信號與系統(tǒng)習(xí)題1

1、1.211.21解)(tx1?012t121?2?)1(?tx1?012t121?3)2(tx?012t121?34)12(?tx23?1?21?0t121?21)24(tx?08t121?101246)()]()([tutxtx??00t131?2)]()()[(2323???tttx??023t)(21?23?)(21?1.27解記憶時變可加齊次因果穩(wěn)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

2、非是G是否是是非非（a）)2()2()(txtxty????①因?yàn)?2()2()0(xxy???，在0?t的輸出與前后時刻的輸入都有關(guān)，所以系統(tǒng)是記憶的。②已知)2()2()(111txtxty????，)2()2()(222txtxty????。當(dāng))()(012ttxtx??時，)2()2()(01012ttxttxty??????，而)2()2()(010101ttxttxtty???????，所以：)()(012ttyty??。因

3、而系統(tǒng)是時變的。③已知)2()2()(111txtxty????，)2()2()(222txtxty????，)2()2()(333txtxty????，當(dāng))()()(213txtxtx??時，)]2()2([)]2()2([)(21213txtxtxtxty????????所以)()()(213tytyty??，因而系統(tǒng)是可加的。當(dāng))()(12taxtx?時，)()2()2()(1112taytaxtaxty?????，因而系統(tǒng)是齊次

4、的。綜合系統(tǒng)的可加性與齊次性，所以系統(tǒng)是線性的。④因?yàn)?2()2()0(xxy???，在0?t的輸出與2?t的輸入也有關(guān)，所以系統(tǒng)是非因果的。⑤若????Btx)(，即輸入有界，則：????????????Btxtxtxtxty2)2()2()2()2()(，即輸出有界。所以系統(tǒng)是穩(wěn)定的。（b）)()].3[cos()(txtty?①可見，在0tt?點(diǎn)的輸出)(0ty僅與0tt?點(diǎn)的輸入有關(guān)?非記憶。②已知)()]3[cos()(11t

5、xtty??，)()]3[cos()(22txtty??。當(dāng))()(012ttxtx??時，)()]3[cos()(012ttxtty???，而)()33cos()(01001ttxtttty?????，)()(012ttyty??，因而系統(tǒng)是時變的。③已知)]()3[cos()(11txtty??，)()]3[cos()(22txtty??，)()]3[cos()(33txtty??，當(dāng))()()(213txtxtx??時，)]()(