How to visualize microphone sound/pressure level in Silverlight 4? - silverlight

I was following this tutorial on how to make a Silverlight audio recorder. I thought it would be great to add a volume bar indicator to provide user with a feedback on what's happening. However, I can't seem to get this to work properly.
OnSamples method of AudioSink class provides raw PCM data as one of the arguments. Also, I set the AudioFrameSize property of AudioCaptureDevice to 40 (1000/40 == 25fps), so OnSamples is triggered every 40ms.
My question is how to extract the sound volume information from PCM data and display it as percentage in a progress bar [0-100]?
This is what I have so far:
double average = 0;
for (int a = 0; a < sampleData.Length; ++a)
{
average += Math.Abs(sampleData[a]);
}
average /= sampleData.Length;
double volume = 20 * Math.Log10(average);
Value of the progress bar is then set to volume:
progressBar.Value = volume;
My code doesn't work, apparently, since the volume value is almost always at the same level.
Any help is appreciated!

try this...this is for (8000,8,1) if you are using 2 channels replace "index+=1" with "index+=2"
for (int index = 0; index < sampleData.Length; index += 1)
{
short sample = (short)((sampleData[index + 1] << 8) | sampleData[index + 0]);
//short sample = (short)(sampleData[index + 0]);
float sample32 = sample / 32768f;
float maxValue = 0;
float minValue = 0;
maxValue = Math.Max(maxValue, sample32);
minValue = Math.Min(minValue, sample32);
float lastPeak = Math.Max(maxValue, Math.Abs(minValue));
this.MicLevel = (100 - (lastPeak * 100)) * 10;
//System.Diagnostics.Debug.WriteLine("Mic Level: " + this.MicLevel.ToString());
}

Related

How can I properly implement zero cross triggering for digital oscilloscope in C?

So I'm doing a simple oscilloscope in C. It reads audio data from the output buffer (and drops buffer write counter when called so the buffer is refreshed). I tried making simple zero-cross triggering since most of the time users will see simple (sine, pulse, saw, triangle) waves but the best result I got with the code below is a wave that jumps back and forth for half of its cycle. What is wrong?
Signal that is fed in goes from -32768 to 32767 so zero is where it should be.
If you didn't understand what I meant you can see the video: click
Upd: Removed the code unrelated to triggering so all function may be understood easier.
extern Mused mused;
void update_oscillscope_view(GfxDomain *dest, const SDL_Rect* area)
{
if (mused.output_buffer_counter >= OSC_SIZE * 12) {
mused.output_buffer_counter = 0;
}
for (int x = 0; x < area->h * 0.5; x++) {
//drawing a black rect so bevel is hidden when it is under oscilloscope
gfx_line(domain,
area->x, area->y + 2 * x,
area->x + area->w - 1, area->y + 2 * x,
colors[COLOR_WAVETABLE_BACKGROUND]);
}
Sint32 sample, last_sample, scaled_sample;
for (int i = 0; i < 2048; i++) {
if (mused.output_buffer[i] < 0 && mused.output_buffer[i - 1] > 0) {
//here comes the part with triggering
if (i < OSC_SIZE * 2) {
for (int x = i; x < area->w + i; ++x) {
last_sample = scaled_sample;
sample = (mused.output_buffer[2 * x] + mused.output_buffer[2 * x + 1]) / 2;
if (sample > OSC_MAX_CLAMP) { sample = OSC_MAX_CLAMP; }
if (sample < -OSC_MAX_CLAMP) { sample = -OSC_MAX_CLAMP; }
if (last_sample > OSC_MAX_CLAMP) { last_sample = OSC_MAX_CLAMP; }
if (last_sample < -OSC_MAX_CLAMP) { last_sample = -OSC_MAX_CLAMP; }
scaled_sample = (sample * OSC_SIZE) / 32768;
if(x != i) {
gfx_line(domain,
area->x + x - i - 1, area->h / 2 + area->y + last_sample,
area->x + x - i, area->h / 2 + area->y + scaled_sample,
colors[COLOR_WAVETABLE_SAMPLE]);
}
}
}
return;
}
}
}
During debugging, I simplified the code until it started working. Thanks Clifford.
I found a trigger index i (let's say it is array index 300). Modified it so that the oscilloscope was drawing lines from [(2 * i) + offset] to [(2 * i + 1) + offset], thus an incorrect picture was formed.
I used (2 * i), because I wanted long waves to fit into oscilloscope. I replaced it with drawing from [i + offset] to [i + 1 + offset] and that solved a problem.
Afterwards, I implemented "horizontal scale 0.5x properly.
The output waveform still jumps a little, but overall it holds it in place.

how to create a simple iir low pass filter with not round errors? (16 bit pcm data)

i have an array of n length fullfilled by 16 bit (int16) pcm raw data,the data is in 44100 sample_rate
and stereo,so i have in my array first 2 bytes left channel then right channel etc...i tried to implement a simple low pass converting my array into floating points -1 1,the low pass works but there are round errors that cause little pops in the sound
now i do simply this :
INT32 left_id = 0;
INT32 right_id = 1;
DOUBLE filtered_l_db = 0.0;
DOUBLE filtered_r_db = 0.0;
DOUBLE last_filtered_left = 0;
DOUBLE last_filtered_right = 0;
DOUBLE l_db = 0.0;
DOUBLE r_db = 0.0;
DOUBLE low_filter = filter_freq(core->audio->low_pass_cut);
for(UINT32 a = 0; a < (buffer_size/2);++a)
{
l_db = ((DOUBLE)input_buffer[left_id]) / (DOUBLE)32768;
r_db = ((DOUBLE)input_buffer[right_id]) / (DOUBLE)32768;
///////////////LOW PASS
filtered_l_db = last_filtered_left +
(low_filter * (l_db -last_filtered_left ));
filtered_r_db = last_filtered_right +
(low_filter * (r_db - last_filtered_right));
last_filtered_left = filtered_l_db;
last_filtered_right = filtered_r_db;
INT16 l = (INT16)(filtered_l_db * (DOUBLE)32768);
INT16 r = (INT16)(filtered_r_db * (DOUBLE)32768);
output_buffer[left_id] = (output_buffer[left_id] + l);
output_buffer[right_id] = (output_buffer[right_id] + r);
left_id +=2;
right_id +=2;
}
PS: the input buffer is an int16 array with the pcm data from -32767 to 32767;
i found this function here
Low Pass filter in C
and was the only one that i could understand xd
DOUBLE filter_freq(DOUBLE cut_freq)
{
DOUBLE a = 1.0/(cut_freq * 2 * PI);
DOUBLE b = 1.0/SAMPLE_RATE;
return b/(a+b);
}
my aim is instead to have absolute precision on the wave,and to directly low pass using only integers
with the cost to lose resolution on the filter(and i'm ok with it)..i saw a lot of examples but i really didnt understand anything...someone of you would be so gentle to explain how this is done like you would explain to a little baby?(in code or pseudo code rapresentation) thank you
Assuming the result of function filter_freq can be written as a fraction m/n your filter calculation basically is
y_new = y_old + (m/n) * (x - y_old);
which can be transformed to
y_new = ((n * y_old) + m * (x - y_old)) / n;
The integer division / n truncates the result towards 0. If you want rounding instead of truncation you can implement it as
y_tmp = ((n * y_old) + m * (x - y_old));
if(y_tmp < 0) y_tmp -= (n / 2);
else y_tmp += (n / 2);
y_new = y_tmp / n
In order to avoid losing precision from dividing the result by n in one step and multiplying it by n in the next step you can save the value y_tmp before the division and use it in the next cycle.
y_tmp = (y_tmp + m * (x - y_old));
if(y_tmp < 0) y_new = y_tmp - (n / 2);
else y_new = y_tmp + (n / 2);
y_new /= n;
If your input data is int16_t I suggest to implement the calculation using int32_t to avoid overflows.
I tried to convert the filter in your code without checking other parts for possible problems.
INT32 left_id = 0;
INT32 right_id = 1;
int32_t filtered_l_out = 0; // output value after division
int32_t filtered_r_out = 0;
int32_t filtered_l_tmp = 0; // used to keep the output value before division
int32_t filtered_r_tmp = 0;
int32_t l_in = 0; // input value
int32_t r_in = 0;
DOUBLE low_filter = filter_freq(core->audio->low_pass_cut);
// define denominator and calculate numerator
// use power of 2 to allow bit-shift instead of division
const uint32_t filter_shift = 16U;
const int32_t filter_n = 1U << filter_shift;
int32_t filter_m = (int32_t)(low_filter * filter_n)
for(UINT32 a = 0; a < (buffer_size/2);++a)
{
l_in = input_buffer[left_id]);
r_in = input_buffer[right_id];
///////////////LOW PASS
filtered_l_tmp = filtered_l_tmp + filter_m * (l_in - filtered_l_out);
if(last_filtered_left < 0) {
filtered_l_out = last_filtered_left - filter_n/2;
} else {
filtered_l_out = last_filtered_left + filter_n/2;
}
//filtered_l_out /= filter_n;
filtered_l_out >>= filter_shift;
/* same calculation for right */
INT16 l = (INT16)(filtered_l_out);
INT16 r = (INT16)(filtered_r_out);
output_buffer[left_id] = (output_buffer[left_id] + l);
output_buffer[right_id] = (output_buffer[right_id] + r);
left_id +=2;
right_id +=2;
}
As your filter is initialized with 0 it may need several samples to follow a possible step to the first input value. Depending on your data it might be better to initialize the filter based on the first input value.

Struggle to program an LFO

based on some tutorial code I've found I coded a little synthesizer with three oscilators and four different waveform. It works well and I want to add an LFO to module the sounds. Since I didn't coded everything on my own I'm a bit confused of how I could fit the LFO formula on my code. This is more or less what I tried in order to implement the LFO formula on a sinewave.(This formula is something like this: sinewaveFormula + 0.5 * Sinefreq * sin(2pi*1) * time)
double normalize(double phase)
{
double cycles = phase/(2.0*pi);
phase -= trunc(cycles) * 2.0 * pi;
if (phase < 0) phase += 2.0*pi;
return phase;
}
double sine(double phase)
{ phase = normalize(phase); return (sin(phase));}
static void build_sine_table(int16_t *data, int wave_length) {
double phase_increment = (2.0f * pi) / (double)wave_length;
double current_phase = 0;
for(int i = 0; i < wave_length; i++) {
int sample = synthOsc(current_phase, oscNum, selectedWave, selectedWave2, selectedWave3, intensity, intensity2, intensity3) + 0.5 * ((current_phase* wave_length) / (2*pi)) * sin(2*pi*(1.0)) * wave_length;
data[i] = (int16_t)sample;
current_phase += phase_increment;
}
}
static void write_samples(int16_t *s_byteStream, long begin, long end, long length) {
if(note > 0) {
double d_sample_rate = sample_rate;
double d_table_length = table_length;
double d_note = note;
// get correct phase increment for note depending on sample rate and table length.
double phase_increment = (get_pitch(d_note) / d_sample_rate) * d_table_length;
// loop through the buffer and write samples.
for (int i = 0; i < length; i+=2) {
phase_double += phase_increment;
phase_int = (int)phase_double;
if(phase_double >= table_length) {
double diff = phase_double - table_length;
phase_double = diff;
phase_int = (int)diff;
}
if(phase_int < table_length && phase_int > -1) {
if(s_byteStream != NULL) {
int16_t sample = sine_waveform_wave[phase_int];
target_amp = update_envelope();
if(smoothing_enabled) {
// move current amp towards target amp for a smoother transition.
if(current_amp < target_amp) {
current_amp += smoothing_amp_speed;
if(current_amp > target_amp) {
current_amp = target_amp;
}
} else if(current_amp > target_amp) {
current_amp -= smoothing_amp_speed;
if(current_amp < target_amp) {
current_amp = target_amp;
}
}
} else {
current_amp = target_amp;
}
sample *= current_amp; // scale volume.
s_byteStream[i+begin] = sample; // left channel
s_byteStream[i+begin+1] = sample; // right channel
}
}
}
}
}
The code compile but there's no LFO on the sine. I don't understand how I could make this formula work with this code.
It may help to get a basic understanding of how a LFO actually works. It is not that difficult - as an LFO is just another oscillator that is mixed to the waveform you want to modulate.
I would suggest to remove your LFO formular from your call of synthOsc(), then you get a clean oscillator signal again. As a next step, create another oscillator signal for which you can use a very low frequency. Mix both signals together and you are done.
Expresssed in simple math, it is like this:
int the_sample_you_want_to_modulate = synthOsc1(...);
int a_sample_with_very_low_frequency = synthOsc2(...);
Mixing two waveforms is done through addition:
int mixed_sample = the_sample_you_want_to_modulate + a_sample_with_very_low_frequency;
The resulting sample will sweep now based on the frequency you have used for synthOsc2().
As you can see, to implement an LFO you actually do not need a separate formular. You already have the formular when you know how to create an oscillator.
Note that if you add two sine oscillators that have the exact same frequency, the resulting signal will just get louder. But when each has a different frequency, you will get a new waveform. For LFOs (which are in fact just ordinary oscillators - like in your build_sine_table() function) you typically set a very low frequency: 1 - 10 Hz is low enough to get an audible sweep. For higher frequencies you get chords as a result.

Detecting and displaying number of players detected using Kinect

I'm trying to display number of players detected by Kinect sensor using an WPF application. In addition to displaying number of player I have also coloured the pixels based on their distance from the Kinect. Original goal was to measure the distance of the pixels and display the distance but I would also like to display how many people are in the frame. Here is the code snippets that I'm using now.
PS I have borrowed the idea from THIS tutorial and I'm using SDK 1.8 with XBOX 360 Kinect (1414)
private void _sensor_AllFramesReady(object sender, AllFramesReadyEventArgs e)
{
using (DepthImageFrame depthFrame = e.OpenDepthImageFrame())
{
if (depthFrame==null)
{
return;
}
byte[] pixels = GenerateColoredBytes(depthFrame);
int stride = depthFrame.Width * 4;
image.Source = BitmapSource.Create(depthFrame.Width, depthFrame.Height,
96, 96, PixelFormats.Bgr32, null, pixels, stride);
}
}
private byte[] GenerateColoredBytes(DepthImageFrame depthFrame)
{
//get the raw data from kinect with the depth for every pixel
short[] rawDepthData = new short[depthFrame.PixelDataLength];
depthFrame.CopyPixelDataTo(rawDepthData);
/*
Use depthFrame to create the image to display on screen
depthFrame contains color information for all pixels in image
*/
//Height * Width *4 (Red, Green, Blue, Empty byte)
Byte[] pixels = new byte[depthFrame.Height * depthFrame.Width * 4];
//Hardcoded loactions for Blue, Green, Red (BGR) index positions
const int BlueIndex = 0;
const int GreenIndex = 1;
const int RedIndex = 2;
//Looping through all distances and picking a RGB colour based on distance
for (int depthIndex = 0, colorIndex = 0;
depthIndex < rawDepthData.Length &&
colorIndex<pixels.Length; depthIndex++, colorIndex+=4)
{
//Getting player
int player = rawDepthData[depthIndex] & DepthImageFrame.PlayerIndexBitmask;
//Getting depth value
int depth =rawDepthData[depthIndex]>>DepthImageFrame.PlayerIndexBitmaskWidth;
//.9M or 2.95'
if (depth <=900 )
{
//Close distance
pixels[colorIndex + BlueIndex] = 0;
pixels[colorIndex + GreenIndex] = 0;
pixels[colorIndex + RedIndex] = 255;
//textBox.Text = "Close Object";
}
//.9M - 2M OR 2.95' - 6.56'
else if (depth >900 && depth<2000)
{
//Bit further away
pixels[colorIndex + BlueIndex] = 255;
pixels[colorIndex + GreenIndex] = 0;
pixels[colorIndex + RedIndex] = 0;
}
else if (depth > 2000)
{
//Far away
pixels[colorIndex + BlueIndex] = 0;
pixels[colorIndex + GreenIndex] = 255;
pixels[colorIndex + RedIndex] = 0;
}
//Coloring all people in Gold
if (player > 0)
{
pixels[colorIndex + BlueIndex] = Colors.Gold.B;
pixels[colorIndex + GreenIndex] = Colors.Gold.G;
pixels[colorIndex + RedIndex] = Colors.Gold.R;
playersValue.Text = player.ToString();
}
}
return pixels;
}
Current goal is to--
Detect total number of players detected and display them in a textBox
Colour them according to the distance logic i.e depth <=900 is red.
With current code I can detect player and color them in Gold but as soon as a player is detected the image freezes and when the player is out of the frame the image unfreezes and acts normal. Is it because of the loop?
Ideas, guidance, recommendation and criticism all are welcome.
Thanks!
Screenshots:
Get a static variable inside your form code
Then set this variable using your video frame routine (dont define it there).
And then update the textbox view, probaply in your _sensor_AllFramesReady
As the arrival of new frames runs in a different thread
I dont see all code maybe to update call textbox.show
the main loop looks a bit strange though, too complex.
basicly you use it to color every pixel in your image.
as the kinect360 has 320x240 pixels so that makes a depth array of size 76800
You might simply create 2 for next loops loops for X and Y and then inside this loop have a variable increase to pick the proper depth value.

PI control algorithm manually implemented in matlab

I'm trying to implement a simple script performing a PI control for a cruise control application, but I'm founding some problems with the integral part. Here is my code:
function [] = PI_cruisecontrol()
clc; close all;
t0 = 0; tfinal = 50; dt = 0.001; % time parameters
r = 10; % reference of 10 m/s
m = 1000; % mass
b = 50; % friction coeff. (depends on v)
yp = zeros(tfinal/dt,1); t = yp; % initialize speed and time array
Ki = 40; % integrarl constant
Kp = 800; % proportional constant
int = 0; % itinialize int error
% CONTROL LOOP (Forward-Euler integrator is used to solve the ODE)
for i=t0+2:tfinal/dt
err = r-yp(i-1); % update error
int = int+err; % integral term
u = (Kp*err)+(Ki*int*dt); % action of control
yp(i) = yp(i-1)+((-b*yp(i)/m) + (u/m))*dt; % solve ode for speed
t(i) = t(i)+dt*i; % log the time
end
% Results
figure(1)
plot(t,yp)
title ('Step Response')
xlabel('Time (seconds)')
ylabel('Amplitud')
axis([0 20 0 12])
hold on
reference = ones(tfinal/dt,1)*10;
plot(t,reference,':')
end
And this is how it should be, using predefinided matlab functions:
function [] = PI_cruisecontrol2()
m = 1000;
b = 50;
r = 10;
s = tf('s');
P_cruise = 1/(m*s + b);
Kp = 800;
Ki = 40;
C = pid(Kp,Ki);
T = feedback(C*P_cruise,1);
t = 0:0.1:20;
step(r*T,t)
axis([0 20 0 12])
end
What am I doing wrong in my code?
Thanks!
I managed to fix the problem, working with float variables instead of arrays. Moreover, I added the derivative term (although for this first order problem was not necessary)
Here I left the code:
function [] = aFortran_PI()
clc; close all;
r = 10; % reference of 10 m/s
m = 1000; % mass
b = 50; % friction coeff. (depends on v)
yp = 0; % init response
Kp = 800; % proportional constant
Ki = 40; % proportional constant
Kd = 0; % derivative term is not necessary in this problem
previous_error = 0;
integral = 0;
dt = 0.001;
% CONTROL LOOP
for i=1:20000
error = r-yp; % update error
integral = integral + error*dt; % integral term
derivative = (error-previous_error)/dt; % derivative term
u = Kp*error+Ki*integral+Kd*derivative; % action of control
yp = yp+(-b*yp/m + u/m)*dt % solve ode for velocity
end
end

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