0195136047.pdf

124 TIME-DEPENDENT CIRCUIT ANALYSIS

I ̄=

V ̄

20 +( 10 /j ω)

V ̄C=

(

10

jω

)

I ̄=

V ̄

1 + 2 jω

Treating each Fourier-series term separately, we have the following.

1. For the dc case,ω=0;VCdc=50 V.


2. For the fundamental component,

v 1 (t)= 63 .7 cos( 20 πt), V ̄ 1 =

63. 7

√

2

 0°

V ̄C 1 =(^63.^7 /

√

2 ) 0°

1 +j( 40 π)

=

0. 51

√

2

 − 89 .5° V

and

vC 1 = 0 .51 cos( 20 πt− 89 .5°)V

3. For the third harmonic (ω= 3 × 20 π) component,

v 3 (t)=− 21 .2 cos( 60 πt), V ̄ 3 =−

21. 2

√

2

0°

V ̄C 3 =−(^21.^2 /

√

2 ) 0°

1 +j 2 ( 60 π)

=−

0. 06

√

2

 − 89 .8° V

and

vC 3 (t)=− 0 .06 cos( 60 πt− 89 .8°)V

4. For the fifth harmonic (ω= 5 × 20 π) component,

v 5 (t)= 12 .73 cos( 100 πt), V ̄ 5 =

12. 73

√

2

 0°

V ̄C 5 =

(

12. 73 /

√

2

)

 0°

1 +j 2 ( 100 π)

=

0. 02

√

2

 − 89 .9° V

and

vC 5 (t)= 0 .02 cos( 100 πt− 89 .9°)V

5. For the seventh harmonic (ω= 7 × 20 π) component,

v 7 (t)=− 9 .1 cos( 140 πt), V ̄ 7 =−

9. 1

√

2

 0°

V ̄C 7 =−(^9.^1 /

√

2 )0°

1 +j 2 ( 140 π)

=−

0. 01

√

2

 − 89 .9° V

and

vC 7 (t)=− 0 .01 cos( 140 πt− 89 .9°)V

Thus,

vC(t)= 50 + 0 .51 cos( 20 πt− 89 .5°)− 0 .06 cos( 60 πt− 89 .8°)

+ 0 .02 cos( 100 πt− 89 .9°)− 0 .01 cos( 140 πt− 89 .9°)V

in which the first five nonzero terms of the Fourier series are shown.