I have RTP packets and I use the pcap.net Packet Object.
I need to get the actual audio data from the packet, without the RTP header - just the payload.
Language: C#
thnaks,
Ofek
RTP header length is 12 bytes usually. Using pcap.net you can create an object of udp.
UdpDatagram udp = null;
Then use
udp.Payload.ToHexadecimal();
So you will get full RTP . In that above hex string remove first 12 byte, that will be the rtp data(ex :voice).
Related
I'm trying to extract audio from a telephone session captured with Wireshark. The capture as send to us from the telephone provider for debugging/analysis. I have 3 files: signalling, and two files with UDP data, one for each direction. After merging two of these files (one direction with signalling), Wireshark provides RTP stream analysis. What I observe (as I do for a second session capture) is that Wireshark isn't able to export RTP stream audio (Payload type: ITU-T G.711 PCMA (8)) for one direction. This happens to be an RTP stream containing "RTF 2833 RTP events" (Payload type: telephone-event (106)). These events seem to transport DTMF tunes out-of-band, for each DTMF tune, there is a section of 7 consecutive RTP events of this type. What Wireshark does is producing an 8 GB *.au file for an audio stream less than two minutes. For the opposite-direction stream I get an audio file that is 2 MB in size.
I have to admit that this is just guesswork: I connect the error with a feature that I can see, I'm a bit confused that Wireshark obviously knows these Events but fails on saving the corresponding audio stream. Do I maybe need some plugin for that?
I tried to search the web for this issue but without success.
This question was previously asked on Network Engineering but turned out to be off-topic there.
You can filter (rtp.p_type != 106) the DTMF events from the wireshark logs (pcap) and then save only the G.711 data in a separate file.
Then do the RTP analysis and save the audio payload in .au/.raw file format.
I am learning about codecs,and I get this question that I didnt understood the answers.
Assuming CODEC G.711 where each datagram carries 20ms of voice, indicate:
3) [E] How many bytes of G.711 encoding does each datagram carry?
A- 20ms/8*0,02=160
4) What is the byte size of each frame carrying G.711 on an Ethernet network?
Note: The dimensions (in bytes) of the base headers of some of the protocols that might be involved in the communication: Ethernet = 18, IP = 20, TCP = 20, UDP = 8, ICMP = 8, RTP = 12
A-18+20+8+12+160=218
I didnt get this math..
g711 codecs pure bandwidth(codec only) is 64kbit, exact
g711 packet length can be 10,20(default),30.. upto 150ms.
So for default settings you have 20ms packet(50 packet/sec) at 64kbit = 160Bytes without udp packet header
Full length of g711(default 20ms) packet is
TPS = 18 bytes+20 bytes+8 bytes+12 bytes+160 bytes
You have 160 bytes of raw data, first you make it rtp packet(timestamp mostly), see packet size at https://en.wikipedia.org/wiki/Real-time_Transport_Protocol
RTP required fore reorder when you get two packet in different order(sometimes happens).
Now you have rtp, BUT it not suitable for send, need know where to send, need address and port. For port part you use UDP packet https://en.wikipedia.org/wiki/User_Datagram_Protocol
For address you use IP packet header, without address it not go destination machine
https://en.wikipedia.org/wiki/Internet_Protocol
Okay, now you have packet. But you still need actually send it. For send you use some hardware level protocol, in this case it is ETHERNET. Ethernet have mac address, which allow fast switching without parse of IP. That is last 18 bytes
https://scialert.net/fulltext/?doi=ajsr.2017.110.115
In some cases you can prefer TCP(when you have packet loss or complex networking), in this case you not use UDP, use TCP instead of it. So you swap 8bytes UDP for 20 bytes TCP.
I'm trying to use the opus Forward Error Correction (FEC) feature.
I have a service which does the encoding with OPUS_SET_INBAND_FEC(1)
and OPUS_SET_PACKET_LOSS_PERC(20) with 10ms packets and sends them over UDP.
I'm not clear on the decoding process though.
When a packet is lost, do I need to call decode with fec=1 ONLY or do I need to call decode with fec=0 after as well on the next packet?
How do I know up front the size of the pcm that I send to decode with fec enabled?
I managed to make it work.
The encoding part stated in the question was correct:
Use the encoder OPUS_SET_INBAND_FEC(1) and OPUS_SET_PACKET_LOSS_PERC(X) where x>0 ans x<100
Send packets with a duration of at least 10ms (for example: 480 samples at 48 kHz)
For the decoding part, when a packet is lost, call the decode function on the next packet first with fec=1 and again with fec=0.
When calling decode with fec=1, the pcm sent will be entirely filled.
If you don't know the length that the pcm should be use on the decoder OPUS_GET_LAST_PACKET_DURATION(x) where x will get the the duration of the last packet.
I'm writing a simple MJPEG streamer in Java.
Since it has very specific application, I had to reverse engineer the rtp packets sent by the GStreamer using the Wireshark sniffer. The GStreamer, in turn, just restreams the mjpeg avi file taken from specific camera.
The first and the most straightforward way is to prepare packets as an array of data and send them to the consumer, particularly VLC player, that uses live555 library. I'm populating the rtp headers well, so the internal payload is hardcoded; in my experiment VLC successfully displays the encoded screen, i.e. it decodes the hardcoded rtp payload well.
So, that is the value of first two packets payload (from total 25) in hex view:
String[] hexArray =
{
"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", //1
"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", //2
According to the spec, the first bytes of the first packet are:
0000000000ffa05a
this is Type-specific, Fragment Offset, Type, Q, Width, Height, MBZ, Precision, Length
the next data starting with 0000008010 and ending with 636363 is Quantization Table 128 bytes long.
So, the question is what is the data starting right after it?
ad8a5a0029
According to specs, no one of it can be recognized as possible JPEG marker, so the question is what is the encoding used and how can I encode the image I like with this particular encoding?
Attempt to encode on the manner of http://www.media.mit.edu/pia/Research/deepview/src/JpegEncoder.java works well for the jpeg files, but is not decoded with VLC's live555.
I asked earlier about H264 at RTP H.264 Packet Depacketizer
My question now is about the audio packets.
I noticed via the RTP packets that audio frames like AAC, G.711, G.726 and others all have the Marker Bit set.
I think frames are independent. am I right?
My question is: Audio is small, but I know that I can have more than one frame per RTP packet. Independent of how many frames I have, they are complete? Or it may be fragmented between RTP packets.
The difference between audio and video is that audio is typically encoded either in individual samples, or in certain [small] frames without reference to previous data. Additionally, amount of data is small. So audio does not typically need complicated fragmentation to be transmitted over RTP. However, for any payload type you should again refer to RFC that describes the details:
AAC - RTP Payload Format for MPEG-4 Audio/Visual Streams
G.711 - RTP Payload Format for ITU-T Recommendation G.711.1
G.726 - RTP Profile for Audio and Video Conferences with Minimal Control
Other