原版847信號與系統(tǒng)歷年考研真題及答案2010

即

?????????????????(????)=???????? ??????(????)=???????????????????? ????(????)=?????????????????(????)=????????(????)+???????? 1 ????????????2????(????)+????????????????(????)+????????(????)= ????????2????(????)+12????????????(????)+36????(????)=????1=????2=

????(????)=(????1????+ ????????(????)=????????????????(????)+????????(????)=2????????????(????)+（20）1????(????)=????????，????????(????)=????????????，試求零輸入響 ?????????????????(????)=???????? ??????(????)=???????????????????? ????(????)=????????

?????????(????)=????????(????)+????????

1 ????????????2????(????)+????????????????(????)+????????(????)=即

????????2????(????)+12????????????(????)+36????(????)==????1=????2=

????(????)=(????1????+????2 ????????(????)=????????????????(????)+????????(????)=2????????????(????)+將將????(0)=5????，????????(0)=10?????

????(0)=????2=????????(0)=2(????1?6????2)+24????2=????=2????(????)=(?25????+（1（10分）????(????)=????(????)?????(????)?????(????)，????(????)與????(????)的波形如題2（1） 1 -1 （2（10-3-3-2-1 12 -3-2-1 123 ????(0)=????2=??????=2????(????)=(?25????+（1（10 1 - -1 2（1）（2（10-3-3-2-1 12 -3-2-1 123 2（2）h(-h(-11-11*②把反褶信號從????②把反褶信號從????=?2到????=2 1 11 11 111 11- 1*)1- 1*1- 1*故卷積結(jié)果為：故卷積結(jié)果為：????(????)=②把反褶信號從????=?4位移????=-3-2-1 12345678n-3-2-1 12 ????=?4位移????=-6-5- -3-2-1 12 -3-2-1 1234567888431-4-3-2-1 12 45678其中????(????)- - -3(????+1)2(????+????(????)=????(????+2)(????+1?2????)(????+1+????3+5????2+7????+ ????2?2?????=????????3+4????2+9????+10=????+????????3+4????2+9????+ 4????? 4(????+ =????+?????5 ?=????+?????5+ ?5× ????+ ????2+2????+ ????+ (????+1)2+ (????+1)2+?(????)=????????(????)+????

?????2????

6??????????cos2?????根據(jù)∫0+?(????)????????可得：????

三（20LTI3∫????+????(????)????????=????.其中????(????)- -

-3±(????+1)2(????+????(????)=????(????+2)(????+1?2????)(????+1+????3+5????2+7????+ ????2?2?????=????????3+4????2+9????+10=????+????????3+4????2+9????+ 4????? 4(????+ =????+?????5+ ?=????+?????5+ + ?5× ????+

????2+2????+

????+

(????+1)2+

(????+1)2+?(????)=????????(????)+?????1?????2????+4??????????sin2?????6??????????cos2????? 根據(jù)∫0+?(????)????????=????????0+(

( 31 4 6

?????=2????

+? + ?sin2???? ?cos2????? 3????3+5????2+7????+ ????(????)?

2

+

+9????+

=2????3????(????)+10????2????(????)+14????????(????)+6????(????)=3????3????(????)+12????2????(????)+27????????(????)+302????(3)(????)+10????′′(????)+14????′(????)+6????(????)=3????(3)(????)+12????′′(????)+27????′(????)+1????1(????)=由此看出極點(diǎn)全位于s平面的左半平面，所系統(tǒng)穩(wěn)定。（15）用PAM????(????)=?????????????????????????????(????)為了補(bǔ)償信4（1）如何設(shè)計????(????)使????(????)和????(????)

?(????)=????0?????????0????????(????????) 3( (

31

6

?????=2????

+2?5

+5

sin2????

cos2?????3????3+5????2+7????+ ????(????)=2????3+4????2+9????+10=?2????3????(????)+10????2????(????)+14????????(????)+6????(????)=3????3????(????)+12????2????(????)+27????????(????)+302????(3)(????)+10????′′(????)+14????′(????)+6????(????)=3????(3)(????)+12????′′(????)+27????′(????)+1由此看出極點(diǎn)全位于s而????<0，?(????)=0，所以?1(????)=0，因此系統(tǒng)因果。（15分）用PAM通信的某一信號其單位沖激響應(yīng)為????(????)=?????????????????????????????(????)為了補(bǔ)償信PAM4（1）如何設(shè)計????(????)使????(????)和????(????)

?(????)=????0?????????0????????(????????)= ????1(????1(????????)=????1(????????)=而????1(????????)=故因此需要設(shè)計????(????)

????(????????)= =????(1+=????(1+????(????)=????????(????) ????′????(????)=????????(????) ????′????????(????+????)=????????(????+????)=????????(????)?????????????(????)+????(????)=????????(????)?????????(????)+??????????????????又由題意知道，系統(tǒng)相位特性為零相位特性，故整個系統(tǒng)為了無失真?zhèn)鬏斝铦M足而????1(????????)=故????(????????)=????1(????????)=????(1+

因此需要設(shè)計????(????)

????(????)=????????(????)+????????′????(????)=????????(????)+????????′K????????(????+????)=????????(????+????)=????????(????)?????????????(????)+????(????)=????????(????)?????????(????)+1 - - -1????1(?????1)=2????1(????)+????2(????)+????(?????????2(?????1)=????(?????1)=????1(????)+????2(????)+2????(?????

?????1(????+1)?= 1)

0?????2(????+ ?2 ????(????)= ?1]?????1(????)?+????2則 ?1 ????+ |?????????????|= ????+

2+2????? ????(????)=????(?????????????)?1????+=

????+ 11 ?1????2+2?????1? ???????11?=????2+2?????1+????2+=????2+2?????Y=(A+D′)(A+B+E′+F′)(A+B+E′+F)(B+C)(B′+D)(A+B+E+F′)(A++E+F)(A+C+D+E+F+1

z-1 z-

-1????1(?????1)=2????1(????)+????2(????)+????(?????????2(?????1)=????1????(?????1)=????1(????)+????2(????)+2????(?????

?=

??????1

0?+

?????2(????+

????(????)= ?1]?????1(????)?+????2則 ?1

????+ |?????????

= ????+

=

+2?????1? ????(????)=????(?????????????)?1????+1=

????+ 1 1?

+2?????1?

=????2+2?????1+????2+=????2+2?????2014Y=(A+D′)(A+B+E′+F′)(A+B+E′+F)(B+C)(B′+D)(A+B+E+F′)(A++E+F)(A+C+D+E+F+（2）（3）略2014

=????????輸入????????與????????8略（1）題9圖中DZ的穩(wěn)定電壓????????=????????，限流電阻????=????????????????，????????=????????????，????????為????????????????和????????????時的輸出電壓????????和穩(wěn)壓管的穩(wěn)定電流????????（2????????????5mAR????????的值。9略

輸入????????與