I am developing an application in which I want to add echo effect in recorded audio files using objective-c.
I am using DIRAC to add other effect e.g. man to women, slow, fast.
now I have to make Robot voice of recorded voice. for robot voice I need to add echo effect
Please help me to do this
Echo is pretty simple. You need a delay line, and little multiplication. Assuming one channel and audio already represented in floating point, a delay line would look something like this (in C-like pseudo-code):
int LENGTH = samplerate * seconds; //seconds is the desired length of the delay in seconds
float buffer[ LENGTH ];
int readIndex = 0, writeIndex = LENGTH - 1;
float delayLine.readNext( float x ) {
float ret = buffer[readIndex];
++readIndex;
if( readIndex >= LENGTH )
readIndex = 0;
return ret;
}
void delayLine.writeNext( float x ) {
buffer[ writeIndex ] = x;
++writeIndex;
if( writeIndex >= LENGTH )
writeIndex = 0;
}
Don't forget to initialize the buffer to all zeros.
So that's your delay line. Basic usage would be this:
float singleDelay( float x ) {
delayLine.writeNext(x);
return delayLine.readNext( x );
}
But you won't hear much difference: it'll just come out later. If you want to hear a single echo, you'll need something like this:
float singleEcho( float x, float g ) {
delayLine.writeNext(x);
return x + g * delayLine.readNext( x );
}
where g is some constant, usually between zero and one.
Now say you want a stream of echos: "HELLO... Hello... hello... h..." like that. You just need to do a bit more work:
float echo( float x, float g ) {
float ret = x + g * delayLine.readNext( x );
delayLine.writeNext( ret );
return ret;
}
Notice how the output of the whole thing is getting fed back into the delay line this time, rather than the input. In this case, it's very important that |g| < 1.
You may run into issues of denormals here. I can't recall if that's an issue on iOS, but I don't think so.
Related
I am implementing this neural network for some classification problem. I initially tried back propagation but it takes longer to converge. So I though of using RPROP. In my test setup RPROP works fine for AND gate simulation but never converges for OR and XOR gate simulation.
How and when should I update bias for RPROP?
Here my weight update logic:
for(int l_index = 1; l_index < _total_layers; l_index++){
Layer* curr_layer = get_layer_at(l_index);
//iterate through each neuron
for (unsigned int n_index = 0; n_index < curr_layer->get_number_of_neurons(); n_index++) {
Neuron* jth_neuron = curr_layer->get_neuron_at(n_index);
double change = jth_neuron->get_change();
double curr_gradient = jth_neuron->get_gradient();
double last_gradient = jth_neuron->get_last_gradient();
int grad_sign = sign(curr_gradient * last_gradient);
//iterate through each weight of the neuron
for(int w_index = 0; w_index < jth_neuron->get_number_of_weights(); w_index++){
double current_weight = jth_neuron->give_weight_at(w_index);
double last_update_value = jth_neuron->give_update_value_at(w_index);
double new_update_value = last_update_value;
if(grad_sign > 0){
new_update_value = min(last_update_value*1.2, 50.0);
change = sign(curr_gradient) * new_update_value;
}else if(grad_sign < 0){
new_update_value = max(last_update_value*0.5, 1e-6);
change = -change;
curr_gradient = 0.0;
}else if(grad_sign == 0){
change = sign(curr_gradient) * new_update_value;
}
//Update neuron values
jth_neuron->set_change(change);
jth_neuron->update_weight_at((current_weight + change), w_index);
jth_neuron->set_last_gradient(curr_gradient);
jth_neuron->update_update_value_at(new_update_value, w_index);
double current_bias = jth_neuron->get_bias();
jth_neuron->set_bias(current_bias + _learning_rate * jth_neuron->get_delta());
}
}
}
In principal you don't treat the bias differently than before when you did backpropagation. It's learning_rate * delta which you seem to be doing.
One source of error may be that the sign of the weight change depends on how you calculate your error. There's different conventions and (t_i-y_i) instead of (y_i - t_i) should result in returning (new_update_value * sgn(grad)) instead of -(new_update_value * sign(grad)) so try switching the sign. I'm also unsure about how you specifically implemented everything since a lot is not shown here. But here's a snippet of mine in a Java implementation that might be of help:
// gradient didn't change sign:
if(weight.previousErrorGradient * errorGradient > 0)
weight.lastUpdateValue = Math.min(weight.lastUpdateValue * step_pos, update_max);
// changed sign:
else if(weight.previousErrorGradient * errorGradient < 0)
{
weight.lastUpdateValue = Math.max(weight.lastUpdateValue * step_neg, update_min);
}
else
weight.lastUpdateValue = weight.lastUpdateValue; // no change
// Depending on language, you should check for NaN here.
// multiply this with -1 depending on your error signal's sign:
return ( weight.lastUpdateValue * Math.signum(errorGradient) );
Also, keep in mind that 50.0, 1e-6 and especially 0.5, 1.2 are empirically gathered values so they might need to be adjusted. You should definitely print out the gradients and weight changes to see if there's something weird going on (e.g. exploding gradients->NaN although you're only testing AND/XOR). Your last_gradient value should also be initialized to 0 at the first timestep.
I am currently in the process of building an application that reads in audio from my iPhone's microphone, and then does some processing and visuals. Of course I am starting with the audio stuff first, but am having one minor problem.
I am defining my sampling rate to be 44100 Hz and defining my buffer to hold 4096 samples. Which is does. However, when I print this data out, copy it into MATLAB to double check accuracy, the sample rate I have to use is half of my iPhone defined rate, or 22050 Hz, for it to be correct.
I think it has something to do with the following code and how it is putting 2 bytes per packet, and when I am looping through the buffer, the buffer is spitting out the whole packet, which my code assumes is a single number. So what I am wondering is how to split up those packets and read them as individual numbers.
- (void)setupAudioFormat {
memset(&dataFormat, 0, sizeof(dataFormat));
dataFormat.mSampleRate = kSampleRate;
dataFormat.mFormatID = kAudioFormatLinearPCM;
dataFormat.mFramesPerPacket = 1;
dataFormat.mChannelsPerFrame = 1;
// dataFormat.mBytesPerFrame = 2;
// dataFormat.mBytesPerPacket = 2;
dataFormat.mBitsPerChannel = 16;
dataFormat.mReserved = 0;
dataFormat.mBytesPerPacket = dataFormat.mBytesPerFrame = (dataFormat.mBitsPerChannel / 8) * dataFormat.mChannelsPerFrame;
dataFormat.mFormatFlags =
kLinearPCMFormatFlagIsSignedInteger |
kLinearPCMFormatFlagIsPacked;
}
If what I described is unclear, please let me know. Thanks!
EDIT
Adding the code that I used to print the data
float *audioFloat = (float *)malloc(numBytes * sizeof(float));
int *temp = (int*)inBuffer->mAudioData;
int i;
float power = pow(2, 31);
for (i = 0;i<numBytes;i++) {
audioFloat[i] = temp[i]/power;
printf("%f ",audioFloat[i]);
}
I found the problem with what I was doing. It was a c pointer issue, and since I have never really programmed in C before, I of course got them wrong.
You can not directly cast inBuffer->mAudioData to an int array. So what I simply did was the following
SInt16 *buffer = malloc(sizeof(SInt16)*kBufferByteSize);
buffer = inBuffer->mAudioData;
This worked out just fine and now my data is of correct length and the data is represented properly.
I saw your answer, there also is an underlying issue which gives wrong sample data bytes which is because of an endian issue of bytes being swapped.
-(void)feedSamplesToEngine:(UInt32)audioDataBytesCapacity audioData:(void *)audioData {
int sampleCount = audioDataBytesCapacity / sizeof(SAMPLE_TYPE);
SAMPLE_TYPE *samples = (SAMPLE_TYPE*)audioData;
//SAMPLE_TYPE *sample_le = (SAMPLE_TYPE *)malloc(sizeof(SAMPLE_TYPE)*sampleCount );//for swapping endians
std::string shorts;
double power = pow(2,10);
for(int i = 0; i < sampleCount; i++)
{
SAMPLE_TYPE sample_le = (0xff00 & (samples[i] << 8)) | (0x00ff & (samples[i] >> 8)) ; //Endianess issue
char dataInterim[30];
sprintf(dataInterim,"%f ", sample_le/power); // normalize it.
shorts.append(dataInterim);
}
This is an interview question I saw on some site.
It was mentioned that the answer involves forming a recurrence of log2() as follows:
double log2(double x )
{
if ( x<=2 ) return 1;
if ( IsSqureNum(x) )
return log2(sqrt(x) ) * 2;
return log2( sqrt(x) ) * 2 + 1; // Why the plus one here.
}
as for the recurrence, clearly the +1 is wrong. Also, the base case is also erroneous.
Does anyone know a better answer?
How is log() and log10() actually implemented in C.
Perhaps I have found the exact answers the interviewers were looking for. From my part, I would say it's little bit difficult to derive this under interview pressure. The idea is, say you want to find log2(13), you can know that it lies between 3 to 4. Also 3 = log2(8) and 4 = log2(16),
from properties of logarithm, we know that log( sqrt( (8*16) ) = (log(8) + log(16))/2 = (3+4)/2 = 3.5
Now, sqrt(8*16) = 11.3137 and log2(11.3137) = 3.5. Since 11.3137<13, we know that our desired log2(13) would lie between 3.5 and 4 and we proceed to locate that. It is easy to notice that this has a Binary Search solution and we iterate up to a point when our value converges to the value whose log2() we wish to find. Code is given below:
double Log2(double val)
{
int lox,hix;
double rval, lval;
hix = 0;
while((1<<hix)<val)
hix++;
lox =hix-1;
lval = (1<<lox) ;
rval = (1<<hix);
double lo=lox,hi=hix;
// cout<<lox<<" "<<hix<<endl;
//cout<<lval<<" "<<rval;
while( fabs(lval-val)>1e-7)
{
double mid = (lo+hi)/2;
double midValue = sqrt(lval*rval);
if ( midValue > val)
{
hi = mid;
rval = midValue;
}
else{
lo=mid;
lval = midValue;
}
}
return lo;
}
It's been a long time since I've written pure C, so here it is in C++ (I think the only difference is the output function, so you should be able to follow it):
#include <iostream>
using namespace std;
const static double CUTOFF = 1e-10;
double log2_aux(double x, double power, double twoToTheMinusN, unsigned int accumulator) {
if (twoToTheMinusN < CUTOFF)
return accumulator * twoToTheMinusN * 2;
else {
int thisBit;
if (x > power) {
thisBit = 1;
x /= power;
}
else
thisBit = 0;
accumulator = (accumulator << 1) + thisBit;
return log2_aux(x, sqrt(power), twoToTheMinusN / 2.0, accumulator);
}
}
double mylog2(double x) {
if (x < 1)
return -mylog2(1.0/x);
else if (x == 1)
return 0;
else if (x > 2.0)
return mylog2(x / 2.0) + 1;
else
return log2_aux(x, 2.0, 1.0, 0);
}
int main() {
cout << "5 " << mylog2(5) << "\n";
cout << "1.25 " << mylog2(1.25) << "\n";
return 0;
}
The function 'mylog2' does some simple log trickery to get a related number which is between 1 and 2, then call log2_aux with that number.
The log2_aux more or less follows the algorithm that Scorpi0 linked to above. At each step, you get 1 bit of the result. When you have enough bits, stop.
If you can get a hold of a copy, the Feynman Lectures on Physics, number 23, starts off with a great explanation of logs and more or less how to do this conversion. Vastly superior to the Wikipedia article.
float jk = 7700; float ck = 8000; - if i do int jk; I get rim=0;
printf (asin(jk/10500)) = 1.57897 - for example
printf (asin(ck/9500)) = 0.87868 - for example
float rim;
rim= asin(jk/10500)+ asin(ck/9500);
printf("\n%f", rim) = nan
Why i get nan?
I don't believe your "for example". Because I don't believe in magic. If you have two valid floats both pretty small, then their sum is not a nan. So, my guess is this:
either |jk| > 10500 or |ck| > 9500. So you make asin with an invalid ( > 1.0 or < -1.0) argument and thus get a nan.
Or you have made another error. Please post a compilable runnable example which will print NAN
There's either something wrong with your code or something seriously wrong with the iphone. The following code:
#include<stdio.h>
#include<math.h>
int main (void) {
printf ("%f\n", asin(1));
printf ("%f\n", asin(0.5));
float rim;
rim = asin(1) + asin (0.5);
printf ("%f\n", rim);
return 0;
}
produces a more sensible:
1.570796
0.523599
2.094395
In other words, both your asin(0.5) and your sum are incorrect.
Are you sure that you didn't actually do something like:
rim = asin(1 + asin (0.5));
That will indeed give you NaN.
Update based on your added info:
Your code still works fine in my environment:
#include<stdio.h>
#include<math.h>
int main (void) {
float jk = 7700;
//jk = 7700/10500;
jk = jk/10500;
printf ("%f\n", asin(jk));
float hg = 8000;
hg = hg / 9500;
printf ("%f\n", asin(hg));
float rim;
rim = asin(jk) + asin (hg);
printf ("%f\n", rim);
return 0;
}
outputting:
0.823212
1.001175
1.824387
You'll notice I changed jk = 7700/10500 to jk = jk/10500 since the latter gave you 0 due to the integer division, but I don't get NaN in either case. You really need to post a complete program which shows the errant behaviour.
#include <stdio.h>
#include <math.h>
main()
{
float jk=7700, ck=8000;
printf ("\n%f",asin(jk/10500));
printf ("\n%f",asin(ck/9500));
float rim;
rim= asin(jk/10500)+ asin(ck/9500);
printf("\n%f", rim);// = nan
}
Output
0.823212
1.001175
1.824387
I think your code is right. The problem is the value of jk and ck.
You kown if |jk/temp|>1 or |ck/temp|>1, the return of asin(jk/temp) will be nan.
so try to make |jk/temp|<=1 and |ck/temp| <=1, I believe that the return will be ok.
I want to check if a floating point value is "nearly" a multiple of 32. E.g. 64.1 is "nearly" divisible by 32, and so is 63.9.
Right now I'm doing this:
#define NEARLY_DIVISIBLE 0.1f
float offset = fmodf( val, 32.0f ) ;
if( offset < NEARLY_DIVISIBLE )
{
// its near from above
}
// if it was 63.9, then the remainder would be large, so add some then and check again
else if( fmodf( val + 2*NEARLY_DIVISIBLE, 32.0f ) < NEARLY_DIVISIBLE )
{
// its near from below
}
Got a better way to do this?
well, you could cut out the second fmodf by just subtracting 32 one more time to get the mod from below.
if( offset < NEARLY_DIVISIBLE )
{
// it's near from above
}
else if( offset-32.0f>-1*NEARLY_DIVISIBLE)
{
// it's near from below
}
In a standard-compliant C implementation, one would use the remainder function instead of fmod:
#define NEARLY_DIVISIBLE 0.1f
float offset = remainderf(val, 32.0f);
if (fabsf(offset) < NEARLY_DIVISIBLE) {
// Stuff
}
If one is on a non-compliant platform (MSVC++, for example), then remainder isn't available, sadly. I think that fastmultiplication's answer is quite reasonable in that case.
You mention that you have to test near-divisibility with 32. The following theory ought to hold true for near-divisibility testing against powers of two:
#define THRESHOLD 0.11
int nearly_divisible(float f) {
// printf(" %f\n", (a - (float)((long) a)));
register long l1, l2;
l1 = (long) (f + THRESHOLD);
l2 = (long) f;
return !(l1 & 31) && (l2 & 31 ? 1 : f - (float) l2 <= THRESHOLD);
}
What we're doing is coercing the float, and float + THRESHOLD to long.
f (long) f (long) (f + THRESHOLD)
63.9 63 64
64 64 64
64.1 64 64
Now we test if (long) f is divisible with 32. Just check the lower five bits, if they are all set to zero, the number is divisible by 32. This leads to a series of false positives: 64.2 to 64.8, when converted to long, are also 64, and would pass the first test. So, we check if the difference between their truncated form and f is less than or equal to THRESHOLD.
This, too, has a problem: f - (float) l2 <= THRESHOLD would hold true for 64 and 64.1, but not for 63.9. So, we add an exception for numbers less than 64 (which, when incremented by THRESHOLD and subsequently coerced to long -- note that the test under discussion has to be inclusive with the first test -- is divisible by 32), by specifying that the lower 5 bits are not zero. This will hold true for 63 (1000000 - 1 == 1 11111).
A combination of these three tests would indicate whether the number is divisible by 32 or not. I hope this is clear, please forgive my weird English.
I just tested the extensibility to other powers of three -- the following program prints numbers between 383.5 and 388.4 that are divisible by 128.
#include <stdio.h>
#define THRESHOLD 0.11
int main(void) {
int nearly_divisible(float);
int i;
float f = 383.5;
for (i=0; i<50; i++) {
printf("%6.1f %s\n", f, (nearly_divisible(f) ? "true" : "false"));
f += 0.1;
}
return 0;
}
int nearly_divisible(float f) {
// printf(" %f\n", (a - (float)((long) a)));
register long l1, l2;
l1 = (long) (f + THRESHOLD);
l2 = (long) f;
return !(l1 & 127) && (l2 & 127 ? 1 : f - (float) l2 <= THRESHOLD);
}
Seems to work well so far!
I think it's right:
bool nearlyDivisible(float num,float div){
float f = num % div;
if(f>div/2.0f){
f=f-div;
}
f=f>0?f:0.0f-f;
return f<0.1f;
}
For what I gather you want to detect if a number is nearly divisible by other, right?
I'd do something like this:
#define NEARLY_DIVISIBLE 0.1f
bool IsNearlyDivisible(float n1, float n2)
{
float remainder = (fmodf(n1, n2) / n2);
remainder = remainder < 0f ? -remainder : remainder;
remainder = remainder > 0.5f ? 1 - remainder : remainder;
return (remainder <= NEARLY_DIVISIBLE);
}
Why wouldn't you just divide by 32, then round and take the difference between the rounded number and the actual result?
Something like (forgive the untested/pseudo code, no time to lookup):
#define NEARLY_DIVISIBLE 0.1f
float result = val / 32.0f;
float nearest_int = nearbyintf(result);
float difference = abs(result - nearest_int);
if( difference < NEARLY_DIVISIBLE )
{
// It's nearly divisible
}
If you still wanted to do checks from above and below, you could remove the abs, and check to see if the difference is >0 or <0.
This is without uing the fmodf twice.
int main(void)
{
#define NEARLY_DIVISIBLE 0.1f
#define DIVISOR 32.0f
#define ARRAY_SIZE 4
double test_var1[ARRAY_SIZE] = {63.9,64.1,65,63.8};
int i = 54;
double rest;
for(i=0;i<ARRAY_SIZE;i++)
{
rest = fmod(test_var1[i] ,DIVISOR);
if(rest < NEARLY_DIVISIBLE)
{
printf("Number %f max %f larger than a factor of the divisor:%f\n",test_var1[i],NEARLY_DIVISIBLE,DIVISOR);
}
else if( -(rest-DIVISOR) < NEARLY_DIVISIBLE)
{
printf("Number %f max %f less than a factor of the divisor:%f\n",test_var1[i],NEARLY_DIVISIBLE,DIVISOR);
}
}
return 0;
}