PCM data flow - part 6: Frames & Periods

http://blog.csdn.net/azloong/article/details/17614859

前面分析了codec、platform、machine驱动的各个组成部分及其注册过程，这三者都是硬件设备相关的，大家应该对音频物理链路有了初步的认知。

往后章节的内容主要集中在pcm native，这是pcm数据流的中间层：

·          往上是与用户态接口的交互，实现音频数据在用户态和内核态之间的拷贝；

·          往下是触发codec、platform、machine的操作函数，实现音频数据在dma_buffer<-> cpu_dai <-> codec之间的传输。

其中会涉及到dma buffer的管理，这需要对音频数据相关概念有一定的了解。因此本章说明下音频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，构成一个完整的声音单元，Frame = Sample * channel；

·          Rate：又称Samplerate，采样率，即每秒的采样次数，针对帧而言；

·          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数据搬运中扮演着非常重要角色。下面引用两段来自alsa官网对Period的解释：

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 thework 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 havemore 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"

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 bean 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).

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 2 x 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_hardwarestructure (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 16bits = 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 via the 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.

Retrieved from "http://alsa-project.org/main/index.php/FramesPeriods"

这里不做翻译了，简单说下Frame和Period要点：

·          Frame：帧，构成一个完整的声音单元，它的大小等于量化位数sample_bits * 声道数channels；

·          Peroid：周期大小，即每次dma硬件中断处理音频数据的帧数。如果周期设定得较大，则单次处理的数据较多，这意味着单位时间内硬件中断的次数较少，CPU也就有更多时间处理其他任务，功耗也更低，但这样也带来一个显著的弊端——数据处理的时延（latency）会增大。后面章节会针对这点介绍Android的deep_buffer和low_latency两种处理方式。

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：

[cpp] view plain copy print?

1. static inline unsigned int  
2. params_period_bytes(const struct snd_pcm_hw_params *p)  
3. {  
4.     return (params_period_size(p) *  
5.         snd_pcm_format_physical_width(params_format(p)) *  
6.         params_channels(p)) / 8;  
7. }