第九章-3信号分析与处理

1、第七章第3讲1第七章第3讲2)()() 1()2()3(0123kkyakyakyakya)()() 1()2()3(00010203kkhakhakhakha0)() 1()2() 3(00010203khakhakhakha0)3()3()2() 1()0(00010203hahahaha0)2()2() 1()0() 1 (00010203hahahaha0) 1() 1()0() 1 ()2(00010203hahahaha1)0()0() 1 ()2() 3(00010203hahahaha301)3(ah301)3(ah第七章第3讲30) 1()2() 1 (000nhhhnanh

2、1)(0nikiikCkh10) 1()()(线性非时变离散系统（零状态）(k)h0(k)(k-n)h0(k-n)A(k-i)Ah0(k-i)第七章第3讲4)()3()2() 1()(00010203kkhakhakhakha1)0() 3()2() 1()0(00010203hahahaha301)0(ah301)0(ah)()3()2() 1()(0123kkyakyakyakya0) 1()2() 1(000nhhhnah1)0(0nikiikCkh10)()()(第七章第3讲50111)()()()(aqaqaqqNqDqNqHnnn)()()(21nqqqqNqnnqqKqqKqqK

3、2211)()()(2211kKKKkhknnkk)()()()(11qDqNqHqHl单位函数响应还可以用单位函数响应还可以用Z变换的方法求取，详见第八章。变换的方法求取，详见第八章。第七章第3讲6)2()()2(61) 1(65)(kkkykyky)()2(61) 1(65)(000kkhkhkh06165231,2121 kkCCkh3122110)(0) 1(, 1)0(00hh )(23)(31210kkhkk)2()()(00khkhkh )2(23)(232312213121kkkkkk第七章第3讲7)2()()2(61) 1(65)(kkkykyky)(111)(3121226

4、11652qqqqqqqH3121312121696)(1)(qqqqqqqqqH31211696)(qqqqqH)()(16)(9)(6)(3121kkkhkk第七章第3讲8 )2(23)(23)(2312213121kkkhkkkk)()(16)(9)(6)(3121kkkhkk) 1()()()2(kkkk )1()()(23)2(23231221231221kkkkkkkk ) 1(0)(6)(23231221kkkkk )(6)(2323)(2312213121kkkhkkkk)()(16)(9)(63121kkkk(1)(2)第七章第3讲91)2(, 0) 1 (00hh)(3)2

5、()(00khkhkh)(3)2()(6) 1(5)2(kfkfkykyky)()(6) 1(5)2(000kkhkhkh2, 321 ) 1(23)(210kCCkhkk ) 1(23) 1(23)(1121310kkkhkkkk ) 1(23 3) 1(231111kkkkkk) 1()() 1() 1(kkkk ) 1(233)1()() 1(23)(1111kkkkkhkkkk ) 1(23332439)(1111kkkkkk ) 1(236)(11kkkk第七章第3讲10)(3)2()(6) 1(5)2(kfkfkykyky36211)3)(2(951653)(22qqqqqqqqq

6、H) 1() 3(6)2()()(11kkkhkk 注：) 1(11kaaqk)(kaaqqk第七章第3讲11) 1()()()() 1()()()(krkrkrkhkkkk00)()()()(iiikhkrikkkikiihkrik00)()()()(对于因果系统：对于因果系统：0)()()(iikikhkr0)(, 0ikkiik时即kiikhkr0)()(第七章第3讲12)()(5 . 01 6)(21kkrk) 1()(5 . 01 6)()(5 . 01 6) 1()()(12121kkkrkrkhkk)()(3)()(1 6)()(5 . 01 6212121kkkkkk第七章第3

7、讲13)()3()2(5 . 0)()(3161kkkhkkkikiikikkiiikikhkr0000)3(31)2(21)(61)()()(321221)()313(21)221(61)(1)(133)(1)(122)(613113121121kkkkkkkkkkkkk第七章第3讲14解：由差分方程可得转移算子) 1(2)(4) 1()(2kekekyky21421224)(qqqqqH)()(4)(2)(21kkkhk)()(46)()(1 8)(2)(1)(14)(2)(4)(2)()(211212112102100kkkkkiihkrkkkkiikikiqqaaSn11第七章第3讲15e(k)=0.25(k)+(k-1)+0.5 (k-2)。解：激励为)2() 1()()(keckebkeaky系统响应为012 3-1ke(k)0.510.25e(k)y(k)05 . 0025. 0125. 0cbbaa8164cba第七章第3讲16已知差分方程为 ，初始状态 。(1) 求系统的零输入响应 yzi(k)