北邮考研信号与系统专业课试卷及答案2003A

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1、A7321ætö1.ft()Fj()fç-+3÷è2ø331æö-j1æöjAFç-÷e2BFç÷e22è2ø2è2øj6-j6C2F(-2)eD2F(-2)eæö2.føt÷F(j)=cosç4+÷f(t)è3ø1éjù1éj+jùAê(t+4)e3úBê(t+4)e3+(t-4)e3ú2êëúû2êëúû1é-j+jù1éj-jùCê(t+4)e3+(t-4)e3úDê(t+4)e3+(t-4)e3ú2êëúû2êëúû¥3.f()t=òu(t-)d01111ABCDS234SSS2t4.f(t)=eu

2、(t)11A:F()S=Re[]S>-2B:F()S=Re[]S<-2S+2S-21A11C:F()S=Re[]S>2D:F()S=Re[]S<2S-2S+2-(s+)Te5.F()s=s+-(t-T)-tA:eu(t-T)B:eu(t-T)-t-(t-)C:eu(t-)D:eu(t-T)næ1ö6.f()n=ç÷u()nZFZ()è3ø-1zz3z3zA:,B:,C:,D:3z-13z-13z-13z+1()j0.2n-j0.3nxn=e+eA10B20C0.2D0.383241.x(n)={3,4,5,

3、6},g(n)=x(2n-1)=2.¥3.T(t)=å(t-nT),n=-¥4.f(t),y(t),H(j)=5.f(t)BHzf(2t)fN=TN=2A2-2ss(1-e)6.F()s=s+1f(0+)=f(¥)=2-1-27.Xn()ZX(Z)=z+1+Z+6Zx(n)x(n)=8.HZ()Z5840ætö1.f(t)fç2-÷è3ø¥f(t)A(1)Otæöæöæö2.x()t=16cosç20t+÷+6cosç30t+÷+4cosç40t+÷è4øè6øè3ø(1)(2)13.8x(t)=22a+t

4、1f(t)2aOt(a)4.ft()AB3Az5.H()z=12z-0.5376518f(n)={2,-1,0}h(n)={-1,2,1,0}y(n)=f(n)*h(n)28f(t)=f1(t)*f2(t)f1(t)f2(t)sintOtOt38h(t)=u(t)h(t)=(t-1)h(t)=-(t)h(t)123-te(t)=eu(t)h(t)1e(t)r(t)h2(t)h1(t)h3(t)484A2()()dytdyt-(k+2)+6y()t=3x()t2dtdt1H(s)2,K-2t58t=0x(t)

5、=2eu(t)t³0(t)ciL1W1H1Ax(t)t=0+1W1F(t)C-615103S1x(n)2y(n)-1Sz7-3S1-14z123710LTIH()=R()+jI()h(t)R()I()1¥I()1¥R()R()=òdI()=-òd-¥--¥-52003(A)73211.D2.D3.C4.C5.B6.C83241.g(n)=x(2n-1)={4,6}2.¥12æ2ö3.T()t=+åcosn1tç1=÷TTn=1èTø-jt4.H(j)=Ke0115.fN=4BHzTN==sfN4B6.f(0

6、+)=1f(¥)=07.x(n)=(n+2)+(n)+(n-1)+6(n-2)8.58401.1ætöfç2-÷è3ø¥A(3)Ot2.1Fnn834362ó4ó3ó2ó4ó3ó2023402341ó32:10()64649944-40-30-20O2030401-a3.8«e22a+taF()aO(b)14.f()t×cos30tÛ[F(+30)+F(-30)]21[f()t×cos30t]*h()tÛ[F(+30)+F(-30)]H()22F[f(t)×cos0t]Y()1122---50-300030

7、50-3000305.jH(e)2jIm[z]23o0.51-2-o23Re[z]76518y(n)={-2,5,0,-1}28ì1[1-cos(t-1)]1£t£2ïïï1[()()]ïcost-2-cost-12£t£3f()t=íï1ï[cos(t-2)-1]3£t£4ïïî038h(t)=h(t)+h(t)*h(t)*h(t)=u(t)-u(t-1)1213-t-tìï1-e0

8、2()s-k+2s+6-t-2t58u(t)=(3e-2e)u(t),C3()(-t-2t)()-tut=2e-eutu(t)=et³0CzsCzi10733615H()z=-1-11-11-z1-z24311y()n-y(n-1)+y(n-2)=x()n+x(n-1)483nné10æ1ö7æ1öùh()n=êç÷-ç÷úu()nêë3è2ø3è4øúû7104