Abstract: Aiming at the problem of sample amount to differential power attack of block cipher, by establishing the SNR model of differential power signal, this paper proposes the expression of the sample amount: . After measuring the parameters σ and ε, the numerical value is got, which is about 8 000. Using the 5 000 samples and 8 000 samples separately to finish the Differential Power Attack(DPA) to Advanced Encryption Standard(AES), and gets the right key when the samples’ amount is 8 000. The result is better than when it is 5 000, so the expression proposed is reasonable.

Key words: block cipher algorithm, Advanced Encryption Standard(AES), Differential Power Attack(DPA), sample amount

摘要： 对分组密码算法差分能量攻击的样本数量选取问题进行研究。通过建立差分能量信号的信噪比模型，推导出样本数量的数学表达式为 ，根据σ和ε计算得到攻击所需样本数量为8 000。分别用5 000组和8 000组随机明文对高级加密标准算法进行差分能量攻击，结果证明，当样本数量为8 000时可以得到正确密钥，效果结果优于5 000组明文的情况。

关键词: 分组密码算法, 高级加密标准, 差分能量攻击, 样本数量

CLC Number: 

• TP309.2