How to calculate the small changes in a signal?

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Susan on 31 Aug 2022

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Commented: Susan on 9 Sep 2022

A.mat

Hi all,

I have an ECG signal and would like to know what minor amplitude changes I can recognize. The goal is to see if I'm able to get the resolution that's provided by the recording system or not. I use the gradient function to calculate the derivative and look at the minimum value to determine the resolution. I would like to know if this is the best approach to take or not. The signal is attached here. Any suggestions would be appreciated.

Fs = 1024; % sampling rate
Ft = gradient(data, 1/Fs)
min(Ft) % the smallest change in the signal's amplitude

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Answer 1

Star Strider on 31 Aug 2022

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Iam not certain what result you want, however to get the smallest amplitude change, using gradient on the original signal is likely not going to produce the result you want, since the minimum gradient is going to be the minimum of the entire record, likely returning the value of the deepest Q- or S-deflection. Taking the gradient of the abolute value instead may do what you want.

Try this —

LD = load(websave('A','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1112930/A.mat'));

data = LD.A;

Fs = 1024;

L = numel(data)

L = 86549

t = linspace(0, L-1, L)/Fs;

Ft = gradient(data, 1/Fs);

[minFt,idx] = min(abs(Ft)) % the smallest change in the signal's amplitude

minFt = 0

idx = 22

minidx = find(Ft == min(abs(Ft)))

minidx = 3223×1

22 35 93 94 95 96 97 98 99 100

Ft2 = gradient(Ft); % Second Derivative

resolution = diff(data);

min_resolution = min(resolution(resolution>0))

min_resolution = 1

figure

subplot(2,1,1)

plot(t, data, 'DisplayName','Data')

hold on

plot(t(minidx), data(minidx), '+r', 'DisplayName','Gradient = 0')

hold off

grid

title('Original')

legend('Location','best')

xlim([0, 10])

subplot(2,1,2)

plot(t, Ft, 'DisplayName','Gradient')

grid

title ('Derivative')

xlim([0, 10])

That returns several values of a flat gradient, all representing inflection points in the EKG trace. Calculating the second derivative will allow the determination of the inflection points being a minimum, maximum, or saddle point. This may be helpful, however I am not certain what you want.

The minimum resolution appears to be equal to 1, however that likely needs to be calibrated with respect to the actual voltage values of the EKG (the R-deflection is normally about 1±0.2 mV) to be useful. It will likely be necessary to check the gain on the instrumentation.

.

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Star Strider on 31 Aug 2022

My pleasure!

I am not certain what sort of ‘interference signal’ you want to use. Gaussian white noise (for example the output of the randn function) could completely obscure the EKG if the amplitude is great enough. A sinusoidal signal creating a varying baseline could preserve the EKG morphology and still cause problems with, for example, machine interpretation.

‘Based on the manual, I should be able to recognize the changes in the amplitude that are less than 20 microvolts.’

If one increment then is 20 µV, then multiply the observed values by 2E-5 to get the values in Volts. Looking at the posted trace, that means that the R-deflection maximum is 3.5E+4 x 2E-5 = 700 mV. (That is unphysiologically high, so check that. A calibration of 2E-6 instead would make that 70 mV and physiologically appropriate.) That also will convert the integer values to floating-point values, making any other calculations easier.

The data in the file appear to be integer data, so it may be necessary to use a calibrated voltage with your instrumentation to determine what the output for the known voltage. (I would use several voltages betweeen -2 mV to +2 mV to get the calibration values and also confirm linearity.) I hope you have some sort of voltage standard available externally or within the instrumentation that will allow you to do this.

What is the ‘NSR signal’? I may have encountered it previously, however likely by a different name.

dpb on 31 Aug 2022

"That returns several values of a flat gradient..."

Which are almost all at the extremes both directions that indicates the signal was clipped in all likelihood.

LD = load(websave('A','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1112930/A.mat'));
data = LD.A;
u=unique(data);
numel(u)
ans = 15528
size(data)
ans = 1×2
       86549           1
[max(data) min(data)]
ans = 1×2
       32767      -32768

Shows the discrete nature of the inputs -- that there are only 15,528 unique values out of the 86,549 total samples and that the min/max values are, indeed at the limit of the 16-bit signed integer.

Since these data are

all(data==fix(data))
ans = logical
   1

all integer-valued (pretty much a given since are reading the 16-bit integers from the data files), the minimum difference in values between samples is, by definition, 1.

There's nothing to be learned here by this exercise until have the absolute scaling to relate the values back to what is the input -- and whether these represent the raw inputs to the device or whether they have been processed before being written out in 16-bit signed integer format.

dpb on 1 Sep 2022

Edited: dpb on 1 Sep 2022

Well, you've kept that hidden -- there's all the data you asked for -- the resolution of the device is that the least significant bit is 19.53 uV.

What that is in output then will depend on the gain you set -- it gives the scaling of mm/V. So, the only thing yet to determine is just what parameter it is that is being written to the data file -- if it were the processed waveforms, you should be able to overlay your plots directly on theirs with some scaling factor; if these are the raw data before any other processing and the waveforms presented are somehow preprocessed, then they won't match identically and you've still the Q? to answer about what happens to the recorded values with known inputs.

@Star Strider probably knows enough about the medical instrumentation field to be able to help as far as what is normally done with raw data collected to present to the trained professional as a polished ECG to give at least some guidance. That may (or may not?) include proprietary algorithms to differentiate one product from another? Or, are there required standards in medical field that everybody has to do the same thing; only user interface/convenience/price differentiates one product from another in market place? I've no idea on that score.

It also confirms the other point made elsewhere that the -3 dB upper frequency is 100 Hz so anything you try to put into it in the kHz or above range thru the input channels simply isn't going to ever be seen.

dpb on 1 Sep 2022

"...should be able to recognize the changes in the amplitude that are less than 20 microvolts."

That would be "greater than", not "less than" 20 uV would be resolvable.

Sure that's what you meant, just correcting the record of what was written.

But, back to the subject of just what your research goal here is regarding the proposed noise injection -- just what is the stated project objective? If it is, indeed, something about detecting stray electrical field noise contaminating a recorded ECG, it would seem you're going to need more sophisticated data collection instrumentation than this commercial product that has that protection from such extraneous sources built into it.

Without a stated research goal/hypothesis we certainly can't design an experimental setup here and it's not the MATLAB forum's place really to do so, but to me it doesn't appear that you can realistically do anything about what you've described with this device, even if you do finally figure out its internals.

You can test this hypothesis with a signal generator on one of the inputs -- do the experiment I outlined before for the DVM -- I'm pretty sure you'll find that if you put a sine wave in at 1 kHz, you'll just get a near zero flat line for that channel recorded, really no different than if you had just grounded that input.

I would suggest you should do these rudimentary tests to see if you've got any chance at all of making this work before investing a great deal more time on other details on the assumption the device will be able to record these presumed corrupted signals when you get there.

Star Strider on 1 Sep 2022

From the .pdf it looks like the full-scale deflection in lead II is 100 mV, and the recording is 10 mm/mV.

This appears to be appropriate —

LD = load(websave('data1','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1113930/data1.mat'))

LD = struct with fields:

data1: [89739×8 double]

VL = compose('V%d',1:6);

VN = {'I', 'II', VL{:}};

EKG8 = array2table(LD.data1, 'VariableNames',VN)

EKG8 = 89739×8 table

I II V1 V2 V3 V4 V5 V6 _____ _____ ____ ____ ____ ____ ____ ____ 808 839 333 265 293 250 304 282 3987 4149 1646 1311 1446 1236 1497 1390 8752 9114 3620 2893 3167 2715 3288 3050 12742 13261 5288 4239 4606 3960 4795 4448 15404 16019 6404 5152 5570 4802 5802 5391 17182 17868 7125 5753 6209 5359 6477 6019 18372 19110 7598 6148 6639 5721 6936 6439 19165 19937 7914 6409 6933 5955 7240 6725 19692 20485 8123 6568 7127 6102 7436 6899 20053 20851 8270 6658 7243 6195 7548 6989 20292 21102 8376 6715 7319 6266 7612 7044 20439 21270 8444 6757 7379 6319 7670 7093 20537 21379 8489 6801 7423 6349 7729 7140 20610 21448 8518 6846 7461 6360 7767 7180 20658 21484 8538 6880 7503 6383 7793 7217 20676 21495 8555 6897 7531 6421 7821 7242

CF = 1E-2/(max(EKG8.II(end-100:end)) - min(EKG8.II(end-100:end))) % Calibration Factor (V/Increment)

CF = 7.7399e-06

EKG8_V = array2table(table2array(EKG8)*CF, 'VariableNames',VN) % Convert Values To Volts

EKG8_V = 89739×8 table

I II V1 V2 V3 V4 V5 V6 _________ _________ _________ _________ _________ _________ _________ _________ 0.0062539 0.0064938 0.0025774 0.0020511 0.0022678 0.001935 0.0023529 0.0021827 0.030859 0.032113 0.01274 0.010147 0.011192 0.0095666 0.011587 0.010759 0.06774 0.070542 0.028019 0.022392 0.024512 0.021014 0.025449 0.023607 0.098622 0.10264 0.040929 0.03281 0.03565 0.03065 0.037113 0.034427 0.11923 0.12399 0.049567 0.039876 0.043111 0.037167 0.044907 0.041726 0.13299 0.1383 0.055147 0.044528 0.048057 0.041478 0.050132 0.046587 0.1422 0.14791 0.058808 0.047585 0.051385 0.04428 0.053684 0.049837 0.14834 0.15431 0.061254 0.049605 0.053661 0.046091 0.056037 0.052051 0.15241 0.15855 0.062872 0.050836 0.055163 0.047229 0.057554 0.053398 0.15521 0.16139 0.064009 0.051533 0.05606 0.047949 0.058421 0.054094 0.15706 0.16333 0.06483 0.051974 0.056649 0.048498 0.058916 0.05452 0.1582 0.16463 0.065356 0.052299 0.057113 0.048909 0.059365 0.054899 0.15896 0.16547 0.065704 0.052639 0.057454 0.049141 0.059822 0.055263 0.15952 0.16601 0.065929 0.052988 0.057748 0.049226 0.060116 0.055573 0.15989 0.16628 0.066084 0.053251 0.058073 0.049404 0.060317 0.055859 0.16003 0.16637 0.066215 0.053382 0.058289 0.049698 0.060534 0.056053

Fs = 1024;

L = size(EKG8,1);

t = linspace(0, L-1, L)/Fs;

figure

plot(t, EKG8_V.II)

grid

xlabel('Time (s)')

ylabel('Amplitude (V)')

xlim([20 30]+0.25)

Each increment then is about 7.74 µV, at least as I calculate it on the basis of the information provided.

No worries about the URL, since I could probably find it online now, however I doubt that’s necessary at this point.

.

Star Strider on 1 Sep 2022

My pleasure!

‘... you mean the peak-to-peak value in the lead II is 100 mV, right?’

In that section of it (after the transients are gone), yes.

‘Why has the calibration factor been calculated based on the lead II?’

That’s the lead I usually calibrate with, since it tends to be the most representative. It also represents the amplitude information in the .pdf file that I’m using as a source of information, so I’m matching those data.

‘It seems that each lead thas different calibration factor, am I right?’

They shouldn’t. The deflections in an EKG are actually projections of an excitation (or repolarization) loop onto each lead. As such, the amplitudes may not be the same as in Lead II.

‘And could you please tell me what is the rational behind calculating the calibration factor in this way?’

It depends on the instrumentation. Since I cannot interact with it to calibrate it, I am using the available information, gleaned from the .pdf file and the data. So it’s largely empirical inb my analysis. My calculation and use of the ‘CF’ variable faithfully reproduces the .pdf file results. That’s the best I can do.

.

dpb on 1 Sep 2022

Edited: dpb on 1 Sep 2022

If the LSB is 20 uV for a 24-bit A/D, but the device is only writing 16-bit signed integers, you can't put 24 bits of resolution into 16 bits -- they don't fit. So

>> 2^24/2^16
ans =
256
>> 

is the ratio of how many bits were digitized over how many were written -- if 24 bits --> 20 uV, then the LSB for 16-bits is 256X that amount. That's for the granularity of the output data waveforms you have available to you to read; the internal software may still have used more, but I'd guess it's probable they're using a 16-bit microprocessor so it fits the word size. If they were using a 32-bit micro, then they could have done all the internal stuff with full precision and then just written the results to the data file as 16-bits.

If it is only actually using an 18-bit A/D instead of 24, then the LSB would be 2^18/2^16 ==> 2^2 --> 4X 20uV or LSB is 80 uV in the actual instrument it would seem then, not 20. The end result would be the same.

The actual resolution in V is then that 1/2^Nbits precision applied to the input range. See a more complete description at <ADC_and_Resolution>

Susan on 1 Sep 2022

@Star Strider No worries at all. I'm glad that I asked you. Then by changing the CF, we got the EKG8_V in mV now (instead of V that we previously got), and it is the same unit as the one in the pdf file, but I'm confused again. You mentioned, "The normal EKG should have an R-deflection amplitude of about 100 mV." Now we get the R-deflection amplitude in the order of 1mV. I remember you previously mentioned that" the R-deflection is normally about 1±0.2 mV". So, there should be a typo here, and I just wanted to ensure I got the correct one since I have no medical background :)

If you don't mind, can I ask you one more question? I collected another data set and generated the pdf file using their software. I thought I would get the identical waveforms as in the pdf file, using your code above. However, it turned out that I was wrong. So, does it mean that for each new data set, I need to figure out what would be the best denominator in the calibration factor to get the same results as the pdf? I attached the data and pdf file here and truly appreciate your input.

Star Strider on 2 Sep 2022

My previous statement ‘1±0.2 mV’ is correct. I don’t know what I was thinking otherwise. (Maybe just fatigued.)

LD = load(websave('data1','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1114355/data1.mat'))

LD = struct with fields:

data1: [72981×8 double]

VL = compose('V%d',1:6);

VN = {'I', 'II', VL{:}};

EKG8 = array2table(LD.data1, 'VariableNames',VN);

CF = 1E-4; % Calibration Factor = 100 µV/Quantization Step

EKG8_V = array2table(table2array(EKG8)*CF, 'VariableNames',VN) % Convert Values To Volts

EKG8_V = 72981×8 table

I II V1 V2 V3 V4 V5 V6 ______ ______ ______ ______ ______ ______ ______ ______ 0.081 0.0836 0.0336 0.0272 0.03 0.0263 0.0314 0.0288 0.4 0.4141 0.1654 0.1346 0.1482 0.1293 0.1546 0.1425 0.8765 0.9111 0.3634 0.2963 0.3254 0.2829 0.3387 0.3129 1.2748 1.3282 0.5297 0.4321 0.4738 0.4111 0.4921 0.4551 1.5401 1.6061 0.64 0.5227 0.5734 0.4963 0.5938 0.5495 1.7172 1.7906 0.7119 0.5824 0.6406 0.5533 0.6617 0.6131 1.836 1.9127 0.7595 0.6219 0.6844 0.591 0.7061 0.656 1.9153 1.9948 0.7908 0.647 0.7111 0.6141 0.7345 0.6827 1.9678 2.0517 0.8107 0.6616 0.7268 0.628 0.7529 0.6972 2.0024 2.0908 0.8243 0.6714 0.7371 0.6378 0.7654 0.7061 2.0243 2.116 0.8345 0.6801 0.7461 0.6455 0.7756 0.7145 2.0386 2.1308 0.8423 0.6866 0.7545 0.6514 0.7844 0.7235 2.0491 2.1397 0.8474 0.6904 0.7614 0.6549 0.79 0.73 2.0563 2.1453 0.8503 0.6946 0.7662 0.6576 0.7926 0.733 2.061 2.1499 0.8512 0.6991 0.7682 0.6595 0.7933 0.734 2.0648 2.1537 0.851 0.7009 0.768 0.6593 0.7926 0.734

Fs = 1024;

L = size(EKG8,1);

t = linspace(0, L-1, L)/Fs;

figure

plot(t, EKG8_V.II,'.', 'MarkerSize',2.5)

grid

xlabel('Time (s)')

ylabel('Amplitude (mV)')

title('II')

xlim([20 21]+0.25)

This one seems to have significantly lower voltage, however the axis is normal (76°) and there is normal R-progression in the precordial leads. The R-peak is about 0.5 The ‘CF’ value here is set at 100 µV/quantization step. This appears to be correct and essentially matches the trace in the .pdf file, however not with the previous data set (follows), since the trace had a Lead II amplitude of only about 0.8 mV while that ‘CF’ value produces an R-amplitude of about 1.1 mV here.

Something is clearly not consistent. I am not certain what that ‘something’ is.

LD = load(websave('data1','https://www.mathworks.com/matlabcentral/answers/uploaded_files/1113930/data1.mat'))

LD = struct with fields:

data1: [89739×8 double]

VL = compose('V%d',1:6);

VN = {'I', 'II', VL{:}};

EKG8 = array2table(LD.data1, 'VariableNames',VN);

CF = 1E-4; % Calibration Factor = 100 µV/Quantization Step

EKG8_V = array2table(table2array(EKG8)*CF, 'VariableNames',VN) % Convert Values To Volts

EKG8_V = 89739×8 table

I II V1 V2 V3 V4 V5 V6 ______ ______ ______ ______ ______ ______ ______ ______ 0.0808 0.0839 0.0333 0.0265 0.0293 0.025 0.0304 0.0282 0.3987 0.4149 0.1646 0.1311 0.1446 0.1236 0.1497 0.139 0.8752 0.9114 0.362 0.2893 0.3167 0.2715 0.3288 0.305 1.2742 1.3261 0.5288 0.4239 0.4606 0.396 0.4795 0.4448 1.5404 1.6019 0.6404 0.5152 0.557 0.4802 0.5802 0.5391 1.7182 1.7868 0.7125 0.5753 0.6209 0.5359 0.6477 0.6019 1.8372 1.911 0.7598 0.6148 0.6639 0.5721 0.6936 0.6439 1.9165 1.9937 0.7914 0.6409 0.6933 0.5955 0.724 0.6725 1.9692 2.0485 0.8123 0.6568 0.7127 0.6102 0.7436 0.6899 2.0053 2.0851 0.827 0.6658 0.7243 0.6195 0.7548 0.6989 2.0292 2.1102 0.8376 0.6715 0.7319 0.6266 0.7612 0.7044 2.0439 2.127 0.8444 0.6757 0.7379 0.6319 0.767 0.7093 2.0537 2.1379 0.8489 0.6801 0.7423 0.6349 0.7729 0.714 2.061 2.1448 0.8518 0.6846 0.7461 0.636 0.7767 0.718 2.0658 2.1484 0.8538 0.688 0.7503 0.6383 0.7793 0.7217 2.0676 2.1495 0.8555 0.6897 0.7531 0.6421 0.7821 0.7242

Fs = 1024;

L = size(EKG8,1);

t = linspace(0, L-1, L)/Fs;

figure

plot(t, EKG8_V.II,'.', 'MarkerSize',2.5)

grid

xlabel('Time (s)')

ylabel('Amplitude (mV)')

xlim([20 21]+0.25)

I am not certain that there is any one value for ‘CF’ that would be uniformly applicable here.

.

Susan on 2 Sep 2022

Edited: Susan on 2 Sep 2022

@dpb It seems to me that there is a linear scaling between the EKG signal amplitude that I can set on the device and the amplitude of the signal that I can plot in Matlab. But since I'm not able to get the same waveform as the one on the software, maybe there is some other post-processing in the data?!

Here is what I got for FS = 1024 and CF = 100 µV/Quantization Step

%                   ~~~ gain = 20mm/mV ~~~          ~~~ gain = 10mm/mV ~~~           ~~~ gain = 5mm/mV ~~~
% 0.05mV Amplitude of ECG Signal--0.0516 mV       -- 0.0254 mV * 2 = 0.0508mV--        --0.0128 mV * 4 = 0.0512
% 0.10mV Amplitude of ECG Signal--0.1003 mV       -- 0.0500 mV * 2 = 0.1000mV--        --0.0250 mV * 4 = 0.1000
% 0.15mV Amplitude of ECG Signal--0.1492 mV       -- 0.0740 mV * 2 = 0.1480mV--        --0.0370 mV * 4 = 0.1480
% 0.20mV Amplitude of ECG Signal--0.1963 mV       -- 0.0978 mV * 2 = 0.1956mV--        --0.0489 mV * 4 = 0.1956        
% 0.25mV Amplitude of ECG Signal--0.2450 mV       -- 0.1219 mV * 2 = 0.2438mV--        --0.0611 mV * 4 = 0.2444
% 0.30mV Amplitude of ECG Signal--0.2945 mV       -- 0.1463 mV * 2 = 0.2928 mV--       --0.0731 mV * 4 = 0.2924
% 0.35mV Amplitude of ECG Signal--0.3420 mV       -- 0.1700 mV * 2 = 0.3400 mV--       --0.0851 mV * 4 = 0.3404
% 0.40mV Amplitude of ECG Signal--0.3892 mV       -- 0.1939 mV * 2 = 0.3878 mV--       --0.0970 mV * 4 = 0.3880
% 0.45mV Amplitude of ECG Signal--0.4373 mV       -- 0.2178 mV * 2 = 0.4356 mV--       --0.1090 mV * 4 = 0.4360
% 0.5mV  Amplitude of ECG Signal--0.4843 mV       -- 0.2416 mV * 2 = 0.4832 mV--       --0.1207 mV * 4 = 0.4828
% 1mV    Amplitude of ECG Signal--0.9557 mV       -- 0.4774 mV * 2 = 0.9548 mV--       --0.2387 mV * 4 = 0.9548
% 1.5mV  Amplitude of ECG Signal--1.4269 mV       -- 0.7130 mV * 2 = 1.4260 mV--       --0.3566 mV * 4 = 1.4264
% 2mV    Amplitude of ECG Signal--1.8964 mV       -- 0.9480 mV * 2 = 1.8962 mV--       --0.4739 mV * 4 = 1.8956
% 2.5mV  Amplitude of ECG Signal--2.3700 mV       -- 1.1848 mV * 2 = 2.3696 mV--       --0.5925 mV * 4 = 2.3700
% 3mV    Amplitude of ECG Signal--2.8296 mV       -- 1.4198 mV * 2 = 2.8396 mV--       --0.7098 mV * 4 = 2.8392
% 3.5mV  Amplitude of ECG Signal--3.2397 mV       -- 1.6554 mV * 2 = 3.3108 mV--       --0.8278 mV * 4 = 3.3112

@Star Strider I don't estimate any parameter; I think I'm looking for a CF value that can match the Matlab plot with the software plot for all collected data sets (like what you did for the two data sets above). Then I can be sure that whatever data set I collect and plot, the waveform is more or less like what I'm going to get from the software for the same collected data, if it makes sense. And I believe non of the parameter that I change, like the amplitude of the EKG signal or the recorder gain, are time-dependent.

Star Strider on 3 Sep 2022

@Susan — Just now looked at the .PNG files. I don’t see anything in them that gives me any insight into the instrumentation. The 50 Hz notch filter just elliminates 50 Hz line noise, although this shouldn’t be necessary using the right leg common reference electrode, since all that should be subtracted out. In any event, there’s not much of interest in the EKG at 50 Hz, so nothing is lost. It seems to be reasonably sophisticated otherwise, marking the E-point at the end of the P-deflection, the J-point at the end of the QRS complex, and the S-T interval. There’s not much to say otherwise about it. I still don’t understand not being able to get the actual recorded EKG information out of it (scaled to produce the actual voltages) rather than only the raw data.

There’s nothing in the .mat files that I can see that hints at any of the settings used to record the signal, scaling factors, or anything else. If those are actually in any of the files the instrumentation produces, it would be extraordinarily helpful to have them.

I still don’t understand the initial decaying exponential transient at the beginning of the recording. That could be a filter transient, although I’ve never seen anything similar in anything I’ve done. (It‘s not due to an anti-aliasing filter, since that occurs before the A/D converter, and in any event is a Bessel filter so there should be no transients due to it.)

dpb on 3 Sep 2022

Edited: dpb on 3 Sep 2022

"I still don’t understand the initial decaying exponential transient at the beginning of the recording"

See the Analog Devices datasheet on their chipset -- they specifically talk about it -- it's from the way the high-pass filter is built into their device...which is part of what makes me think the sensor @Susan has is essentially identical.

" that hints at any of the settings used to record the signal, scaling factors, or anything else"

There's an associated .inf file with a few hints but nothing that really helps more than what little already know; the previous long thread discussed those at some length, including trying to decipher the 512-byte .dat-file header. So far, we've been unable to actually figure out what most of the data in it are...

Part of the exercise @Susan was going to do was to see if could build a table of those header files with varying controllable parameters -- the .inf files I've seen from that exercise had no discernible changes; I'm not sure if she was able to decipher anything else from the headers or not.

Star Strider on 7 Sep 2022

There does not appear to be anything in the ‘EVT.DAT’ or ‘LEADINFO.DAT’ files, however it is possible to read the last two (fourth and fifth) files in the cell arrays. The 'int16' precison appears to be correct for the EKG file.

If ‘Record Gain=2’ translates to 2mV full-scale gain, the voltages are still really strange.

Uz3 = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118395/RecordGain2mVECGamp1mV.zip')

Uz3 = 1×5 cell array

{'RecordGain2mVECGamp1mV/'} {'RecordGain2mVECGamp1mV/EVT.DAT'} {'RecordGain2mVECGamp1mV/LEADINFO.DAT'} {'RecordGain2mVECGamp1mV/SE8ECG4E.DAT'} {'RecordGain2mVECGamp1mV/SE8ECG4E.INF'}

Uz4 = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118560/In.zip')

Uz4 = 1×5 cell array

{'In/'} {'In/EVT.DAT'} {'In/LEADINFO.DAT'} {'In/SE8ECG4E.DAT'} {'In/SE8ECG4E.INF'}

Uz5 = unzip('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1118565/Out.zip')

Uz5 = 1×2 cell array

{'Out/'} {'Out/RG2PS1-30082022-000751.arc'}

Uz33 = Uz3{3}

Uz33 = 'RecordGain2mVECGamp1mV/LEADINFO.DAT'

type(Uz33) % Empty?

Uz42 = Uz4{2}

Uz42 = 'In/EVT.DAT'

type(Uz42) % Empty?

Uz43 = Uz4{3}

Uz43 = 'In/LEADINFO.DAT'

type(Uz43) % Empty?

fidi = fopen(Uz4{4})

fidi = 3

v4 = fread(fidi,'int16');

fclose(fidi);

FullScaleGain = 0.002; % Volts (From 'Uz4{5}')

vs = v4 * FullScaleGain;

L = numel(vs);

Fs = 1024;

t4 = linspace(0, L-1, L)/Fs;

figure

plot(t4, vs)

grid

xlim([2 2.5]*1E+2)

Uz45 = Uz4{5}

Uz45 = 'In/SE8ECG4E.INF'

type(Uz45)

Model=SE-2012V1 Firmware Version=V1.31 Hardware Version=V1.20 Updated Date=2018-01-25 Recorder SN=B41741881 Model Number=IO-SD Record Gain=2 Sample Rate=1024 HP Filter=0.05 LP Filter=60 Start Delay= Record Hours=24H Record Date=2022-08-30 Record Time=00:07:51 Finish Reasons=Manual Stop Date=2022-08-30 Stop Time=00:09:02 Time Passed=00:01:11 EDAN Key=160

fidi2 = fopen(Uz5{2})

fidi2 = 3

v5 = fread(fidi2,'uint8'); % A Binary File (How To Read It?)

fclose(fidi2);

char(v5).'

ans =

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Q î ñ a M P @ V P î ð a L O > U O î ï b L N < S N í î b L M : Q M í í c L M 8 O K ì í c M L 8 M J ë ì c M L 7 K H ë ì d M L 7 H G ê ì d M L 7 F E é ì e M M 8 D D é ì e N N : C C è í f N P = B C ç î f P S @ A C ç ð g Q V E B D æ ò g S Z K D E æ õ g T ^ Q F G æ ø g V c X H J æ û g X i ` L M ç ÿ h [ o i P P ç h ] w s V U è i ` } ] Z é i d e _ ê j g m e ì k k w k í k n £ « q î l r · x ð $l v · Ã ò *m y Á Ï ô /m } Ë Û ¬ ö 4m Õ æ ¸ ø 8m Ý ð Ä ú <m å ø Ð ¦ ü @m ê ÿ Ú þ Cl ï ä ³ Ej ò í ¸ Gh ó ö ½ He ó ü Á Ha { ð þ Â G] t é û Á DY l à õ ¾ @U d Öï º =Q \ Ê ç ¶ 8M S ¿ ÷ ß ° 4I J ³ î × « ÿ 1D A § ä Ï ¥ ý -? 8 Û È ü ); / Ò À ú %7 & È ¹ ù 3 x ¾ ² ÷ / l ´ ª ö + ` © ¢ õ & T ó " úÿH z ñ ðÿ< t ï çÿ0 o î Ýÿ# u z i ì ý Ôÿ j r d ë ù Ëÿ ` j ^ é õ ÂÿÿÿV b Y è ñ ºÿóÿK Z T æ í ²ÿèÿA R O å ê þÿÿàÿ9 L J å è þÿ¬ÿÜÿ3 G G ä ç ÿÿ®ÿÛÿ/ E D ä æ ²ÿÞÿ- C C ä æ ¹ÿâÿ- C C ä æ Áÿèÿ. C D ä ç Êÿïÿ/ D D å ç Õÿ÷ÿ0 E E å è ßÿÿÿ2 G F å è % ê 3 H F å é + õÿ 5 I G å é 1 ÿÿ 7 J H å ê 7 9 L I å ë > ) ; M K å ì D 1 = O L å í K * : ? Q M æ í Q 4 B A R N æ î U < H C S O æ î X A L D T O æ î Z E O E T O æ î [ G P E U O æ î \ H Q D U O æ î ] I R D U O æ î ^ J S E U O æ î ^ K S E V P æ î _ L T E V P æ î _ L T E V P æ î _ K T E V P æ î ^ K T E V O å î ^ K T D U O å î ^ K T D U O å î ^ K T E U O å î _ K T E U P å î _ K T E U P å î _ K T E U O å î _ K T E U P å í _ L T E V P å í _ L T E U P å í ^ K T E U P ä í ^ K T E U P ä í ^ K T E U P ä í ^ K T E U O ä ì ^ K T E U O ä ì ^ K T E U O ä ì _ J T E U P ä ì _ J T F U P ä ì _ J T F U P ä ì ^ J T F U P ä ì ^ J U F T P ä ì ^ J U F T P ä ì ^ J U E T P ä ì ^ J U E T P ä ì ^ J U E U P ä ì ^ J U E U P ä ì ] J U E U P ä ì ] J V E U P ä ì ] J V E T O ä ì ] J V D T O ä ì ^ J V D T O ã ì ^ J V D T O ã ì ^ J V D T O ã ì ^ J V D U O ã ì ^ J W E U O ã ì ^ J W E U O ã ë ] J W E U O ã ë ] J W D U O â ë ] J W D T O ã ë ] J X D T O â ë ^ J X E U O â ê ^ J X E U O â ë ^ K X E U O â ë ^ K X E U O ã ê ] K Y D U O â ê ] J Y D T O â ê ] J Y D T O â ê ] J Y D T O â ê ] J Y D T O â ê ] J Y E T O á ê ] K Y E T O á ê ^ K Z E T O á ê ^ K Z E T O â é ^ K Z E T O â é ] K Z E T O â ê ] K Z E T O â ê ] K Z E T O â ê ] K Z E T O á ê ] K Z F T O á ê ] K [ F U O á ê ] K [ F U O á ê ] K [ F V P á ê ] K [ G V P á ë \ L [ G V P á ë \ L [ F V O á ê \ L [ F V O á ê \ L [ G V O á ê \ L [ G V O á ê \ L [ G V O á ê \ L \ H V O á ê \ L \ H V O á ë \ L \ H W P á ë \ L \ H W P á ë [ L \ H V P á ë [ L \ H V P á ë [ M \ H W P á ë [ M ] H W P á ë [ N ] I W Q á ë [ N ] I X Q á ê \ O ^ J X Q á ê \ O ^ J X Q á ê \ O ^ J X Q á ê \ N ^ J X Q à ê \ N ^ K X Q à ê [ N ^ K Y Q à ê [ O ^ K Y Q à ê [ O _ K Y Q à ê [ O _ K Y R à ê [ O _ L Y R à ë [ O _ L Y R à ë [ O _ L Y R à ê [ O _ L Y R à ê Z O ` L Y R à ë Z O _ L Y R à ë Z O ` L Z R à ê Z O ` M Z S á ê Z P ` M Z S á ê Z P ` M Z S à ê Z Q ` M Z S à ê Z Q ` N [ S ß ê Z Q a N [ S ß ë Z Q ` N [ S ß ë Z Q a N [ S ß ë Y Q a N [ S à ë Y Q a N [ S à ë Y Q a N \ S à ë Y Q a N \ S à ë Y R a O \ T à ë Y S b O \ T ß ê Z S b O \ T ß ê Z R a O \ S ß ê Y R b O \ S ß ê Y S b P \ S ß ê Y S b P \ S ß ë Y S b P ] T ß ê Y S b P ] T ß ê Y S c P ] T Þ ê Y S c Q ] U Þ ê Y S c Q ] U Þ ê Y S c Q ] T Þ ë X S c Q ] T Þ ë X S c Q ] T ß ë X T d Q ^ T ß ë X T d R ^ T ß ë Y T d R ^ U ß ë Y T d R ^ U ß ê Y T d R ^ U ß ê Y T d R ^ U ß ê X T d S ^ U ß ê X T d R ^ U ß ê X T d S ^ U ß ê X T e S ^ U ß ë W U e S _ V ß ê X U e S _ V ß ê X U e S _ V ß ë X U e T _ V ß ë X U e T _ V ß ë X U e S _ V ß ë W U e S _ V ß ë W U e T _ V ß ë W U e T _ V ß ë W U e T ` V ß ë W U e U ` V ß ì W U e U ` V ß ì W V f V ` V ß ë X V f V ` W Þ ë X V g V ` W Þ ì X V g V a W Þ ì Y W h V a W Þ ì Y W h W a X Þ ì Y X i W a X Þ ì Y Y i X a X ß ì Z Y j Y b X ß ì Z Z j Y b Y ß ì Z Z k Y c Y ß ì Z Z k Y c X Þ ì Z Y l Y c Y Þ í [ Y l Z c Y Þ í [ Z m Z d Y Þ í \ Z m [ d Y Þ í \ [ n \ e Z Þ í ] [ o \ e Z Þ í ] \ o ] e Z Þ í ] \ o ] e Z Þ í ^ \ p ] f Z Þ í ^ ] p ] f Z Þ í ^ ] q ] f Z Þ í ^ ] r ^ g Z Þ í ^ ^ s ^ g [ Þ í _ ^ s _ g [ Þ í _ _ s _ g [ Þ í _ _ s ` g \ Þ í ` _ s ` h \ Þ î ` _ t ` h \ Þ î ` _ t ` h \ Þ î ` _ t a i \ Þ î ` ` u a i ] Þ ï a ` u b j ] Þ î a a v c j ] Þ î b a v c j ] Þ î b a v c j ] Þ î b a v c j ^ Þ ï b a v c j ^ Þ ï b ` v d k ^ Ý ï a ` v d k ^ Þ ï b a w e l ^ Þ ï b a w f l _ Þ ï b a x f m _ Þ ï b a x f m _ Þ ï b b x g m ` Þ ï c c x g m _ Þ ï c d y g m _ Ý ð d e z g m _ Ý ð e f { g m _ Þ ï e f | g n _ Þ ð e g | g n _ Þ ï f g | g n ` Ý ï f g | g n ` Ý ï f g | g n ` Ý ï f g { g m _ Ý ï e f { f m _ Ü î e e z f l ^ Ü î e d z e l ^ Ü í e d y e l ^ Ü í d d y e l ^ Û í d d x d k ^ Û í d d w c k ^ Û ì d d w c j ^ Û ì d c v c i ] Û ì d b u b i ] Û ì d b t a i \ Û ë c a s ` i \ Û ë c a s ` h \ Û ë b a r _ h \ Û ë b a r _ g \ Û ê b ` r ^ g [ Ú ê b _ q ] f Z Ú ê b ^ o \ e Y Ú é a ] m [ d Y Ú é a \ l Z c X Ù è a [ j X b W Ù è a Z h W b W Ù ç a Y g V a V Ù æ ` Y e U ` U Ù å _ X d T _ U Ù å _ W b S ^ T Ù ä _ U a Q ] S Ø ä ^ T _ P \ S Ø ã ] S ^ N [ R Ø ã ] R \ M Z Q × â ] R [ L Y Q × â ] Q Y K X P × á \ P X J W P Ö á \ O V I V O Ö à [ O U H V O Ö à [ O T H U N Õ à [ N S G T N Õ ß [ M R F S M Õ ß [ L R E S L Ö Þ Z K Q E R L Ö Þ Z K P D R L Ö Þ Z K P D R L Õ Ý Z J P D R L Õ Ý Z J O D Q L Õ Ý Z J O C Q L Ô Ý Y J N C Q L Ô Ý Y I M B Q K Ô Ü Y I L A P J Ô Ü Y H L @ P J Ô Ü Y H L ? O J Ô Ü Y H L ? O J Ô Ü X H K ? O J Ô Ü X H K ? O J Ô Ü X H K ? O J Ô Ü X G K ? O J Ô Ü X G J ? N I Ô Ü X G J > N I Ô Ü X G J > N I Ô Ü X F J > N I Ô Ü X F J > M I Õ Û X F J > M I Ô Û Y F J > N J Ô Û Y G J ? N J Ô Û Y G J ? N J Ô Û X G K ? N I Ô Û X G J ? N I Ô Û X G J ? N I Ô Û X G J ? N H Ô Û X G J ? N H Ô Û X G J ? M H Ô Û X G J ? M I Ô Û W G J ? N I Ó Û W G J ? N I Ó Û W G J ? N I Ó Û W F J ? M I Ó Û W F J ? M H Ó Ú W F J ? M H Ó Ú W F J ? N H Ó Ú W G J ? N I Ó Ú X G J ? N I Ò Ú X G J ? N I Ò Ù X G J ? N I Ò Ù X G J ? N I Ò Ù X G J ? M H Ò Ù X G J ? M H Ò Ù X G J ? M H Ò Ù X G J ? M I Ò Ù X G J ? N I Ò Ù X G J ? N I Ò Ù X G J ? N I Ñ Ù X F J ? N I Ñ Ù W F I ? M I Ñ Ù W F J ? M I Ñ Ø W F J ? M H Ò Ø W F J ? N H Ò Ø W G J ? N H Ò Ù W G J ? N H Ñ Ù W G J ? M H Ñ Ù W G J ? M H Ñ Ù W G J ? M H Ñ Ù W G J ? M H Ñ Ù W F J > M H Ñ Ù W F I > M G Ñ Ù W F I > M H Ñ Ù W F I > N H Ñ Ù W F I > N H Ñ Ù W F I > N H Ñ Ù W F I > N H Ñ Ø W F I > M H Ñ Ø W F J > M H Ð Ø V F J > M H Ð Ø V F J > M H Ð Ø V F J = M G Ñ Ø W F I > M G Ñ Ø W E J > M H Ð Ø W E J > M H Ð × W F J ? M H Ð × W F J > M H Ð × W F I > M H Ð × W F I > M H Ð × W F I > M H Ð Ö W F I > M H Ð Ö W F I > M G Ð Ö W F I > L G Ð × W F I > M G Ï × W F I > M G Ï Ö W F I > M G Ï Ö W F I > M G Ï Ö W F I > M G Ï Ö V F I = L G Ï Ö V F I = L G Ï Ö V F I = L G Ï Ö W F I = L G Ï Ö W F I > L G Ï Ö V F I > L G Ï Ö V F I > L G Ï Ö U F I > L G Ï Ö U F I > L G Î Ö U E I > L G Î Ö U E H > L G Î Ö U E H > L G Î Ö U E H = L G Ï Ö V E H = L G Ï Ö V E I > L G Î Ö V E I > L G Î Ö V F I > L G Î Ö V E I > K G Î Ö U E I > K G Î Ö U E H > K F Î Ö U E I > L F Î Ö U E I > L G Î Õ V E I > L G Î Õ V E I > L G Î Õ U F I > L G Î Õ U F I > L G Î Õ V F H > L G Í Õ U F H = L F Í Õ U F H = K F Í Õ U F H = L G Í Õ U E H = L G Í Õ U E H = L G Î Õ U E H = L G Í Õ U E H = L G Í Õ U E H = L G Í Ô U D H = K H Í Ô U D H > K G Í Ô U D H = K G Ì Ô U D H = K G Ì Ô U E H = K G Ì Ô U E H = L G Ì Ô V E H > L G Ì Ô V E H > L G Ì Ô V E H > L G Ì Ô V E H > L G Ì Ó U E H = L G Ì Ó U E H = K G Ì Ó U E H = K G Ì Ó U E H = K G Ì Ó U E H > L G Ì Ó U E H = L G Ì Ó V E H > L G Ì Ó V E H > L G Ë Ó V E H = L G Ë Ó U E H = K G Ë Ó U E H = K F Ë Ó U E H = K F Ë Ó U E H < K F Ë Ó U E H < K F Ì Ó U E H < K F Ë Ó U E H = K F Ë Ò U E H = K G Ë Ò U E H = K G Ë Ò U D H = K F Ë Ò U D G < K F Ë Ò U D G < K F Ë Ò U D G < K G Ë Ò U D G < K G Ë Ò U E G = K G Ë Ò U E G = K G Ë Ò U E H = K G Ë Ò U E H = K G Ë Ò T D H = K G Ê Ò T D H = K F Ê Ñ T D H < K G Ê Ò T D G < K G Ê Ò T D G < K G Ê Ò T D G < K F Ê Ò T D G < K F Ê Ò T D G < K F Ê Ò T D G < K F Ê Ñ T D G < K F Ê Ñ T D G < J F Ê Ñ T C G < J F Ë Ñ T C G = K F Ë Ñ T C G = K E Ê Ñ T C G = K E Ê Ñ T C G = K F Ê Ñ T D H > K F É Ñ T D H = K F É Ñ T D G = K F É Ñ S D G < K F É Ñ S C G < J F É Ñ T C G < J F É Ñ T D G < J F É Ð T D G = J F É Ð T D G = K F É Ð T E G = K F É Ð T D G = K F É Ð T D G = K F É Ð S C G < K F É Ð S C G < J F É Ð T C G < J F É Ð T D G < K E É Ï T D F < K F É Ï U D F < K F É Ï U D F < K F É Ï U D F < J F É Ï T D F < J F È Ï T D F < J F È Ï S C F < I E È Ð S C F < J E È Ð S C F < J E È Ï S C F < J E È Ï T C F = J E È Ï T C F = J E È Ï S C G = J E È Ï T C G < J E È Ï T C G < J E È Ï S C F < J E È Ï S C F ; J E È Ï S C F ; J E È Ï S C F < J E È Ï S C F < J E Ç Ï S C G < J E Ç Ï S C G ; J E Ç Ï S C F ; J E Ç Ï S C F ; I E Ç Î S C F ; I E Ç Î R C F ; I D Ç Î S B F < I D Ç Î S B F < I D Ç Î S B F < I D Ç Î S B F ; I D Ç Î S B F < I E Ç Î S C F < I E Ç Î S C E < I F Ç Î S C E ; I F Ç Î S C E ; I E Ç Î S C E ; I E Ç Î S C E ; I E Ç Î S C E ; I E Ç Î S C E ; I E Æ Î R C E ; I E Ç Î R C E ; I D Ç Í R B F ; I D Æ Í R B F ; I D Æ Í R B E ; I D Æ Í R B E ; I D Æ Í R B F ; I E Æ Í R C F ; I E Æ Í R C F ; I D Æ Í R C F ; I D Æ Ì R C F ; I D Å Ì R C F ; I E Å Ì R C F ; I E Å Ì R B F ; I E Å Ì R B F ; I E Å Ì R B F ; I E Å Ì R B F ; I E Å Ì R B F ; I E Æ Ì R B F ; H E Æ Ì R B E ; H D Æ Ì R B E ; H D Æ Ì R B E ; H D Å Ì R B E ; H D Å Ì R B E ; H D Å Ì R B E ; I D Å Ì R B E ; I D Å Ì R B E ; I D Å Ì R C E ; I D Å Ì R C E ; I D Å Ì R C E ; I D Å Ì R C E ; H D Ä Ì R C E : H D Ä Ì R B E : H D Ä Ì R B E : H D Ä Ì R B E ; H D Ä Ë R B E ; I D Ä Ë R B E ; I D Ä Ë R B E ; I D Ä Ë R B E ; H C Ä Ë R B E ; H C Ã Ë Q B E : H C Ã Ë Q B E : H C Ã Ë Q B E ; H D Ã Ë Q B E ; H D Ä Ë Q B F ; I D Ã Ë R C F ; I D Ã Ë R C G J E Ä Ì R E J ? K F Ä Ì R F K @ L F Ä Í S G L A M F Ä Í S H M A M G Ä Í T H N B N H Ä Í T I O C O H Ä Í U I Q D O H Ä Í U J R D P H Ä Î U J S E P I Ä Ï U K S F Q I Ä Ï U L T G Q J Ä Ï U L U H R K Ä Ï U M V I S K Ä Ð U M W I S K Ä Ð V M W J T L Å Ð V M X J T L Å Ð V N Y K T L Å Ð V N Y K U L Å Ñ V O Z L U M Å Ñ V O [ M V N Å Ñ W P \ N V N Å Ò W Q ] N W O Å Ò W Q ] N W O Å Ò W Q ^ O W O Å Ò W R _ O X O Å Ò W R _ P X O Å Ò W S ` P X O Å Ò W S a P X O Å Ò X S a Q Y O Å Ò X T b Q Z O Å Ò X T b R Z P Å Ñ X T b R Z Q Å Ò X T b R Z Q Ä Ò X T b R Z P Ä Ò X T b Q Y P Ä Ò X T b Q Y P Å Ò X T b Q X O Å Ò X T b Q X O Å Ò X T b Q Y O Ä Ò X T b Q Y O Ä Ò X S a Q Y O Ä Ò X S a Q Y O Ä Ñ W S a P Y O Ä Ñ W R a P X O Ä Ð W R a P X O Ä Ð W R ` P X O Ä Ñ W R ` P X O Ä Ñ W Q ` O X O Ä Ñ W Q _ O X O Ä Ð W Q ^ O W O Ã Ð W Q ^ N W N Ã Ï V P ] N V N Ã Ï V P \ M V M Ã Ï V O Z L U M Â Î V N Z K T L Â Î V N Y J T L Â Í V M X J T L Â Í U M X I T L Â Ì U L W I S L Á Ì U K V H R K Á Ì T K U H R J Á Ì T J T G Q J Á Ë S J S F P I Á Ë S I R E O I Á Ê S I Q D O I Á Ê S H Q D N H À É S H P C N H À É S G O C M G À É S F N B M G À É R E L A L F À È R D K ? K E À È Q D J > J E ¿ Ç Q C H = I D ¿ Ç Q B G < H D ¿ Æ P A F ; H C ¿ Æ P A E : G C ¿ Æ P @ D 9 G B ¿ Æ P @ D 9 G B ¿ Æ P @ C 9 G B ¿ Æ P @ C 8 G B ¾ Æ P @ C 8 G B ¾ Æ O @ C 8 F B ¿ Æ O @ C 8 F B ¿ Æ O @ C 8 F B ¿ Å O @ C 9 F B ¿ Å O @ C 9 F B ¿ Ä P @ C 9 F B ¿ Ä P @ C 9 F B ¾ Ä P @ C 9 F B ¾ Ä P @ C 8 F B ½ Ä P ? C 8 F B ½ Å O ? C 8 F A ½ Å O ? C 8 F B ½ Å P ? B 8 F B ½ Å P ? B 9 F B ½ Ä P @ C 9 F A ½ Ä P @ C 9 F A ½ Ä P @ C 8 F A ½ Ä O @ B 8 E A ½ Ä O ? B 8 E A ½ Ä O ? B 8 E A ½ Ä O @ C 8 E A ½ Ä O @ C 8 F B ½ Ä O @ C 8 F B ½ Ä O @ C 7 F B ½ Ä O @ C 7 F A ½ Ä O @ B 7 F A ½ Ä O @ B 7 E A ½ Ä O @ B 7 E @ ½ Ä N @ B 8 E @ ½ Ã N @ B 8 E @ ½ Ã N @ B 8 E @ ½ Ã N @ B 8 E @ ½ Ä N ? A 8 E @ ½ Ä N ? A 8 E @ ½ Ä N ? A 8 E A ¼ Ä N ? A 7 E A ¼ Ã N ? B 7 E A ¼ Ã N ? B 7 E A ¼ Ã N ? B 8 F A ¼ Ã N ? B 8 F A ¼ Ã N ? B 8 E A ¼ Ã N ? B 8 F A ¼ Ã N ? B 8 F A ¼ Ã N ? B 8 F A ¼ Ã N ? B 7 E A ¼ Ã M ? B 7 E A ¼ Ã M ? B 8 E A ¼ Ã N ? B 8 E A ¼ Â N ? B 8 E A » Â N ? B 8 E A » Â N ? B 9 E B » Â N ? B 9 E B » Â N ? B 9 E A » Â N ? B 8 E A » Â N ? B 8 E @ » Â N ? B 7 E @ » Â O ? B 7 E @ » Â O ? B 7 E A » Á O ? B 8 F A » Á O ? B 8 F A » Á O ? B 8 F A » Á O ? B 8 E A º Â N ? B 7 E A º Â N ? B 7 D A º Â N ? B 7 D A º Â N ? B 7 E A º Â N ? A 7 E A º Á N ? A 7 E A º Á N ? B 7 E A º Á N ? B 7 E A º Á N ? B 7 E A » À N > B 7 E @ » À M > A 6 E @ » À M = A 5 D @ ¼ ¿ M = ? 3 D ? ¼ ¾ M < > 0 C > ¼ ½ M < = . B = ¼ ¼ M ; < - A < ¼ » N ; ; + ? ; ¼ º N ; : ) = : » ¹ N ; 9 ' : 9 º ¸ O ; 9 & 8 7 ¹ · O ; 8 % 6 6 ¹ · P < 8 % 3 4 ¸ · Q < 8 % 1 3 ¸ · Q < 9 ' 0 2 · ¸ R = : ) / 2 ¶ ¹ R > < + . 2 ¶ º S ? ? / - 2 µ ¼ S @ B 3 . 2 µ ¾ S A F 9 / 3 µ À S C J ? 1 4 ¶ Ã T E O G 4 7 ¶ Æ T G U O 7 : ¶ Ê U I \ W < > · Î U L c a B B · Ò V O k k I G ¸ × V R s u P L ¹ Ü V U | X Q º á W X a X » ç W ] k _ ½ ì X a § v f ¾ ò Y e ¤ ³ n À ÷ Z i ® ¿ v Â ü Z l · Ê } Ä Z o ¿ Ó £ Ç Z q Ç Ü ® ÉY s Î ä ¹ Ë X t Õ ë Å Í W t Û ó Ð Ï U t ß ù Û § Ñ S r â ü ä ¬ Ò Q o á þ ê ° Ó M j Ü ê ° Ô I c Ô ÿ æ ® Ó E [ Ê û à ª Ò A R À ô Ù ¦ Ñ< J µ í Ò ¢ Ð 8 A « å Ë Ï 3 8 Ý Å Î þ / / Ô ¾ Ì ù + ' Ê ¶ Ë õ ' | À ® É ñ # p µ ¥ È ì d ª ~ Æ è X x Å ä úÿL r Ã ß ñÿ? m Â Û çÿ3 } h À × Þÿ' x u b ¿ Ó Õÿ m m ] ½ Ï Íÿ c e W » Ê ýÿÄÿ X ] R º Æ ùÿ¼ÿøÿN U L ¸ Â õÿ´ÿíÿC M G · ¾ ñÿªÿáÿ8 E A µ º ìÿ¡ÿÕÿ. = < ´ ¶ éÿÿËÿ& 7 7 ´ ´ éÿÿÇÿ! 2 4 ´ ² ëÿÿÇÿ 0 2 ´ ² ïÿ¡ÿÊÿ / 1 ´ ² óÿ©ÿÏÿ / 1 ´ ² ùÿ²ÿÖÿ / 1 ´ ³ ÿÿ»ÿÝÿ 0 1 ´ ³ Åÿåÿ 1 2 ´ ´ Ïÿíÿ 2 2 ´ ´ Ùÿõÿ" 4 3 ´ µ äÿýÿ$ 5 5 ´ ¶ îÿ & 7 6 µ · # ùÿ ( 8 7 µ · * * : 8 µ ¸ 1 , < : µ ¹ 7 ' . = ; µ º = $ . 0 ? < µ º B , 4 1 @ < µ º E 1 8 2 @ = ¶ » G 5 ; 2 A = µ » H 7 = 3 A > µ » I 8 > 3 B > µ º J 9 ? 4 B > ´ º J : @ 5 B > ´ º K : A 5 B > ´ º K : A 4 B > µ º J : @ 4 B = µ º J : @ 4 B = µ º K : @ 4 B = µ º K ; @ 4 B > µ º K ; @ 5 B > µ º K ; @ 5 B > µ º K ; A 4 B > µ º K ; A 4 B > µ º K ; A 4 B > µ º K ; @ 4 A = µ º K ; @ 4 B = ´ º K ; @ 5 B > ´ º L ; A 5 B > ´ º L ; A 5 A > ´ º L ; A 5 A > ´ º L ; @ 5 A = ´ º L ; @ 5 A = ´ º L ; @ 5 A = ³ º K ; A 4 A = ³ º K ; A 4 A = ³ º L ; A 4 A > ³ º K ; B 4 A > ³ º K ; B 5 A > ² º K ; B 5 B > ³ ¹ K ; B 5 A > ³ ¹ K : B 4 A > ³ ¹ J : B 4 A = ³ ¹ J : B 4 A = ³ ¹ I : B 4 A = ³ ¹ I : B 4 A = ´ ¹ I : C 4 A = ´ ¹ I : C 4 A = ´ ¹ J : D 4 A > ´ ¹ J : D 4 B > ³ ¹ J : D 4 B > ³ ¸ J ; D 4 A > ³ ¸ J ; D 3 A > ² ¹ J : D 3 A = ³ ¹ J : D 3 A = ³ ¹ J : D 3 A = ´ ¹ J : D 4 A = ³ ¹ J : D 4 B = ³ ¸ J ; E 4 B = ³ ¸ J ; E 4 B = ² ¸ J ; F 4 A = ² ¸ J : F 4 A = ² ¸ J : F 4 A = ² ¸ J ; F 4 A = ² ¸ J ; G 4 A = ² ¸ J ; G 4 B > ² · J ; G 4 B > ² · J ; G 4 A > ² · J ; G 4 A > ² · J ; G 4 A = ² · J : G 4 A = ± · I : G 4 A = ² ¸ I : H 4 A = ² ¸ I : H 4 A = ² ¸ I : H 4 A = ² ¸ I : H 5 A = ² ¸ I : H 5 B = ² ¸ I ; H 5 B > ² ¸ J ; H 6 B > ± ¸ J ; H 6 B > ± ¸ I < H 6 B > ± ¸ I ± ¸ I ± ¸ I ² ¸ I = I 6 C > ² ¸ I = I 6 C > ² ¸ I = I 7 C > ² ¸ I = I 7 C > ± ¸ I = J 7 C ? ± ¸ I = J 7 C ? ± ¸ I = J 7 D ? ± ¸ I = J 8 D ? ± ¸ I = J 8 E ? ² ¸ I = J 8 E ? ² ¸ I > J 9 E ? ± ¸ I > J 9 E ? ± ¸ I > J 9 E ? ± ¸ I > K 8 E ? ± ¸ I > K 8 E ? ± ¸ H > K 9 E ? ± ¸ H > K : E @ ± ¸ H > L : E @ ± ¸ H > K : E @ ± ¹ H > K : E @ ± ¹ H ? K : E @ ± ¹ H ? K : E @ ± ¹ G ? K : F @ ± ¹ G ? K : E @ ° ¹ G ? L ; F @ ° ¸ H @ L ; F @ ° ¸ H @ L < G A ° ¹ H @ L < G A ± ¹ H @ L < G A ± ¹ H @ M < G A ± ¹ G @ M < G A ± ¹ G @ M < F A ± ¹ G @ M < F @ ± ¹ G @ M < F @ ± ¹ F @ M < F A ± ¹ G A M = G A ± ¹ G A M > G A ± ¹ G A M > H A ± ¹ G A M > H A ± ¹ G A N > H B ° ¹ G A N > H B ° ¹ G A N > I B ° ¹ G B N > I B ° ¹ G B O > I B ± ¹ F B O > I B ± º G B O > I C ± º G B O ? I C ± º F B O ? I C ± º F B O ? I B ± º F B O ? I B ± º F B O ? I B ± º F B O @ I B ± º F C P @ I C ° º F C P @ I C ° º F C P @ J C ° º F C P @ J C ± º E D P A J C ± º E D P A J C ± º E D P A J C ± º E D Q A J C ± º E D Q A J C ° º E E Q A K D ° º E E R A K D ° º E E R B K D ° º E E R B K D ° º E E R C K D ° º E E R C K D ° º E E R C L D ° º E E R C L D ° º E E R C L D ° º D E S C L D ° º D E S C L D ° º D E S D L E ° º E F T D L E ° º E F T D M F ° º E F T E M F ¯ º E G T E M F ¯ º E G T E N F ¯ º E G T E N F ¯ » E G T F N F ¯ » E G U F N F ° » E H V G O F ° » F H V G O F ° » F H V G O F ° » F I W G O F ° » F I W H P F ° ¼ G I W H O G ° ¼ G I X H O G ° ¼ H J X I P G ° ¼ H J Y J P G ° ¼ I K Z J Q H ° ¼ I K Z K Q H ° ¼ I K [ L Q H ° ¼ I K [ L R H ° ¼ I L [ L R H ° ¼ I L [ L R H ° ¼ I L \ M R H ° ¼ J M ] M S I ° ¼ J M ] N S J ° ¼ J N ^ N T J ° ½ K N ^ N T J ° ½ K N _ N T J ° ½ L O _ O T J ° ½ L O ` O T J ° ½ L O ` O T J ¯ ½ L O a P T J ¯ ½ M P a P U K ° ¾ M P b Q U K ° ¾ M P b R V K ° ¾ N P c R V L ° ¾ N Q c R W L ° ¾ N Q d R W L ° ¾ N Q d R X L ° ¾ N Q c S X L ° ¾ N Q c S X L ° ¾ O Q d T X L ± ¾ O Q d T Y M ± ¾ O R d T Y M ± ¾ O R d U Y M ° ¾ O R d U Y N ° ¾ O R d U Z O ° ¾ O R e V Z O ° ¾ P R e V Z O ¯ ¿ P R e V Z O ° ¿ P R f W Z O ° ¿ Q T g X [ O ° ¿ R U h X [ O ° ¿ R V i X [ O ° ¿ S V j X [ O ° ¿ S W j W [ O ° ¿ S W j W Z O ° ¾ S V i W Z N ° ¿ S V i W Z N ° ¿ S V i W Z M ¯ ¾ S V i W Z M ¯ ¾ S W h W Z M ¯ ¾ S V g V Y M ¯ ¾ R V f U Y M ¯ ¾ R U f T X L ¯ ¾ R T e S X L ¯ ¾ Q S d S W K ¯ ½ Q S d R W K ® ½ Q S c R W K ® ¼ Q S c Q W K ® ¼ Q R b Q V K ® ¼ Q R b P V K ® » P R a P V K » P Q ` O U J » O P _ N U J º O O ^ N T I º O N ] M S I º O N \ L R H º N M [ K Q H ¹ O L Z J P H ¬ ¸ N L X I P G ¬ · N K W H O G « · N J U G N F « ¶ M I T E M E « ¶ M H R C L D « µ L G P B K C « µ L F O A J B « ´ K E M ? I A ª ´ K D L > H A ª ³ J C J = G A ª ³ J B I < F @ ª ² J A G ; E @ © ² I @ F 9 D ? © ± I @ D 8 C > © ± I ? C 7 B = © ° H > B 7 A < © ° H > A 6 A < © ¯ H > A 6 A < © ¯ H = @ 5 A < © ¯ H = ? 5 A < © ¯ H < ? 5 @ ; ¨ ¯ H < > 5 @ ; ¨ ® H ; > 4 ? : ¨ ® G ; = 4 ? : ¨ ® G : < 3 > : ¨ ® G : ; 2 > 9 ¨ G 9 : 2 > 9 ¨ G 9 : 1 = 9 § G 9 : 1 = 9 § ¬ F 9 9 0 = 9 § ¬ F 9 9 0 = 9 ¨ ¬ F 9 9 0 = 9 § F 8 9 / < 8 § ¬ E 8 8 / < 8 § ¬ E 9 8 / < 8 § ¬ F 9 8 / < 8 § E 9 8 / ; 8 ¨ E 8 8 / ; 8 ¨ E 8 9 / ; 8 ¨ E 8 9 / ; 8 § F 8 9 / ; 8 § F 8 9 / ; 8 § ¬ F 8 9 . < 8 § ¬ F 8 8 / < 8 § ¬ F 8 8 / < 9 ¦ ¬ F 8 8 / < 9 ¦ ¬ G 9 8 / < 9 ¦ ¬ G 9 9 0 < 9 ¦ « G 8 9 0 < 9 ¦ « F 8 9 / < 9 ¦ « F 8 8 / < 9 ¦ « E 8 8 / < 8 ¦ « F 7 8 / < 8 ¦ « F 7 8 / < 8 ¦ « F 7 8 / < 8 ¦ « F 8 8 / < 8 § « F 8 8 / < 8 ¦ ¬ F 8 8 / ; 8 ¦ ¬ F 8 8 / < 8 ¦ « F 8 8 / ; 8 ¥ « F 8 8 / ; 8 ¥ « F 7 8 / ; 8 ¦ « F 7 8 / ; 8 ¦ « F 7 8 / ; 8 ¦ « F 7 8 / ; 8 ¦ « F 7 8 / ; 8 ¦ « E 8 8 / < 8 ¦ « F 7 8 / ; 8 ¦ « E 7 8 / ; 8 ¦ « E 7 8 / ; 8 ¦ « E 7 8 / ; 8 ¦ ª E 7 8 / : 8 ¥ ª E 7 8 / ; 8 ¥ ª E 8 9 . ; 8 ¥ ª E 8 9 / < 8 ¥ ª F 8 9 / < 8 ¥ © F 8 9 / < 8 ¤ © F 8 9 / < 9 ¤ © F 8 9 / < 8 ¥ © F 7 8 / < 8 ¥ ª F 7 8 / ; 8 ¥ ª E 7 8 . ; 8 ¥ ª E 7 8 . < 8 ¥ ª E 7 8 . < 8 ¥ ª F 7 8 / ; 8 ¥ ª E 7 8 / ; 8 ¤ ª E 7 8 . : 8 ¤ ª E 7 8 . : 8 ¤ © E 7 8 . : 8 ¤ © E 7 8 . ; 8 ¤ ª E 7 8 . ; 8 ¤ ª E 7 8 . ; 8 ¤ © E 7 8 . ; 8 ¤ © E 8 8 . ; 8 ¤ © E 7 8 . ; 8 ¤ © E 7 8 . ; 8 ¤ © E 7 8 . ; 8 ¤ © E 7 8 . : 8 ¤ © E 7 8 . ; 8 ¤ © E 7 8 / ; 8 ¤ © E 7 8 / ; 8 ¤ © E 7 8 . ; 8 ¤ ¨ E 7 8 . ; 8 ¤ ¨ E 7 8 . ; 8 £ © E 7 8 . ; 8 £ © E 7 8 . ; 7 £ © E 7 8 / ; 7 £ © E 7 8 / : 7 £ © E 7 8 / : 7 £ © E 7 7 / : 7 £ © E 7 7 / : 8 £ © E 7 7 / : 8 £ © E 7 7 / ; 8 £ © D 7 7 / : 8 £ © D 7 7 . : 7 £ © D 6 7 . : 7 £ © D 6 7 . : 7 £ © D 6 7 . : 7 £ ¨ D 6 7 . : 6 £ ¨ D 6 7 . : 6 £ ¨ E 7 7 . : 7 £ ¨ E 7 7 . : 7 £ ¨ E 7 7 . : 7 £ ¨ E 6 7 . : 7 £ ¨ D 6 7 . : 7 £ ¨ D 6 7 . : 7 £ ¨ D 6 7 . : 7 £ ¨ D 7 7 . : 7 £ § D 7 7 . : 7 £ § D 7 7 . : 8 £ § D 6 7 / : 8 ¢ ¨ D 6 7 / : 7 ¢ ¨ C 6 7 . : 7 ¢ ¨ C 6 7 . 9 7 ¢ ¨ C 6 6 . 9 6 ¢ ¨ C 6 6 . 9 6 ¢ § C 6 6 . 9 6 ¢ § D 7 7 . : 7 ¢ § D 7 7 . : 7 ¢ § D 7 7 . : 7 ¢ § D 6 7 . : 7 ¢ § C 6 7 . : 7 ¢ § C 6 7 . : 7 ¢ § C 6 7 - 9 7 ¡ § C 6 7 - 9 7 ¡ § C 6 7 . : 7 ¡ § D 6 7 . : 7 ¡ § E 6 7 . : 7 ¡ § E 6 7 . : 8 ¡ § E 6 7 . : 7 ¡ § D 6 7 - : 7 ¡ § D 6 7 - 9 7 ¡ ¦ D 6 7 - : 7 ¡ ¦ C 6 7 - : 6 ¡ ¦ C 6 7 - : 7 ¡ ¦ C 6 7 - : 7 ¡ ¦ C 6 7 . : 7 ¡ ¦ C 6 7 . : 7 ¡ ¦ C 6 7 . : 7 ¡ ¦ C 6 7 - : 6 ¡ ¦ C 6 7 - : 6 ¡ ¦ C 6 7 - : 6 ¦ D 6 7 - : 6 ¡ ¦ D 6 7 - : 6 ¡ ¦ D 6 7 - : 7 ¡ ¦ D 6 7 . : 6 ¦ D 6 7 . : 6 ¦ D 6 7 . : 6 ¦ C 6 7 . : 6 ¦ C 6 7 . : 6 ¦ C 6 7 - : 6 ¦ C 6 7 . : 6 ¦ D 6 7 . : 6 ¥ D 6 7 . : 7 ¥ D 6 7 . : 7 ¥ D 6 7 . : 7 ¥ C 6 8 . ; 7 ¥ C 6 8 . : 7 ¥ C 6 7 . : 6 ¥ C 6 7 - 9 6 ¥ C 5 7 - 9 6 ¥ C 5 6 - 9 6 ¥ B 5 6 - 9 6 ¥ B 5 6 - 9 6 ¥ B 5 7 - 9 6 ¥ C 5 7 . 9 6 ¥ C 5 6 . 9 6 ¥ C 5 6 - 9 6 ¥ C 5 6 - 8 6 ¥ C 5 6 - 8 6 ¥ C 5 6 - 9 6 ¥ C 5 6 - 8 6 ¤ C 6 6 - 9 6 ¤ C 6 7 . 9 7 ¤ C 6 7 . : 7 ¤ C 6 7 . : 7 ¤ C 6 6 . : 7 ¤ C 6 6 - 9 7 ¤ C 6 6 - 9 6 ¤ C 6 6 - 9 6 ¤ C 6 6 - 9 6 ¤ B 6 6 - 9 6 ¤ C 6 6 - 9 6 ¤ C 6 6 - 9 6 ¤ C 6 6 - 9 6 ¤ B 5 6 - 9 6 ¤ B 5 5 - 9 6 ¤ B 5 5 - 9 6 ¤ B 5 5 - 9 6 £ B 5 5 - 9 5 ¤ B 5 6 - 8 6 ¤ B 5 6 - 8 5 ¤ B 5 6 - 9 5 ¤ B 5 6 - 9 5 ¤ B 5 6 - 9 5 £ B 4 6 - 9 5 £ B 5 6 , 8 5 £ B 5 6 , 8 5 £ C 5 6 , 8 5 £ C 5 6 - 9 5 £ C 5 6 - 9 5 £ C 5 5 - 9 5 £ C 5 5 - 9 5 £ C 5 5 , 9 5 £ C 5 6 , 9 5 £ C 5 6 , 9 5 £ C 5 6 , 9 5 £ B 5 6 - 9 5 ¢ B 5 6 - 9 5 ¢ B 4 6 - 9 5 ¢ B 4 6 - 9 6 ¢ B 4 6 - 9 6 ¢ A 4 6 , 9 6 ¢ A 4 5 , 8 5 ¢ A 4 5 , 7 5 ¢ A 4 5 , 8 5 ¢ A 5 5 , 8 5 ¢ B 5 5 , 8 6 ¡ B 5 5 , 8 6 ¡ A 5 6 , 8 5 ¢ A 5 5 , 8 5 ¢ A 5 5 , 8 5 ¢ A 4 5 , 8 5 ¢ A 4 5 , 8 5 ¢ A 4 5 - 8 5 ¢ B 4 5 - 8 5 ¢ B 4 5 - 8 5 ¢ B 4 5 - 8 5 ¢ B 4 5 - 8 5 ¢ B 5 5 - 8 5 ¢ B 4 5 , 8 5 ¡ B 4 5 , 8 5 ¡ B 4 5 , 8 5 ¡ B 4 5 , 7 5 ¡ B 5 5 , 8 5 ¡ A 5 5 , 8 5 ¡ A 5 5 , 8 5 ¡ A 5 5 , 8 5 ¡ A 4 5 , 8 5 ¡ A 4 5 , 8 4 ¡ A 4 5 , 8 4 ¡ A 4 5 , 8 4 ¡ B 4 5 + 8 4 ¡ B 4 5 + 8 4 ¡ B 4 6 , 8 5 ¡ B 4 6 , 8 5 ¡ C 4 5 , 8 5 ¡ B 4 5 - 8 5 ¡ B 4 5 - 8 5 ¡ B 4 5 - 8 5 ¡ B 4 5 , 8 5 ¡ B 4 5 , 8 5 ¡ A 4 5 , 8 5 ¡ A 4 5 , 8 5 A 4 5 , 8 4 A 4 5 , 8 4 A 4 5 , 8 4 A 4 5 - 8 4 A 4 5 , 7 4 A 4 5 , 7 4 @ 4 5 , 7 4 @ 4 5 , 7 3 @ 4 5 , 7 3 A 4 5 , 8 4 A 4 5 , 8 4 A 4 5 , 8 4 @ 4 4 , 8 4 @ 4 4 + 7 4 @ 4 4 + 7 4 @ 4 4 , 7 4 A 5 5 , 7 4 A 5 5 - 8 4 A 4 5 - 8 4 A 4 5 , 8 4 A 4 5 , 8 4 A 4 5 , 7 4 A 4 5 , 7 4 A 5 6 , 8 4 A 5 6 - 8 5 A 6 8 . 9 6 A 6 9 / : 6 ¡ B 7 : 0 ; 7 ¡ B 7 : 1 ; 7 ¡ C 8 ; 2 < 7 ¡ C 8 < 2 < 8 ¡ C 9 = 3 = 8 ¡ C : > 3 > 9 ¢ C ; @ 4 > : ¢ D < A 5 ? : ¢ D = B 6 @ : £ D = C 7 A ; £ D > D 8 A ; £ D > E 8 A ; £ E > E 9 B < £ E > F : B < ¤ E > G : B < ¤ E ? G : C < ¤ E ? H ; C < ¥ F @ I < D < ¥ F A J = E = ¥ F A K > E > ¦ F B K > F > ¦ F B L ? F > ¦ F C M ? G ? ¦ G C N @ G ? ¦ G D N @ G ? § G E O A H ? § H E P A H @ § H E P B H @ ¦ H E P B H @ ¦ H F P C I A ¦ H F Q C I A § G E Q C I @ § G E Q C H @ § F E R C H @ § G E R C I @ § G F R C I @ § G F S D I @ ¦ G F R C I @ ¦ G E R C I @ ¦ G E R C I @ ¦ G E Q C I @ ¦ G D Q B I @ ¦ F D Q B I @ ¦ F E Q B I A ¦ F E P B H A ¦ F E P B H @ ¦ F D O B H @ ¦ E D O A H ? ¦ E D N A H ? ¦ E C N @ G ? ¥ E C N @ G > ¥ F C M ? F > ¤ F B M ? F > ¤ E B L ? E > £ E A K > E > £ E @ J = D = ¢ E @ I = D = ¢ E ? I < C = ¢ D ? H < C < ¢ D > G ; B < ¡ C > F : A ; ¡ C > E : A ; ¡ C = E 9 @ ; ¡ C = C 8 @ : C 8 B 9 > 4 = 8 A 9 = 3 = 7 A 8 < 2 < 7 A 7 : 1 ; 7 A 6 9 1 : 7 A 6 8 0 : 6 @ 5 6 . 9 6 @ 4 5 - 8 5 @ 3 4 , 7 5 ? 3 3 + 6 4 ? 2 3 + 6 3 ? 2 3 * 5 3 ? 2 3 * 6 3 @ 2 3 * 6 3 @ 2 3 + 6 3 @ 2 2 + 6 3 ? 2 2 + 6 3 ? 2 2 * 6 3 ? 2 2 * 5 3 ? 2 2 ) 5 2 ? 2 2 ) 5 2 ? 2 2 ) 6 2 ? 2 2 * 6 2 ? 2 2 * 6 3 ? 2 2 * 6 3 ? 2 2 * 6 3 ? 2 2 * 6 3 ? 2 2 ) 6 2 ? 2 2 ) 5 2 ? 1 2 ) 5 2 ? 1 2 ) 5 2 ? 2 2 ) 5 2 ? 2 2 ) 5 2 ? 1 2 ) 5 2 ? 1 2 ) 5 2 ? 1 2 ) 5 2 ? 0 2 ) 5 2 ? 1 2 ) 5 2 ? 1 2 ) 5 2 > 1 2 ) 5 2 > 1 2 ) 5 2 > 2 2 * 5 2 ? 2 2 * 5 2 > 2 2 * 5 3 > 2 2 * 5 3 > 1 2 * 5 3 > 1 2 * 5 2 > 1 2 * 5 2 > 2 2 * 5 2 > 2 2 * 5 3 > 2 2 * 5 3 > 1 3 * 5 3 > 1 3 * 5 3 > 1 2 * 5 2 = 1 2 * 4 2 = 1 2 ) 4 1 > 1 2 ) 4 1 > 1 2 ) 4 1 > 1 2 ) 4 1 > 1 2 ) 5 1 > 1 2 ) 5 2 > 1 2 * 5 2 > 1 1 * 5 2 > 1 1 ) 5 1 = 1 1 ) 5 1 = 1 1 ( 4 1 > 1 1 ) 4 1 > 2 1 ) 5 2 ? 2 2 * 5 2 ? 2 2 * 5 2 ? 2 2 * 5 2 > 2 2 * 5 2 > 2 2 * 5 2 > 2 2 * 5 1 = 1 1 * 5 1 = 1 1 * 5 1 = 1 1 ) 5 1 = 1 1 ) 5 2 > 1 1 * 5 2 > 1 2 * 5 2 > 1 2 * 5 2 > 1 1 * 5 2 > 1 1 * 5 2 > 0 1 ) 5 1 > 0 1 ) 5 1 > 0 1 ) 5 1 > 0 1 ( 4 1 > 0 0 ' 4 2 > 0 0 % 4 2 = / . # 3 1 = . - ! 2 0 = . , 1 / = . * . - = . ) , , > . ) * * ? . 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Any meaningful information on how to scale them appears to be missing.

.

dpb on 7 Sep 2022

The previous datasheet you posted has available output scalings "Gains" of 5, 10, 20 mm/mV. The user-settable gain factor is 0.5, 1, 2 to provide those scale factors; it's that user setting that is recorded in the .dat header and the .inf file.

If you can introduce identically the same input signals into the device at the different gains, the results should be scaled in that ratio as to their output amplitudes.

Again, we're back to if one could simply input known sine waves from a signal generator of a given amplitude and freguency one would be able to have a know reference from which to work.

I'm still led to wonder if there's any real point in continuing to try to reverse-engineer this product, however. If the purpose of the experiment is to determine the effects of the supposed external RF/EMI on the results; can't that just as well be determined by comparing the results with/without the interference and simply observing the difference? That again, of course, presupposes one can actually manage to introduce this interference into the device and get it by the lowpass filtering and actually manage to corrupt the data signal in the first place...

It would seem a device more along the lines of the Analog Devices evaluation board would be far more amenable to experimentation of this purpose unless the research grant is specific about the use of a given commercial device.

$0.02, imo, ymmv, etc., etc., etc., ...

Star Strider on 8 Sep 2022

‘But, as a clinician, does it really matter all that much what the actual scale is; aren't ECGs read by pattern recognition of the general waveform and not by any absolute magnitude (other than frequencies)?’

Voltages matter. For example, a 1 mm (0.1 mV) or greater Q deflection is significant, usually indicating a transmural myocardial infarction. Also, an R deflection of 12 mm (1.2 mV) or greater in lead II generally is interpreted as left ventricular hypertrophy (usually attributed to hypertension or a stenotic — narrowed — aortic valve). The absolute amplitude of ST-T interval voltages, positive or negative, are also important, for different reasons and with different interpretations. The segment from the end of the P deflection to the beginning of the QRS complex is generally considered to be isoelectric, and for that reason a reference.

The information I was able to recover listed the sampling frequency as 1024 Hz. When I used that to create a time vector, and plotted it with the EKG trace, the intervals and heart rate were appropriate.

I still can’t figure out how to translate the EKG data to absolute voltages.

Star Strider on 9 Sep 2022

As always, my pleasure!

It’ill be interesting to see the results, and I hope they’ll lead to some insight into this. I suggest recording the same signal with different instrumentation parameters in different runs in order to understand how they relate to the recorded signal. (I wish I’d thought of that a few days ago!)

This may be the information we need. If not, well, at least we considered that approach.

Susan on 9 Sep 2022

@Star Strider Sure thing!

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Answer 2

dpb on 31 Aug 2022

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https://nl.mathworks.com/matlabcentral/answers/1791605-how-to-calculate-the-small-changes-in-a-signal#answer_1038730

Edited: dpb on 31 Aug 2022

We've elsewhere determined the data are recorded as signed 16-bit integers which span +/-32K which is a 64K range. A default double has approximately 15 decimal digits of precision (53 bits in the significand) so it can easily hold all the precision available from the instrument without any loss.

Given the number of possible inputs from the device is only the 64K unique values, any more than that number of unique values in a processed input signal will have come only as artifacts during the calculations as long as the calculations are on a per channel basis.

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dpb on 31 Aug 2022

Edited: dpb on 31 Aug 2022

As @Star Strider says, that's all dependent upon the scaling in terms of absolute resolution.

As we discussed before and he brings up again, the answer to that Q? is still dependent upon figuring out "whatcomesoutta" with a given known "whatgoesinna" signal.

As far as "interference", that also will depend wholly upon just what that interference is in amplitude and frequency content as far as what it will mean in interpretation.

What is the source of this supposed interference; where is it entering the signal chain? If it's external RF or somesuch modifying the inputs of the instrument, it'll have the same gains and filtering applied to it as to the real data and its relative effect will be exactly proportional to the signal as is its magnitude in comparison to the input signal at the point it is picked up.

IOW, you can't tell the difference once you've sampled the waveform; if it's significant, it'll undoubtedly wreak havoc on the interpretation, but unless you have some way to characterize it, it'll be difficult at best to know what it is. I don't see that some definition of "resolution" in the instrument has any real bearing on the problem of identifying signal corruption; as above, the instrument itself has an N-bit A/D and is outputting 16-bit signed integers of the scaled magnitude of the inputs. As of yet, we here don't yet know what has been applied to those values -- are they the raw outputs of the A/D scaled to some input voltage range or the actual trace magnitudes scaled back to a 16-bit output range? We dunno until have the results of the tests or you can find (so far missing) documentation of the real content of the data files.

dpb on 1 Sep 2022

Edited: dpb on 1 Sep 2022

Again, it will depend upon how this interference is actually introduced into the system and where that occurs -- if it is an induced voltage in the leads before it gets to the analog input anti-alias and low-pass filtering modules, then it should be pretty-much taken care of -- that's their purpose. Since those kinds of signals are in the MHz and higher frequencies and this device has analog low pass filtering somewhere around 60 Hz, I'd not expect it (the device) to have any appreciable sensitivity in those frequencies. (You can illustrate the effect with an ordinary digital volt-meter and a signal generator -- input the output of a signal generator sine wave at 1V RMS into the DVM on AC volts at 60 Hz, say, then raise the frequency slowly. By the time you get to several kHz and beyond, the meter will have dropped in its shown reading significantly, just how much depending on the specific meter, but by probably 20 kHz it'll read essentially 0 V. The input circuitry simply doesn't have the bandwidth to pass higher frequencies; the device is designed for low frequency inputs. An oscilliscope, otoh, depending on the model, may have bandwidths up into the GHz range.)

I would strongly suspect you'll see the same effect with this device but at much lower frequency inputs -- it simply won't "know" the high-frequency components of the signal are there by the time the signal gets to the A/D to be digitized.

But, if you somehow manage to corrupt the data aftewards, you've got the problem before that you can't really tell anything about what is/is not noise from some "resolution" number; it'll all be mixed up together and present itself in some form of distortion of a waveform that otherwise would have resulted.

As to the Q? you raised about resolution -- the resolution of the sampled data is 1X the input scalling factor; the inferred scaling if they are raw waveforms would seem to be of the right order of magnitude based on @Star Strider's physiological knowledge presuming the digitized waveforms represent those signals and haven't been otherwise normalized internally first.

Again, until can determine just what those data are in relationship to a known input, it's not possible to know with absolute certainty.

dpb on 2 Sep 2022

Edited: dpb on 2 Sep 2022

Since there is now a veritable plethora of personal monitoring devices on the market, it dawned on me somebody must be making chipsets -- the first I found was (almost of course) from Analog Devices. They even have an evaluation board kit. Something of this sort might be more suitable for the kind of project you've embarked upon. <AD8232 Product Evaluation Kit> is a link to the documentation set for the chipset and the evaluation board plus links to other useful information for data acquisition. The Designer's Guide for instrumentation amplifiers is a gold mine of "gotcha's" and "how-to's" to get through the snake pit of low-level signal acquisition.

The data sheet for the AD8232 chipset does, in fact, have a section on the issues with RFI and it has built into it right on the board filtering specifically for the problem (as I'd wager will the product you're using) .

While this is "just" a three-lead setup, from the description I'd venture the sensors for the Holter device are very much similar if not based on the specific chipset -- I did not do an extensive search for more channel devices but they're bound to be out there. It wouldn't seem that there would be any need of having more channels for your purposes of investigation of noise contamination of signals.

Good luck...

dpb on 2 Sep 2022

Edited: dpb on 2 Sep 2022

Unless the device gets saturated at/on the sensor board circuitry itself, I can't see how the analog lowpass filters can't help but prevent such high frequency components from ever showing up.

I don't know for absolute certainty, but from the AD datasheet theory section,the bias voltage simply provides a common reference "ground" on which the actual signal (the output of the differential amplifier @Star Strider mentioned above) "rides" -- it simply moves the overal signal to be sampled by the A/D converter to be in the mid-range of its scale; it's not directly related to the signal level about that reference other than to ensure the pk-pk amplitudes are within the sampling range of the A/D scale. Study the AD datasheet; I'll bet $$ to donuts your device uses virtually the identical circuitry given the descriptions do have, it might even be built around the given chipset or its bigger brother/cousin.

One thing that I didn't understand before is now clear to me though -- the initial transients are real and simply artificacts of the low cutoff frequency of the highpass filter after the leads are connected and the device comes out of saturation. Initially I had figured those artifacts would not be present in that one wouldn't record data until a steady-state base level had been reached. Hence, "back in the beginnng" of this exercise I was doubtful we were actually reading the data file correctly. Clearly, that isn't so; we're pulling the data as it is; there's still some uncertainty as to whether it was written before, during, or after all the internal processing to create the plotted outputs. That's going to be difficult to do really well if you can't find a way to digitize them or get to the actual plot data itself.

dpb on 2 Sep 2022

Edited: dpb on 3 Sep 2022

It has certainly been through the anti-aliasing filters already, yes.

I don't know that we yet know for sure about whether what is written is raw, partially processed or finally processed waveforms, though, do we?

All (at least I) know is they are writing out 16-bit signed integers that reflect the general shape/pattern of the processed results you get for the same dataset -- but you've not yet been able to prove that you can reproduce those by a consistent, unique linear transformation of the those values. But, then again, you haven't yet demonstrated that can't be done, either.

We're back to the original suggestions that need to put the known voltages and waveforms in and see if get identically those voltages out to really be able to determine if those are raw data values or whether they may have been processed and then the output waveforms of the plots converted to 16-bits and written out -- or somewhere in between. One would presume it is either the input or the final output and not someplace on the way between home and the forum they chose to stop and write, but "ya' never know!" what they may have been thinking.

I don't suppose the user manual says anything at all more about what the actual data are, does it? The very short blurb you posted previously only had a (very incomplete) description of the header and that it was 16-bit signed integer, but what those integers represent wasn't mentioned.

dpb on 9 Sep 2022

As @Star Strider commented above, start simple with sine wave of known amplitude and frequency. What you have for test gear will control the "how" and what can be done easily, of course. I'm such an old guy and been out of consulting gig for 20+ years now and what I had then was easily 10 years out of being current I really don't have a handle on any of the current stuff on the market; either from the traditional vendors like Tektronix nor any of the now-ubiquitous miniature stuff built around the PC or a standalone device.

The signal generator doesn't even have to be a calibrated source, precisely, if you have a scope or a DVM -- you can tee the signal into it as well as into the device to measure the voltage...

Susan on 9 Sep 2022

@dpb Thank you so much for your input. Got it.

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