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AES encryption algorithmConsider the following plain text SE - 3 3 0 - FINALEXAM 1 - Convert this Plaintext to a state matrix:Given the ASCIl

AES encryption algorithmConsider the following plain text SE-330-FINALEXAM
1- Convert this Plaintext to a state matrix:Given the ASCIl values of the plaintext: 83,69,45,51,51,48,45,70,73,78,65,76,69,88,65,77,32, and the encryption key ADVANCEDENCRYPTI converted to ASCIl as 65,68,86,65,78,67,69,68,69,78,67,82,89,80,84,73,32:a.What is the length of the key used in this algorithm in bits?b.Describe how the AddRoundKey step is performed without showing the output (results). Use the provided plain text and encryption key2- If the output of SubBytes is 6568866578676968697867828980847332Perform the ShiftRows stage

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