answer:序列{1, 1, 1, 1}的8点DFT可以表示为： X[k] = 1/8 * (x[0] * W^0k + x[1] * W^1k + x[2] * W^2k + x[3] * W^3k + x[4] * W^4k + x[5] * W^5k + x[6] * W^6k + x[7] * W^7k) 其中，W代表旋转因子，用复数表示为： W = cos(2π/N) – j sin(2π/N) 其中j表示虚数单位，N表示DFT的长度。对于8点DFT，N=8。 因此，计算序列{1, 1, 1, 1}的8点DFT，可得： X[0] = 1/8 * (1 * 1 + 1 * 1 + 1 * 1 + 1 * 1 + 1 * 1 + 1 * 1 + 1 * 1 + 1 * 1) = 1 X[1] = 1/8 * (1 * 1 + cos(2π/8) – j sin(2π/8) + 1 * 1 + cos(2π/8) – j sin(2π/8) + 1 * 1 + cos(2π/8) – j sin(2π/8) + 1 * 1 + cos(2π/8) – j sin(2π/8)) = 0 X[2] = 1/8 * (1 * 1 + cos(4π/8) – j sin(4π/8) + 1 * 1 + cos(4π/8) – j sin(4π/8) + 1 * 1 + cos(4π/8) – j sin(4π/8) + 1 * 1 + cos(4π/8) – j sin(4π/8)) = 0 X[3] = 1/8 * (1 * 1 + cos(6π/8) – j sin(6π/8) + 1 * 1 + cos(6π/8) – j sin(6π/8) + 1 * 1 + cos(6π/8) – j sin(6π/8) + 1 * 1 + cos(6π/8) – j sin(6π/8)) = 0 X[4] = 1/8 * (1 * 1 + 1 * 1 + 1 * 1 + 1 * 1 + 1 * 1 + 1 * 1 + 1 * 1 + 1 * 1) = 1 X[5] = 1/8 * (1 * 1 + cos(2π/8) + j sin(2π/8) + 1 * 1 + cos(2π/8) + j sin(2π/8) + 1 * 1 + cos(2π/8) + j sin(2π/8) + 1 * 1 + cos(2π/8) + j sin(2π/8)) = 0 X[6] = 1/8 * (1 * 1 + cos(4π/8) + j sin(4π/8) + 1 * 1 + cos(4π/8) + j sin(4π/8) + 1 * 1 + cos(4π/8) + j sin(4π/8) + 1 * 1 + cos(4π/8) + j sin(4π/8)) = 0 X[7] = 1/8 * (1 * 1 + cos(6π/8) + j sin(6π/8) + 1 * 1 + cos(6π/8) + j sin(6π/8) + 1 * 1 + cos(6π/8) + j sin(6π/8) + 1 * 1 + cos(6π/8) + j sin(6π/8)) = 0 因此，序列{1, 1, 1, 1}的8点DFT为{1, 0, 0, 0, 1, 0, 0, 0}。

answer:序列{1.1}的2点DFT可以表示为： X[k] = x[0] * W^0k + x[1] * W^1k 其中，W代表旋转因子，用复数表示为： W = cos(2π/N) – j sin(2π/N) 其中j表示虚数单位，N表示DFT的长度。对于2点DFT，N=2。 因此，计算序列{1.1}的2点DFT，可得： X[0] = 1.1 * 1 + 0 * 0 = 1.1 X[1] = 1.1 * cos(π/2) – j sin(π/2) = – j * 1.1 因此，序列{1.1}的2点DFT为{1.1, – j * 1.1}。

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