MPEG2 Presentation Time Stamps (PTS) calculation

Question

I have an MPEG2 TS file and now I am interested in extracting PTS information from each picture frame. I know that PTS is described in 33 bits including 3 marker bits. But I don't know how this bitfield can be converted to more understandable form (seconds, milliseconds). Can anybody help me please

Scarito

Answer 1

The MPEG2 transport stream clocks (PCR, PTS, DTS) all have units of 1/90000 second. The PTS and DTS have three marker bits which you need to skip over. The pattern is always (from most to least significant bit) 3 bits, marker, 15 bits, marker, 15 bits, marker. The markers must be equal to 1. In C, removing the markers would work like this:

uint64_t v; // this is a 64bit integer, lowest 36 bits contain a timestamp with markers
uint64_t pts = 0;
pts |= (v >> 3) & (0x0007 << 30); // top 3 bits, shifted left by 3, other bits zeroed out
pts |= (v >> 2) & (0x7fff << 15); // middle 15 bits
pts |= (v >> 1) & (0x7fff <<  0); // bottom 15 bits
// pts now has correct timestamp without markers in lowest 33 bits 

They also have an extension field of 9bits, forming a 42bit integer in which the extension is the least significant bits. The units for the base+extension are 1/27000000 second. Many implementations leave the extension as all zeros.

Whelk

Hi Alex I, Long shot here; do you know why the data is stored in this way? I've looked around and I haven't been able to find an answer. – Assimilative

@Assimilative I suspect the markers are there for one of two reasons: (1) to catch all-zero timestamps as being invalid, and (2) to catch bitstreams that are being read or decoded a bit at a time as being invalid when there is a bit offset. The units of 90000 are chosen so that both 30 fps and 29.97 fps are exact fractions. – Whelk

I see, those are both good points, thank you very much! To me it just seemed like it made the spec more complicated for no good reason. – Assimilative

Actually extension to 27MHz has PCR only. PTS and DTS have units in 90KHz only. NOTE: only [0,299] values used from 9bit of extension (not full range [0,511]). – Naphthyl

Answer 2

24hours/day * 60min/hr * 60secs/min *90k/sec (clock) = 7962624000, which needs 33 bits to be represented; You can extract your time from the clock using this info;

Proudman

Nice remark. I think the "k" is for 1e3 (not "K"). so 7776000000 still 33bits though. – Shrill

does it mean I can get absolute time of day at encoder side from PTS? Just using 00:00:00 as the base? – Griseofulvin

@Griseofulvin No, unfortunately it doesn't mean that. PTS starts at an arbitrary value, it's not specified in the standard. Most programs/devices that produce transport streams start at 0 for the beginning of each stream. They could start at the time of day expressed in PTS, but in practice none do that. – Whelk

PTS could be combined with PCR to get actual clock at the encoder side, though it is not guaranteed in any way to have any relation to the wall clock :). – Substation