信号与系统课后答案5

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1、第五章章章习习习题题题5.1求下列各时间函数f(t)的像函数F(s)。()(-at)()(1)ft=1-eUt(2)f(t)=sin(wt+f)U(t)()1(-at)()()-at()()ft=1-eUt(3)ft=e1-atUt(4)a()2()(5)ft=tUt(6)f(t)=(t+2)U(t)+3d(t)()()-at(7)ft=tcoswtUt(8)f(t)=(e+at-1)U(t)答案11aF(s)=-=(1)ss+as(s+a)ssiny+wcosyF(s)=2(2)s+wsF(s)=2(3)(s+a)1a1F(s)=´=(4)as(s+a)s(s+a)2F(s)=2(5)s

2、2123s+2s+1F(s)=++3=22(6)sss(7)1jtw-jtw1jtw1-jtwFs()=Lt[coswt]=Lte(+e)=Lte+Lte=222221111s-w´+´=222222(s-jw)2(s+jw)(s+w)(8)2-at-aaaFs()=Le[+at-1]=L--(1e)+L[at]=+=22ss(+a)sss(+a)5.2求下列各像函数F(s)的原函数f(t)。()()2s+1s+32s+16(1)F()s=(2)F()s=()()2ss+2s+4(s+5s+6)(s+12)232s+9s+9s(3)F()s=(4)F()

3、s=22s+3s+2(s+3s+2)s答案(1)KKK123F(s)=++ss+2s+4(s+1)(s+)33K=´s=1s(s+2)(s+)48s=0(s+1)(s+)31K=(s+)2=2s(s+2)(s+)44s=-2(s+1)(s+)33K=(s+)4=3s(s+2)(s+)48s=-4313F(s)=8+4+8ss+2s+431*2t3-4tf(t)=(+e+e)U(t)848(2)1234152-K1K2K35945F(s)=++=++s+2s+3s+12s+2s+3s+1212-2t34-3t152-12tf(t)=(e-e+e)U(t)5945(3)3s+521F(s)=2

4、+=2++(s+1)(s+)2s+1s+2-t-2tf(t)=2d(t)+2(e+e)U(t)(4)2s3s+214F(s)==1-=1+-2s+3s+2(s+1)(s+)2s+1s+2-t-2tf(t)=d(t)+(e-4e)U(t)5.3求下列各像函数F(s)的原函数f(t)。32s+6s+6s1(1)F()s=(2)F()s=222s+6s+8s(s+1)答案(1)2-4F(s)=s++s+2s+4-2t-4tf(t)=d¢(t)+2(e-4e)U(t)(2)KKKKK1231-31112132122F(s)=++++=++++322322(s+)1(s+)1s+1ss(s+)1(s

5、+)2s+1ss12-t-t-t12-tf(t)=(te+2te+3e+t-)3U(t)=(t+t+)3eU(t)+(t-)3U(t)225.4求下列各像函数F(s)的原函数f(t)。-(s-1)2+e1(1)F()s=(2)F()s=2(-s)(s-1)+4s1-e2-s1-e(3)F()s=s答案(1)-(s-)1212eF(s)=+´2222因(s-)1+22(s-)1+22sin2tU(t)«2又因有s+42-ssin(2t-)1U(t-)1«e2故由时移性有s+4又由复频移性有t2-(s-)1esin(2t-)1U(t-)1«e2(s-)1+4t1tf(t)=esin

6、2tU(t)+esin(2t-)1U(t-)1故2(2)11F(s)=´-ss1-e¥¥f(t)=U(t)*∑d(t-K)=∑U(t-K)故n=0k=0,KÎN(3)-s-s1-e1-eF(s)=´2s1-sU(t)-U(t-)1«1(-e)因有s故f(t)=[U(t)-U(t-)1]*[U(t)-U(t-)1]=tU(t)-(2t-)1U(t-)1+(t-)2U(t-)225.5用留数法求像函数()4s+17s+16f(t)。Fs=的原函数()(2)s+2s+3答案2F(s)的分母(s+)2(s+)3=0，得到一个单极点s=-3令1和一个二重极点s=-22。下面求各极点上的留数。2[st

7、]4s+17s+16st-3tRes=F(s)(s+)3e=e=e1s=-32(s+)2s=-31d[st]Res2=F(s)(s+)2e=2(-1)!dts=-22d4s+17s+16st(e)=dss+3s=-2224s+24s+35st4s+17s+16st2e+te=s+6s+9s+3s=-2-2t-2t-2t3e+(-)2te=3(-2t)e[-3t-2t]故f(