[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"$fCbBi-A2ljW-05QzKug2qUS6G4ndfndFI-6ZwQ-UQKWc":3},{"answer":4,"createTime":5,"id":6,"options":7,"origin":11,"question":18,"related":19,"source":28,"type":29},[],"2024-05-18 16:37:10",144418263,[8,9,10],"根轨迹增益","动态增益","静态增益",{"count":12,"courseId":13,"courseImg":14,"courseName":15,"workId":16,"workName":17},39,"32884ecc8078bd664dc5b804879faac5","https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F320bee5fb96e260ebc3971940bcbdee9.jpg","自动控制原理","work_35111090","Unit 3","尾一多项式形式的传递函数可以分解因式写成基本环节乘积的形式,对应的增益称为(),表示系统在阶跃输入下输出稳态值变化量与输入阶跃幅度的比值",[20,30,39,46,49,57,66,75,83,94],{"answer":21,"createTime":22,"id":23,"options":24,"question":27,"source":28,"type":29},[],"2024-05-18 16:37:09",144418260,[25,26],"静","动","控制所以关注的稳定性、快速性属于对系统 ()态的要求","v1",0,{"answer":31,"createTime":22,"id":32,"options":33,"question":38,"source":28,"type":29},[],144418261,[34,35,36,37],"不确定","线性定常","离散","非线性","传递函数适用于()系统",{"answer":40,"createTime":5,"id":41,"options":42,"question":45,"source":28,"type":29},[],144418262,[43,44],"零极点表达式","基本环节乘积表达式","首一多项式形式的传递函数可以分解因式写成(),对应的增益称为根轨迹增益",{"answer":47,"createTime":5,"id":6,"options":48,"question":18,"source":28,"type":29},[],[8,9,10],{"answer":50,"createTime":5,"id":51,"options":52,"question":56,"source":28,"type":29},[],144418264,[53,54,55],"严格真有理分式","真有理分式","多项式","一个实际的即物理上可实现的线性系统,其传递函数必然是 ()",{"answer":58,"createTime":5,"id":59,"options":60,"question":65,"source":28,"type":29},[],144418265,[61,62,63,64],"阶跃","脉冲","斜坡","加速度","系统传递函数反拉普拉斯变换得到系统的()响应",{"answer":67,"createTime":5,"id":68,"options":69,"question":74,"source":28,"type":29},[],144418266,[70,71,72,73],"微分","积分","拉普拉斯变换","拉普拉斯反变换","零初始条件下,线性定常系统的单位阶跃响应是其单位脉冲响应的 ()",{"answer":76,"createTime":5,"id":77,"options":78,"question":82,"source":28,"type":29},[],144418267,[79,79,80,81],"","s+2","s--2","已知系统在零初始条件下的脉冲响应为 exp(-2t) ,则系统的传递函数为()",{"answer":84,"createTime":85,"id":86,"options":87,"question":92,"source":28,"type":93},[],"2024-05-19 21:07:23",144678205,[88,89,90,91],"机理","拟合","实验","化简","建立控制系统的数学模型的方法有 ()方法和()方法",1,{"answer":95,"createTime":85,"id":96,"options":97,"question":102,"source":28,"type":93},[],144678206,[98,99,100,101],"形式","精度","复杂度","种类","在系统的建模过程中,模型的()和()之间存在矛盾,所以需要应根据系统的实际结构参数及要求的计算精度,略去一些次要因素,使模型既能反映系统的动态本质,又能简化分析计算的工作"]