EE 461 - Section 4c
Numerically Controlled Oscillators
EE 461
Section 4c - Numerically Controlled Oscillators
1
Amplitude Noise for ܰ௥ = ܰ஺ + 1
• In the notes, ஺ ଶ (஺ being the amplitude noise due to
truncating ) was calculated from 2 ଶ ௘ ଶ :
௦మ
௘ ଶ = ଵଶ =
஺ ଶ =
ଶషమಿಲ
ଵଶ
(under the assumption ܰ௥ ≫ ܰ஺ )
ଶ ିଶே
2 ಲ
6
• Find the exact values for ௘ ଶ and ஺ ଶ in terms of ஺
when ௥ = ஺ + 1
EE 461
Section 4c - Numerically Controlled Oscillators
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1
Amplitude Noise for ܰ௥ = ܰ஺ + 1 (cont.)
• is an ஺ + 1 bit unsigned fraction
• ் is an ஺ bit unsigned fraction
• ௘ = − ் :
• Is zero if the LSB (Least Significant Bit) of is zero
• Is 2ିேೝ =
ଶషಿಲ
ଶ
if the LSB of is one
௥ bit has value
2ିேೝ or 2ି ேಲାଵ
஺ bits
߮݊
଴ ଵ
EE 461
⋯
Section 4c - Numerically Controlled Oscillators
3
Amplitude Noise for ܰ௥ = ܰ஺ + 1 (cont.)
• The probability density function for ௘ :
1
2
−2ିேಲ −2ିேೝ
0
Probability
௘
Therefore:
1
1
௘ = ௘ = −2ିேೝ × + 0 ×
2
2
=
௘ = ௘ =
EE 461
−2ିேೝ −2ି ேಲ ାଵ
=
2
2
2ିேಲ
4
Section 4c - Numerically Controlled Oscillators
4
2
Amplitude Noise for ܰ௥ = ܰ஺ + 1 (cont.)
௘ ଶ = −2ିேೝ
௘ ଶ =
ଶ
1
1 2ିଶேೝ
× + 0ଶ × =
2
2
2
2ିଶ ேಲ ାଵ
2ିଶேಲ
=
2
8
௘ ଶ = ௘ ଶ − ௘ ௘
ଶ
2ିேಲ
2ିଶேಲ
=
−
8
4
௘ ଶ =
ଶ
ଶ
2ିଶேಲ
16
EE 461
Section 4c - Numerically Controlled Oscillators
5
Amplitude Noise for ܰ௥ = ܰ஺ + 1 (cont.)
஺ ଶ = 2 ଶ ௘ ଶ ஺ ଶ =
ܴܵܰ =
ܴܵܰ =
EE 461
ଶ ିଶே
2 ಲ
8
ܲୱ୧୥୬ୟ୪
‫ݍ‬஺ ଶ ݊ + ‫ݍ‬஽ ଶ ݊
=
1ൗ
2
ߨଶ
8
2ିଶேಲ +
2ିଶேವ
3
1
ߨ ଶ ିଶேಲ 2 ିଶேವ
2
+ 2
3
4
Section 4c - Numerically Controlled Oscillators
6
3
߮ ݊−1
NCO Input
∑
∆݂ ݊
Phase
Accumulator
FM Modulators
߮݊
∑
ROM
‫݊ݕ‬
݂௢ + ∆݂ ݊
݂௢
Instantaneous Frequency
= − 1 + ௢ + ∆ EE 461
Section 4c - Numerically Controlled Oscillators
7
FM Modulators (cont.)
Let 0 = ௢
= 0 + ௢ + ∆ 1 + ௢ + ∆ 2 + ⋯ + ௢ + ∆ ௡
= 0 + ௢ + ∆ ௞ୀଵ
௡
= 0 + ௢ + ∆ ௞ୀଵ
∆ = cos 2௢ + 2௢ + 2∆ Carrier Phase
EE 461
Phase Deviation
Section 4c - Numerically Controlled Oscillators
8
4
FM Modulators (cont.)
• The signal component is ∆ • It is always much smaller in magnitude than ௢ :
∆݂ ݊ ≪ ݂௢
• Usually ∆ is a signed fraction but it could be an unsigned
fraction
• In practical systems, ∆ is always small and ௢ + ∆ is
ଵ
always less than ଶ
• Since ௢ ≫ ∆ , ௢ + ∆ is always greater than zero
• This means that ௢ + ∆ can be represented as an unsigned
fraction
EE 461
Section 4c - Numerically Controlled Oscillators
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FM Modulators (cont.)
signed fraction
signed fraction
∆݂ ݊
݂௢ + ∆݂ ݊
∑
ܰ௥ + 1
Note:
The adder has ௥ + 1
inputs
ܰ௥ + 1 bit signed fraction
append sign
0
bit
ܰ௥
݂௢ unsigned fraction
• Since ௢ + ∆ is always positive, there is no reason to
compute the sign bit
• So an ௥ bit adder can be used
EE 461
Section 4c - Numerically Controlled Oscillators
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5
FM Modulators (cont.)
Sign bit is discarded
unsigned fraction
∆݂ ݊
݂௢ + ∆݂ ݊
∑
ܰ௥
signed fraction
ܰ௥
ܰ௥
signed fraction with
the sign bit dropped
݂௢ unsigned fraction
• The adder is an ܰ௥ bit input adder with the carry out ignored
• The carry out can be safely ignored since the system is always
ଵ
designed so that ௢ + ∆ is always less than
ଶ
EE 461
Section 4c - Numerically Controlled Oscillators
11
Scaling the Input
• Two of the key parameters of a frequency modulation (FM)
modulator is the centre frequency ௢ and the peak deviation
from the centre frequency, which is denoted ∆୮ୣୟ୩
• For commercial FM:
•
•
•
•
௢ = 92.1 to 107.9 MHz in 200 kHz increments
Carrier Stability ±2000 Hz
∆୮ୣୟ୩ = 75kHz
Bandwidth of ∆ = 15kHz
• TV Aural (FM):
• ௢ = 4.5 MHz
• ∆୮ୣୟ୩ = 25kHz
• Bandwidth of ∆ = 15kHz
EE 461
Section 4c - Numerically Controlled Oscillators
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6
Scaling the Input (cont.)
Audio
Input
Preemphasis
Filter
ADC
m݊
Message signal
to be transmitted
signed fractions
m݊
×
g×m ݊
ܰ௥ + 1
∆݂ ݊
truncated to have
ܰ௥ fraction bits
unsigned
݃ fraction
• The gain is chosen to make the peak value of ∆݂ ݊ equal to
∆݂୮ୣୟ୩
• If ݉ ݊ is an AC signal with peak value ݉୮ୣୟ୩ , then the gain is given
by:
݃=
∆݂୮ୣୟ୩
݉୮ୣୟ୩
EE 461
Section 4c - Numerically Controlled Oscillators
13
Scaling the Input (cont.)
• Notice that the product × is truncated to ௥ bits
• This introduces both AC and DC noise:
• The DC component will simply modify ݂௢ :
݂௢ becomes ݂௢ + ߤ = ݂௢ −
ଶషమಿೝ
ଶ
• The AC component has power
ଶషమಿೝ
ଵଶ
• This means ∆ has an AC noise component and therefore an
SNR
• The SNR due to the quantization of × is:
ܴܵܰ௥ =
EE 461
∆݂ ଶ ݊
= 12 2ଶேೝ ∆݂ ଶ ݊
2ିଶேೝ
12
Section 4c - Numerically Controlled Oscillators
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7
Scaling the Input (cont.)
• Depending on the signal being transmitted, the ratio of ௣௘௔௞
to rms value of will vary
• If is a sinusoid, then:
௣௘௔௞ ଶ
ଶ =
• The ratio of
1
2
௠೛೐ೌೖ
is sometimes called headroom
௠మ ௡
• Since
∆௙೛೐ೌೖ
=
௠೛೐ೌೖ
∆௙మ ௡
2ଶேೝ =
௠మ ௡
, the equation for ௥ can be written:
݉௣௘௔௞
ܴܵܰ௥
12∆݂௣௘௔௞
EE 461
ଶ
ଶ
݉ଶ ݊
This ratio will depend on
the application and will
be given
Section 4c - Numerically Controlled Oscillators
15
Scaling the Input (cont.)
2ேೝ =
ܴܵܰ௥ 1
12 ∆݂௣௘௔௞
log
ܰ௥ ≥
EE 461
݉௣௘௔௞
݉ଶ ݊
ܴܵܰ௥ 1
12 ∆݂௣௘௔௞
This will be given
݉௣௘௔௞
݉ଶ ݊
log 2
Section 4c - Numerically Controlled Oscillators
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8
Scaling the Input (cont.)
• The centre frequency of the NCO will be ௢ + ∆ ∆݂ ݊
∑
݂௢ + ∆݂ ݊
݂௢
௢ =
DC component of this will
be the centre frequency
௢
௢
2ିேೝ
௢ 2ିேೝ
− ∆ = − −
= +
2
௦
௦
௦
2
where ௢ is the desired centre frequency in Hz
EE 461
Section 4c - Numerically Controlled Oscillators
17
Example 4.7
• Design an NCO for a commercial FM modulator
• Assume ஽ = ∞ and ஺ = ∞
• System Parameters:
• ௢ = 96.3MHz
• ௣௘௔௞ = 1.0
•
௠೛೐ೌೖ
=4
௠మ ௡
• ௥ > 60dB
• ௌ = 200Msamples/second
EE 461
Section 4c - Numerically Controlled Oscillators
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9
Example 4.7 Solution
• First find ∆௣௘௔௞ :
∆௣௘௔௞
0.075
=
௦
200Msample/second
= 0.375 × 10ିଷ cycles/sample
∆௣௘௔௞ =
• Next find :
=
∆௣௘௔௞
= 0.375 × 10ିଷ
௣௘௔௞
• Next find ௥ :
•
In this case, there are two inequalities that must be satisfied:
–
One for the accuracy of ௢
–
The second for the SNR due to quantizing × EE 461
Section 4c - Numerically Controlled Oscillators
19
Example 4.7 Solution (cont.)
• The accuracy of ௢ must be ±2000Hz (carrier stability):
−log 2∆݂
ܰ௥ >
=
log 2
> 15.6
Therefore ܰ௥ ≥ 16
−log 2
2000
‫ܨ‬௦
=
log 2
−log 2 × 10ିହ
log 2
• Also:
log
௥ ≥
௥ 1
12 ∆௣௘௔௞
log 2
௣௘௔௞
ଶ =
10଺
1
12 0.375 ×
10ିଷ
4
≥ 21.55
Therefore ܰ௥ = 22
EE 461
Section 4c - Numerically Controlled Oscillators
20
10
Example 4.7 Solution (cont.)
• Finally find ௢ :
௢ =
௢ 2ିேೝ 96.3 2ିଶଶ
+
=
+
௦
2
200
2
We want to find the integer such that ଶಿೝ ≅ ݂௢
௞
= round ௢ 2ேೝ
= round
96.3 2ିଶଶ ଶଶ
+
2
200
2
= 2019558
2019558 111101101000011100110
௢ =
=
2ଶଶ
2ଶଶ
= .0111101101000011100110
EE 461
Section 4c - Numerically Controlled Oscillators
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11