Calculating the number of samples in a WAV file - audio

I am currently looking at online examples and here is a WAV file contents in bytes
52 49 46 46 24 08 00 00 57 41 56 45 66 6d 74 20 10 00 00 00 01 00 02 00
22 56 00 00 88 58 01 00 04 00 10 00 64 61 74 61 00 08 00 00 00 00 00 00
24 17 1e f3 3c 13 3c 14 16 f9 18 f9 34 e7 23 a6 3c f2 24 f2 11 ce 1a 0d
and here is the visual; representation:
So according to the Subchunk2Size there is 2048 bytes in the data. The formula to calculate the number of samples in a WAV is given as:
Subchunk2Size /(NumChannels * BitsPerSample/8 ) = NumSamples
If I plugin numbers and according to the information given I get NumSamples = 512. But in the diagram the sample rate is 22050. How can the total number fo samples be less than a single second of samples?

For those wondering, here is a link to the source.
I suspect they are just using a bad example where the duration of the wav file would be less than a second. Their formula makes sense and we can use it to verify the data size of a one second wav file.
If our sample rate is 22050 samples/sec and our wav file is one second, then numSamples = 22050. We know that Subchunk2Size is the number of bytes in the data and can be calculated using this formula: Subchunk2Size = numSamples * numChannels * bitsPerSample / 8 , so, assuming numChannels = 2 and bitsPerSample = 16, we know that a one second wav file should be (22050 * 2 * 16 / 8) bytes which is 88200 bytes, so it would make sense that if Subchunk2Size is 2048 bytes, as per the website's example, then the duration of the wav file would be less than a second and thus, numSamples would be less than 22050.

Related

Converting string to bytes and writing to file in Lua

I'm trying to convert and write string data into the file as bytes.
I have already tried something to, but instead of seeing 00 inside hexdump, im seeing 0x30 inside file which is hexadecimal value for character 0.
Here is what I wrote:
local data = "000000010000000100000004000000080000000100000000"
for i=1,#data,2 do
file:write(tonumber(data:sub(i,i+1)))
end
io.close(file)
When I do hexdump of the file I'm getting this:
0000000 30 30 30 31 30 30 30 31 30 30 30 34 30 30 30 38
0000010 30 30 30 31 30 30 30 30
0000018
Expected is:
0000000 00 00 00 01 00 00 00 01 00 00 00 04 00 00 00 08
0000010 00 00 00 01 00 00 00 00
0000018
You want to use string.char in one way:
local data = "000000010000000100000004000000080000000100000000"
for i=1,#data,2 do
file:write(string.char(tonumber(data:sub(i,i+1), 16)))
end
io.close(file)
or another:
local data = string.char(0,0,0,1,0,0,0,1,0,0,0,4,0,0,0,8,0,0,0,1,0,0,0,0)
file:write(data)
io.close(file)
Note that strings in Lua may contain any bytes you want including null bytes. See Values and Types.
Hint: Use string.char to convert numbers to bytes:
file:write(string.char(tonumber(data:sub(i,i+1))))
If the strings contains hexadecimal, use tonumber(...,16).

How to convert Big Endian Hex to ASCII in Node?

I'm trying to interpret a HEX message that is transmitted in network byte order (big endian), how should I proceed to achieve conversion to ASCII?
server.on('message', function (message, remote) {
//I receive message via UDP in HEX.
});
The message comes in this format:
2B 41 43 4B 19 EF 24 10 01 02 03 02 56 50 22 00 0A 02 3B 01 00 00 4E 07 DD 02 17 11 21 20 46 AD 4E 1E 0D 0A
With that being said, each byte has a parameter attached, let's say I have 4 bytes to parameter1, 2 bytes to parameter2 and 8 bytes to parameter 3, how would I interpret it?

Unknown Bittorrent Message

I have been receiving an odd/unknown message while attempting to communicate with some bittorrent peers. In this particular case I am in the middle of downloading pieces and all of a sudden this new/odd message pops up in front of a piece response.The message is odd because it doesn't appear to follow the protocol, all messages are supposed to look like this
'<length prefix><message ID><payload>'
length prefix is 4 bytes, message id is 1 byte and the payload. I am including a capture to show what I mean, on line 509 of the capture you will
see a request for a piece, on line 510 you will see the beginning of the response.
The first 4 bytes of the response are 00 00 00 00, ie 0 length message (Which is causing me issues), the next 4 bytes are the actual length of the message which is 30. The actual response to the piece request starts on line 513, so I get the piece I was requesting but this new/odd message is messing me up. I'm certain I can find a workaround but I would really like to understand what this means.
Also, I have no idea what the actual message means, and cannot find any information about it anywhere.
Here is the Wireshark capture.
https://1drv.ms/u/s!Agj06pa-wu0tnFqsYn_KnHmVz3x2
Data from packet 510:
0000 00 00 00 00 00 00 00 1e 14 01 64 35 3a 61 64 64 ..........d5:add
0010 65 64 36 3a 63 f2 7a 48 17 f4 37 3a 64 72 6f 70 ed6:c.zH..7:drop
0020 70 65 64 30 3a 65 ped0:e
00 00 00 00 4 bytes keep-alive message
00 00 00 1e message length 30 bytes
14 message type extended message (BEP10)
01 extended message ID = 1 as specified by the previous extension handshake: ut_pex
64 35 3a 61 64 64 65 64 36 3a 63 f2 7a 48 17 f4 37 3a 64 72 6f 70 70 65 64 30 3a 65
d5:added6:c.zH..7:dropped0:e
ut_pex message data (bencoded)
d
5:added
6:c.zH..
7:dropped
0:
e
ut_pex message data (bencoded with added white space)
The first 4 bytes of the response are 00 00 00 00, ie 0 length message (Which is causing me issues)
The bittorrent spec says
Messages of length zero are keepalives, and ignored.

How do you decode an Ethernet Frame without things like Wireshark?

For example: How would one decode the following ethernet frame?
00 26 b9 e8 7e f1 00 12 f2 21 da 00 08 00 45 00 05 dc e3 cd 20 10 35 06 25 eb 0a 0a 0a 02 c0 a8 01 03 c3 9e 0f 40 00 00 10 00 00 00 14 00 70 10 00 5c 59 99 00 00 02 04 05 b4 01 03 03 06 00 00 01 98 64 34 e8 90 84 98 20 12 18 19 04 85 80 00
I know that the first 6 bytes are the MAC destination address : 00 26 b9 e8 7e f1 The next 6 bytes are the source MAC address : 00 12 f2 21 da 00 The next 2 bytes show the ethernet type : 08 00 The next 4 bytes are : 45 00... Ipv4... "5" the number of bytes in the header.. and "00" means there are no differentiated services.
What I don't know is what anything after that is or how to read it.
Anyone help?
Rearranging a bit your packet, we have:
00 26 b9 e8 7e f1 00 12 f2 21 da 00 08 00 45 00
05 dc e3 cd 20 10 35 06 25 eb 0a 0a 0a 02 c0 a8
01 03 c3 9e 0f 40 00 00 10 00 00 00 14 00 70 10
00 5c 59 99 00 00 02 04 05 b4 01 03 03 06 00 00
01 98 64 34 e8 90 84 98 20 12 18 19 04 85 80 00
If you know that the first 6 octets form the destination mac address, that means that it is an Ethernet layer 2 packet.
According to IEEE 802.3, $3.1.1:
First 6 octets are the destination mac address (00 26 b9 e8 7e f1)
Next 6 octets are the source mac address (00 12 f2 21 da 00)
Next 4 octets are, optionally the 802.1Q tag (present, 08 00 45 00)
Next 2 octets are either:
Maximum payload size - aka MTU (if <= 1500, which is the case, 05 dc is 1500)
Ethernet 2 frame (if >= 1536)
Next is the payload ranging from 46 octets (if the 802.1Q tag is absent) or 42 octets (if the 802.1Q tag is present) to up to 1500 octets (starts at e3 cd 20 10 ..., ends either at 20 12 18 19 or at 03 06 00 00, depends on the 7th item)
Last 4 octets form the CRC32 code (either 01 98 64 34 or 04 85 80 00, depending on the 7th item)
There is also 12 octets used for padding (random - not so random - bytes), that may or may not be inserted in this packet. (if inserted, the padding is e8 90 84 98 20 12 18 19 04 85 80 00)

What are those 32 bits near the start of a FLAC file?

I am trying to write a parser to extract information from the following FLAC file:
$ hd audio.flac | head -n 6
00000000 66 4c 61 43 00 00 00 22 12 00 12 00 00 00 00 00 |fLaC..."........|
00000010 00 00 0a c4 42 f0 00 78 9f 30 00 00 00 00 00 00 |....B..x.0......|
00000020 00 00 00 00 00 00 00 00 00 00 84 00 02 64 1f 00 |.............d..|
00000030 00 00 47 53 74 72 65 61 6d 65 72 20 65 6e 63 6f |..GStreamer enco|
00000040 64 65 64 20 76 6f 72 62 69 73 63 6f 6d 6d 65 6e |ded vorbiscommen|
00000050 74 10 00 00 00 12 00 00 00 54 49 54 4c 45 3d 52 |t........TITLE=R|
Now, according to the specification, the format should be as follow (numbers are in bits):
<32> "fLaC", the FLAC stream marker in ASCII
<16> The minimum block size (in samples) used in the stream.
<16> The maximum block size (in samples) used in the stream.
<24> The minimum frame size (in bytes) used in the stream.
<24> The maximum frame size (in bytes) used in the stream.
<20> Sample rate in Hz.
<3> (number of channels)-1. FLAC supports from 1 to 8 channels
<5> (bits per sample)-1. FLAC supports from 4 to 32 bits per sample.
<36> Total samples in stream.
<128> MD5 signature of the unencoded audio data.
So, I start to write my parser and, while testing, get very strange results. So I test with a "real" metadata extractor:
$ metaflac --list audio.flac
METADATA block #0
type: 0 (STREAMINFO)
is last: false
length: 34
minimum blocksize: 4608 samples
maximum blocksize: 4608 samples
minimum framesize: 0 bytes
maximum framesize: 0 bytes
sample_rate: 44100 Hz
channels: 2
bits-per-sample: 16
total samples: 7905072
MD5 signature: 00000000000000000000000000000000
From the numbers, I can deduce the following:
66 4c 61 43 00 00 00 22 12 00 12 00 00 00 00 00
~~~~~~~~~~~ ~~~~~~~~~~~ ~~~~~ ~~~~~ ~~~~~~~~ ~~
^ ^ ^ ^ ^ ^
| | | | | |
| | | | | + Etc.
| | | | + Minimum frame size
| | | + Maximum block size
| | + Minimum block size
| + What is that ?!?
+ FLAC stream marker
Where does those 32 bits come from? I see they represent the length of the header, but isn't it against the standard to put it there (Taking into account that we already know the length: (32+16+16+24+20+3+5+36+128)/8)?
The 0x22 (34) is indeed the header block size in bytes as part of the METADATA_BLOCK_HEADER which follows the fLaC marker in the stream. Of the first 8 bits (00), bit 7 indicates that there are more metadatablocks to follow, the next 7 bits indicate that it's a STREAMINFO block. The following 3 bytes (00 00 22) is the length of the contents of the block;
16 + 16 + 24 + 24 + 20 + 3 + 5 + 36 + 128 = 272 bits
272 bits / 8 = 34 (0x22) bytes.

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