某信号的Z变换结果是$X(z)=\frac{1-az^{-1}}{z^{-1}-a}(a不等于0）$，则$x[0]=?$
A. $a$
B. $-a$
C. $\frac{1}{a}$
D. $-\frac{1}{a}$
查看答案
当$0.5
A. $(\frac{1}{2})^nu[n]+2^nu[-n-1]$
B. $(\frac{1}{2})^nu[-n-1]+2^nu[n]$
C. $2^nu[n]-(\frac{1}{2})^nu[-n-1]$
D. $2^nu[n]+(-\frac{1}{2})^nu[-n-1]$
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. $[2^n+(-\frac{1}{2})^n]u[-n-1]$
已知某信号的Z变换结果是$X(z)=\frac{-3z^{-1}}{2-5z^{-1}+2z^{-2}}$，当$|z|>2$时右边序列$x[n]$为
A. $[(\frac{1}{2})^n-2^n]u[n]$
B. $[(\frac{1}{2})^n+2^n]u[n]$
C. $[2^n-(\frac{1}{2})^n]u[n]$
D. $[2^n+(-\frac{1}{2})^n]u[n]$
已知某信号的Z变换结果是$X(z)=\frac{1-0.5z^{-1}}{1+\frac{3}{4}z^{-1}+\frac{1}{8}z^{-2}}$，则信号为
A. $x[n]=[2^{n+2}-3(\frac{1}{4})^n]u[n]$
B. $x[n]=[4(-\frac{1}{2})^n-3(\frac{1}{4})^n]u[n]$
C. $x[n]=[4(-\frac{1}{2})^n-3(-\frac{1}{4})^n]u[n]$
D. $x[n]=[2^{n+2}-3(-\frac{1}{4})^n]u[n]$
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