I am trying to use the webusb api to connect my USB device which basically is a digitizer. the usb version is 2.0.
I have successed to request the device in chrome but failed to open the device with the error Access Deny.
request device
the usb device has 3 interface: 2 HID and 1 USB (winusb loaded). here is the information shown in xusb:
Using libusb v1.0.23.11397
Opening device 2D80:1010...
libusb: warning [hid_open] could not open HID device in R/W mode (keyboard or mouse?) - trying without
Device properties:
bus number: 1
port path: 14 (from root hub)
speed: 12 Mbit/s (USB FullSpeed)
Reading device descriptor:
length: 18
device class: 0
S/N: 3
VID:PID: 2D80:1010
bcdDevice: 0100
iMan:iProd:iSer: 1:2:3
nb confs: 1
Reading BOS descriptor: 1 caps
Unknown BOS device capability 05:
Reading first configuration descriptor:
nb interfaces: 3
interface[0]: id = 0
interface[0].altsetting[0]: num endpoints = 1
Class.SubClass.Protocol: 03.01.01
endpoint[0].address: 81
max packet size: 0040
polling interval: 01
interface[1]: id = 1
interface[1].altsetting[0]: num endpoints = 1
Class.SubClass.Protocol: 03.01.01
endpoint[0].address: 82
max packet size: 0040
polling interval: 01
interface[2]: id = 2
interface[2].altsetting[0]: num endpoints = 1
Class.SubClass.Protocol: FF.00.00
endpoint[0].address: 83
max packet size: 0040
polling interval: 00
Claiming interface 0...
Claiming interface 1...
Claiming interface 2...
Reading string descriptors:
String (0x01): ""
String (0x02): ""
String (0x03): "******"
Reading Extended Compat ID OS Feature Descriptor (wIndex = 0x0004):
libusb: warning [hid_submit_control_transfer] unsupported HID control request
libusb: warning [hid_submit_control_transfer] unsupported HID control request
00000000 28 00 00 00 00 01 04 00 01 00 00 00 00 00 00 00 (...............
00000010 02 01 57 49 4e 55 53 42 00 00 00 00 00 00 00 00 ..WINUSB........
00000020 00 00 00 00 00 00 00 00 ........
Reading Extended Properties OS Feature Descriptor (wIndex = 0x0005):
libusb: warning [hid_submit_control_transfer] unsupported HID control request
libusb: warning [hid_submit_control_transfer] unsupported HID control request
00000000 8e 00 00 00 00 01 05 00 01 00 84 00 00 00 01 00 ................
00000010 00 00 28 00 44 00 65 00 76 00 69 00 63 00 65 00 ..(.D.e.v.i.c.e.
00000020 49 00 6e 00 74 00 65 00 72 00 66 00 61 00 63 00 I.n.t.e.r.f.a.c.
00000030 65 00 47 00 55 00 49 00 44 00 00 00 4e 00 00 00 e.G.U.I.D...N...
00000040 7b 00 31 00 44 00 34 00 42 00 32 00 33 00 36 00 {.1.D.4.B.2.3.6.
00000050 35 00 2d 00 34 00 37 00 34 00 39 00 2d 00 34 00 5.-.4.7.4.9.-.4.
00000060 38 00 45 00 41 00 2d 00 42 00 33 00 38 00 41 00 8.E.A.-.B.3.8.A.
00000070 2d 00 37 00 43 00 36 00 46 00 44 00 44 00 44 00 -.7.C.6.F.D.D.D.
00000080 44 00 37 00 45 00 32 00 36 00 7d 00 00 00 D.7.E.2.6.}...
Releasing interface 0...
Releasing interface 1...
Releasing interface 2...
Closing device...
the first hid interface is used to report digitizer which i think it is already claimed by the system. I guess that is the issue because I have tried to remove the two hid interface and repeat the process. this time the device was opened successfully.
however the same issue is not presented on Mac OS.
so i am wonder if anyone has any suggestion to solve the issue on windows?
It is possible that this is a Chrome issue. I have been working on a new backend for connecting to USB devices on Windows that should be more reliable, especially for complex composite devices.
Please install Chrome canary-channel (at least version 84.0.4110.2) from https://www.google.com/chrome/canary/ and try turning on the “Enable new USB backend” in flag in chrome://flags.
Let me know if this resolves the issue.
I am currently using C# to retrieve frames from a borescope (via the FFMPEG library). However, I came across a problem weeks ago and I can't solve it.
The images are returned in JPEG format (since the borescope stream is MJPEG).
Some images come without quality problems, but others come with a strange line in the middle
followed by random staining. (At the end of the question there is an example of a normal image and one with problems).
Analyzing the structure of the files, I realized that there are some differences, but I don't really understand JPEG's binary structure very well, and I can't tell what is corrupted.
Getting to know what is corrupted in the image, which culminates in the quality problem, is very important to me because, through this, I can discard the frame using C#. However, without understanding this problem, I can't even discard the frame, much less fix it.
So, having the image without quality problems as a reference, what is the problem with the binary structure of the image with quality problems?
Examples:
JPEG 1: Image without quality problems
Image's preview (just to see the quality, do not download from here)
JPEG 2: Image with quality problems
Image's preview (just to see the quality, do not download from here)
It's possible to look into binary structure of images through online HEX editors like: Online hex editor, Hexed or Hex-works.
Thank you for reading and have a nice day.
There are at least 2 issues with the file.
The first I can detect with ImageMagick by running this command:
magick identify -verbose image.jpg
and it tells me that the data segment ends prematurely.
Image: outExemplo0169.jpeg
Format: JPEG (Joint Photographic Experts Group JFIF format)
Mime type: image/jpeg
Class: DirectClass
Geometry: 640x480+0+0
Units: Undefined
Colorspace: sRGB
Type: TrueColor
Base type: Undefined
Endianess: Undefined
Depth: 8-bit
Channel depth:
Red: 8-bit
Green: 8-bit
Blue: 8-bit
Channel statistics:
Pixels: 307200
Red:
min: 0 (0)
max: 255 (1)
mean: 107.234 (0.420527)
standard deviation: 66.7721 (0.261851)
kurtosis: -0.67934
skewness: 0.577494
entropy: 0.92876
Green:
min: 0 (0)
:2020-02-26T18:59:19+00:00 0:00.057 0.070u 7.0.9 Resource identify[80956]: resource.c/RelinquishMagickResource/1067/Resource
Memory: 3686400B/0B/32GiB
identify: Corrupt JPEG data: premature end of data segment `outExemplo0169.jpeg' # warning/jpeg.c/JPEGWarningHandler/399.
The second I can see with exiftool when I run this command:
exiftool -v -v -v outExemplo0169.jpeg
ExifToolVersion = 11.11
FileName = outExemplo0169.jpeg
Directory = .
FileSize = 66214
FileModifyDate = 1582743337
FileAccessDate = 1582743559
FileInodeChangeDate = 1582743337
FilePermissions = 33188
FileType = JPEG
FileTypeExtension = JPG
MIMEType = image/jpeg
JPEG APP0 (14 bytes):
0006: 4a 46 49 46 00 01 01 00 00 01 00 01 00 00 [JFIF..........]
+ [BinaryData directory, 9 bytes]
| JFIFVersion = 1 1
| - Tag 0x0000 (2 bytes, int8u[2]):
| 000b: 01 01 [..]
| ResolutionUnit = 0
| - Tag 0x0002 (1 bytes, int8u[1]):
| 000d: 00 [.]
| XResolution = 1
| - Tag 0x0003 (2 bytes, int16u[1]):
| 000e: 00 01 [..]
| YResolution = 1
| - Tag 0x0005 (2 bytes, int16u[1]):
| 0010: 00 01 [..]
| ThumbnailWidth = 0
| - Tag 0x0007 (1 bytes, int8u[1]):
| 0012: 00 [.]
| ThumbnailHeight = 0
| - Tag 0x0008 (1 bytes, int8u[1]):
| 0013: 00 [.]
JPEG SOF0 (15 bytes):
0018: 08 01 e0 02 80 03 01 21 00 02 11 01 03 11 01 [.......!.......]
ImageWidth = 640
ImageHeight = 480
EncodingProcess = 0
BitsPerSample = 8
ColorComponents = 3
JPEG DQT (130 bytes):
002b: 00 03 03 03 03 03 03 04 03 03 03 04 04 04 05 06 [................]
003b: 09 06 06 05 05 06 0c 08 09 07 09 0e 0c 0e 0e 0d [................]
004b: 0c 0d 0d 0f 11 15 12 0f 10 14 10 0d 0d 13 19 13 [................]
005b: 14 16 17 18 18 18 0f 12 1a 1c 1a 17 1c 15 17 18 [................]
006b: 17 01 04 04 04 06 05 06 0b 06 06 0b 17 0f 0d 0f [................]
007b: 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 [................]
008b: 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 [................]
[snip 18 bytes]
JPEG DHT (416 bytes):
00b1: 00 00 01 05 01 01 01 01 01 01 00 00 00 00 00 00 [................]
00c1: 00 00 01 02 03 04 05 06 07 08 09 0a 0b 10 00 02 [................]
00d1: 01 03 03 02 04 03 05 05 04 04 00 00 01 7d 01 02 [.............}..]
00e1: 03 00 04 11 05 12 21 31 41 06 13 51 61 07 22 71 [......!1A..Qa."q]
00f1: 14 32 81 91 a1 08 23 42 b1 c1 15 52 d1 f0 24 33 [.2....#B...R..$3]
0101: 62 72 82 09 0a 16 17 18 19 1a 25 26 27 28 29 2a [br........%&'()*]
0111: 34 35 36 37 38 39 3a 43 44 45 46 47 48 49 4a 53 [456789:CDEFGHIJS]
[snip 304 bytes]
JPEG SOS
JPEG EOI
Unknown trailer (50 bytes at offset 0x10274):
10274: 42 6f 75 6e 64 61 72 79 45 42 6f 75 6e 64 61 72 [BoundaryEBoundar]
10284: 79 53 00 00 01 00 90 08 01 00 fb 4b db 6a 2a 22 [yS.........K.j*"]
10294: 00 00 2a 22 00 00 01 00 01 00 80 02 00 00 e0 01 [..*"............]
102a4: 00 00
So there are 50 extraneous bytes at the end including the text string "BoundaryEBoundaryS" which may be recognisable to you as coming from somewhere else in your processing chain?
One test you could do for JPEG quality is check the last 2 bytes are a valid EOI which means it should end in FF D9 - see here.
I am communicating with a servo via RS232 serial. The built-in functions that came with my servo are too slow (25 ms for a simple 54 byte message on a 57,600 baud port), so I am trying to write my own communication functions, however the built-in functions are not documented. I have used a port monitor to determine what information is being sent to the servo and I need help deciphering the results.
I used the built-in functions to command the servo to "goto" incrementally increasing steps (1, 2, 3, etc.). This resulted 5 packets being sent to the servo for each "goto" command. The first 4 packets are identical for each "goto" command. I have attached about 50 hex packet below (1 per line). If you need more, post, and we can work something out.
10 13 04 20 00 01 B6 24 E9 68
10 13 04 20 00 00 AE 24 54 82
10 13 04 20 00 00 B5 24 8B 0B
10 13 04 20 00 01 43 01 71 9B
The 5th packet varies based on the step the motor is being commanded to move to. I have included 1 packet here as an example. I have attached a file with about 1000 of these packets (1 per line).
10 13 08 20 03 01 11 25 0A 00 00 00 81 CF
The first 8 bytes of this packet (10 13 08 20 03 01 11 25) appear to be the actual "goto" command. They remain the same no matter what step is specified.
The last 6 bytes (0A 00 00 00 81 CF) change based upon the step that is requested. In the file I attached, I instructed the servo to initially goto step "0", then "1", "2", etc. The first 4 bytes appear to be a little-endian integer corresponding to the number of steps (i.e. the sample command I showed above instructs the servo to goto step 10 decimal).
My question regards the last 2 bytes of the command. They appear to vary randomly, but whenever the specified step is the same they match. This leads me to believe that these 2 bytes are a checksum of some kind. My question to you is: how is the checksum calculated?
I have already tried xor'ing all the bytes, both singly and in 2 byte pairs, and I tried Fletcher's checksum, and a simple checksum (sum of all bytes). I also checked the 2's complement of each of these methods (though I certainly wouldn't mind someone checking to make sure I didn't make a mistakes in the calculations). Does anyone have any ideas?
10 13 08 20 03 01 11 25 00 00 00 00 E9 64
10 13 08 20 03 01 11 25 01 00 00 00 9F D0
10 13 08 20 03 01 11 25 02 00 00 00 04 0C
10 13 08 20 03 01 11 25 04 00 00 00 23 95
10 13 08 20 03 01 11 25 05 00 00 00 55 21
10 13 08 20 03 01 11 25 06 00 00 00 CE FD
10 13 08 20 03 01 11 25 07 00 00 00 B8 49
10 13 08 20 03 01 11 25 08 00 00 00 6C A7
10 13 08 20 03 01 11 25 09 00 00 00 1A 13
10 13 08 20 03 01 11 25 0A 00 00 00 81 CF
10 13 08 20 03 01 11 25 0C 00 00 00 A6 56
10 13 08 20 03 01 11 25 0D 00 00 00 D0 E2
10 13 08 20 03 01 11 25 0F 00 00 00 3D 8A
10 13 08 20 03 01 11 25 10 10 00 00 00 17 FA
10 13 08 20 03 01 11 25 11 00 00 00 84 77
10 13 08 20 03 01 11 25 12 00 00 00 1F AB
10 13 08 20 03 01 11 25 13 00 00 00 69 1F
10 13 08 20 03 01 11 25 14 00 00 00 38 32
10 13 08 20 03 01 11 25 15 00 00 00 4E 86
10 13 08 20 03 01 11 25 16 00 00 00 D5 5A
10 13 08 20 03 01 11 25 17 00 00 00 A3 EE
10 13 08 20 03 01 11 25 18 00 00 00 77 00
10 13 08 20 03 01 11 25 19 00 00 00 01 B4
10 13 08 20 03 01 11 25 1A 00 00 00 9A 68
10 13 08 20 03 01 11 25 1B 00 00 00 EC DC
10 13 08 20 03 01 11 25 1C 00 00 00 BD F1
10 13 08 20 03 01 11 25 1D 00 00 00 CB 45
10 13 08 20 03 01 11 25 1E 00 00 00 50 99
10 13 08 20 03 01 11 25 1F 00 00 00 26 2D
10 13 08 20 03 01 11 25 20 00 00 00 DE 2A
10 13 08 20 03 01 11 25 21 00 00 00 A8 9E
10 13 08 20 03 01 11 25 22 00 00 00 33 42
10 13 08 20 03 01 11 25 24 00 00 00 14 DB
10 13 08 20 03 01 11 25 25 00 00 00 62 6F
10 13 08 20 03 01 11 25 26 00 00 00 F9 B3
10 13 08 20 03 01 11 25 27 00 00 00 8F 07
10 13 08 20 03 01 11 25 28 00 00 00 5B E9
10 13 08 20 03 01 11 25 29 00 00 00 2D 5D
10 13 08 20 03 01 11 25 2A 00 00 00 B6 81
10 13 08 20 03 01 11 25 2B 00 00 00 C0 35
10 13 08 20 03 01 11 25 2C 00 00 00 91 18
10 13 08 20 03 01 11 25 2D 00 00 00 E7 AC
10 13 08 20 03 01 11 25 2E 00 00 00 7C 70
10 13 08 20 03 01 11 25 2F 00 00 00 0A C4
10 13 08 20 03 01 11 25 30 00 00 00 C5 8D
10 13 08 20 03 01 11 25 31 00 00 00 B3 39
10 13 08 20 03 01 11 25 32 00 00 00 28 E5
10 13 08 20 03 01 11 25 33 00 00 00 5E 51
10 13 08 20 03 01 11 25 34 00 00 00 0F 7C
10 13 08 20 03 01 11 25 35 00 00 00 79 C8
10 13 08 20 03 01 11 25 36 00 00 00 E2 14
10 13 08 20 03 01 11 25 37 00 00 00 94 A0
10 13 08 20 03 01 11 25 38 00 00 00 40 4E
10 13 08 20 03 01 11 25 39 00 00 00 36 FA
10 13 08 20 03 01 11 25 3A 00 00 00 AD 26
10 13 08 20 03 01 11 25 3B 00 00 00 DB 92
10 13 08 20 03 01 11 25 3C 00 00 00 8A BF
10 13 08 20 03 01 11 25 3D 00 00 00 FC 0B
10 13 08 20 03 01 11 25 3E 00 00 00 67 D7
10 13 08 20 03 01 11 25 3F 00 00 00 11 63
this is a late answer, but hopefully this can help for other CRC re-engineering tasks:
Your CRC is a derivation of the so-called "16 bit width CRC as designated by CCITT", but with "init value zero".
The CRC is calculated from byte position 3 to byte position 12 of your example data. e.g.
08 20 03 01 11 25 00 00 00 00
The full CRC specification according to our CRC specification overview is:
CRC:16,1021,0000,0000,No,No
The problem was not only to find the right CRC polynomial, but finding the following answers:
Which part of the data is included in the CRC calculation, and which is not.
Which init value to use? Apply final xor value?
Does this algorithm expect reflected input or output values?
Again, see our manual description or the Boost CRC library on what this means.
What I did is running a brute-force script that simply tries out several popular 16 bit CRC polynomials with all kinds of combinations of start/end positions, initial values, reflected versions. Here is how the processing output looked:
Finding CRC for test message (HEX): 10 13 08 20 03 01 11 25 00 00 00 00 E9 64
Trying CRC spec : CRC:16,1021,FFFF,0000,No,No
Trying CRC spec : CRC:16,8005,0000,0000,No,No
Trying CRC spec : CRC:16,8005,FFFF,0000,No,No
Trying CRC spec : CRC:16,1021,FFFF,FFFF,No,No
Trying CRC spec : CRC:16,1021,0000,FFFF,No,No
Trying CRC spec : CRC:16,1021,0000,0000,No,No
Found it!
Relevant sequence for checksum from startpos=3 to endpos=12
08 20 03 01 11 25 00 00 00 00
CRC spec: CRC:16,1021,0000,0000,No,No
CRC result: E9 64 (Integer = 59748)
With the result I could re-calculate the checksum of your example telegrams correctly
19.09.2016 12:18:12.764 [TX] - 10 13 08 20 03 01 11 25 00 00 00 00 E9 64
19.09.2016 12:18:14.606 [TX] - 10 13 08 20 03 01 11 25 01 00 00 00 9F D0
19.09.2016 12:18:16.030 [TX] - 10 13 08 20 03 01 11 25 02 00 00 00 04 0C
I uploaded the documented CRC finder example script which works with the free Docklight Scripting V2.2 evaluation. I assume this can be very useful for other CRC re-engineering puzzles, too.
The example also helped to solve Stackoverflow question 22219796
I am developing an application which reads NFC card from the reader.
I know the code for reading binary block like this:
FF B0 00 04 10
04 for the block 4 and 10 for 16 bytes data. My card has the data "TEST009996".
I run 5 code for read binary blocks from 4-8 like this:
FF B0 00 04 10
FF B0 00 05 10
FF B0 00 06 10
FF B0 00 07 10
FF B0 00 08 10
I got the following results:
T☻enTEÉ ☺
T☻enTEST00É
T☻enTEST009996É
enTEST009996■ 6É
ST009996■ 6 É
or in hexadecimal:
01 03 A0 10 44 03 11 D1 01 0D 54 02 65 6E 48 43 90 00
44 03 11 D1 01 0D 54 02 65 6E 48 43 49 44 30 30 90 00
01 0D 54 02 65 6E 48 43 49 44 30 30 39 39 39 36 90 00
65 6E 48 43 49 44 30 30 39 39 39 36 FE 00 00 36 90 00
49 44 30 30 39 39 39 36 FE 00 00 36 00 00 00 00 90 00
Should I create an algorithm to cut the result to get the data? Are there any better ways?
Source:
http://downloads.acs.com.hk/drivers/en/API-ACR122U-2.02.pdf
It seems that your tag is an NFC Forum Type 2 Tag (find the NFC Forum Type 2 Tag Operation specification on the NFC Forum website). As you mention MIFARE this could, for instance, be a MIFARE Ultralight, MIFARE Ultralight C or NTAG tag.
A block on a Type 2 Tag consists of 4 bytes. The read command reads 4 blocks at a time. So the read command gives you 4 blocks (4 bytes each) starting at a given block offset plus a status word for the read command (0x9000 for success). In your case you get:
Read(4, 16): 0103A010 440311D1 010D5402 656E4843 9000
Read(5, 16): 440311D1 010D5402 656E4843 49443030 9000
Read(6, 16): 010D5402 656E4843 49443030 39393936 9000
Read(7, 16): 656E4843 49443030 39393936 FE000036 9000
Read(8, 16): 49443030 39393936 FE000036 00000000 9000
Consequently, the memory of your tag looks like this:
0103A010
440311D1
010D5402
656E4843
49443030
39393936
FE000036
00000000
A Type 2 Tag (btw. in order to make sure that this tag actually conforms to the Type 2 Tag Operation Specification you would also need to read the capability container which is located in block 3) contains a series of tag-length-value (TLV) structures:
01 (Tag: Lock Control TLV)
03 (Length: 3 bytes)
A0 10 44 (Value: Information on position and function of lock bytes)
03 (Tag: NDEF Message TLV)
11 (Length: 17 bytes)
D1010D5402656E48434944303039393936 (Value: NDEF message)
FE (Tag: Terminator TLV; has no length field)
So your tag contains the NDEF message
D1010D5402656E48434944303039393936
This translates to
D1 (Header byte of record 1)
- Message begin is set (= first record of an NDEF message)
- Message end is set (= last record of an NDEF message)
- Short record flag is set (= Payload length field consists of 1 byte only)
- Type Name Format = 0x1 (= Type field contains an NFC Forum well-known type)
01 (Type length: 1 byte)
0D (Payload length: 13 bytes)
54 (Type: "T")
02656E48434944303039393936 (Payload field)
The payload field of a NFC Forum Text record decodes like this:
02 (Status byte: Text is UTF-8 encoded, Language code has a length of 2 bytes)
656E (Language code: "en")
48434944303039393936 (Text: "TEST009996")
I am writing a code to parse MFT of NTFS. I`m trying analyse Data Run of non residental $INDEX_ALLOCATION attrib:
11 01 2C 11 02 FE 11 00
9F 0B 21 01 DB 00 21 01
D9 00 21 01 E0 00 21 01
F6 00 21 01 10 01 00 F1
After regroup I see problem in Data Run No 3:
DataRun 1: 11 01 2C
DataRun 2: 11 02 FE
DataRun 3: 11 00 9F <- what does mean "00" ?
I tried analyse it using Active Disk Editor 3 and this software decompose it to:
DataRun 3: 11 00 9F 0B
In my opinion header of DataRun 3 ("11") mean 1 length and 1 offset so there should be 2 bytes after header, but there are 3 bytes.
Any idea?