信号与系统吴大正第四章作业第一章作业

?0,t?0?t,0?t?1(2)f(t)=r(t)-2r(t-1)+r(t-2)= ? ??2?t,1?t?2??0,t?2F(t) 1 0 1 2 t

(3) f(t)=ε(t)r(2-t) 解：f(t)=??2?t,0?t?2

?0,t?0或t?2y 2 0 2 x

?0,t?0?(4)f(t)=r(t)?(2-t)= ?T,0?t?2

?0,t?2?F(t) 2 0 1 2 t

(5) f(t)=r(2t)ε(2-t)

?2t,0?t?2解：f(t)=2t·ε(2t)ε(2-t)=?

0,t?2或t?0?y 4 0 2 x

(6)f(t)=sin(?t)[? (t)-?(t-1)]

?0,t?0? = ?sin(?t),0?t?1

?0,t?1?F (t) 1 0 1 t (7) f(t)=sinπ(t-1)[ε(2-t)-ε(-t)]

Y 0 1 2 X

?0,k?0?(8) f(k)=k[?(k)-?(k-5)]= ?k,0?k?5

?0,k?5?F(k) 3 1 k 0 1 2 3

（9）f(k)=2-kε(k)

?k??2,k?0解:由题意可得f(k)= ?

??0,k?0

Y 1 1/2 1/4 1/8 0 1 2 3 X

(10) f(k)=2-(k-2)?(k-2)=

F(k)

?2?(k?2),k?2 ??0,k?21 1 21 40 1 3 4 k (11) f(k)=sin(

k?)[ε(k)-ε(k-7)] 6

Y 1 1/2 32 0 1 2 3 4 5 x

k??2,0?k?3(12) f(k)=2[?(3-k)-?(-k)]=?

??0,k?3或k?0 k

F(k) 4 2 0 1 2 3 k

1.3 解：(1) f(t)=2?(t+1)- ?(t-1)- ?(t-2)

(2)f(t)=(t+1)[?(t+1) ?(t-1)]-(t-3)[ ?(t-1)-?(t-3)] (3) f(t)=10sin(?t)[?(t)-?(t-1)]

(4) f(t) =?(-t-2)+()2t+5[?(t+2)-?(t-1)]+(t+1)[ ?(t+1)-?(t-1) ]?(t-1) 1.4 解: (a) f(k)= ?(k+2) (b) f(k)= ?(k-3)-?(k-7)

(c) f(k) =?(-k+2)

(d) f(k)=(-1)k?(k) 1.5 解：(1) 由于2 ?3?3?10=为有理数，所以f1(k)cos(k)是周期序列，

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