题目内容

函数$\frac{2e^{-s}-e^{-2s}}{s}$的Laplace逆变换$L^{-1}[\frac{2e^{-s}-e^{-2s}}{s}]=?$

A. $u(t-2)-2u(t-1)$
B. $u(t-1)-2u(t-2)$
C. $2u(t-1)-u(t-2)$
D. $2u(t-2)-u(t-1)$

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信号$f(t)=e^{-t}u(t-2)$的拉氏变换为

A. $\frac{1}{s-2}e^{-s-1}$
B. $\frac{1}{s+1}e^{-2(s+1)}$
C. $\frac{1}{s-2}e^{-2(s-1)}$
D. $\frac{1}{s+1}e^{-(s+1)}$

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. $-\frac{1}{a}$

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. $[2^n+(-\frac{1}{2})^n]u[-n-1]$

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