21解)(tx1012t1212)1(tx1012t1213)2(tx012t12134)12(tx231210t12121)2/4(tx08t121101246)()]()([tutxtx00t1312)]()()[(2323tttx023t)(2123)(2110123432121211n][nx4121
22解0123412121211n]4[nx567811012121211n]3[nx23456781102121n]3[nx1821011n]13[nx212101231n]2[]2[nnx012343211n][)1(][2121nxnxn4120123n])1[(2nx101234312121211n]3[][nunx4121
27解记忆时变可加齐次因果稳定A是是是是非是B非是是是是是C是是是是非非D是是是是是是E是否是非是是F是是是是非是G是否是是非非(a))2()2()(txtxty①因为)2()2()0(xxy,在0t的输出与前后时刻的输入都有关,所以系统是记忆的
②已知)2()2()(111txtxty,)2()2()(222txtxty
当)()(012ttxtx时,)2()2()(01012ttxttxty,而)2()2()(010101ttxttxtty,所以:)()(012ttyty
因而系统是时变的
③已知)2()2()(111txtxty,)2()2()(222txtxty,)2()2()(333txtxty,当)()()(213txtxtx时,)]2()2([)]2()2([)(21213txtxtxtxty