ஽ = 9 + 1 = 10 and ஺ = 10 + 1 = 11:
=
1
ଶ
3
2ିଶ
ଵଵ
2
+ 2ିଶ
3
= 58.5dB
ଵ଴
SNR meets the specification
EE 461
Section 4b - Numerically Controlled Oscillators
23
Example 4.6 (cont.)
• Therefore:
• A ROM with ஺ = 11 (11 address lines) and ஽ = 10 (word
length of 10 bits) will work
• Size of ROM (in bits) = 2ேಲ words × ஽ bits⁄word
= 2ଵଵ words × 10 bits⁄word
= 20,480bits
• ஽ = 10 and ஺ = 11 are sufficient but may not yield the
smallest ROM
EE 461
Section 4b - Numerically Controlled Oscillators
24
12
Example 4.6 (cont.)
• From the inequality for noise due to truncating :
ௗ஻ ≤ −5.17dB + 6.02
dB
bit ஺
with equality holding when ஽ = ∞. With ஺ = 10 and ஽ =
∞, SNR = 55.03dB
• Now find the minimum value for ஽ to make the SNR 55 dB
with ஺ = 10:
Trials for ஽
SNR
9 + 1 = 10 bits
54.3 dB
9 + 2 = 11 bits
54.82 dB
9 + 3 = 12 bits
54.98 dB
9 + 4 = 13 bits
55.02 dB
EE 461
Section 4b - Numerically Controlled Oscillators
25
Example 4.6 (cont.)
• With ஺ = 10 and ஽ = 13 (which also meets SNR spec):
• Size of ROM (in bits) = 2ேಲ words × ஽ bits⁄word
= 2ଵ଴ words × 13 bits⁄word
= 13,312bits
• Therefore answer is:
஺ = 10, ஽ = 13, and ௥ = 27
• Comments:
• In practice, there is usually an upper limit on ஽ , for example, it
could be the size of the DAC
• The ROM can be shrunk to ଵ⁄ସ of its size by storing the
coefficients for ଵ⁄ସ cycle and using a logic circuit to recompute
the address of the OM
EE 461
Section 4b - Numerically Controlled Oscillators
26
13