Due to the discovery of the leaked default HomeMatic AES-key my interest in the AES signing-process was once again renewed, so I decided to take a fresh look at it.
First we take a look at the messages involved in a signed command-execution:
Initial command-message from the initiator to the device (m-frame):
0E 03 A0 11 68EA14 1ED017 02 01C80000 len | flg | sender recvr | \params/ cnt type subtype
Signing request (challenge) from the device to the initiator (c-frame):
11 03 A0 02 1ED017 68EA14 04 D962D9FB2B03 00 len | flg | sender recvr | \challenge?/ kno cnt type sig.req.
Signing response from the initiator to the device (r-frame):
19 03 A0 03 68EA14 1ED017 344305154D33A8766DBAE938311FA514 len | flg | sender recvr \---------AES-payload----------/ cnt type
ACK-message from the device to the initiator, containig device-authentication (a-frame):
12 03 80 02 1ED017 68EA14 0101C80016 8B0C277F len | flg | sender recvr \ack-data/ \-auth-/ cnt type
The messages were captured when switching on the output of a HM-LC-Sw1-PI-2 plug with a HM-CFG-USB-2 using the default key.
The obvious approach -- decrypting the AES-payload with the key -- did not result in anything which could be resembled to the input-messages, so I started playing around with the key and the challenge, replacing and XORing them together.
XORing the first 6 bytes of the key with the challenge and then decrypting the payload with the modified key produced the following "plaintext":
Comparing this to the a-frame, we can see that two bytes from
the auth-part are identical (
12 03 80 02 1ED017 68EA14 0101C80016 8B0C
Repeating this process with different messages confirmed that the last 2 or 3 bytes of the a-frame were included in the decrypted plaintext.
Replacing the two "wrong" bytes with the "correct" bytes from the a-frame results in this:
If this is then decrypted again, the following plaintext is produced:
This contains the first 10 bytes (excluding length) of the m-frame in bytes 7 to 16 and what seems to be a timestamp or counter in bytes 1 to 6, so it seems to be decrypted correctly.
But how are the two magic bytes
8b0c produced? XORing them to the
bytes found in their place produces:
8b0c ^ 8ac4 =
These two bytes are the non-null parameters of the m-frame:
0E 03 A0 11 68EA14 1ED017 02
Testing with different messages confirmed that the parameters of
the m-frame (padded with
0x00 at the end to fill 16 bytes) are
XORed to the result of the first decryption before the second
decryption, sort of like an IV used in CBC-mode.
Based on these findings, the algorithm to verify the signature is:
A temporary key is built by XORing the first 6 bytes of the key with the challenge (bytes 12-17 of the c-frame).
The parameters of the m-frame are used as an IV, padded with
0x00 at the end to fill 16 bytes.
The AES-payload P of the r-frame is decrypted with the temporary key, producing Pd.
The IV is XORed to the decrypted Pd, resulting in Pd^. The first 4 bytes of Pd^ will be sent in the ACK-message.
The resulting Pd^ is decrypted (again) with the temporary key, producing Pd^d.
The end-result Pd^d should contain the beginning of the
m-frame (starting with the counter) which caused the
challenge-response-exchange starting at byte 7.
Bytes 1 to 6 seem to be the local timestamp or a counter
of the device generating the response.
RPTED-flag should be masked out in both the original
and the decrypted message before comparing, as the sender can't
know the value of the flag at the recipient (
flg & 0xbf)!
No HomeMatic-devices were harmed in this process.
2015-06-07, Michael Gernoth < gernoth at cs.fau.de >