Effect of sampling jitter on signal tracking in a direct sampling dual ...
Input Signal and Sampling Frequencies Requirements for ...
Transcript of Input Signal and Sampling Frequencies Requirements for ...
Input Signal and Sampling Frequencies
Requirements for Efficient ADC Testing with
Histogram Method
Yujie Zhao, A. Kuwana, S. Yamamoto, Y. Sasaki
H. Kobayashi, S. Katayama, J. Wei
T. Nakatani, K. Hatayama
K. Sato, T. Ishida,
T. Okamoto, T. Ichikawa
Division of Electronics and Informatics
Gunma University
ROHM Semiconductor
June 28,2021
The 36th International Technical Conference on Circuits/Systems, Computers and Communications
2/27
Outline
• Objective
• ADC Test with Histogram Method
• Input Sine Wave and Sampling Frequencies
Relationship in ADC Histogram Test Method
➢Sine Wave Histogram and Waveform Missing
➢Golden Ratio Sampling
➢Metallic Ratio Sampling
➢Prime Number Ratio Sampling
• Conclusion
3/27
Outline
• Objective
• ADC Test with Histogram Method
• Input Sine Wave and Sampling Frequency
Relationship in ADC Histogram Test Method
➢Sine Wave Histogram and Waveform Missing
➢Golden Ratio Sampling
➢Metallic Ratio Sampling
➢Prime Number Ratio Sampling
• Conclusion
4/27
Background
IoT era is coming !
High quality & Low cost ADC test is required
ADC is a key component
◆Analog signal ◆Digital signalSampling
A/D
Conversion
01010101110
10101010111
11101101010
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Research Objective & Approach
SAR ADC linearity test takes a long time
● low-speed sampling
● high-resolution
ADC linearity test with histogram method:Investigation of "high efficiency relationship" between input and sampling frequencies
This Work
Test cost is proportional to test time
6/27
Outline
• Objective
• ADC Test with Histogram Method
• Input Sine Wave and Sampling Frequency
Relationship in ADC Histogram Test Method
➢Sine Wave Histogram and Waveform Missing
➢Golden Ratio Sampling
➢Metallic Ratio Sampling
➢Prime Number Ratio Sampling
• Conclusion
7/27
■Histogram method (Ramp wave input)
7
⚫ Highly linear ramp signal generation is difficult
(limitation up to 14-bit ADC)
⚫ ADC output histograms for all bins are equal
if ADC is perfectly linear
t
Ramp wave
ADC
DUT
001010011100101110111
Output
Code
Number
of Samples
INL
DNL
Conventional Linearity Testing 1
DNL
INL
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Conventional Linearity Testing 2
8
■Histogram method(Single sine wave input)
⚫ Low distortion sine using an analog filter
⚫ Number of samples is small around the middle of
output range Many samples required (long test time)
t
Sine wave
Sine wave
GeneratorBPF ADC
DUT
Output
Code
Number
of Samples
INL
DNL
f
Remove
DNL
INL
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DNL & INL
9
001 010 011 100 101 110 111
Output
Code
Number
of Samples
DNL
⚫ Important ADC testing items
DNL : Difference between
actual step width and ideal value
INL : Deviation from ideal conversion line
=
=k
i
iDNLkINL1
)()(
10/27
Outline
• Objective
• ADC Test with Histogram Method
• Input Sine Wave and Sampling Frequencies
Relationship in ADC Histogram Test Method
➢Sine Wave Histogram and Waveform Missing
➢Golden Ratio Sampling
➢Metallic Ratio Sampling
➢Prime Number Ratio Sampling
• Conclusion
11/27
Repetitive waveform sampled asynchronously
Sampling CLK
Measured waveform
Reconstruct a 1-period waveform0
50
100
150
200
250
0 128 256 384 512 640 768 896 1024N
um
ber
of
Sam
ple
s
Output Code
PDF:
Probability Distribution Function
𝑝 𝑣 =1
𝜋𝐴2−𝑣2
- Sampled histogram
is compared with PDF.
- Histogram is obtained. -
DNL , INL are calculated.
Sine Wave Histogram
12/27
A large amount of data is required to reconstruct the waveform Test time: long
Waveform Missing
𝑻𝑪𝑳𝑲
TSIG
Waveform missing occurs
at special ratio (TCLK / Tsig)
Sampling CLK
Measured waveform
Repetitive waveform sampled asynchronously
Reconstruct a 1-period waveform
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𝛼 = 1,1
2,1
3,2
3,⋯ ,
1
6,⋯
Τ1 6
1
Τ1 1024
CLK
sig
Waveform Missing
𝑓𝐶𝐿𝐾 ≈1
𝛼𝑓𝑠𝑖𝑔𝑓𝐶𝐿𝐾 ≫ 𝑓𝑠𝑖𝑔
𝑓𝐶𝐿𝐾 ≈ 𝑓𝑠𝑖𝑔
Yuto Sasaki, Yujie Zhao, Anna Kuwana and Haruo Kobayashi, "Highly Efficient Waveform Acquisition Condition in
Equivalent-Time Sampling System", 27th IEEE Asian Test Symposium, Hefei, Anhui, China (Oct. 2018)
Special ratio
TCLK and Tsig , 𝑓𝐶𝐿𝐾 and 𝑓𝑠𝑖𝑔
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Waveform Missing for Saw Signal
Normal situation Waveform Missing
15/27
Outline
• Objective
• ADC Test with Histogram Method
• Input Sine Wave and Sampling Frequency
Relationship in ADC Histogram Test Method
➢Sine Wave Histogram and Waveform Missing
➢Golden Ratio Sampling
➢Metallic Ratio Sampling
➢Prime Number Ratio Sampling
• Conclusion
16/27
Golden Ratio
Golden Ratio: 𝐥𝐢𝐦𝒏→∞
𝑭𝒏
𝑭𝒏−𝟏= 𝟏. 𝟔𝟏𝟖𝟎𝟑𝟑𝟗𝟖𝟖𝟕𝟒𝟗𝟖𝟗𝟓 = 𝝋
The most beautiful ratio
17/27
Yuto Sasaki, Yujie Zhao, Anna Kuwana and Haruo Kobayashi, "Highly Efficient Waveform
Acquisition Condition in Equivalent-Time Sampling System", 27th IEEE Asian Test Symposium,
Hefei, Anhui, China (Oct. 2018)
Τ1 𝜑
CLK
Golden Ratio 𝝋
𝒇𝑪𝑳𝑲 = 𝝋 × 𝒇𝒔𝒊𝒈𝝋 = 1.6180339887…
Golden Ratio Sampling
Proposal of sampling conditions for the highest waveform acquisition efficiency
18/27
Outline
• Objective
• ADC Test with Histogram Method
• Input Sine Wave and Sampling Frequency
Relationship in ADC Histogram Test Method
➢Sine Wave Histogram and Waveform Missing
➢Golden Ratio Sampling
➢Metallic Ratio Sampling
➢Prime Number Ratio Sampling
• Conclusion
19/27
Metallic Ratio
Golden Ratio: 𝐥𝐢𝐦𝒏→∞
𝑭𝒏
𝑭𝒏−𝟏= 𝟏. 𝟔𝟏𝟖𝟎𝟑𝟑𝟗𝟖𝟖𝟕𝟒𝟗𝟖𝟗. . = 𝝋
n Decimal
0 1
11 + 5
21.6180339887… Golden Ratio
2 1 + 2 2.4142135623… Silver Ratio
33 + 13
23.3027756377… Bronze Ratio
4 2 + 5 4.2360679774…
… …
n𝑛 + 𝑛2 + 4
2
Generalization of Golden Ratio
20/27
Histogram of Saw Signal
𝑇𝑠𝑖𝑔
ℎ𝑖 𝑘 =𝑀
𝑁, 𝑘 = 1,2,3, . . . , 𝑁ideal value error 𝑒 𝑘 =
𝑁 ∙ ℎ 𝑘
𝑀− 1
ideal value ℎ𝑖 𝑘 =𝑀
𝑁
Output Code
Number of
Samples
Total number of samples: M
ADC resolution :N.
ℎ 𝑘
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RMS Error Calculation
Root mean square error
between actual and ideal histograms 𝑅𝑀𝑆 =σ ⅇ 𝑘
2
𝑁
Total number of samples: M=65536
0
10
20
30
40
50
60
0 32 64 96 128 160 192 224 256
RM
S
N
golden ratio sliver ratio bronze ratio
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Outline
• Objective
• ADC Test with Histogram Method
• Input Sine Wave and Sampling Frequency
Relationship in ADC Histogram Test Method
➢Sine Wave Histogram and Waveform Missing
➢Golden Ratio Sampling
➢Metallic Ratio Sampling
➢Prime Number Ratio Sampling
• Conclusion
23/27
RMS of Prime Number Sampling
RATIO with two large prime numbers ⇒ Small RMS
In contrast, Golden Ratio has a smaller RMS
𝒇𝑪𝑳𝑲 = 𝒇𝒔𝒊𝒈 × 𝑹𝑨𝑻𝑰𝑶 Total number of samples: M=65536
24/27
RMS of Prime Number Sampling: Big Number Case
𝒇𝑪𝑳𝑲 :Big prime number 997
𝒇𝑺𝑰𝑮 :Some prime numbers 389,599,751,853,907,953,991
→ almost the same
Total number of samples: M=65536
25/27
RMS ComparisonTotal number of samples: M=65536
Increasing N to 16384,
Bronze ratio result is better (RMS range is smaller)
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Outline
• Objective
• ADC Test with Histogram Method
• Input Sine Wave and Sampling Frequency
Relationship in ADC Histogram Test Method
➢Sine Wave Histogram and Waveform Missing
➢Golden Ratio Sampling
➢Metallic Ratio Sampling
➢Prime Number Ratio Sampling
• Conclusion
27/27
Conclusion
27
⚫ Like the golden ratio
Find conditions for efficient sampling
at a specific location
(ADC resolution N=256,512,1024,2048,4096)
Next work
Golden Ratio sampling
Efficiency: high
Sampling frequency: low
Metallic ratio sampling
Efficiency: high
Sampling frequency: high
Prime number ratio sampling
Efficiency:Not good
Sampling frequency: low
Thanks for your attention.