百度试题 题目离散系统的差分方程为y(n)=x(n)+ay(n-1),则系统的频率响应( )。 A. 当|a| B. 当|a|>1时,系统呈低通特性 C. 当0 D. 当-1 相关知识点: 试题来源: 解析 C.当0 反馈 收藏
【答案】：方程两端进行傅里叶变换，得到Y(ejω)-ae-jωY(ejω)=X(ejω)[知识点分析]主要考察由离散系统的差分方程求解系统的频响特性。[逻辑推理]首先将差分方程两边进行傅里叶变换，然后根据系统的频响特性定义即可求得。
令x(n) = &n ; 则y(n) = h(n),即为单位脉冲响应，原式写为1.5h(n - 1) = &(n) -&(n-1)。令n = n - 1，1.5h(n ) = &(n + 1) -&(n)，h(n ) =3[ &(n + 1) -&(n) ] /2。方程中输入输出均为一次关系项，所以系统是线性的，若输入输出序列前的系数是...
设系统的差分方程为 y(n)一ay(n一1)=x(n)其中,x(n)为输入,y(n)为输出。 (1)当边界条件分别为y(0)=0,Y(一1)=0时,试判断系统是否为因果的
The domain of YB-1 required for nucleic acid binding is the CSD, composed of RNP1 and RNP2: the structure of this domain is known, both from X-ray crystal- lography and NMR studies,62 but the interactions with RNA or DNA are not detailed and modeling exercises could not provide ...
解:令输入x(n)=,则输出y(n)=h(n) 又已知,y(-1)=0 由y(n)=ay(n-1)+x(n)得: y(0)=ay(-1)+x(0)=1 y(1)=ay(0)+x(1)=a y(2)=ay(1)+x(2)= y(3)=ay(2)+x(3)= …… y(n)=,n≥0; 由y(n-1)= [y(n)- x(n)]得: y(-2)= [y(-1)-x(-1)]...
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X-autosome translocation (XY1Y2) has been reported in distinct groups of vertebrates suggesting that the rise of a multiple sex system within a species may act as a reproductive barrier and lead to speciation. The viability of this system has been linked
设系统差分方程 y(n)=ay(n-1)+x(n)其中x(n)为输入,y(n)为输出。当边界条件选为y(—1)=0时,是判断系统是否线性的、移不变的.《数字信号处理
={1,1,0,0}，做 DFT 的结果如下：(1) 首先计算出奇偶对称序列：偶对称序列 Xe(n)： {1, 0, 1, 0}奇对称序列 Xo(n)： {0, 1, 0, -1}(2) 计算 X(k)：X(k) = Σ[j=0, N-1] x(j)*exp(-2πijk/N), k = 0, 1, ..., N-1其中 N = 4，代入 x(n) ={1...