what is the proper way to calculate the size in hex bytes of a code segment. I am given:
IP = 0848 CS = 1488 DS = 1808 SS = 1C80 ES = 1F88
The practice exercise I am working on asks what is the size (in hex bytes) of the code segment and gives these choices:
A. 3800 B. 1488 C. 0830 D. 0380 E. none of the above
The correct answer is A. 3800, but I haven't a clue as to how to calculate this.
How to calculate the length:
Note CS. Find the segment register that's nearest to it, but greater.
Take the difference between the two, and multiply by 0x10 (read: tack on a 0).
In your example, DS is closest. 1808 - 1488 == 380. And 380 x 10 = 3800.
BTW, this only works on the 8086 and other, similarly boneheaded CPUs, and in real mode on x86. In protected mode on x86 (which is to say, unless you're writing a boot sector or a simple DOS program), the value of the segment register has very little to do with the size of the segment, and thus the stuff above simply doesn't apply.
Related
I'm configuring the internal temperature sensor of a STM32F4 MCU, and when reading the documentation, I came across some "duplicated" but divergent definitions.
Take a look in the image below:
The temperature sensor is connected to the ADC1, channel 16. When reading the ADC value inside my room, I always get values around ~920;
The values for the calibrations (read from MCU memory) are the following:
TS_CAL1 = 941
TS_CAL2 = 1199
It seems to me that calculating the final temperature using the relationship shown on Table 69 leads to different results from when using the relationship from Table 70.
Can anyone help me in this topic? What's the difference between the data of Tables 69 and 70? What is the purpose of each one? How to calculate the temperature correctly?
As Clifford explained in the comments, the information in table 69 tells you the typical behaviour of any device from this family, whereas the pointers in table 70 give you the address of the calibration data for your particular device which were measured in the factory.
If you told me that some device of this type gave a reading of 920, I would estimate the temperature as follows:
ADC voltage = 920/4096 * 3.3V = 741mV
Voltage offset from V(25C) = 741mV - 760mV = -19mV
Temperature offset from 25C = -19/2.5 = 7.6C
Temperature = 25 - 7.6 = 17.4 degrees C
For your calibrated device I would estimate the temperature like this:
Slope = (1199 - 941) / (110 - 30) = 3.225 LSB/degree
ADC offset from ADC(30C) = (920 - 941) = -21 LSB
Temperature offset from 30C = -21 / 3.225 = 17.775 C
Temperature = 30 - 17.775 = 12.2 degrees C
It is important to note however that although this second number is "calibrated", it is done so using calibration data from much higher temperatures. To use it below 30 degrees requires to extrapolate in a way which may not be physically valid.
Ask yourself, was the room closer to 17 degrees or 12 degrees? Bear in mind that the internal temperature sensor is probably subject to a certain amount of self-heating from the high performance processor.
If you want to use the internal temperature sensor to measure low temperatures outside the calibration range like this it might be appropriate to use the offset from the lower calibration point, but then use the typical slope from the datasheet.
Note also that many STM32 evaluation boards run at 3.0V not 3.3V, so all the calculation will have to be changed if that is the case.
I can not see the table but in theory there are following points need to considers when working with sensor:
Offset: value that is different with the actual value. You may check by feed a constant voltage to system and compare with ADC value measured.
Error of sensor: you might need to measure at known value. For example, for temperature, it is used to measure at 0 temperature which is a steady state.
I've started editing the RaspiStillYUV.c code. I eventually want to process the image I receive, but for now, I'm just working to understand it. Why am I working with YUV instead of RGB? So I can learn something new. I've made minor changes to the function camera_buffer_callback. All I am doing is the following:
fprintf(stderr, "GREAT SUCCESS! %d\n", buffer->length);
The line this is replacing:
bytes_written = fwrite(buffer->data, 1, buffer->length, pData->file_handle);
Now, the dimensions should be 2592 x 1944 (w x h) as set in the code. Working off of Wikipedia (YUV420) I have come to the conclusion that the file size should be w * h * 1.5. Since the Y component has 1 byte of data for each pixel and the U and V components have 1 byte of data for every 4 pixels (1 + 1/4 + 1/4 = 1.5). Great. Doing the math in Python:
>>> 2592 * 1944 * 1.5
7558272.0
Unfortunately, this does not line up with the output of my program:
GREAT SUCCESS! 7589376
That leaves a difference of 31104 bytes.
I figure that the buffer is allocated in fixed size chunks (the output size is evenly divisible by 512). While I would like to understand that mystery, I'm fine with the fixed size chunk explanation.
My question is if I am missing something. Are the extra bytes beyond the expected size meaningful in this format? Should they be ignored? Are my calculations off?
The documentation at this location supports your theory on padding: http://www.raspberrypi.org/wp-content/uploads/2013/07/RaspiCam-Documentation.pdf
Specifically:
Note that the image buffers saved in raspistillyuv are padded to a
horizontal size divisible by 16 (so there may be unused bytes at the
end of each line to made the width divisible by 16). Buffers are also
padded vertically to be divisible by 16, and in the YUV mode, each
plane of Y,U,V is padded in this way.
So my interpretation of this is the following.
The width is 2592 (divisible by 16 so this is ok).
The height is 1944 which is 8 short of being divisible by 16 so an extra 8*2592 are added (also multiplied by 1.5) thus giving your 31104 extra bytes.
Although this kindof helps with the size of the file, it doesn't explain the structure of the YUV output properly. I am having a look at this description to see if this provides a hint to start with: http://en.wikipedia.org/wiki/YUV#Y.27UV420p_.28and_Y.27V12_or_YV12.29_to_RGB888_conversion
From this I believe it is as follows:
Y Channel:
2592 * (1944+8) = 5059584
U Channel:
1296 * (972+4) = 1264896
V Channel:
1296 * (972+4) = 1264896
Giving a sum of :
5059584 + 2*1264896 = 7589376
This makes the numbers add up so only thing left is to confirm if this interpretation is correct.
I am also trying to do the YUV decode (for image comparisons) so if you can confirm if this actually does correspond to what you are reading in the YUV file this would be much appreciated.
You have to read the manual carefully. Buffers are padded to multiples of 16, but colour data is half-size, so your image size needs to be in multiples of 32 to avoid problems with padding breaking external software.
I have a 256 x 256 boolean array. These array is constantly changing and set bits are practically randomly distributed.
I need to send a current list of the set bits to many clients as they request them.
Following numbers are approximations.
If I send the coordinates for each set bit:
set bits data transfer (bytes)
0 0
100 200
300 600
500 1000
1000 2000
If I send the distance (scanning from left to right) to the next set bit:
set bits data transfer (bytes)
0 0
100 256
300 300
500 500
1000 1000
The typical number of bits that are set in this sparse array is around 300-500, so the second solution is better.
Is there a way I can do better than this without much added processing overhead?
Since you say "practically randomly distributed", let's assume that each location is a Bernoulli trial with probability p. p is chosen to get the fill rate you expect. You can think of the length of a "run" (your option 2) as the number of Bernoulli trials necessary to get a success. It turns out this number of trials follows the Geometric distribution (with probability p). http://en.wikipedia.org/wiki/Geometric_distribution
What you've done so far in option #2 is to recognize the maximum length of the run in each case of p, and reserve that many bits to send all of them. Note that this maximum length is still just a probability, and the scheme will fail if you get REALLY REALLY unlucky, and all your bits are clustered at the beginning and end.
As #Mike Dunlavey recommends in the comment, Huffman coding, or some other form of entropy coding, can redistribute the bits spent according to the frequency of the length. That is, short runs are much more common, so use fewer bits to send those lengths. The theoretical limit for this encoding efficiency is the "entropy" of the distribution, which you can look up on that Wikipedia page, and evaluate for different probabilities. In your case, this entropy ranges from 7.5 bits per run (for 1000 entries) to 10.8 bits per run (for 100).
Actually, this means you can't do much better than you're currently doing for the 1000 entry case. 8 bits = 1 byte per value. For the case of 100 entries, you're currently spending 20.5 bits per run instead of the theoretically possible 10.8, so that end has the highest chance for improvement. And in the case of 300: I think you haven't reserved enough bits to represent these sequences. The entropy comes out to 9.23 bits per pixel, and you're currently sending 8. You will find many cases where the space between true exceeds 256, which will overflow your representation.
All of this, of course, assumes that things really are random. If they're not, you need a different entropy calculation. You can always compute the entropy right out of your data with a histogram, and decide if it's worth pursuing a more complicated option.
Finally, also note that real-life entropy coders only approximate the entropy. Huffman coding, for example, has to assign an integer number of bits to each run length. Arithmetic coding can assign fractional bits.
Closed. This question is off-topic. It is not currently accepting answers.
Want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 9 years ago.
Improve this question
Let's say I have a 12-bit Analog to Digital Converter (4096 bins). And let's say I have a signal from 0 to 5 Volts.
What is the proper conversion formula to convert ADC bins into Volts?
V = ADC / 4096 * 5
or
V = ADC / 4095 * 5
Do I divide by 4096 because there are 4096 bins in the ADC?
Or do I divide by 4095 because that is the highest value that the ADC returns?
V = ADC / 4096 * 5
is the correct formula for converting the digital value back to (an approximation of) the analog voltage.
This is according to The Data Conversion Handbook, edited by Walt Kester (Newnes, 2005),
and available at (as of 2018/10/18) at:
https://www.analog.com/en/education/education-library/data-conversion-handbook.html
See in particular Figures 2.4 and 2.5 in Chapter 2:
In your case, FS would be 5 V. (And of course you're using a 12-bit ADC, not a 3-bit one.) Note that even if the ADC value is the maximum possible value (4095 in your case), the corresponding analog voltage will be slightly less than the "full-scale" voltage (5 V in your case).
Brian's suggestion about checking ADC datasheet is ideal. BUT! Assuming your maximum voltage (5V) equates to the maximum ADC input (12-bits = 4095), the following conversion should work for you:
const float maxAdcBits = 4095.0f; // Using Float for clarity
const float maxVolts = 5.0f; // Using Float for clarity
const float voltsPerBit = (maxVolts / maxAdcBits);
float yourVoltage = ADCReading * voltsPerBit;
A quick inspection of the math with Excel leads me to believe this is correct.
How picky do you want to get? If you want to really get picky, then you should also consider that each "bin" represents a small range of values (about 1.2 mV in your case). So when you convert to a voltage value, do you want to return the voltage value of the middle of the bin, or the lower edge of the bin? That is, do you want to effectively "truncate" or "round" in the value you report?
Also, the ADC's steps are probably even (linear), but take care as to what the ADC does with the bins at the two extremities of the range. Those bins may be possibly half the width of the others. It depends on the ADC, so check the spec.
Whether this concern matters at all depends on your application.
The spec for the ADC should identify how 5V is represented in terms of your 12 bits.
I would suspect that 4095 corresponds to 5V and thus your second solution is correct. Otherwise you would never be able to identify a signal of 5V correctly.
For a 12-bit value the maximum representable value is 4095, but of course there are 4096 values total (including zero). Assuming that your ADC is linear then yes, 4095 is equivalent to full scale. This is not necessarily 5V but whatever your reference voltage is equivalent to OR a value exceeding that voltage (of course).
Trying to understand an fft (Fast Fourier Transform) routine I'm using (stealing)(recycling)
Input is an array of 512 data points which are a sample waveform.
Test data is generated into this array. fft transforms this array into frequency domain.
Trying to understand relationship between freq, period, sample rate and position in fft array. I'll illustrate with examples:
========================================
Sample rate is 1000 samples/s.
Generate a set of samples at 10Hz.
Input array has peak values at arr(28), arr(128), arr(228) ...
period = 100 sample points
peak value in fft array is at index 6 (excluding a huge value at 0)
========================================
Sample rate is 8000 samples/s
Generate set of samples at 440Hz
Input array peak values include arr(7), arr(25), arr(43), arr(61) ...
period = 18 sample points
peak value in fft array is at index 29 (excluding a huge value at 0)
========================================
How do I relate the index of the peak in the fft array to frequency ?
If you ignore the imaginary part, the frequency distribution is linear across bins:
Frequency#i = (Sampling rate/2)*(i/Nbins).
So for your first example, assumming you had 256 bins, the largest bin corresponds to a frequency of 1000/2 * 6/256 = 11.7 Hz.
Since your input was 10Hz, I'd guess that bin 5 (9.7Hz) also had a big component.
To get better accuracy, you need to take more samples, to get smaller bins.
Your second example gives 8000/2*29/256 = 453Hz. Again, close, but you need more bins.
Your resolution here is only 4000/256 = 15.6Hz.
It would be helpful if you were to provide your sample dataset.
My guess would be that you have what are called sampling artifacts. The strong signal at DC ( frequency 0 ) suggests that this is the case.
You should always ensure that the average value in your input data is zero - find the average and subtract it from each sample point before invoking the fft is good practice.
Along the same lines, you have to be careful about the sampling window artifact. It is important that the first and last data point are close to zero because otherwise the "step" from outside to inside the sampling window has the effect of injecting a whole lot of energy at different frequencies.
The bottom line is that doing an fft analysis requires more care than simply recycling a fft routine found somewhere.
Here are the first 100 sample points of a 10Hz signal as described in the question, massaged to avoid sampling artifacts
> sinx[1:100]
[1] 0.000000e+00 6.279052e-02 1.253332e-01 1.873813e-01 2.486899e-01 3.090170e-01 3.681246e-01 4.257793e-01 4.817537e-01 5.358268e-01
[11] 5.877853e-01 6.374240e-01 6.845471e-01 7.289686e-01 7.705132e-01 8.090170e-01 8.443279e-01 8.763067e-01 9.048271e-01 9.297765e-01
[21] 9.510565e-01 9.685832e-01 9.822873e-01 9.921147e-01 9.980267e-01 1.000000e+00 9.980267e-01 9.921147e-01 9.822873e-01 9.685832e-01
[31] 9.510565e-01 9.297765e-01 9.048271e-01 8.763067e-01 8.443279e-01 8.090170e-01 7.705132e-01 7.289686e-01 6.845471e-01 6.374240e-01
[41] 5.877853e-01 5.358268e-01 4.817537e-01 4.257793e-01 3.681246e-01 3.090170e-01 2.486899e-01 1.873813e-01 1.253332e-01 6.279052e-02
[51] -2.542075e-15 -6.279052e-02 -1.253332e-01 -1.873813e-01 -2.486899e-01 -3.090170e-01 -3.681246e-01 -4.257793e-01 -4.817537e-01 -5.358268e-01
[61] -5.877853e-01 -6.374240e-01 -6.845471e-01 -7.289686e-01 -7.705132e-01 -8.090170e-01 -8.443279e-01 -8.763067e-01 -9.048271e-01 -9.297765e-01
[71] -9.510565e-01 -9.685832e-01 -9.822873e-01 -9.921147e-01 -9.980267e-01 -1.000000e+00 -9.980267e-01 -9.921147e-01 -9.822873e-01 -9.685832e-01
[81] -9.510565e-01 -9.297765e-01 -9.048271e-01 -8.763067e-01 -8.443279e-01 -8.090170e-01 -7.705132e-01 -7.289686e-01 -6.845471e-01 -6.374240e-01
[91] -5.877853e-01 -5.358268e-01 -4.817537e-01 -4.257793e-01 -3.681246e-01 -3.090170e-01 -2.486899e-01 -1.873813e-01 -1.253332e-01 -6.279052e-02
And here is the resulting absolute values of the fft frequency domain
[1] 7.160038e-13 1.008741e-01 2.080408e-01 3.291725e-01 4.753899e-01 6.653660e-01 9.352601e-01 1.368212e+00 2.211653e+00 4.691243e+00 5.001674e+02
[12] 5.293086e+00 2.742218e+00 1.891330e+00 1.462830e+00 1.203175e+00 1.028079e+00 9.014559e-01 8.052577e-01 7.294489e-01
I'm a little rusty too on math and signal processing but with the additional info I can give it a shot.
If you want to know the signal energy per bin you need the magnitude of the complex output. So just looking at the real output is not enough. Even when the input is only real numbers. For every bin the magnitude of the output is sqrt(real^2 + imag^2), just like pythagoras :-)
bins 0 to 449 are positive frequencies from 0 Hz to 500 Hz. bins 500 to 1000 are negative frequencies and should be the same as the positive for a real signal. If you process one buffer every second frequencies and array indices line up nicely. So the peak at index 6 corresponds with 6Hz so that's a bit strange. This might be because you're only looking at the real output data and the real and imaginary data combine to give an expected peak at index 10. The frequencies should map linearly to the bins.
The peaks at 0 indicates a DC offset.
It's been some time since I've done FFT's but here's what I remember
FFT usually takes complex numbers as input and output. So I'm not really sure how the real and imaginary part of the input and output map to the arrays.
I don't really understand what you're doing. In the first example you say you process sample buffers at 10Hz for a sample rate of 1000 Hz? So you should have 10 buffers per second with 100 samples each. I don't get how your input array can be at least 228 samples long.
Usually the first half of the output buffer are frequency bins from 0 frequency (=dc offset) to 1/2 sample rate. and the 2nd half are negative frequencies. if your input is only real data with 0 for the imaginary signal positive and negative frequencies are the same. The relationship of real/imaginary signal on the output contains phase information from your input signal.
The frequency for bin i is i * (samplerate / n), where n is the number of samples in the FFT's input window.
If you're handling audio, since pitch is proportional to log of frequency, the pitch resolution of the bins increases as the frequency does -- it's hard to resolve low frequency signals accurately. To do so you need to use larger FFT windows, which reduces time resolution. There is a tradeoff of frequency against time resolution for a given sample rate.
You mention a bin with a large value at 0 -- this is the bin with frequency 0, i.e. the DC component. If this is large, then presumably your values are generally positive. Bin n/2 (in your case 256) is the Nyquist frequency, half the sample rate, which is the highest frequency that can be resolved in the sampled signal at this rate.
If the signal is real, then bins n/2+1 to n-1 will contain the complex conjugates of bins n/2-1 to 1, respectively. The DC value only appears once.
The samples are, as others have said, equally spaced in the frequency domain (not logarithmic).
For example 1, you should get this:
alt text http://home.comcast.net/~kootsoop/images/SINE1.jpg
For the other example you should get
alt text http://home.comcast.net/~kootsoop/images/SINE2.jpg
So your answers both appear to be correct regarding the peak location.
What I'm not getting is the large DC component. Are you sure you are generating a sine wave as the input? Does the input go negative? For a sinewave, the DC should be close to zero provided you get enough cycles.
Another avenue is to craft a Goertzel's Algorithm of each note center frequency you are looking for.
Once you get one implementation of the algorithm working you can make it such that it takes parameters to set it's center frequency. With that you could easily run 88 of them or what ever you need in a collection and scan for the peak value.
The Goertzel Algorithm is basically a single bin FFT. Using this method you can place your bins logarithmically as musical notes naturally go.
Some pseudo code from Wikipedia:
s_prev = 0
s_prev2 = 0
coeff = 2*cos(2*PI*normalized_frequency);
for each sample, x[n],
s = x[n] + coeff*s_prev - s_prev2;
s_prev2 = s_prev;
s_prev = s;
end
power = s_prev2*s_prev2 + s_prev*s_prev - coeff*s_prev2*s_prev;
The two variables representing the previous two samples are maintained for the next iteration. This can be then used in a streaming application. I thinks perhaps the power calculation should be inside the loop as well. (However it is not depicted as such in the Wiki article.)
In the tone detection case there would be 88 different coeficients, 88 pairs of previous samples and would result in 88 power output samples indicating the relative level in that frequency bin.
WaveyDavey says that he's capturing sound from a mic, thru the audio hardware of his computer, BUT that his results are not zero-centered. This sounds like a problem with the hardware. It SHOULD BE zero-centered.
When the room is quiet, the stream of values coming from the sound API should be very close to 0 amplitude, with slight +- variations for ambient noise. If a vibratory sound is present in the room (e.g. a piano, a flute, a voice) the data stream should show a fundamentally sinusoidal-based wave that goes both positive and negative, and averages near zero. If this is not the case, the system has some funk going on!
-Rick