I'm trying to measure the frequency of a simulated digital signal per SOLVED: measure frequency of digital random signal. I don't need an averaging window because my signal will be consistent and not random.
CODE:
This is my code, which is identical to the code from the solution, minus the averaging window:
QUESTION:
My simulated signal is a 0-5 VDC square wave with a frequency of 100 Hz and a sampling rate of 1000 Hz. Why is it outputting that my frequency is 125 Hz? (NOTE: Regardless of what value I enter for frequency, I measure a higher value than the true value)
It was a stupid logic error on my end. I forgot to add dt to Period one last time before enqueuing onto the Periods queue.
Periods holds the length of time of each recorded half period of the signal (hence the divide by two at the end before outputting to the graph). In this example, the signal frequency was 100 Hz (period 0.01 sec), so every half cycle is 0.005 sec long. By not adding the last sampling period before enqueuing, I was saying every half cycle was 0.004 sec. This is why the frequency was 125 Hz ( 0.5 cycles / 0.004 sec = 125 Hz).
Therefore, since we should add dt to Period regardless of whether a zero-crossing has occurred, we fix the code by moving the addition from the False case of the select statement to before the select statement, and simply pass the queue and sum through the False case (see below).
IF True:
IF False:
NOTE:
One could use the Express >> Signal Analysis >> Timing and Transitions VI to measure frequency, but it wasn't working for me when I started out for some reason, and more importantly I wanted to learn how to program this myself.
Related
I have an output of a spectrogram with the shape of (8193, 2110). This means that I have 2110 time frames and 8193 frequency bins. I want to convert time bins to real time. This spectrogram is related to a one day long file, so instead on time frame I need the exact real time like : 12:32:45.
A single FFT in your spectrogram represents a time window of duration T = N / Fs, where N = FFT size and Fs = sample rate. If you have no overlap between successive FFTs then the i’th FFT represents a time window of duration T, starting at time i*T.
If you have overlap between successive FFTs, e.g. k = 0.5 (50% overlap), then i*T obviously becomes (1-k)*i*T.
I have retrieved some signal in my Abaqus simulation for verification purpose. The true signal shall be a perfect sinusoid at 300kHz and I performed fft on the sampled signal using scipy.fftpack.fft.
But I got a strange spectrum as shown below (sorry that I am too lazy to scale the x-axis of the spectrum to the correct frequency). In the same figure, I sliced the signal into pieces and plotted in the time domain. I also repeated the same process for a pure sine wave.
This totally surprises me. As indicated below in the code, sampling frequency is 16.66x of the frequency of the signal. At the moment, I think it is due to the very little error in the sampling period. In theory, Abaqus shall sample it in a regular time interval. As you can see, there is some little error so that the dots in my signal appear to be thicker than the perfect signal. But does such a small error give a striking difference in the frequency spectrum? Otherwise, why is the frequency spectrum like that?
FYI1: This is the magnified fft spectrum of my signal:
FYI2: This is the python code that was used to produce the above figures
def myfft(x, k, label):
plt.plot(np.abs(fft(x))[0:k], label = label)
plt.legend()
plt.subplot(4,1,1)
for i in range(149800//200):
plt.plot(mysignal[200*i:200*(i+1)], 'bo')
plt.subplot(4,1,2)
myfft(mysignal,150000//2, 'fft of my signal')
plt.subplot(4,1,3)
[Fs,f, sample] = [5e6,300000, 150000]
x = np.arange(sample)
y = np.sin(2 * np.pi * f * x / Fs)
for i in range(149800//200):
plt.plot(y[200*i:200*(i+1)], 'bo')
plt.subplot(4,1,4)
myfft(y,150000//2, 'fft of a perfect signal')
plt.subplots_adjust(top = 2, right = 2)
FYI3: Here is my signal in .npy and .txt format. The signal is pretty long. It has 150001 points. The .txt one is the raw file from Abaqus. The .npy format is what I used to produce the above plot - (1) the time vector is removed and (2) the data is in half precision and normalized.
Any standard FFT algorithm you use operates on the assumption that the signal you provide is uniformly sampled. Uniform in this context means equally spaced in time. Your signal is clearly not uniformly sampled, therefore the FFT does not "see" a perfect sine but a distorted version. As a consequence you see all these additional spectral components the FFT computes to map your distorted signal to the frequency domain. You have two options now. Resample your signal i.e. it is uniformly sampled and use your off the shelf FFT or take a non-uniform FFT to get your spectrum. Here is one library you could use to calculate your non-uniform FFT.
I was going thru this book to understand wavelets. Its a beautifully written not much technical document.
web.iitd.ac.in/~sumeet/WaveletTutorial.pdf
But in its very first chapter it describes below figure with explanation:
The frequency is measured in cycles/second, or with a more common
name, in "Hertz". For example the electric power we use in our daily
life in the US is 60 Hz (50 Hz elsewhere in the world). This means
that if you try to plot the electric current, it will be a sine wave
passing through the same point 50 times in 1 second. Now, look at the
following figures. The first one is a sine wave at 3 Hz, the second
one at 10 Hz, and the third one at 50 Hz. Compare them
But I am unable to understand what X and Y axis values represents. The X values range is in between [1,-1] so I am assuming it is value of the signal while Y axis is representing the time in milliseconds (1000ms = 1 sec). But then the document goes on further to state the representation of same signal in frequency-amplitude domain:
So how do we measure frequency, or how do we find the frequency
content of a signal? The answer is FOURIER TRANSFORM (FT). If the FT
of a signal in time domain is taken, the frequency-amplitude
representation of that signal is obtained. In other words, we now have
a plot with one axis being the frequency and the other being the
amplitude. This plot tells us how much of each frequency exists in our
signal.
But I am not able to understand what does in the upper graph X and Y axis values represents - shouldn't is be Frequency (X Axis) and Amplitude (Y axis) - if I am correct then why does Y axis has values ranked as 0,200 and 400 - shouldn't it be between range [1,-1] or rather [0,1]?
For the time domain signals the X axis is time and the Y axis is amplitude.
For the frequency domain equivalents the X axis is frequency and the Y axis is magnitude.
Note that when using most FFTs there is a scaling factor of N, where N is the number of points, so the magnitude values in the frequency domain plots are much greater than amplitude of the original time domain signal.
As Paul R wrote above, in the first image the horizontal X-axis represents time with the units ms.
The time interval has the length 1000ms.
The vertical Y-axis represents the amplitude of the signal. However, in the diagram the unit is not Volt, but it is normalized to amplitude 1.
If you perform a Fourier Transformation on that time signal, you will get a frequency spectrum.
If you use a DFT (Discrete Fourier Transformation) or a FFT (Fast Fourier Transformation),
the result depends on the implementation of the algorithm.
a) If the algorithm delivers a normalized result, the amplitude of your frequency line is 0.5 (if the amplitude of your input signal is 1).
b) If the algorithm delivers a non normalized result, the amplitude of your frequency line is half the value of the number of DFT/FFT input values.
Your frequency line has the value of 500, which means the algorithm does not use normalization and the number of input samples was 1000.
Now, what is represented by the horizontal X-axes in the frequency domain?
In the time domain, the length of your time input interval is T = 1000ms = 1s.
Therefore the distance between the frequency lines in the frequency domain is df = 1/s = 1Hz.
As we know from the amplitude in the frequency domain, the input signal in time domain had 1000 samples. This means the sampling time was dt = T/1000 = 1s/1000 = 1ms.
Therefore the total frequency interval F = (fmin, ..., fmax) in frequency domain is 1/dt = 1/1ms = 1kHz.
However, the range does NOT start at fmin = 0 Hz and ends at 1kHz, as one could assume inspecting the upper diagram in the second image. The spectrum calculated by a DFT/FFT contains a positive and a negative frequency range. This means you get a frequency range: (-500Hz, -499Hz, -498Hz, ... -1Hz, 0Hz, 1Hz, 2Hz, ..., 498Hz, 499Hz). The value 500Hz does not exist!
However, for the user's convenience the spectrum is not output in this order, but it is shifted by 500Hz (F/2). This means the spectrum starts with the DC value:
0Hz, 1Hz, 2Hz, ..., 498Hz, 499Hz, -500Hz, -499Hz, -498Hz, ..., -2Hz, -1HZ.
Because the spectrum of a real input function is hermitian Y(f) == Y(-f)*, the positive band carries the complete information. So, you can cut off the negative side band.
The upper diagram in the second image shows two peaks. The first peak appears at f = 50Hz and the second peak is shown at f=950Hz. However, this is not correct. The labels of the horizontal axes are wrong. The second peak appears at f = -50Hz.
In the lower diagram the frequency range ends at 500Hz (499Hz would be correct)a). The range of the negative frequencies is cut off.
I'm newbie in cepstrum analysis. So that's the question.
I have signal with the length 4096 and sample rate 8000 Hz. I make FFT and get the array with the length 4096*2 (2*i position is for cosinus coeff, 2*i+1 position is for sinus coeff). Frequency step is (sampleRate/signalLength == 8000/4096). So, I can calculate frequency at i position this way: i*sampleRate/signalLength.
Then, I make the cepstrum transformation. I can't understand how to find quefrency step and how to find frequency for given quefrency.
The bin number of an FFT result is inversely proportional to the length of the period of a sinusoidal component in the time domain. The bin number of a quefrency result is also inversely proportional to the distance between partials in a series of overtones in the frequency domain (this distance often the same as a root or fundamental pitch). Thus quefrency bin number would be proportional to period or repeat lag (autocorrelation peak) of a harmonically rich periodic signal in the time domain.
So I am writing a piece of code where I have to detect different movement gestures using Accelerometer and gyroscope in iOS 4.3 above.
Q1: Is there any existing opensource code which has achieved any movement/gesture detection?
If not
Q2: For now I want to detect if the iPhone is dropped.
What I have achieved so far:
CoreMotion API gives userAcceleration, which (afaik) is the acceleration that the user is giving to the device or the acceleration of device in some direction (x, y or z) with out considering the gravity so what I can do is: store, let's say, previous 5-6 values of acceleration parameters and can check where any of them hits large negative value, which basically represents the sudden deceleration.
But this solution is not very optimal, I think I need to somehow detect the freefall/downwards motion of the device too first. Any idea how to approach this problem?
UPDATE:
Thanks Misch for sharing your approach. I was not at all thinking about total acceleration. I did exactly what you said:
"You would however have to test yourself, what means "total acceleration corresponds
approximately to earth acceleration" and "for some time" in your case."
The acceleration value is actually in G's so I tested the "total acceleration" values with in range of 0.9 - 1.1. And I checked them over some time, initially I checked four consecutive values when updateInterval is set to 1/20.0. For now I have relaxed the condition of consecutiveness a little.
Here's a sample output:
Acceleration = 0.090868
Acceleration = 0.074473
Acceleration = 0.159797
Acceleration = 1.157513
Acceleration = 1.224588
Acceleration = 1.036272
Acceleration = 0.914698
Acceleration = 0.904093
Acceleration = 0.941516
Acceleration = 0.046362
Acceleration = 0.045109
Acceleration = 0.060045
I think, I still have to keep on testing and adjusting values. If you have any optimization in mind kindly do share, I know in order to help you'd need to see the many samples of freefall acceleration values. For now I am thinking to:
round off the values of acceleration to 3 decimal places and play with the acceleration range I am using.
May be check if, after the freefall condition is met, total acceleration value suddenly goes down.
Do you think the acceleration values I have quoted are a bit noisy?
To Q2:
Checking for large negative values does not tell you whether the phone is being dropped.
First, the user could just move the phone with a rapid gesture, this would also result in a large (maybe negative) value.
Secondly, the phone could fall in another direction than you imagine, and therefore, the acceleration could be positive, although the phone is moving towards the ground.
You could just calculate the total acceleration (a = sqrt(ax^2 + ay^2 + az^2)) and check whether this total acceleration corresponds approximately to earth acceleration (9.81). The phone is falling if the acceleration corresponds to earth acceleration for some time.
You would however have to test yourself, what means "total acceleration corresponds approximately to earth acceleration" and "for some time" in your case.
The physics behind this:
Let's assume you drop your phone in a way, that the y axis of the phone shows upwards. Then the x and z accelerations will be 0 and the y acceleration will be like this:
The acceleration will be 0 in the beginning, will then reach -9.81 the moment, you release your phone. Then it will hit the ground, which you see in the small acceleration peak, and then the acceleration is zero again.
However you can't use only the acceleration in y direction, because your phone may fall in a different angle.
So you have to watch the total acceleration of your phone:
Here you don't see a negative acceleration anymore. However, here it doesn't matter anymore in which direction your phone is facing, as a free fall will always be characterized by an acceleration of 9.81.
To your edits:
1. Why would you want to round off the values of acceleration to 3 decimal places? If you test 0.9 < x < 1.1, it doesn't matter if x is 0.914698 or 0.915.
2. What if someone drops the phone but then catches it again, so the total acceleration does not necessarily have to go down again. Additionally, there should be a large acceleration value (a sudden deceleration) the moment the phone hits the floor. Maybe the reason one does not see this in your values is that it is so short, that it falls between two consecutive measurements. However this could be measured, so don't suppose that immediately after the free fall, the acceleration should decrease again.