700+ câu Trắc nghiệm Xử lý số liệu tín hiệu đáp án - Phần 7

$X\left(Z\right)=\frac{1}{\left(1-a\cdot Z{}^{-1}\right)}\left|Z\right|\ge \left|a\right|$

$X\left(Z\right)=\frac{1}{\left(1+a\cdot Z{}^{-1}\right)}\left|Z\right|\ge \left|a\right|$

$X\left(Z\right)=\frac{1}{\left(1-a\cdot Z\right)}\left|Z\right|\ge \left|a\right|$

$X\left(Z\right)=\frac{Z}{\left(1-a\cdot Z{}^{-1}\right)}\left|Z\right|\ge \left|a\right|$

Miền hội tụ $\left|Z\right|>1$

Miền hội tụ với mọi $Z$

Miền hội tụ $\left|Z\right|<1$

Không tồn tại biến đổi $Z$

Có. Hai điểm cực tại $\text{Z}=0$ và $\text{Z}=-1$Không có điểm không

Có. Một điểm cực tại $Z=0$Một điểm không tại $\text{Z}=-1$

Có. Hai điểm không tại $\text{Z}=0$ và $\text{Z}=-1$Không có điểm không

Có. Một điểm cực tại $\text{Z}=0$Không có điểm không

$x\left(n\right)=u\left(n\right)\cdot \left[a\cdot 2{}^n+a{}^n\right]$

$x\left(n\right)=u\left(n-1\right)\cdot \left[a\cdot 2{}^{n-1}+a{}^{n-1}\right]$

$x\left(n\right)=-u\left(n-1\right)\cdot \left[a\cdot 2{}^{n-1}+a{}^{n-1}\right]$

$x\left(n\right)=-u\left(n-1\right)\cdot \left[a\cdot 2{}^n+a{}^n\right]$

$\text{X}\left(\text{Z}\right)=\frac{2}{(Z-1){}^2}+\frac{1}{Z-1}+\frac{1}{Z+1}$

$\text{X}\left(\text{Z}\right)=\frac{1}{(Z-1){}^2}-\frac{1}{Z-1}-\frac{1}{Z+1}$

$X\left(Z\right)=\frac{2}{(Z-1){}^2}+\frac{1}{Z-1}-\frac{1}{Z+1}$

$X\left(Z\right)=\frac{1}{(Z-1){}^2}-\frac{2}{Z-1}-\frac{1}{Z+1}$

$\text{X}\left(\text{Z}\right)=\frac{1}{Z-1}-\frac{1}{Z+1}-\frac{2}{(Z+1){}^2}+\frac{4}{(Z+1){}^3}$

$\text{X}\left(\text{Z}\right)=\frac{1}{Z-1}-\frac{1}{Z+1}+\frac{2}{(Z+1){}^2}+\frac{2}{(Z+1){}^3}$

$\text{X}\left(\text{Z}\right)=\frac{1}{Z-1}-\frac{2}{Z+1}-\frac{2}{(Z+1){}^2}+\frac{2}{(Z+1){}^3}$

$\text{X}\left(\text{Z}\right)=\frac{1}{Z-1}-\frac{2}{Z+1}+\frac{2}{(Z+1){}^2}+\frac{4}{(Z+1){}^3}$

$\text{X}\left(\text{Z}\right)=\frac{\frac{1}{a}}{Z}+\frac{\frac{1}{a}}{\left(Z-a\right)}-\frac{1}{(Z-a){}^2}$

$\frac{1}{aZ}-\frac{1}{a\left(Z-a\right)}+\frac{a}{(Z-a){}^2}$

$\text{X}\left(\text{Z}\right)=\frac{\frac{1}{a}}{Z}-\frac{\frac{1}{a}}{\left(Z-a\right)}-\frac{a}{(Z-a){}^2}$

$X\left(Z\right)=\frac{1}{aZ}-\frac{1}{a\left(Z-a\right)}+\frac{1}{(Z-a){}^2}$

$X\left(Z\right)=\frac{2}{Z-2}-\frac{1}{Z-1}$

$\text{X}\left(\text{Z}\right)=\frac{2.Z{}^5}{Z-2}-\frac{Z{}^5}{Z-1}$

$X\left(Z\right)=Z{}^4\left|\frac{2}{Z-2}-\frac{1}{Z-1}\right|$

$\text{X}\left(\text{Z}\right)=\frac{Z{}^5}{Z-2}+\frac{Z{}^5}{Z-1}$

$x\left(n\right)=\delta \left(n-1\right)-a\cdot \delta \left(n-2\right)$

$x\left(n\right)=a{}^n\cdot u\left(n-2\right)$

$x\left(n\right)=\delta \left(n-1\right)-a{}^n.u\left(n\right)$

$x\left(n\right)=\delta \left(n-2\right)\cdot a{}^n\cdot u\left(n\right)$

$x\left(n\right)=3{}^{n-3}\cdot u\left(n+3\right)-2{}^{n-3}\cdot u\left(n+3\right)$

$x\left(n\right)=3{}^{n+3}\cdot u\left(n+3\right)-2{}^{n+3}\cdot u\left(n+3\right)$

$x\left(n\right)=3{}^{n+3}\cdot u\left(n+3\right)+2{}^{n+3}\cdot u\left(n+3\right)$

$x\left(n\right)=3{}^{n+4}\cdot u\left(n+3\right)-2{}^{n+4}\cdot u\left(n+3\right)$

$\text{H}\left(\text{Z}\right)=1+\beta Z{}^{-1}-\alpha Z{}^{-1}$

$\text{H}\left(\text{Z}\right)=1+\beta Z{}^{-1}+\alpha Z{}^{-1}$

$H\left(Z\right)=1+\beta Z{}^{-1}+\frac{1}{1+\alpha Z{}^{-1}}$

$H\left(Z\right)=\frac{1}{1+\beta Z{}^{-1}}+\alpha Z{}^{-1}$

$x\left(n\right)=2{}^n\cdot u\left(n\right)+3{}^n\cdot u\left(n\right)\left|z\right|<3$

$\text{x}\left(\text{n}\right)=2{}^{\text{n}}\cdot \text{u}\left(\text{n}\right)+3{}^{\text{n}}$

$x\left(n\right)=2{}^n\cdot u\left(n\right)+3{}^n\cdot u\left(n\right)2<\left|z\right|<3$

$\text{x}\left(\text{n}\right)=2{}^{\text{n}}\cdot \text{u}\left(\text{n}\right)+3{}^{\text{n}}\cdot \text{u}\left(\text{n}\right)\left|\text{z}\right|>3$

$H\left(Z\right)=\frac{1}{1-2Z{}^{-1}}+\frac{1}{1-3Z{}^{-1}}$

$H\left(Z\right)=\frac{1}{1+2Z{}^{-1}}+\frac{1}{1+3Z{}^{-1}}$

$H\left(Z\right)=\frac{1}{1-2Z}+\frac{1}{1-3Z}$

$H\left(Z\right)=\frac{2}{1-2Z{}^{-1}}+\frac{3}{1-3Z{}^{-1}}$

$h\left(n\right)=\left[\left(\frac{1}{2}\right){}^n+2{}^n\right]u\left(n\right)\left|z\right|>2$

$h\left(n\right)=\left[\left(\frac{1}{2}\right){}^n-2{}^n\right]u\left(n\right)\left|z\right|>2$

$h\left(n\right)=\left[\left(\frac{1}{2}\right){}^n+2{}^n\right]u\left(n\right)\left|z\right|<2$

$h\left(n\right)=\left[\left(\frac{1}{2}\right){}^n-2{}^n\right]u\left(n\right)\left|z\right|<2$

$Y\left(Z\right)=-\frac{1}{2}\cdot \frac{1}{\left(1-Z{}^{-1}\right)}+\frac{1}{2}\frac{1}{\left(1-3Z{}^{-1}\right)}$

$Y\left(Z\right)=-\frac{1}{2}\cdot \frac{1}{\left(1-Z{}^{-1}\right)}-\frac{1}{2}\frac{1}{\left(1-3Z{}^{-1}\right)}$

$Y\left(Z\right)=-\frac{1}{2}\cdot \frac{1}{\left(1+Z{}^{-1}\right)}+\frac{1}{2}\frac{1}{\left(1-3Z{}^{-1}\right)}$

$Y\left(Z\right)=-\frac{1}{2}\cdot \frac{1}{\left(1-Z{}^{-1}\right)}+\frac{1}{2}\frac{1}{\left(1+3Z{}^{-1}\right)}$

$y\left(n\right)=\frac{1}{3}\left[3{}^n-1\right]\cdot u\left(n\right)$

$y\left(n\right)=\frac{1}{3}{3}^{″}+1\cdot u\left(n\right)$

$y\left(n\right)=\frac{1}{2}\left[3{}^n+1\right]\cdot u\left(n\right)$

$y\left(n\right)=\frac{1}{2}\left[3{}^n-1\right]\cdot u\left(n\right)$