PCM data flow之二：Frames and Periods

在开始之前，我们先了解下关于PCM数据的几个重要概念：

Sample：样本长度，音频数据最基本的单位，常见的有8位和16位。

Channel：声道数，分为单声道mono和立体声stereo。

Frame：帧，构成一个声音单元，Frame = Sample * channel。

Rate：又称Sample rate，采样率，即每秒的采样次数，针对帧而言。

Interleaved：交错模式，一种音频数据的记录方式，在交错模式下，数据以连续桢的形式存放，即首先记录完桢1的左声道样本和右声道样本（假设为立体声），再开始桢2的记录。而在非交错模式下，首先记录的是一个周期内所有桢的左声道样本，再记录右声道样本，数据是以连续通道的方式存储。多数情况下使用交错模式。

Period size：周期，每次硬件中断处理音频数据的帧数，对于音频设备的数据读写，以此为单位。

Buffer size：数据缓冲区大小，这里特指runtime的buffer size，而不是snd_pcm_hardware定义的buffer_bytes_max。一般来说Buffer size = period_size * period_count，period_count相当于处理完一个buffer数据所需的硬件中断次数。

下面一张图直观的表示buffer/period/frame/sample之间的关系：

敏感的读者会察觉到Period和Buffer size在PCM数据搬运中占据着非常重要角色。下面引用两段来自alsa官网对Period的解释，英文不作翻译。

Period

The interval between interrupts from the hardware. This defines the input latency, since the CPU will not have any idea that there is data waiting until the audio interface interrupts it.

The audio interface has a "pointer" that marks the current position for read/write in its h/w buffer. The pointer circles around the buffer as long as the interface is running.

Typically, there are an integral number of periods per traversal of the h/w buffer, but not always. There is at least one card (ymfpci)
that generates interrupts at a fixed rate indepedent of the buffer size (which can be changed), resulting in some "odd" effects compared to more traditional designs.

Note: h/w generally defines the interrupt in frames, though not always.

Alsa's period size setting will affect how much work the CPU does. if you set the period size low, there will be more interrupts and the work that is done every interrupt will be done more often. So, if you don't care about low latency,
set the period size large as possible and you'll have more CPU cycles for other things. The defaults that ALSA provides are in the middle of the range, typically.

(from an old AlsaDevel thread[1], quoting Paul
Davis)

Retrieved from "http://alsa.opensrc.org/Period"

来自：http://alsa.opensrc.org/Period

FramesPeriods

A frame is equivalent of one sample being played, irrespective of the number of channels or the number of bits. e.g.
  * 1 frame of a Stereo 48khz 16bit PCM stream is 4 bytes.
  * 1 frame of a 5.1 48khz 16bit PCM stream is 12 bytes.
A period is the number of frames in between each hardware interrupt. The poll() will return once a period.
The buffer is a ring buffer. The buffer size always has to be greater than one period size. Commonly this is 2*period size, but some hardware can do 8 periods per buffer. It is also possible for the buffer size to not be an integer multiple of the period size.
Now, if the hardware has been set to 48000Hz , 2 periods, of 1024 frames each, making a buffer size of 2048 frames. The hardware will interrupt 2 times per buffer. ALSA will endeavor to keep the buffer as full as possible. Once the first period of samples has
been played, the third period of samples is transfered into the space the first one occupied while the second period of samples is being played. (normal ring buffer behaviour).

Additional example

Here is an alternative example for the above discussion.
Say we want to work with a stereo, 16-bit, 44.1 KHz stream, one-way (meaning, either in playback or in capture direction). Then we have:
  * 'stereo' = number of channels: 2
  * 1 analog sample is represented with 16 bits = 2 bytes
  * 1 frame represents 1 analog sample from all channels; here we have 2 channels, and so:
      * 1 frame = (num_channels) * (1 sample in bytes) = (2 channels) * (2 bytes (16 bits) per sample) = 4 bytes (32 bits)
  * To sustain 2x 44.1 KHz analog rate - the system must be capable of data transfer rate, in Bytes/sec:
      * Bps_rate = (num_channels) * (1 sample in bytes) * (analog_rate) = (1 frame) * (analog_rate) = ( 2 channels ) * (2 bytes/sample) * (44100 samples/sec) = 2*2*44100 = 176400 Bytes/sec
Now, if ALSA would interrupt each second, asking for bytes - we'd need to have 176400 bytes ready for it (at end of each second), in order to sustain analog 16-bit stereo @ 44.1Khz.
  * If it would interrupt each half a second, correspondingly for the same stream we'd need 176400/2 = 88200 bytes ready, at each interrupt;
  * if the interrupt hits each 100 ms, we'd need to have 176400*(0.1/1) = 17640 bytes ready, at each interrupt.
We can control when this PCM interrupt is generated, by setting a period size, which is set in frames.
  * Thus, if we set 16-bit stereo @ 44.1Khz, and the period_size to 4410 frames => (for 16-bit stereo @ 44.1Khz, 1 frame equals 4 bytes - so 4410 frames equal 4410*4 = 17640 bytes) => an interrupt will be generated each 17640 bytes - that is, each 100 ms.
  * Correspondingly, buffer_size should be at least 2*period_size = 2*4410 = 8820 frames (or 8820*4 = 35280 bytes).
It seems (writing-an-alsa-driver.pdf), however, that it is the ALSA runtime that decides on the actual buffer_size and period_size, depending on: the requested number of channels, and their respective properties (rate and sampling resolution) - as well as the
parameters set in the snd_pcm_hardware structure (in the driver).
Also, the following quote may be relevant, from http://mailman.alsa-project.org/pipermail/alsa-devel/2007-April/000474.html:

> > The "frame" represents the unit, 1 frame = # channels x sample_bytes.
> > In your case, 1 frame corresponds to 2 channels x 16 bits = 4 bytes.
> >
> > The periods is the number of periods in a ring-buffer.  In OSS, called
> > as "fragments".
> >
> > So,
> >  - buffer_size = period_size * periods
> >  - period_bytes = period_size * bytes_per_frame
> >  - bytes_per_frame = channels * bytes_per_sample
> >
> 
> I still don't understand what 'period_size' and a 'period' is?

The "period" defines the frequency to update the status, usually viathe invokation of interrupts.  The "period_size" defines the frame sizes corresponding to the "period time".  This term corresponds to the "fragment size" on OSS.  On major sound hardwares,
a ring-buffer is divided to several parts and an irq is issued on each boundary. The period_size defines the size of this chunk.

On some hardwares, the irq is controlled on the basis of a timer.  In this case, the period is defined as the timer frequency to invoke an irq.

来自：http://alsa-project.org/main/index.php/FramesPeriods

再说说period bytes，对于dma处理来说，它关心的是数据大小，而不管period size和period count，因此有个转换关系：

period_bytes = period_size * sample_bits * channels / 8

代码如下：

static inline unsigned int
params_period_bytes(const struct snd_pcm_hw_params *p)
{
	return (params_period_size(p) *
		snd_pcm_format_physical_width(params_format(p)) *
		params_channels(p)) / 8;
}