kandi X-RAY | Waveform Summary
kandi X-RAY | Waveform Summary
Top functions reviewed by kandi - BETA
- Show wave datas
- Show lines
- Set the width of a wave line
- Shows the specified lines
- Show lines in normal range
- Region Drawable
- Draw the grid
- Draw path
- Draw the wave line
- Draws the wave line
- Show wave show data
- Show a line
- Destroy the wavelet
- Stop the timer
- Sets the background color
- Initialize the attributes
- Set the grid line line color
- Set the color of the wave line
Waveform Key Features
Waveform Examples and Code Snippets
Trending Discussions on Waveform
I'm trying to simulate a pulse width modulate (PMW) waveform generator and getting a syntax error in ISE. Checked fuse.xmsgs and found out it's near counter. Can someone point out the syntax error, please?...
ANSWERAnswered 2021-Jun-10 at 13:14
counter count := 0;
This is illegal syntax, as you didnt declare the object class (signal, constant, variable). You need to use the format:
signal count : counter := 0
This is also illegal, as you are comparing an integer to a
std_logic_vector that you havent included a package for. You need to convert the slv to an
if (count <= unsigned(width)) then
reset is missing from the sensitivity list
I am currently trying to run a windowed CRQA in R using heart and respiration waveform signals. I have to run 94 windowed CRQA, where each signal has 20000 - 50000 data points. Hence the computational load is relatively high and takes forever. Therefore I am trying to get R to increase the memory size and use multiple cores. Both this does not seem to work with the folllowing code:...
ANSWERAnswered 2021-Jun-10 at 09:47
You need to use
clusterMap() or another function from the
parallel package rather than
Look those options up with:
You are also currently mixing things up by introducing
registerDoParallel(cl), which is from the
foreach package. That would require you to then use
foreach() and its helper
%dopar%. If you do not use
foreach() then you do not need
The relevant part of your code would look something like this with
clusterMap(). I've tidied it a bit but I can't test it on my machine.
I need to render moving audio waveform like iOS voice memo app. Here I maintain waveform:[Int] rms amplitude of waves.
Now when new waves come add it in waveform[Int] and I add new
UIBezierPath line at right of
CAShapeLayer and translate whole
CAShapeLayer by 5 points.
But translation animation is not so smooth. Could you please suggest any better approach for it?
My current implementation:...
ANSWERAnswered 2021-May-29 at 12:59
Don't try to do your animation in
draw(rect:). That puts all the work on the CPU, not the GPU, and does not take advantage of the hardware-accelerated animation in iOS.
I would suggest instead using a
CAShapeLayer and a
CABasicAnimation to animate your path.
CGPath from your
UIBezierPath into the
CAShapeLayer, and then create a
CABasicAnimation that changes the path property of the shape layer.
The trick to getting smooth shape animations is to have the same number of control points for every step in the animation. Thus you should not add more and more points to your path, but rather create a new path that contains a graph of the last n points of your waveform.
I would suggest keeping a ring buffer of the n points you want to graph, and building a GCPath/UIBezierPath out of that ring buffer. As you add more points, the older points would "age out" of the ring buffer and you'd always graph the same number of points.Edit:
Ok, you need something simpler than a ring buffer: Let's call it a lastNElementsBuffer. It should let you add items, discarding the oldest element, and then always return the most recent elements added.
Here is a simple implementation:
I have a 1024 samples and I chucked it into 32 chunks in order to perform FFT on it, below is the output from FFT:...
ANSWERAnswered 2021-May-26 at 22:27
Your signal is a sine wave. You chop it up. Each segment will have the same frequency components, just a different phase (shift). The FFT gives you both the magnitude and phase for each frequency component, but after
abs only the magnitude remains. These magnitudes are necessarily the same for all your chunks.
I am looking to render audio waveform on screen like iOS voice memo app, so I am using AVAudioEngine and installed Tap on input node. But it gives fastest call back on 0.1sec frequency , I need to get buffer data on faster frequency , so that can draw more and more waveform on screen for smooth and realtime visualisation.
Could you please sugger better approach for it ?
let inputNode = self.audioEngine!.inputNode let format = inputNode.inputFormat(forBus: 0)...
ANSWERAnswered 2021-May-26 at 13:13
I think you misunderstand how to handle this. Instead of getting very small amounts of new data at a high frequency, you should factor your code to get a larger chunk of data on a longer time interval.
You can then set up your display code to start an animation to animate in a graph of the new data over the time interval you are given. 1/10th of a second is actually a pretty short time interval for an animation.
I would suggest averaging the amplitude of the data for that 1/10th of a second time interval and animating that new average data into place.
BTW, see the update I posted to the answer to your question about creating an animated graph of the data.
The animation I came up with looks like this:
I am using EDA Playground with Aldec Riviera simulator, and I have this module here:...
ANSWERAnswered 2021-May-23 at 14:27
In the testbench, you declared the
result signal, but it is not connected to anything. You probably intended it to be driven by the
alu output of the same name. In that case, you should connect it to the instance:
while drawing a GL_LINE_STRIP I noticed the drawing is outside of window bounds, I can't see half of the drawing.
I want to center it within the window and automatically resize it to fit it. can someone redirect me to how can I achieve that?
this is what I'm seeing:
I want it to be in that size & position:
This is my code:...
ANSWERAnswered 2021-May-21 at 11:06
To solve the isseu, you need to know the bound of your curve. You have to change the the orthographic projection:
glOrtho(-6, 6, -6, 6, -1, 1);
In Orthographic Projection, the view space coordinates are linearly mapped to the clip space coordinates. The viewing volume is defined by 6 distances (left, right, bottom, top, near, far). The values for left, right, bottom, top, near and far define a cuboid (box).
If you want to center the view on the curve, the center of the projection must be the center of the curve. e.g.:
This is more of a math problem than a programming problem, but it has now come up for me twice in two different programming projects.
I would like to create a function,
w_sweep(t, f0, t0, f1, t1), that samples at time
t a sine wave that sweeps through frequencies from
t == t0 to
t == t1. (You can assume frequencies are in Hz and times are in seconds. I don't care what happens when
t is outside the range
I started by writing a simpler function that uses a constant frequency:...
ANSWERAnswered 2021-May-20 at 14:29
It can be wise to look at the reverse, the change in periods. For a frequency f0, you have a period T0 = 1/f0. For a frequency f1, you have a period T1=1/f1.
Now, when you evaluate
w_sweep(t), you're not using one period T0, then a period
lerp(T0,T1,...) etcetera. Instead, you determine a single period
T = 1.0/std::lerp(f0, f1, (t - t0)/(t1 - t0)), and then calculate
sin(2*pi*t/T) which is of course
I think what you want (not entirely sure) is
f=1/lerp(1/f0, 1/f1, (t - t0)/(t1 - t0))
I am trying to build a Finite State Machine vending machine which consists of Datapath Unit and Control Unit. The attached links is the Control unit which consists the input of EQ(Equal), GT(Greater) and product. When product is "1" and either EQ or GT is "1", the output will be out=product. However, in my problem, the verilog code shows correct for GT but not EQ. It seem that the output cannot response to EQ when it is high.
My design of the state diagram. State Diagram
My Verilog code. Verilog code
The result. Result Waveform...
ANSWERAnswered 2021-May-19 at 16:00
The answer is very simple. Add
ns to your simulation charts and you will understand why.
At the beginning of your simulation you are in state
S0. When the
product is bigger than
0 (first yellow mark) you go to state
S1. Then you are waiting for
EQ fall down one clock cycle ago so next
GT one clock cycle later.
I'm trying to figure out how to play a raw 8-bit (signed) waveform using Java, and struggling because it seems like every API example I find assumes media playback from a file. Here's what I have so far (shortened for clarity):...
ANSWERAnswered 2021-May-18 at 23:23
Call the constructor directly like this:
No vulnerabilities reported
You can use Waveform like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the Waveform component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
Reuse Trending Solutions
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page