哈工大電路答案第11章

章用時可以刪除0s0s0s20s2一-一一一一FsasFsf(0)]+bF(s)+c[F(s)/s]=(as+b+c/s)F(s)=(as2+bs+c)A/T11s2(s+1/T)設(shè)F(s)=L{f(t)}=5,F則f(t)與f(t)的卷積為120030300L{f(t)*f(t)}=F(s)F(s)。1212A=2s+1|=-3，A=2s+1|=-31s+3s=-21s+3s=-2F(s)=3=(3/5)〉5s2+2s+6(s+1)2+(5)2f(t)=e-tsin(5t)5(b)畫出運算電路如圖(c)所示+++U(s)2UsUs)1__下((1+s)U(s)-U(s)=I(s)(1)212由式(2)解得代入式(1)得i2s+126+U(s)U(s)S4I(s)+I(s)L3sU(s)2L11/(21/(2s)_bL3s1LLLI(s)=U(s)1=1+2/sU(s)24+1/(2s)4+0.5/sLU(s)=2[I(s)+I(s)]+U(s)=[2+1+2+1+2/s]U(s)SL213ss4+0.5/sL化簡得S3s(8s+1)LtetU(s)=2+2sU(s)=(2/3)s=A1+A21s+2s=1/315A=(2/3)s|=0.8Vs2s+1/3s=2SI(s)S+I(I(s)SU(s)C3I(s)=3I(s)=3Cs+3+1/(0.5s)Ss2+3s+216AAAs16AAAA=6|=3Vs1(s+1)(s+2)s=0A=6|=6Vs2s(s+2)s=1A=6|=3Vs3s(s+1)s=2CCuUViUR5A1L12++_(b)sUs/s+1/s+5/242445s/625s/6s45/s5s/6U(s)=s(s2+5s+6)ss+2s+3A=s2+6.25s+6|=1.25Vs2s(s+3)s=2sLssI(s)=L{e3t(t)}=1s11+sC2U(s)U(s2S22I(s)222U(s)=9Ss由節(jié)點電壓法得221SU(s)=2.7u(t)=L1{U(s)}=2.7e0.1tε(t)V2222iiA，i(0)==1A10210s4s4sI(s)+U(s)-2+U(s)-2ssss+4i(t)=L1{I(s)}=(0.50.3e4t)A(t>0)1122+C2一ss(s+5)s=一6CI(s)sMI(s)121R1**sLsLRs12212((R+sL)I(s)+sMI(s)=1/s1222I(s)=10(s+10)=1+0.5+0.5ssssss+20I(s)=5=0.5+0.520.75s2+20s+100s+20/3s+2012a初始值i(0)=U2RsL1I(s)2I(s)2L22〈1s(s+200)2s(s+200)21U(s)sLs+10s+40s-j314s+j314+2_3I(s)1+U(s)1_I(s)1ls11s1121ls11s1121將I(s)=(5/s)-U(s)2代入上式化簡解得A=(s+1)2|=1.667Vs2s+0.6s=0A=|=(s+1)(s+0.2)|=0.556Vs3dss=0(s+0.6)2s=01i(0)=10V=2A,u(0)=4i(0)=8V,u(0)=0L(4+1)C1LC21s24U(s)2s2s+1442s+11/(0.2s)U(s)=1.6s2+3.6s+102s(s2+1.5s+1.5)22+22s+U(s)CsC0.52s得Us10Cs(s2+4s+1)2,32ZZ(s)U(s)RocSoc111+1/ss+1未畫)列節(jié)點電壓方程(1111U(s)(1111U(s)〈H(s)=U(s)rI(s)rI(s)1I(s)I(s)22slss2m1|-20I(s)+(20+30)I(lss2m1I(mm1m1,2m12m121*om*+1*om+1**o*+1mm_故當(dāng)SSo