I'm looking to generate a wave form generated by a cycle of numbers that increase and then decrease on a given rate. The frequency can vary between 1 to 40 per minute and the amplitude varies between 100 and 3000. The idea is to form a breathing like pattern for "breaths per minute" (1-40) and an inhaled volume per breath (100-3000).
I'm new here and I can only find random generators. I have looked at NSTimer and UIGraphs from the Ios-Developer Tesla tutorial app.
Could anyone point me in the right direction.
Many Thanks.
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I've been given some digitized sound recordings and asked to plot the sound pressure level per Hz.
The signal is sampled at 40KHz and the units for the y axis are simply volts.
I've been asked to produce a graph of the SPL as dB/Hz vs Hz.
EDIT: The input units are voltage vs time.
Does this make sense? I though SPL was a time domain measure?
If it does make sense how would I go about producing this graph? Apply the dB formula (20 * log10(x) IIRC) and do an FFT on that or...?
What you're describing is a Power Spectral Density. Matlab, for example, has a pwelch function that does literally what you're asking for. To scale to dBSPL/Hz, simply apply 10*log10([psd]) where psd is the output of pwelch. Let me know if you need help with the function inputs.
If you're working with a different framework, let me know which, 100% sure they'll have a version of this function, possibly with a different output format in which case the scaling might be different.
I want to process the value from InfluxDB on Grafana.
The final demand is to show how many miles the current vehicle has traveled in a certain time frame.
You can use the formula: average velocity * time.
Do the seniors have any good methods?
So what I'm thinking is: I've got the mean function for the average speed over a fixed period of time and the corresponding mileage, and then I want to add all the mileage together. How do I do that?
What if you only use SQL?
1.) InfluxDB uses InfluxQL, not a SQL
2.) Your approach average velocity * time is innacurate
3.) Use suitable InfluxDB functions, I would say INTEGRAL() is the best function for this case + some basic arithmetic. Don't expect the 100% accuracy. Accuracy depends heavily on the metric sampling, e.g. 1 minute sampling - but what if vehicle is driving 59 seconds and it is not moving for that second when sampling is happening. So don't be supprised, when even 10 sec sampling will be inacurrate.
A current source is exciting a Loud by an AC current of ±5mA. The voltage through the loud is measured using the NI data acquisition. The resistance of the loud changes with time, the peak-to-peak amplitude of the voltage signal changes accordingly. How to define the relationship between the loud's resistance and the voltage peak-to-peak amplitude?! in other words, how can I plot the graph of signal's peak-to-peak amplitude as a function of time in LabView?
measure with appropriate Sample-Rate, at least 10 times higher than the max. frequency of your signal.
use a DAQ, that has "synchronous sampling"
measure current and voltage (synchronously with high sample rate). You can use either a shunt or a current transducer for current measurement
Sample in "Blocks". This means: let the DAQ device store e.g. 10k Values (at a Sample rate of 100kHz) in it's internal memory, and read that buffer every 100 ms. Go to the Example finder (Help -> Find exampes) and look for "continous analog measurement" examples.
calculate the RMS-Value of both signals of each block and plot that in a graph. If you want it simple, feed both signals into a "Chart".
if the current is constant (which should be a strait line in the graph!), the voltage should rise over time, when the inner resistance of the loudspeaker rises ...
Note: be aware, that with the example numbers above (100kHz SR, Blocksizte 10k) calculating the RMS value will produce wrong results, when your signal main frequency is below 10 Hz!
How do you interpret "frames per second" when it comes to computer vision models?
If my model can analyze each test image in .50 seconds, does that mean it's 2 frames per second since it can analyze 2 images in one second?
Yes, you are right. At the end it only depends on where you use it. Online/offline, video or photo. The construct of frames per seconds processing speed (FPS) is a way to express how fast the method is. Often you can bump it to the term "real time" which is a way to express that the method can process a video sequence "online" with the asumption of the video having a frame rate of 25 (FPS). The term "real time" is often disliked, because in and of itself it does not state anything.
According To This Article about Throughput and Latency H
"When You Go To Buy a Water Pipe, There Are Two Completely Independent Parameters That You Look At: The Diameter of the Pipe and Its Length"
But I Think These Two Parameters Are Related. Throughput Is Measured As Per Unit Time, So A Long Latency Will Affect Throughput, Say, If The Droplet Is Fast, More Of Them Will Pass The Pipe In One Second,
Can Any One Help Me Understand This?
EDIT:
the confusion is originated from counting queuing time as part of latency which we should not. Once a request is handled, the latency is independent of throughput.
Let me give you another anology...Think of a car travelling on a single lane road from location A to location B..time taken by that car to travel from A to B is your latency...and the number of cars travelling at an interval, maintaining the latency is your throughput.
The factors that affect here is your medium of travel ie by road and no of lanes on the road.
You're thinking about frequency. Say you have a window into the water pipe at some given point, and you send water droplets at some constant interval (say 1 droplet ever second). You count how often you see a single droplet pass by, and take the inverse (1/seconds). So if you count 1 second of elapsed time between droplets being observed, then you have a frequency of 1Hz.
Now say that you keep this frequency constant (1Hz), but you elongate the pipe. You send one droplet down and count how much time elapses before it reaches the end of the pipe. So say it takes 2 seconds for a single drop to travel from the beginning to the end of the pipe, then you have a latency of 2 seconds.
Now say that you widen the diameter of the pipe, and now you are able to send 2 droplets with a frequency of 1Hz. At the end of the pipe you will count 2 droplets coming out every second. So your throughput will be 2 droplets per second.
Here is my bit in a language which I can understand
When you go to buy a water pipe, there are two completely independent parameters that you look at: the diameter of the pipe and its length. The diameter determines the throughput of the pipe and the length determines the latency, i.e., the time it will take for a water droplet to travel across the pipe. Key point to note is that the length and diameter are independent, thus, so are are latency and throughput of a communication channel.
More formally, Throughput is defined as the amount of water entering or leaving the pipe every second and latency is the average time required to for a droplet to travel from one end of the pipe to the other.
Let’s do some math:
For simplicity, assume that our pipe is a 4inch x 4inch square and its length is 12inches. Now assume that each water droplet is a 0.1in x 0.1in x 0.1in cube. Thus, in one cross section of the pipe, I will be able to fit 1600 water droplets. Now assume that water droplets travel at a rate of 1 inch/second.
Throughput: Each set of droplets will move into the pipe in 0.1 seconds. Thus, 10 sets will move in 1 second, i.e., 16000 droplets will enter the pipe per second. Note that this is independent of the length of the pipe. Latency: At one inch/second, it will take 12 seconds for droplet A to get from one end of the pipe to the other regardless of pipe’s diameter. Hence the latency will be 12 seconds.