HW6 - ShanghaiTechsist.shanghaitech.edu.cn/faculty/zhoupq/Teaching/Fall15/HW6... · Electric Circuits Fall 2015 Solution

Electric Circuits Fall 2015 Solution

1

HW6

RULES:

Please try to work on your own. Discussion is permissible, but identical submissions are

unacceptable!

Please show all intermediate steps: a correct solution without an explanation will get zero

credit.

Please submit on time. NO late submission will be accepted.

Please prepare your submission in English only. No Chinese submission will be accepted.

1. [6%] In the circuit of Fig. 1, find:

(a) v(0+) and i(0+)

(b) dv(0+)/dt and di(0+)/dt

(c) v(∞) and i(∞)

Fig. 1

Solution:

(a) At t = 0-

, 1u t and 0u t so that,

0 4 / 3 5 0.5 Ai , and 0 5 0 2.5 Vv i .

Hence, 0 0 0.5 Ai i

0 0 2.5 Vv v

(b) C

dvi C

dt or 0 / 0 /Cdv dt i C

For t = 0+,

4u(t) = 4 and 4u(-t) = 0.

Since i and v cannot change abruptly,

0 0 / 5 2.5 / 5 0.5ARi v ,

For KCL, 0 4 0 0C Ri i i

Hence, 0 . 5 4 0 0 . 5Ci , which leads to 0 4Ci


2

0 / 0 / 4 / 0.1 /Cdv dt i C 40 V s

Similarly, L

div L

dt which leads to 0 / 0 /Ldi dt v L

3 0 0 0 0Li v v

1.5 0 2.5 0 0 4L Lv or v

0 / 0 / 4 / 0.25 /Ldi dt v L 16 A s

(c) 5 4 / 3 5 2 . 5 Ai

5 4 - 2.5 7.5 Vv

2. [8%] The circuit shown in Fig. 2 has been in operation for a long time. At t = 0, the source voltage

suddenly jumps to 250 V.

Find vo(t) for t 0.

Fig. 2

Solution:

Initial condition: (0 ) (0 ) (0 ) 0AC L Li i i , (0 ) (0 ) 50VC Cv v

As a series RLC circuit, 3

800025000rad/s

2 2 160 10

R

L

2 6

0 3 9

1 1625 10

160 10 10 10LC

We find 2 2 , so the circuit is critical damping

Then, 25000 25000

0 1 2( ) t t

fv t V Ate A e

When t = ∞，we get 250VfV

When t = 0, apply Initial condition,

0 2(0) 250 50v A , so 2 200A

01 2(0) 25000 0

dvA A

dt ,so

6

1 5 10 VA

Above all, 6 25000 25000

0( ) 250 5 10 200 Vt tv t te e


3

3. [8%] The natural response for the circuit shown in Fig. 3 is known to be

v(t) = -11e-100t + 20e-400t V, t 0

If C = 2 μF and L = 12.5 H, find iL(0+) in milliamperes.

Fig. 3

Solution:

From equation,400 100 500

= = = 2502 2

Then,

61 101000

2 2 2 250R

C

When t = 0+, 0 11 20 9Vv ,

So, 9

(0 ) 9mA1000

Ri

Also we can get, 100 4001100 8000t tdve e

dt

When t = 0+, (0 )

1100 8000 6900V/sdv

dt

,

6(0 )(0 ) 2 10 ( 6900) 13.8mAC

dvi C

dt

For KCL, we can get (0 ) [ (0 ) (0 )] (9 13.8) 4.8mAL R Ci i i

4. [14%] In the circuit in Fig. 4, the switch has been in position 2 for all t < 0, no initial energy in L

and C. At t = 0, the switch is moved to position 1 (and remains there for t > 0).

a) Find vC(t) for t > 0, determine which damped case it is.

b) Let R = 20 Ω, find vC(t) again and determine the damped case.

c) Let R = 0 Ω, find vC(t) again and determine the damped case.

d) Sketch vC(t) at these 3 cases on 1 graph with MATLAB. (Your MATLAB script should be

attached, containing XY axis and grid!)


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Fig. 4

Solution:

a) Initial condition: (0 ) (0 ) 0C Cv v , (0 ) (0 ) (0 ) 0CL L

dvi i C

dt

,

For t>0, the circuit is a series RLC, apply KVL

LS L C

diV Ri L v

dt (1)

As CL

dvi C

dt , substitute this into (1)

2

2

C C C Sd v dv v VR

dt L dt LC LC , gives 2

0

1,

2

R

L LC

For R=40 Ω, 0

40 120 10

2 0.01 1

, overdamped

So, 20 10 3 20 10 3

1 2( )t t

C fv t V Ae A e

, 2fV ,

Hence, 20 10 3 20 10 3

1 2( ) 2t t

Cv t Ae A e

,

20 10 3 20 10 3

1 2( ) 20 10 3 20 10 3t tCdv

t Ae A edt

,

Substitute initial condition:

1 20 2 A A and 1 20 20 10 3 20 10 3A A

Get 1 2

2 3 2 31 , 1

3 3A A ,so

20 10 3 20 10 32 3 2 3( ) 2 ( 1) ( 1) V

3 3

t t

Cv t e e

b) For R=20 Ω, 0

20 110 10

2 0.01 1

, critically damped

Hence, 10 10

1 2( ) 2 t t

Cv t Ate A e ,

Similarly, 1 220, 2A A ,

So, 10 10( ) 2 20 2t t

Cv t te e


5

c) For R=0 Ω, 0 10 , under damped.

Hence, 1 2( ) 2 cos10 sin10Cv t A t A t ,

Similarly, 1 22, 0A A ,

So, ( ) 2 2cos10Cv t t

d) (4%)

5. [10%] The switch in the circuit of Fig. 5 has been in position a for a long time. At t = 0 the switch

moves instantaneously to position b. Find

a) vo(0+)

b) dvo(0+)/dt

c) vo(t) for t 0.

Solution:

a) t < 0:

2 0 4 0 6 0

( 0 ) ( 0 ) 6 m A4 0 0 0 6 0 0 0 1 0 0 0 0

o oi i

( 0 ) ( 0 ) 2 0 6 0 0 0 0 . 0 0 6 1 6 VC Cv v

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

3

3.5

4

时间t/s

电压

/V

overdamped

criticaldamped

underdamped


6

t = 0+:

(0 ) (0 ) 6 || 3 16 0.006 2000 16 4Vo ov i ,

(0 ) 20 (0 ) 24VL ov v

b) For KVL, ( ) 2000o o Cv t i v

( ) 2 0 0 0o o Cd v d i d vt

d t d t d t

So,

3

9

(0 )(0 ) 24 6 10(0 ) 2000 (0 ) (0 ) 2000 2000 103680V/s

0.5 781.25 10

o o C oLdv di dv iv

dt dt dt L C

c) 6

0

12.56 10 1600rad/s

LC ， 2000rad/s

2

R

L

For 2 2

0 , circuit is overdamped.

2 2

1,2 2000 2000 1600 2000 1200s

So, 800 3200

1 2( ) t t

o fv t V Ae A e

( ) 20Vf oV v ,

Apply initial condition: 1 220 4A A ,

1 2800 3200 103680A A

So, 1 211.2, 35.2A A

800 3200( ) 20 11.2 35.2 Vt t

ov t e e , t >0+


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6. [10%] Given the network in Fig. 6, find v(t) for t > 0.

Fig. 6

Solution:

For t = 0-, 0 0 4 12 8Vv v

For t > 0, the circuit becomes that shown in Figure below after source transformation.

So, 1/ 1/ 1 1/ 25 5o LC

/ 2 6 / 2 3R L

2 2

1,2 0 3 4s j

Thus, 3 4 4 t

sv t V Acos t Bsin t e , 12VsV

0 8 12 v A or 4A

dv

i Cdt

, so,

3 3/ / 3 4 4 4 4 4t ti C dv dt Acos t Bsin t e Asin t Bcos t e

0 3 4 3i A B B

Finally, v t 3t

12 4cos4t 3sin4t e A , t > 0

7. [10%] The switch in the circuit shown in Fig. 7 has been closed for a long time before it is opened

at t = 0. Assume that the circuit parameters are such that the response is underdamped.

a) Derive the expression for vo(t) as a function of Vg, ,d, C, and R for t 0.

b) Derive the expression for the value of t when the magnitude of vo is maximum.


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Fig. 7

Solution:

a) Let i be the current in the direction of the voltage drop vo(t).

Then by hypothesis, circuit is underdamped

1 2cos sint t

f d di i Ae t A e t

Easy to find, ( ) 0fi i , 1(0)gV

i AR

Therefore, 1 2cos sint t

d di Ae t A e t

For (0)

0,di

Ldt

2 1 2 1[( )cos ( )sin ] t

d d d d

diA A t A A t e

dt

So, 2 1 2 10,g

d

d d

VA A A A

R

Therefore,

So, ( ) sin Vg t

o d

d

Vv t e t

RC

, t>0


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b) ( )

( cos sin )g to

d d d

d

Vdv te t t

dt RC

,

When ( )

0odv t

dt , get cos sin 0d d dt t

So, 1

arctan d

d

t

8. [10%] Obtain i1 and i2 for t > 0 in the circuit of Fig. 8.

Fig. 8

Solution:

At t = 0-, 1 1 2 2(0 ) (0 ) 0, (0 ) (0 ) 0i i i i (1)

Apply KCL, 11 2

14

2

dii i

dt (2)

Also, 1 22 ( ) 3

di dii

dt dt (3)

Take the derivative of (2),

2

1 1 2

20 2 2

d i di di

dt dt dt (4)

From (2) and (3), 2 1 12 13 2.5 12 3

di di dii i

dt dt dt

Subsituting this into (4), get

2

1 112

7 6 24d i di

idt dt

, which gives 2 7 6 0 ( 1)( 6)s s s s

Thus, 6

1( ) [ ], 24 / 6 4At t

s si t I Ae Be I

6

1( ) 4 [ ]t ti t Ae Be , and 1(0) 4i A B (5)

6 6 612 1

14 4 4 [ ] 0.5[ 6 ] 0.5 2

2

t t t t t tdii i Ae Be Ae Be Ae Be

dt

For 2(0) 0.5 2 0i A B , combine with (5)


10

Get 3.2, 0.8A B , so

6

1( ) 4 3.2 0.8t ti t e e , 6

2( ) 1.6 1.6t ti t e e

9. [10%] The circuit in in Fig. 9 is the electrical analogue of a temperature control system.

Fig. 9

Assuming CA = 1 F, CB = 4 F, RA = 1 Ω, RB = 4 Ω.

iS = K(V0 − vB)2 where K = 25 A/V2, V0 = 1.1 V.

a) Write dynamical equations for this network in state form. Use vA and vB as state variables.

(As a check on your state equations, the stable steady-state value of vB is 1 V. That is, you should

have dvA/dt = dvB/dt = 0 for vB = 1 V.)

b) Now assume vA = VA + va and vB = VB + vb, where VA and VB are the steady-state values and va

and vb are small variations. Determine and draw a small-signal linear circuit model in which va and

vb are the state variables.

c) Is the zero-input response of the small-signal circuit under-damped, over-damped, or critically-

damped?

Solution:

a) Use KCL for two node equations:

2

0( )A A BA B

A

dv v vC K V v

dt R

,

B B A BB

B A

dv v v vC

dt R R

_

b) Since 2

0S Bi K V v , then the following small-signal approximation is valid:

0 02 ( ) 2 ( )s SB B

b B

i diK V v K V V

v dv ,

02 ( )s B bi K V V v .

See figure below for a small-signal model.


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c) First, write two new state equations using the small-signal model:

02 ( )a a bA B b

A

dv v vC K V v v

dt R

,

b b a bB

B A

dv v v vC

dt R R

We substitute in the numerical values given, and get the following:

Get 1VBV from(a), then

5aa b b

dvv v v

dt ,

4 0.25bb a b

dvv v v

dt .

Eliminate va, getting the following:

2

216 21 21 0b b

b

d v dvv

dt dt

This has the following characteristic equation:

216 21 21 0s s

Since 221 4 21 16 0 , the system is underdamped.

10. [4%] Draw the dual of the circuit in Fig. 10.

Fig. 10

Solution:


12

Construct,

Redraw:

11. [10%] a) Derive the differential equation that relates the output voltage to the input voltage for the

circuit shown in Fig. 11.

Fig. 11

b) Compare the result with Eq. 1 when R1C1 = R2C2 = RC in Fig. 12.

2

2

1 1 2 2

1 1og

d vv

dt R C R C (1)


13

Fig. 12

c) What is the advantage of the circuit shown in Fig. 11?

Solution:

a) For KCL on Va, V- and Vb

2 02

a g ga a a av v vdv v dv v

Cdt R R dt CR CR

(1)

(0 ) 00 0b a b ad v v dv v

Cdt R dt CR

(2)

( ) 20

2

b b o b b b odv d v v v dv v dvC C

dt dt R dt CR dt

(3)

From (2), ba

dvv RC

dt and

2

2

a bdv d vRC

dt dt

Substitude into (1), get

2 2

2 2 2 2

1

2 2

g gb b b bv vd v dv d v dv

RCdt dt RC dt RC dt R C

(4)

Differentiate (3), get

2 2

2 2

1 1

2

b b od v dv d v

dt CR dt dt (5)

Substitute (4) into (5),

We get

2

2 2 2

govd v

dt R C


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b) When 1 1 2 2R C R C RC ,

2

2 2 2

govd v

dt R C

The two equations are the same except for a reversal in algebraic sign.

c) Two integrations of the input signal with one operational amplifier.

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HW6 - ShanghaiTechsist.shanghaitech.edu.cn/faculty/zhoupq/Teaching/Fall15/HW6... · Electric Circuits Fall 2015 Solution