Основы теории цепей

ЗАДАНИЕ №1

1.     Определение переходной и импульсной характеристик цепи

Цепь:                                       i1               i2   

                                                                  + I(t)

              R1                                                                                    R2        U2      

                                         C       

Исходные данные:

Топологические уравнения:

I0=i1+i2

I2R2=I1R1+Uc

(I0-I1)R2= I1R1+Uc

I1(R1+R2)+Uc=I0R2

Дифференциальное уравнение:

(С (R1+R2)/R2)dUc/dt+Uc/R2=I(t)

Характеристическое уравнение:

          (С (R1+R2)/R2)p+1/R2=0

 Начальные условия нулевые :

          p=-1/С(R1+R2)=-1/t

                           t                                                                                                                                                                                    t

          Uc(t)=e-t/tò (I(t)1(t)*R2/ С(R1+R2))et/t dt=(I0*R2/ С(R1+R2))tцe-t/tòet/t dt =I0*R2e-t/tet/t ½ =

0                                                                                                               0

=I0*R2e-t/t[ et/t-1]= I0*R2 [1-e-t/t]

          I1(t)=CdUc/dt=(IoCR21/tц) e-t/t =(IoR2/(R1+R2)) e-t/t

          I2(t)=Io[1-R2/(R1+R2)) e-t/t]

          U2=I2*R2= Io[R2-(R22/(R1+R2)) e-t/t]

Переходная характеристика:

           hI2=1-R2/(R1+R2)) e-t/t=1-0.67 e-t/t     

hU2=R2[1-(R2/(R1+R2)) e-t/t]1(t)   tц=C(R1+R2)=0.6 10-6

hU2=200[1-0,67 e-t/t]1(t)

Импульсная характеристика:

       gI= R2/(R1+R2)2C)e-t/t+[1-R2/(R1+R2)) ]e-t/td(t)=1.1*106 e-t/t+0.33d(0)

          gU2=d hU2/dt=(R2*R2/(R1+R2)tц e-t/t)) 1(t)+ R2[1-(R2/(R1+R2)) e-t/t]d(t)

          gU2=0,22*109e-t/t1(t)+66d(0)

2. Определение отклика цепи:

Входное воздействие:

I(t)=2*(1-e-t/0,610  )1(t)

hI2=1-(R2/(R1+R2)) e-t/t1(t)

                          t               

Iвых=I(0)hI2(t)+ ò I’(y) hI2(t-y)dy

                         0

I(0)hI2(t)= 2*(1-e0/0,610  ) hU2=0

I’(t)=(2/0.6 10-6) e-t/0.6 10

t

 ò(2/0.6 10-6 )e-y/0.6 10[1-0,67 e-(t-y)/0.6 10]dy

0

            t                                                                                                                             t

1)   ò(2/0.6 10-6)e-y/0.6 10dy= -(0.6 10-62/0.6 10-6) e-y/0.6 10½=-2[e-t/0.6 10-1]= 2[1-e-t/0.6 10]

0                                                                                                                                                                                            0

                                   t                                                                                                  t                

2) -(2*0,67/0.6 10-6 ) òe-y/0.6 10 ey/0.6 10 e-t/0.6 10dy=(2,23 106)e-t/0.6 10ò1dy=

                                                      0                                                                                                      0

=-2,23 106 te-t/0.6 10=-2,23 106 te-t/0.6 10

I(t)2=-2,23*106 te-t/0.6 10-2e-t/0.6 10+2=2-2,23*106*te-t/0.6 10-2e-t/0.6 10

U2= I(t)2*R2

Выходное напряжение:

U2(t)=400-446*106 te-t/0.6 10-400e-t/0.6 10

3.Опредиление АЧХ, ФЧХ :

К(jw)=Iвых/Iвх= (U2/R2)/(U2/Zэ)= Zэ/ R2

Zэ=(R2(R1-j/wC))/((R1+R2)-j/wC)

К(jw)=(R1-j/wC)/((R1+R2)-j/wC)=Ö(R12+(1/Cw)2)/ (((R1+R2))2+(1/wC)2)      *

*e-jarctg(1/wCR1)+ jarctg(1/wC(R1+R2)) =

=Ö((R1w)2+(1/C)2)/ ((R1+R2)w)2+(1/C)2) *e-jarctg(1/wCR1)+ jarctg(1/wC(R1+R2)) =

К(jw)=Ö(10000*w2+0,25 1018)/(90000*w2+0,25 1018)  * e-jarctg(10/0,2w)+ jarctg(10/w0,6)

АЧХ(w)=Ö(10000*w2+0,25 1018)/(90000*w2+0,25 1018)

ФЧХ(w)=-arctg(106/0,2w)+ arctg(106/w0,6)

ЗАДАНИЕ №2

1.Определить параметры цепи : Q0, r

 Цепь:                             Rг

1	2	3