Instrumentation and Measurement Technology Conference-IMTC 2007 Warsaw,Poland, May 1-3, 2007
Selecting the Optimal Resolution and Conversion Frequency for A/D and D/A Converters
Camilo Quintans Grafna
Departmentof Electronic Technology, University of Vigo CampusUniversitario,36310Vigo, Spain Phone: +34 986 812 088, Fax: +34 986 811 987,Email:
Abstract -When a signal is converted by means of a converter circuit as an analog to digital converter (ADC) or a digital to analog converter (DAC) the quantization and sampling effects appeared, which causes a quality decrease of the signal that is usually indicated by the signal to noise ratio (SNR). ThisSNRlost decreases when the resolution and conversion frequency of the converter is increased Nevertheless, in this way the cost and the complexity of the circuit are also unnecessarily increased. In this work, a methodtoselect the optimal resolution and theconversion frequency is proposed. This method, which is based on the dynamic characteristic of the converterand thesignal, utilizes the desirable resulting signal quality as the constraint.Fromthesignal'sSNRand the quantity ofitthatcanbe lost the minimumnumber of bitsNis obtained. Then, from N the minimum conversion frequency is calculated. A mathematical analysis anda simulation test of the methodarecarriedout. Resultscanbedirectly appliedtoselect the maindynamics characteristics of the ADC, theDACor, ingeneral, ofaDAS.
Keywords -Data acquisition, data conversion, digital-analog conversion, signal sampling, signalresolution, white noise, random noise.
I. INTRODUCTION
When anADC (Analog to Digital Converter) [1]-[3] or a DAC (Digital to Analog Converter) has to be selected, or even aDAS (Data Acquisition System) [4] [5], the question of howmanybits arenecessaryand how fast the signalmust besampled for the applicationinquestion is formulated.
As a first approach, the number of bits needed may be specified taking into account the minimal variation of the input signaltobedetected, which is calledonecode binorQ.
This coefficient is calculatedbydividing the Full Scale input Range (FSR) by the number of code transition levels in the discrete transfer function (2N_1) [6] [7]. This is the most common method used but it is only suitable from a static point of view. Therefore, it isnotenough when the lost signal qualitymustbeknown; this is thedynamic point of view.
On the one hand, the characteristic defining the signal quality from adynamic point of view is itsSNR. Onthe other hand the conversionprocess, sampling and quantization, has two negative effects over the signal's SNR. The sampling effectdepends onthe frequencyfsand thequantization effect dependsonthe resolutionorthe number of bitsN.
Therefore the SNR decreases dependingonthe fsy N. To aid their selection it isnecessary torelate thembymeansofa rule or an equation which are easy to apply. That is to say what is the SNR of the signal and what its tolerated loss is?
Thus, firstly the initial signal's SNR and its quantity sacrificed arerelatedtothe minimum N. Secondly the value obtained ofNis utilizedtocalculate the minimum value offs.
Inthis way, the dimension and theworking frequency of the converter areoptimized.
On study this relation it has been found that when the ADC resolution N increased, the SNR of the signal also improved, but only up to a certain N, above whichpointno more improvement is achieved. Therefore, it is pointless to selectanADCwithabetter resolution.As for the conversion frequency, it is deduced that to optimize the value ofN to obtain acertainsignal quality, it isnecessaryto use acriteria in order to establish a minimum conversion frequency.
Therefore the introduced criteria mustbe taken into account, which is strongerthan theNyquist criteria [7] but guarantees that withnoperiodic signals the same desirable qualitymay be achieved. From that analysis, considering that this signal always has some quality loss, albeit it small (for instance0.1 dB),twoequationsareobtained.
To validate the previous analysis, using a random noise model, uniformly distributed for the signal and also for the ADCmodel, simulations have been made.
These resultscanbedirectly appliedas acriteriontoselect the dynamic characteristic ofan ADC, a DAC or a DAS.
Moreover, it can be usefulto answer another question. That is, if the SNR of an analog signal is unknown and it is converter to digital format, could that SNR be deduced?
Well, iftheoriginal characteristic of the signal is known, if theADC SNRis also known andifthe SNRof the converted signal is calculated, then, theSNRof the original signalmay be deduced.
In the next section a detailed problem definition and its solution is presented. In section III a simulation example is presented and in section IV experimental results, which utilize a real DAS including an ADC with its conditioning circuit and the interface with the PC, are shown in orderto prove the main contributions of this work. Finally, conclusionsaresummarizedinsectionV.
II. PROPOSED SOLUTION
The problem definition follows the Fig. 1 diagram. The starting data is the SNR of the signal and the maximum accepted SNRloss. From this data the minimum number of bits is obtained. Finally, with this value and with the input signal frequency, the minimum conversion frequency is obtained.
Signal quality (SNRsignal) Maximum input
+ signal
Lost signal quality (SNRIost) frequency (fi) Minimum numberof bits (N)
Minimumsampledfrequency (fs)
each twoblocks, is obtained from (3). This equationmaybe generalizedto more blocksbymeans of(4). Commonly, the signal to noise ratio is expressed in a logarithmic scale according toexpression (5) and theacronym SNRis used. In thisway, the total SNR of thetwo blocks is represented by (6). This equation is specified, by the following example depictedinFig. 1,by (7).
snr = x
Vr2+ 2
x . x
rj= n r2=
snrl snr2
snr = x
fx2 x2
x 2 + x
2 2
Snr; snr2
(1)
(2)
snrlf
snr2(3)
2 2
~snrj
+snr2
Fig. 1. Input and output information for the problem.
A. Problem analysis
To analyze the problem the diagram depicted in Fig. 2 is followed. The diagram consists of two blocks, B1 and B2, which are characterized by their SNR and are connected in serial.
B1 constitutes the source of the original signal S1 that is composed by the x information and by the r1 noise, so the outputsignal of theB1 isSI=x±+r. This signal crossesB2 and becomes noisier since this block introduces it ther2noise. To do the analysis the two following suppositions are made.
First, signal information x is maintained through the two blocks. Second, noises r1 and r2 are considered as independent random variables. Then, the total noise in the signalattheB2block outputisequaltothequadrature sumof thetwonoises
(rj±+r22)'
2 [8] and theoutputinformation isx.With thesesuppositions the final signal is
S=x+(r±2+r2j2)'
2 Todistinguish how the signaltonoise ratio isexpressed, if it is in a logarithmic scale (dB) the capital letters are used (SNR) andifit isperunit the lowercaselettersareused(snr).Then, the snr ofBI is snrl=x1r and for B2 is given by snr2=xlr2. When this relation is expressed in a logarithmic scale we have the equations SNR =20log(xrl) and SNR2=20log(xlr2)both of themexpressedin(dB).
B1
SI=
1 2 +r2(snzr,,r,) 1| (snr2,r2)
Fig.2.Diagram of the signalsource(B1)and theprocess(B2)handling the signal.
The total signal tonoise ratio is given by (1), and taking into account the noise of each block given by (2), the total signaltonoiseratio, depending onthe signaltonoise ratio of
fJ
snrsnr = m=1 Zsn"22 m=1
SNR
SNR=20 logsnr ; snr=10 20
2 SNR1
flOI 20
SNR=20.log ml
ml 20
m=1
2 SNR. SNR1+SNR2
=l 201
SNR=20-log lOm=I
=20
logS]2 SNRm 2 SNR1 SNR2
YI0
10 10 10 +10 10m=1
(4)
(5)
(6)
(7)
B.ApplicationtotheADCnumberofbits selection
The previous analysis may be directly applied to the problem of calculating the minimum number of bits needed tolimit the maximum lossinthesignal quality. The block B1 is substituted by a generator representing the process that provides the signaltobe converted andB2 ischanged by the ADC (Fig. 3). To simplify, the terminology of Table I is followed.
N
SNRT SNRs SNRADC ~SINAD
Fig.3.Diagram of thetwoblocks connectedinserial.
Table I. Terminology.
Term Definition
N Number of bits
snrs
InputsignalSNR SnrADC ADC SNRsnrL SNR lost of thesignal snrT Total SNR in the output
SNRS
Inputsignal SNR in dB SNRADC ADC SNRindB SNRL SNR lost ofthesignal SNRT Total SNR in the output in dBnecessary
follows. to adapt the previously presented calculus as
TotalSNR 80
75 70 65
m 60 sr 55 zc, 50 45 40 35 30
-ADCSNR -Signal SNR -Total SNR
By substituting in (7) the terms defined in Table expression of the total SNRis obtained (8). In this equ the value of the SNRADC may be substituted by (9). In way,theSNRTdependingontheSNRsandNis obtained
SNRs+SNRADC
10 20
SNk =20 log SNRs SNRADC
10 10 +10 10 SNRADC =6.02 N+1.76
I the [ation
i this (10).
(8)
(9)
,.1IY
SNRs+6.02 N+1.76
In 20
SNRT =20 log |SNRS 6.02N+1.76
1U)
10 10 +10 10
According to (10) Fig. 4 shows how the quality of the sampled signal (indicated by SNRs) improves when the number of bits(N) oftheADCincreases. Inthisexample, the input signal has an SNR (SNRs) equal to 50 dB. The total SNR(SNRT) increases significantlyuptocertainN(around9- 10bits). After this,evenif Nis increased, the total SNRdoes not improve. This performance is important because it may be takenas arule. Since, ifanADC with aSNR with 10 dB above the SNR ofthesignal isselected, then the signalis minimally worsened (about 0.1 dB). Therefore, ifan ADC with more bits of resolution is selected, it will be more expensive but the performance achieved will not be better.
Moreover, if more bits than those needed areused, not only thecostoftheADCisgreaterbut also thecostofdigitalparts (the bus width, the memory size, etc.), the power consumption and the conversion time are unnecessarily increased.
Fromthispoint of view, it is possibletoknow what is the SNR of the ADC needed and then to deduce from it the number of bits. But, ifwhat is interesting is to be able to select N from the acceptable loss in the SNR, it will be
5 6 7 8 9 10
Numberof bits
11 12 13
Fig.4.TotalSNR versustheADCnumber of bits (N).
The starting point is the loss acceptable inthe SNRof the signal. This loss in the SNR is, according to (11), the difference between the signal SNR and the total SNRinthe output. Bysubstituting theSNRvalues(12) and transforming it to a linear scale the equation (13) is obtained. In this equation the value ofsnrTis substitutedby its value from the general expression obtained in (4). Thus, the value ofsnrADC is obtained from(9) and equalizedtothe expression (14) and then, the expression (15) is achieved. Finally, from (15),Nis deduced and withsomecalculus thegeneral expression forN, which depends on the SNRs and the value of SNRL, is obtained.
SNRT=
SNRs-SNRL
(1 1)20 log
snr,
=20 logsnrs-20 logsnrLsnr5 snr5
log
snr,
=log snrT= SsnrL snrL
SflrS - nSf snrADC* ; 2 snrS2+ 2
snrL
snri
+snrA snrLSf2rADC *(- ) 2 2
snrA rsnsnr5
I
snr s
snrADC = i- B/ 2
(12) (13)
(14)
6.02 N+1.76=20 log
snr2-I Sflr
20 N*log2+ 20 log 6 =20 log
sn,
g g ~~~2
snr,2-
log 2N 2. =log X
snrL 22
~6
snrsni--1IN62
snr -IN= -log, 22snrs ) 3 snrL -1)
= 1.Llog2 +log02 snrS -log2(snr2 -1)
2
Log2 +1og2i10
iU -1092 10 )1= 2F[1°g2 3S+NRSS Rlo
N 19 +S R *1og210 -1og2
2 3
After the minimum number of bits has been obtained the
(15)
optimalconversion frequency must be calculated. This is the frequency at which all code bins have almost one sample and whenthe signal has the maximum slope only one sample. In thisway, the loss of some code, due the amplitude change of the signal between two consecutive samples is greater than the corresponding increment between two consecutive codes (Fig. 6), is avoided. Moreover, obtaining several samples of the signal corresponding to the same code is also avoided (Fig. 7). The optimal frequency corresponds to Fig. 8, in whichwe can see how, in the portion of the signal with the greatest slope, there is the minimum code bin, in other words only one sample per code bin that guarantees that there is almostonecode bin per sample.111 (16)
Ifthe SNR ofthe input signal to be converted and the SNR of the converter system are both known depending on N.
Therefore, it will be easy to calculate N from the loss of the signal quality which canbe assumed. To do this, it is enough toapply (16). Finally, it is important to point out that in real systems the ENOB (Effective Number of Bits) [8] is used instead ofN. Therefore, equation (16) may be changed to (17).
1
Flg2 SNRs5 10NR
iENOB=
2o2 1+10 lo210-lo2
(10 -1 ( 7Fig. 5 presents the value of N that should be taken when the SNRof the signal and the SNR to loss are known. This figure has threetraces correspondingto three different losses ofSNR(0.1, 1.0and3.0dB).
Min. numberof bits
Q z
425 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105
SignalSNR(dB)
Fig.5.Minimum number of bitsneeded versus the SNR lost in a certain input sampled signal withacertainSNR.
110
a 101
:t I
E 1OQ
& 011
*-0 1
o)
010.001
000
Fig. 6.
111
110
0 a, 010
a)
9
There isn'tone sample of everycode bin. Codes"101" and
"011"are loss.
Discrete time
Signal sampledat a tooslow frequency (fs =1I1 f ).
I mr)rtIul st-, Li lCithnnQrlnUl It,
001
sample.These extrasamplesareredundant.
000
Discrete
timeFig.7.Signal sampledat a toofast frequency(fs=33 f ).
0 P-
fs (2 -_1).,.j
2fs =
(23
-1)
t;rf
=21.98.f,
V
LSB.I
Nocodebinsare
loss.There is Almostonesample
for eachcodebin.
Discretetime
/ dv 1LSB
/ dt~ 1
* * kI
!,4 ~~k
1/fs
Fig. 9. Maximum slewrateof the signal's voltage.
Fig. 8. Signalsampledattheoptimal frequency ( fs=22 ).
To obtain that optimum frequency value the maximum slew rate of the signal must be taken into account. If the signal is composed of a set of harmonic components, the componentwithgreaterfrequencymustbeconsidered. Either
way, the maximum slewrate must be considered. To do the analysis it is supposed that the signalrepresented by (18) has the greatest frequency f and its amplitude corresponds with AO=FSR/2. Differentiating (18)twotimes and equalizing itto
zero,thepoint of maximum slope is obtained (20), this is the
zerocrossing point. Substituting this pointin(19) it gives the expression of thegreatestslope (21).
v(t)=
A,
sin(w. t)dv(t)=A0 co cos(w. t) dt
dtV
=AO- 2sin(ot)=O
; M t=0,z,2yz..dt
dv(t)
=-A c=A2a zf-dtdtmax 0
2N-1 FSR
5FSR 2 l
III. SIMULATION RESULTS
Fig. 10 shows a signal with a SNR of 15 dB that is sampled at a frequencyfSi=5f, by an ADC of 4 bits. The ADC has a SNRequalto25 dB therefore it has 10 dBmore
than the signal, accordingtothe presented rule the lost SNR of thesignal becomes 0.137 dB, which isanacceptable result becauseit isnearthe theoreticalvalue of 0.1 dB.
1111
1110 1101 1100
(18) 1010
n 1001
1000
(19) - 0111
( 0110
0101
0100
(20) 0011
0010 0001
(21) 0000
(22)
0 10 20 30 40 50
Sample
Fig. 10.Example ofanoisy sampled signal with 15 dB of SNR.
Now, the slope ofthesignal is comparedtothe maximum slopethat thesampledprocess candetect(Fig. 9). This value mustbe greater than or equaltothe expression (21), thus the general equation (22)is obtained, whichcan be simplifiedto (23). Inthe example of Fig. 6toFig. 8, the number of bits is 3, and the number ofsamples persignal period is 11, 33 and 22respectively.
Applying the equation (23) the relation (24) is obtained.
This resultdemonstrates, aswould be expected, that the value of 22 samplesperperiodis theoptimumnumber.
IV. EXPERIMENTAL RESULTS
In ordertoprovethesimulation results which validatethe rulespresentedinthiswork,twoexperimentsarepresented in this section. To do this an ADC including conditioning and converters circuits is used to acquire the data and then the spreadsheet in the PC is utilizedtoanalyze them.
In the firstexperiment a setofnoisy signals formed by a
fixed 24kHz sinusoidalcomponentandbyavariable random noiseareacquired by the ADC and then analyzed (Fig. 11).
il
110
-0 101
;D E
-2 011
a) .@ 010
n 001 000
(23) (24)
...*...*...
,.#,,.
Inthe second experiment the acquired noise signal is fixed at25dB (Fig. 12) and theSNRof theADCis varied from20 to 42dB by connecting thenecessarynumber of bits(from3 to8) (Fig. 13). Each time the number of bits is changed the ADC SNRis measuredby acquiring the sinusoidal signal but disconnecting the noise. After this, the noise is addedtothe sinusoidal signal and the result is acquired in order to be analyzed by the spread sheet (Fig. 13).
55,0- 53,0- 51,o- 49,0- 47,0- 45,0- 43,0-
m7 41,0-
z 39,0
CO37,0- 35,0- 33,0- 31,0-
29,0-
27,0- 25,0-
-TotalSNR X ADC SNR
LO r- m O; M LOr-:IpusignaLO SR [ LO
CN C4 C04 C0 C0 C0 C0 tU)ULO LO
Input signalSNR[dB]
Fig.11. Result from sampling, withafixedADC SNRof42dB, signals with different SNR.
44 42 40 38 m.3634
0 32 Z 30
C') 288 26 24 22 20 18
TotalSNR --> - Signal SNR
m-L Lu L L lo I- _-lo I-_
I1 4 -k-W t 1- ~L
18 20 22 24 26 28 30 32 34 36 38 40 42 44 ADCSNR [dB]
Fig. 13. Result from sampling the 25 dBsignal varying theADC SNR.
V.CONCLUSIONS
When the resolution of an ADC must be selected in an optimalwayfollowing rulescanbe useful:
* Select theADCwithaSINADof10dB above the signal'sSNR.
* For acertain maximumSNR tobe lostinthe conversion process,selectan ADCwithanENOB greaterthan the onegiven by the expression(17).
* Finally, after selecting theADCresolution, select the sampled frequency given byexpression (21).
These applied results have been initially proved by simulation and afterby real experiments.
ACKNOWLEDGMENT
Measure P1l:freq(Cl) P2:rms(Cl) P3:rms(C2) P4:pkpk(Cl)
value 24.024024 kHz 2.S265 V 162 mV S 6 V
status v1 v1 v v
mm = =:ItIlT.gger~~~~~~~Ilm
Fig. 12.Oscilloscopescreenwith theinput sinusoidal (C1), the noisetobe added(C2) and theoutputnoised sinusoidal of25dB ofSNR.
In both experiments the sampledrate is maintained at 24 MHiz and, accordingto(23) andtaking intoaccountthat the maximum number of bits in the experiments is 8 b, the size Mof the data record has been established at 1 k samples and sothefrequency ofthe signalto24kHz.
Fig. 11 shows that the total SNRdoes not improve when theinput signal SNRis 10dB overtheADC SNR. Therefore, theADCperformance limits the resultingSNR.
Reciprocally, Fig. 13 shows that when theADC SNR gets over in 10 dB the input signal SNR, in this caseequal to25 dB(8 Vpp sinusoidal plus 162 mVrmswhitenoise), the final SNRdoes not improve and thus, sampling the signal with a major resolution is pointless.
The author wish to express their gratitude to the Autonomous Government of the "Xunta de Galicia" (Spain) which has funded this work through the research grant:
PGIDIT04DPI340004CT.
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