I am doing some simulation in SUMO at high speed 100 Km/h, the space between vehicles is large and i would like to narrow. I think this space is coming because of the high speed. Does there exist any command to control the maximum Gap between vehicles in SUMO same as it exist for the minimum gap "minGap.
This gap is controlled by the time gap parameter tau which can be modified just like the minGap (but has seconds as unit). The default is 1 and commonly used values for automated driving are 0.8 or 0.5 but even down to 0.2. You need to make sure though that your simulation step size (--step-length) is at most the time gap.
I am trying to generate pulse waves with a width of 2 milliseconds and frequency of approximately 100 Hz as shown below:
According to this website: http://www.ni.com/white-paper/2991/en#toc1
under the section "Introduction to Pulse Width Modulation" it describes the duty cycle to be 20% if pulse width is 2 ms with a frequency of 100 Hz (or 10 milliseconds).
As you can see in the diagram above the "duty cycle %" indicator correctly computes a percentage close to 20%.
If I perform the calculations correctly, why am I getting a waveform of pulses that have a width of 3 ms instead of 2 ms shown below?
Following is the back panel diagram containing the logic I am using to generate the waveform:
Your generation frequency is 1 kHz, so you are at the minimum resolution.
Your pulse is 2 ms high. I would advise you to make the samplerate higher.
I need to calculate the FFT of audiodata in an Android Project and I use jTransforms to achieve this.
The samples of the audiodata are a few seconds long and are recorded with a samplerate of 11025 Hertz.
I am not sure how to set the length of the FFT in jTransforms.
I do not really need high frequency resolution, so a size of 1024 would be enough.
But from what I have understood learning about the FFT, if I decrease the FFT size F and use a sample with a lenght of N > F, only the first F values of the original sample are transformed.
Is that true or did I understand something wrong?
If it is true, is there an efficient way to tranform the whole signal and decreasing the FFT-Size afterwards?
I need this to classify different signals using Support Vector Machines, and FFT-Sizes > 1024 would give me too much features as output, so I would have to reduce the result of the FFT to a more compact vector.
If you only want the FFT magnitude results, then use the FFT repeatedly on successive 1024 chunk lengths of data, and vector sum all the successive magnitude results to get an estimate for the entire much longer signal.
See Welch's Method on estimating spectral density for an explanation of why this might be a useful technique.
Im not familiar with the jTransform library, but do you really set the size of the transform before calculating it? Amplitude values of the time-domain signal and the sampling frequency (11.025 kHz) is enough to calculate the FFT (note that the FFT assumes constant sampling rate)
The resolution in frequency domain will be determined by Nyquist's theorem; the maximum resolvable frequency in your signal will be equal to half your sampling rate. In other words, sampling your signal with 11.025 kHz, you can expect your frequency graph to contain frequency values (and corresponding amplitudes) between 0 Hz - 5.5125 kHz.
UPDATE:
The resolution of the FFT (the narrowness of the frequency bins) will increase/improve if your input signal is longer, thus 1024 samples might not be a long sequence enough if you need to distinguish between very small changes in frequency. If thats not a problem for you application, and the nature of your data is not variying quickly, and you have the processing time, then taking an average of 3-4 FFT estimates will greatly reduce noise and improve estimates.
So I am studying for an up and coming exam, one of the questions involves calculating various disk drive properties. I have spent a fair while researching sample questions and formula but because I'm a bit unsure on what I have come up with I was wondering could you possibly help confirm my formulas / answers?
Information Provided:
Rotation Speed = 6000 RPM
Surfaces = 6
Sector Size = 512 bytes
Sectors / Track = 500 (average)
Tracks / Surfaces = 1,000
Average Seek Time = 8ms
One Track Seek Time = 0.4 ms
Maximum Seek Time = 10ms
Questions:
Calculate the following
(i) The capacity of the disk
(ii) The maximum transfer rate for a single track
(iii) Calculate the amount of cylinder skew needed (in sectors)
(iv) The Maximum transfer rate (in bytes) across cylinders (with cylinder skew)
My Answers:
(i) Sector Size x Sectors per Track x Tracks per Surface x No. of surfaces
512 x 500 x 1000 x 6 = 1,536,000,000 bytes
(ii) Sectors per Track x Sector Size x Rotation Speed per sec
500 x 512 x (6000/60) = 25,600,000 bytes per sec
(iii) (Track to Track seek time / Time for 1 Rotation) x Sectors per Track + 4
(0.4 / 0.1) x 500 + 4 = 24
(iv) Really unsure about this one to be honest, any tips or help would be much appreciated.
I fairly sure a similar question will appear on my paper so it really would be a great help if any of you guys could confirm my formulas and derived answers for this sample question. Also if anyone could provide a bit of help on that last question it would be great.
Thanks.
(iv) The Maximum transfer rate (in bytes) across cylinders (with cylinder skew)
500 s/t (1 rpm = 500 sectors) x 512 bytes/sector x 6 (reading across all 6 heads maximum)
1 rotation yields 1536000 bytes across 6 heads
you are doing 6000 rpm so that is 6000/60 or 100 rotations per second
so, 153,600,000 bytes per second (divide by 1 million is 153.6 megabytes per second)
takes 1/100th of a second or 10ms to read in a track
then you need a .4ms shift of the heads to then read the next track.
10.0/10.4 gives you a 96.2 percent effective read rate moving the heads perfectly.
you would be able to read at 96% of the 153.6 or 147.5 Mb/s optimally after the first seek.
where 1 Mb = 1,000,000 bytes
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