700+ câu Trắc nghiệm Xử lý số liệu tín hiệu đáp án - Phần 16

Bộ lọc thông cao

Bộ lọc chắn dải

Bộ lọc thông dải

Bộ lọc thông thấp

Bộ lọc thông cao

Bộ lọc chắn dải

Bộ lọc thông dải

Bộ lọc thông thấp

Bộ lọc thông cao lý tưởng

Bộ lọc chắn dải lý tưởng

Bộ lọc thông dải lý tường

Bộ lọc thông thấp lý tưởng

Bộ lọc thông cao lý tưởng

Bộ lọc chắn dải lý tưởng

Bộ lọc thông dải lý tường

Bộ lọc thông thấp lý tưởng

$X{}_R\left(t\right)=A\text{s}\text{i}\text{n}\left(\omega {}_{0}t+\phi {}_0\right)-A\text{c}\text{o}\text{s}\left(\omega {}_{0}t+\phi {}_0\right)$

$X{}_R\left(t\right)=A\text{c}\text{o}\text{s}\left(\omega {}_{0}t+\phi {}_0\right)+A\text{s}\text{i}\text{n}\left(\omega {}_{0}t+\phi {}_0\right)$

$X{}_R\left(t\right)=A\text{s}\text{i}\text{n}\left(\omega {}_{0}t+\phi {}_0\right)$

$X{}_R\left(t\right)=A\text{c}\text{o}\text{s}\left(\omega {}_{0}t+\phi {}_0\right)$

$X{}_I\left(t\right)=A\text{s}\text{i}\text{n}\left(\omega {}_{0}t+\phi {}_0\right)-A\text{c}\text{o}\text{s}\left(\omega {}_{0}t+\phi {}_0\right)$

$X{}_I\left(t\right)=A\text{c}\text{o}\text{s}\left(\omega {}_{0}t+\phi {}_0\right)+A\text{s}\text{i}\text{n}\left(\omega {}_{0}t+\phi {}_0\right)$

$X{}_I\left(t\right)=A\text{s}\text{i}\text{n}\left(\omega {}_{0}t+\phi {}_0\right)$

$X{}_I\left(t\right)=A\text{c}\text{o}\text{s}\left(\omega {}_{0}t+\phi {}_0\right)$

$\text{a}\text{r}\text{g}\left[X\left(e{}^{j\omega }\right)\right]=-\frac{\omega }{2}+\left\{2k+\frac{1}{2}\left[1-\frac{\text{s}\text{i}\text{n}3\omega }{\left|\text{s}\text{i}\text{n}3\omega \right|}\right]\right\}\pi$

$\text{a}\text{r}\text{g}\left[X\left(e{}^{j\omega }\right)\right]=-\frac{\omega }{2}+\text{a}\text{r}\text{g}\left[\text{s}\text{i}\text{n}3\omega \right]$

$\text{a}\text{r}\text{g}\left[X\left(e{}^{j\omega }\right)\right]=\text{a}\text{r}\text{c}\text{t}\text{g}\frac{\text{I}\text{m}\left[X\left(e{}^{j\omega }\right)\right]}{\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]}$

Cả 3 phương án trên đều đúng

$\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]=\text{s}\text{i}\text{n}3\omega \text{c}\text{o}\text{s}\frac{\omega }{2}$

$\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]=-\text{s}\text{i}\text{n}3\omega \text{s}\text{i}\text{n}\frac{\omega }{2}$

$\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]=\text{s}\text{i}\text{n}3\omega \text{s}\text{i}\text{n}\frac{\omega }{2}$

$\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]=-\text{s}\text{i}\text{n}3\omega \text{c}\text{o}\text{s}\frac{\omega }{2}$

$\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]=\text{s}\text{i}\text{n}3\omega \text{c}\text{o}\text{s}\frac{\omega }{2}$

$\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]=-\text{s}\text{i}\text{n}3\omega \text{s}\text{i}\text{n}\frac{\omega }{2}$

$\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]=\text{s}\text{i}\text{n}3\omega \text{s}\text{i}\text{n}\frac{\omega }{2}$

$\text{R}\text{e}\left[X\left(e{}^{j\omega }\right)\right]=-\text{s}\text{i}\text{n}3\omega \text{c}\text{o}\text{s}\frac{\omega }{2}$

Bộ lọc thông thấp không lý tưởng

Bộ lọc thông cao không lý tưởng

Bộ lọc thông thấp $1/2$ băng tần lý tưởng

Bộ lọc thông cao $1/2$ băng tần lý tưởng

Sử dụng để làm suy giảm biên độ tín hiệu

Sử dụng để di pha tín hiệu đi 1 góc bằng $\pi$ hay $180{}^{\circ }$

Sử dụng để di pha tín hiệu đi $1\text{g}\text{ó}\text{c}$ bằng $\pi /2$ hay $90{}^{\circ }$

Sử dụng để vi phân tín hiệu

Vì nếu ta không rời rạc hoá tần số, ta

Ta phải rời rạc hoá tần số vì phân

Rời rạc hoá tần số sẽ cho kết quả tính sẽ không thể phân tích tính toán hệ toán $\text{H}\left(\text{e}{}^{\text{j}\omega }\right)$ chính xác hơn tần số liên tục thống

Đối với tín hiệu số, nếu không rời tích tín hiệu bằng tần số số hiệu quả hơn rạc hoá $\omega$ ta sẽ không thể tính $\text{H}\left(\text{e}{}^{\text{j}\omega }\right)$ với phân tích theo tần số liên tục mọi trị số của $\omega$ vì như vậy sẽ có vô hạn các trị số

$\omega {}_k=\frac{2\pi }{N}kk\left\{0,1,2,\dots N-1\right\}$

$\omega {}_{k.n}=\frac{2\pi }{N}k\cdot nk\left\{0,1,2,\dots N-1\right\}$

$\omega {}_k=\frac{1-e{}^{j\frac{2\pi }{k}k}}{1-e{}^{j\frac{2\pi }{k}}}k\left|0,1,2,\dots N-1\right|$

$\omega {}_k=\frac{1}{1-\text{W}{}_{\text{N}}}k\left\{0,1,2,\dots N-1\right\}$

$\tilde{X}\left(k\right)=e{}^{-j\frac{2k\pi }{5}}$

$\tilde{X}\left(k\right)=\frac{\text{s}\text{i}\text{n}\frac{k\pi }{2}}{\text{s}\text{i}\text{n}\frac{k\pi }{10}}e{}^{-j\frac{2k\pi }{5}}$

$\tilde{X}\left(k\right)=e{}^{-jk\pi }$

$\tilde{X}\left(k\right)=5\cdot \frac{\text{s}\text{i}\text{n}\frac{k\pi }{2}}{\text{s}\text{i}\text{n}\frac{k\pi }{10}}e{}^{-j\frac{2k\pi }{5}}$

${\displaystyle \sum }_{n=0}\mathrm{Unsupported\; }e{}^{j\frac{2n}{N}kn}=\{\begin{array}{c}N\\ k=r\pi \\ 0\\ k\ne 0\end{array}$voi $r=0,1,2\dots \infty$

${\displaystyle \sum }_{n=0}\mathrm{Unsupported\; }e{}^{j\frac{2n}{N}kn}=0$

${\displaystyle \sum }_{n=0}\mathrm{Unsupported\; }e{}^{j{}_N^{2n}kn}=\{\begin{array}{c}N\\ k=r\pi \\ 0\\ k\ne 0\end{array}$ voi r nguyen

${\displaystyle \sum }_{n=0}\mathrm{Unsupported\; }e{}^{j\frac{2n}{N}hn}=kN$ voi $\forall k$

$\tilde{X}\left(k\right)={\displaystyle \sum }_{n=0}\mathrm{Unsupported\; }x\left(n\right)\cdot W{}_N^{kn}$

$\tilde{X}\left(k\right)={\displaystyle \sum }_{n=-\infty }\mathrm{Unsupported\; }x\left(n\right)\cdot W{}_N^{kn}$

$\tilde{X}\left(k\right)={\displaystyle \sum }_{n=0}\mathrm{Unsupported\; }x\left(n\right).W{}_N^{-kn}$

$\tilde{X}\left(k\right)=\frac{1}{N}{\displaystyle \sum }_{n=0}\mathrm{Unsupported\; }x\left(n\right)\cdot W{}_N^{kn}$

$\tilde{x}\left(n\right)=\frac{1}{N}{\displaystyle \sum }_{k=-\infty }\mathrm{Unsupported\; }\tilde{X}\left(k\right)\cdot \text{W}{}_{\text{N}}^{-\text{k}\text{n}}$

$\tilde{x}\left(n\right)={\displaystyle \sum }_{k=0}\mathrm{Unsupported\; }\tilde{X}\left(k\right)\cdot \text{W}{}_{\text{N}}^{\text{k}\text{n}}$

$\tilde{x}\left(n\right)={\displaystyle \sum }_{k=0}\mathrm{Unsupported\; }\tilde{X}\left(k\right)\cdot \text{W}{}_{\text{N}}^{-\text{k}\text{n}}$

$\tilde{x}\left(n\right)=\frac{1}{N}{\displaystyle \sum }_{k=0}\mathrm{Unsupported\; }\tilde{X}\left(k\right)\cdot \text{W}{}_{\text{N}}^{-\text{k}\text{n}}$

$\tilde{X}\left(k\right)=a\cdot b\cdot {\tilde{X}}_1\left(k\right)\cdot {\tilde{X}}_2\left(k\right)$

$\tilde{X}\left(k\right)=a\cdot {\tilde{X}}_1(k){}^{\text{*}}b\cdot {\tilde{X}}_2\left(k\right)$

$\tilde{X}\left(k\right)=a\cdot {\tilde{X}}_1\left(k\right)+b\cdot {\tilde{X}}_2\left(k\right)$

$\tilde{X}\left(k\right)=a\cdot {\tilde{X}}_1\left(k\right)+b\cdot {\tilde{X}}_2\left(k\right)+ab\cdot {\tilde{X}}_1\left(k\right)\cdot {\tilde{X}}_2\left(k\right)$

$\text{D}\text{F}\text{T}\left[\tilde{x}\left(n-n{}_0\right)\right]=W{}_N^{-kn{}_0}\cdot \tilde{X}\left(k\right)$

$\text{D}\text{F}\text{T}\left[\tilde{x}\left(n-n{}_0\right)\right]=W{}_N^{kn{}_0}\cdot \tilde{X}\left(k\right)$

$\text{D}\text{F}\text{T}\left[\tilde{x}\left(n-n{}_0\right)\right]=W{}_N^{n{}_0}\cdot \tilde{X}\left(k\right)$

$\text{D}\text{F}\text{T}\left[\tilde{x}\left(n-n{}_0\right)\right]=W{}_N^{n{}_0}\cdot \tilde{X}\left(-k\right)$

$\text{D}\text{F}\text{T}\left[{\tilde{x}}^{\text{*}}\left(n\right)\right]={\tilde{X}}^{\text{*}}\left(-k\right)$$\text{D}\text{F}\text{T}\left[\tilde{x}\left(-n\right)\right]={\tilde{X}}^{\text{*}}\left(k\right)$

$\text{D}\text{F}\text{T}\left[\tilde{x}\left(-n\right)\right]=\frac{1}{2}\left[{\tilde{X}}^{\text{*}}\left(k\right)-\tilde{X}\left(-k\right)\right]$$\text{D}\text{F}\text{T}\left[{\tilde{x}}^{\text{*}}\left(n\right)\right]=\frac{1}{2}\left[{\tilde{X}}^{\text{*}}\left(k\right)+\tilde{X}\left(-k\right)\right]$

$\text{D}\text{F}\text{T}\left[{\tilde{x}}^{\text{*}}\left(n\right)\right]={\tilde{X}}^{\text{*}}\left(k\right)$$\text{D}\text{F}\text{T}\left[{\tilde{x}}^{\text{*}}\left(n\right)\right]=\frac{1}{2}\left[{\tilde{X}}^{\text{*}}\left(k\right)+\tilde{X}\left(k\right)\right]$

$\text{D}\text{F}\text{T}\left[\tilde{x}\left(-n\right)\right]=\frac{1}{2}\left[{\tilde{X}}^{\text{*}}\left(k\right)-\tilde{X}\left(k\right)\right]$

$\tilde{X}\left(k\right)=\tilde{X}{}_1(k){}_N\cdot \tilde{X}{}_2(-k){}_N$

$\tilde{X}\left(k\right)={\tilde{X}}_1(k){}_N(\text{*}){}_N{\tilde{X}}_2(k){}_N$

$\tilde{X}\left(k\right)=\tilde{X}{}_1(k){}_N(\text{*}){}_N\tilde{X}{}_2(-k){}_N$

$\tilde{X}\left(k\right)={\tilde{X}}_1(k){}_N\cdot {\tilde{X}}_2(k){}_N$