What is the smallest audio buffer needed to produce Tone sound without distotions with WaveOUT API - audio

Question

Does the WaveOut API has some internal limitation of the size for the current piece of buffer played ? I mean if I provide a very small buffer does it affects somehow the sound played to the speakers. I am experiencing very strange noise when I am generating and playing the sinus wave with small buffer. Something like a peak, or "BUMP".
The complete Story:
I made a program that can generate Sinus sound signal in real time.
The variable parameters are Frequency and Volume. The project requirement was to have a maximum latency of 50 ms. So the program must be able to produce Sinus signals with manually adjustable frequency of audio signal in real time.
I used Windows WaveOut API, C# and P/invoke to access the API.
Everything works fine when the sound buffer is 1000 ms large. If I minimize the buffer to 50 ms as per latency requirement then for certain frequencies I am experiencing at the end of every buffer, a noise or "BUMP". I do not understand if the sound generated is malformed ( I checked and is not) or something happens with the Audio chip, or some delay in initializing and playing.
When I save the produced audio to .wav file everything is perfect.
This means the must be some bug in my code or the audio subsystem has a limitation to the buffer chunks sent to it.
For those who doesn't know WaveOut must be initialized at first time and then must be prepared with audio headers for each buffer that are containing the number of bytes that needs to be played and the pointer to a memory that contains the audio that needs to be player.
UPDATE
Noise happens with the following combinations 44100 SamplingRate, 16 Bits, 2 channels, 50 ms buffer and generated Sinus audio signal of 201Hz, 202Hz, 203Hz, 204Hz, 205Hz ... 219Hz,
220Hz, 240 Hz, is ok
Why is this difference of 20, I do not know.

Answer 1

There are a few things to keep in mind when you need to output audio smoothly:
waveOutXxxx API is a legacy/compatibility layer on top of lower level API and as such it has greater overhead and is not recommended when you are to reach minimal latency. Note that this is unlikely to be your primary problem, but this is a piece of general knowledge helpful for understanding
because Windows is not real time OS and its audio subsystem is not realtime either you don't have control over random latency involved between you queue audio data for output and the data is really played back, the key is to keep certain level of buffer fullness which protects you from playback underflows and delivers smooth playback
with waveOutXxxx you are no limited to having single buffer, you can allocate multiple reusable buffers and recycle them
All in all, waveOutXxxx, DirectSound, DirectShow APIs work well with latencies 50 ms and up. With WASAPI exclusive mode streams you can get 5 ms latencies and even lower.
EDIT: I seem to have said too early about 20 ms latencies. To compensate for this, here is a simple tool LowLatencyWaveOutPlay (Win32, x64) to estimate the latency you can achieve. With sufficient buffering playback is smooth, otherwise you hear stuttering.
My understanding is that buffers might be returned late and the optimal design in terms of smallest latency lies along the line of having more smaller buffers so that you are given them back as early as possible. For example, 10 buffers 3 ms/buffer rather than 3 buffers 10 ms/buffer.
D:\>LowLatencyWaveOutPlay.exe 48000 10 3
Format: 48000 Hz, 1 channels, 16 bits per sample
Buffer Count: 10
Buffer Length: 3 ms (288 bytes)
Signal Frequency: 1000 Hz
^C

Answer 2

So I came here because I wanted to find the basic latency of waveoutwrite() as well. I got around 25-26ms of latency before I got to the smooth sine tone.
This is for:
AMD Phenom(tm) 9850 Quad-Core Processor 2.51 GHz
4.00 GB ram
64-bit operating system, x64-based processor
Windows 10 Enterprise N
The code follows. It is a modfied version of Petzold's sine wave program, refactored to run on the command line. I also changed the polling of buffers to use of a callback on buffer complete with the idea that this would make the program more efficient, but it didn't make a difference.
It also has a setup for elapsed timing, which I used to probe various timings for operations on the buffers. Using those I get:
Sine wave output program
Channels: 2
Sample rate: 44100
Bytes per second: 176400
Block align: 4
Bits per sample: 16
Time per buffer: 0.025850
Total time prepare header: 87.5000000000 usec
Total time to fill: 327.9000000000 usec
Total time for waveOutWrite: 90.8000000000 usec
Program:
/*******************************************************************************
WaveOut example program
Based on C. Petzold's sine wave example, outputs a sine wave via the waveOut
API in Win32.
*******************************************************************************/
#include <stdio.h>
#include <windows.h>
#include <math.h>
#include <limits.h>
#include <unistd.h>
#define SAMPLE_RATE 44100
#define FREQ_INIT 440
#define OUT_BUFFER_SIZE 570*4
#define PI 3.14159
#define CHANNELS 2
#define BITS 16
#define MAXTIM 1000000000
double fAngle;
LARGE_INTEGER perffreq;
PWAVEHDR pWaveHdr1, pWaveHdr2;
int iFreq = FREQ_INIT;
VOID FillBuffer (short* pBuffer, int iFreq)
{
int i;
int c;
for (i = 0 ; i < OUT_BUFFER_SIZE ; i += CHANNELS) {
for (c = 0; c < CHANNELS; c++)
pBuffer[i+c] = (short)(SHRT_MAX*sin (fAngle));
fAngle += 2*PI*iFreq/SAMPLE_RATE;
if (fAngle > 2 * PI) fAngle -= 2*PI;
}
}
double elapsed(LARGE_INTEGER t)
{
LARGE_INTEGER rt;
long tt;
QueryPerformanceCounter(&rt);
tt = rt.QuadPart-t.QuadPart;
return (tt*(1.0/(double)perffreq.QuadPart));
}
void CALLBACK waveOutProc(HWAVEOUT hwo, UINT uMsg, DWORD_PTR dwInstance, DWORD_PTR dwParam1, DWORD_PTR dwParam2)
{
if (uMsg == WOM_DONE) {
if (pWaveHdr1->dwFlags & WHDR_DONE) {
FillBuffer((short*)pWaveHdr1->lpData, iFreq);
waveOutWrite(hwo, pWaveHdr1, sizeof(WAVEHDR));
}
if (pWaveHdr2->dwFlags & WHDR_DONE) {
FillBuffer((short*)pWaveHdr2->lpData, iFreq);
waveOutWrite(hwo, pWaveHdr2, sizeof(WAVEHDR));
}
}
}
int main()
{
HWAVEOUT hWaveOut ;
short* pBuffer1;
short* pBuffer2;
short* pBuffer3;
WAVEFORMATEX waveformat;
UINT wReturn;
int bytes;
long t;
LARGE_INTEGER rt;
double timprep;
double filtim;
double waveouttim;
printf("Sine wave output program\n");
fAngle = 0; /* start sine angle */
QueryPerformanceFrequency(&perffreq);
pWaveHdr1 = malloc (sizeof (WAVEHDR));
pWaveHdr2 = malloc (sizeof (WAVEHDR));
pBuffer1 = malloc (OUT_BUFFER_SIZE*sizeof(short));
pBuffer2 = malloc (OUT_BUFFER_SIZE*sizeof(short));
pBuffer3 = malloc (OUT_BUFFER_SIZE*sizeof(short));
if (!pWaveHdr1 || !pWaveHdr2 || !pBuffer1 || !pBuffer2) {
if (!pWaveHdr1) free (pWaveHdr1) ;
if (!pWaveHdr2) free (pWaveHdr2) ;
if (!pBuffer1) free (pBuffer1) ;
if (!pBuffer2) free (pBuffer2) ;
fprintf(stderr, "*** Error: No memory\n");
exit(1);
}
// Load prime parameters to format
waveformat.wFormatTag = WAVE_FORMAT_PCM;
waveformat.nChannels = CHANNELS;
waveformat.nSamplesPerSec = SAMPLE_RATE;
waveformat.wBitsPerSample = BITS;
waveformat.cbSize = 0;
// Calculate other parameters
bytes = waveformat.wBitsPerSample/8; /* find bytes per sample */
if (waveformat.wBitsPerSample&8) bytes++; /* round up */
bytes *= waveformat.nChannels; /* find total channels size */
waveformat.nBlockAlign = bytes; /* set block align */
/* find average bytes/sec */
waveformat.nAvgBytesPerSec = bytes*waveformat.nSamplesPerSec;
printf("Channels: %d\n", waveformat.nChannels);
printf("Sample rate: %d\n", waveformat.nSamplesPerSec);
printf("Bytes per second: %d\n", waveformat.nAvgBytesPerSec);
printf("Block align: %d\n", waveformat.nBlockAlign);
printf("Bits per sample: %d\n", waveformat.wBitsPerSample);
printf("Time per buffer: %f\n",
OUT_BUFFER_SIZE*sizeof(short)/(double)waveformat.nAvgBytesPerSec);
if (waveOutOpen (&hWaveOut, WAVE_MAPPER, &waveformat, (DWORD_PTR)waveOutProc, 0, CALLBACK_FUNCTION)
!= MMSYSERR_NOERROR) {
free (pWaveHdr1) ;
free (pWaveHdr2) ;
free (pBuffer1) ;
free (pBuffer2) ;
hWaveOut = NULL ;
fprintf(stderr, "*** Error: No memory\n");
exit(1);
}
// Set up headers and prepare them
pWaveHdr1->lpData = (LPSTR)pBuffer1;
pWaveHdr1->dwBufferLength = OUT_BUFFER_SIZE*sizeof(short);
pWaveHdr1->dwBytesRecorded = 0;
pWaveHdr1->dwUser = 0;
pWaveHdr1->dwFlags = WHDR_DONE;
pWaveHdr1->dwLoops = 1;
pWaveHdr1->lpNext = NULL;
pWaveHdr1->reserved = 0;
QueryPerformanceCounter(&rt);
waveOutPrepareHeader(hWaveOut, pWaveHdr1, sizeof (WAVEHDR));
timprep = elapsed(rt);
pWaveHdr2->lpData = (LPSTR)pBuffer2;
pWaveHdr2->dwBufferLength = OUT_BUFFER_SIZE*sizeof(short);
pWaveHdr2->dwBytesRecorded = 0;
pWaveHdr2->dwUser = 0;
pWaveHdr2->dwFlags = WHDR_DONE;
pWaveHdr2->dwLoops = 1;
pWaveHdr2->lpNext = NULL;
pWaveHdr2->reserved = 0;
waveOutPrepareHeader(hWaveOut, pWaveHdr2, sizeof (WAVEHDR));
// Send two buffers to waveform output device
QueryPerformanceCounter(&rt);
FillBuffer (pBuffer1, iFreq);
filtim = elapsed(rt);
QueryPerformanceCounter(&rt);
waveOutWrite (hWaveOut, pWaveHdr1, sizeof (WAVEHDR));
waveouttim = elapsed(rt);
FillBuffer (pBuffer2, iFreq);
waveOutWrite (hWaveOut, pWaveHdr2, sizeof (WAVEHDR));
// Run waveform loop
sleep(10);
printf("Total time prepare header: %.10f usec\n", timprep*1000000);
printf("Total time to fill: %.10f usec\n", filtim*1000000);
printf("Total time for waveOutWrite: %.10f usec\n", waveouttim*1000000);
waveOutUnprepareHeader(hWaveOut, pWaveHdr1, sizeof (WAVEHDR));
waveOutUnprepareHeader(hWaveOut, pWaveHdr2, sizeof (WAVEHDR));
// Close waveform file
free (pWaveHdr1) ;
free (pWaveHdr2) ;
free (pBuffer1) ;
free (pBuffer2) ;
}