当$|z|<0.5$时左边序列$x[n]$为
A: $[(\frac{1}{2})^n-2^n]u[-n-1]$
B: $[(\frac{1}{2})^n+2^n]u[-n-1]$
C: $[2^n-(\frac{1}{2})^n]u[-n-1]$
D: $[2^n+(-\frac{1}{2})^n]u[-n-1]$
A: $[(\frac{1}{2})^n-2^n]u[-n-1]$
B: $[(\frac{1}{2})^n+2^n]u[-n-1]$
C: $[2^n-(\frac{1}{2})^n]u[-n-1]$
D: $[2^n+(-\frac{1}{2})^n]u[-n-1]$
举一反三
- 下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x>1) $
- 设`\n`阶方阵`\A`满足`\|A| = 2`，则`\|A^TA| = ,|A^{ - 1}| = ,| A^ ** | = ,| (A^ ** )^ ** | = ,|(A^ ** )^{ - 1} + A| = ,| A^{ - 1}(A^ ** + A^{ - 1})A| = `分别等于（ ） A: \[4,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] B: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n + 1)}^2}}},2{(\frac{3}{2})^n},\frac{{{3^n}}}{2}\] C: \[4,\frac{1}{2},{2^{n + 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\] D: \[2,\frac{1}{2},{2^{n - 1}},{2^{{{(n - 1)}^2}}},2{(\frac{3}{2})^{n - 1}},\frac{{{3^n}}}{2}\]
- 已知信号$x[n]=2^nu(n)+\frac{1}{2^n}u(-n-1)$，其Z变换结果为 A: $\frac{z}{z-2}-\frac{z}{z-0.5}$ B: $\frac{1}{z-2}-\frac{1}{z-0.5}$ C: $\frac{z}{z-2}+\frac{z}{z-0.5}$ D: 不存在
- 1．下列数列中，收敛但极限不为$1$的是 A: ${{(2+\frac{1}{n})}^{\frac{1}{n}}}$ B: ${{n}^{\frac{1}{n}}}$ C: $\frac{1}{{{n}^{2}}+1}+\frac{2}{{{n}^{2}}+2}+\cdots +\frac{n}{{{n}^{2}}+n}$ D: $\frac{{{(n!)}^{2}}}{{{n}^{n}}}$
- ${X_1},{X_2},...,{X_n}$是来自均匀分布X~U(-a,a)的样本，用矩估计法估计参数a为() A: ${(\frac{3}{n}\sum\limits_{k = 1}^n {x_k^2} )^{\frac{1}{2}}}$ B: ${(\frac{2}{n}\sum\limits_{k = 1}^n {x_k^2} )^{\frac{1}{2}}}$ C: ${(\frac{3}{n}\sum\limits_{k = 1}^n {x_k} )^{\frac{1}{2}}}$ D: ${(\frac{2}{n}\sum\limits_{k = 1}^n {x_k} )^{\frac{1}{2}}}$