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SECOND ORDERSECOND ORDER

Examples Examples

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SECOND ORDERSECOND ORDERWhat Circuit do we require and why? What Circuit do we require and why? 1. Circuit at t = 01. Circuit at t = 0--

This circuit is required to find the initial values of i(0-) and/or v(0-). These values are used to form the first equation of finding A1 and A2.

3. Circuit for t > 0 3. Circuit for t > 0

This circuit is required to find the initial values of di(0+)/dt and/or dv(0+)/dt. These values are used to form the second equation of finding A1 and A2.

2. Circuit at t = 02. Circuit at t = 0++

a) Source free circuitWe can divide this circuit into two:

This circuit is used to find the natural response of the circuit.

b) Circuit at t = infinityThis circuit is used to find the steady-state response of the circuit.

Superposition theorem allows us to solve separately and then combine the solution in (a) and (b) together and to get the complete solution easily.

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Example 1

(i) Find v(t) for t > 0.(ii) Find i(t) for t >0.

vi

24 V

R = 6 1 H

0.25 F

+

-

1

t = 0

The switch has been closed for a long time and it is open at t = 0.

4

v

i

24V

5

1 H

0.25 F

+

-

1

t = 0

Example 2

t = 0

(i) Find i(t) for t > 0.(ii) Find v(t) for t >0.

The switches have been opened for a long time and they are closed at t = 0.

5

(i) Find i(t) for t > 0.(ii) Find iR(t) for t >0.

Example 3

vi

24 V

R = 1 1 H

0.25 F

+

-

1

t = 0

iR

The switch has been opened for a long time and it is closed at t = 0.

6

Example 1 : Solution

(i) Find v(t) for t > 0.(ii) Find i(t) for t >0.

vi

24 V

R = 6 1 H

0.25 F

+

-

1

t = 0

The switch has been closed for a long time and it is open at t = 0.

7

1. Draw the circuit for t = 0-

This statement means that the circuit is in steady state at t = 0-. Therefore, C is opened and L is shorted.

“The switch has been closed for a long time and it is open at t = 0”

24 V

6

1

i(0) = 3.429 A

v (0) = 3.429 V

+

-

Find i(0) and v(0)

Steps to solve this problem

8

2. Draw the circuit for t = 0+

This is a starting point for the circuit to experience transient. Therefore, C is not opened and L is not shorted.

We know that i(0-) = i(0+) and v(0-) = v(0+)

24 V

6

i(0+) = 3.429 A0.25 F

+

-

v(0+) = 3.429 V

Find dv(0+)/dt or/and di(0+)/dt

1 H

9

24 V

6

i(0+) = 3.429 A0.25 F

+

-

v(0+) = 3.429 V

Find dv(0+)/dt or/and di(0+)/dt

1 H

sVC

idt

dv /714.1325.0

429.3)0()0(

Lv

dtdi L )0()0(

+ -vL(0+)

-24+ 6(3.429)+vL(0+)+3.429 = 0vL(0+) = 0

KVL around the loop

0)0()0(

Lv

dtdi L

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3. Draw the circuit for t = ∞At t = ∞ the circuit has reached the steady state again. Therefore, C is open and L is shorted.

24 V

6

+

-

Find v(∞) or/and i(∞)

v(∞) = 24 V

i(∞) = 0

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4. Draw the source free circuit for t >0

1H 6

+

-

Voltage source is shorted and current source is opened.

iv0.25F

Generally for this step, we have to find the differential equation for the source free circuit, then its characteristic equation. Since the circuit is a series RLC, we can directly write down its characteristic equation, and determine the type of the response.

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For second order circuit:

02 22 oss

012 LC

sLRs

For second order circuit, its characteristics equation is

General form

Series RLC circuit

3)1(2

62

L

R 225.01

11

LCo rad/srad/s

o the response is overdamped

the roots are real and distincts1

2o

2 s1 0.764

s2 2 o2 s2 5.236

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5. Write the general solution for the circuit for t > 0.

24 V

6

i0.25 F

+

-

v

1H

tsts eAeAvtv 2121)()( tsts eBeBiti 21

21)()( or

tsts eAeA 212124 tsts eBeB 21

210

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6. Find A1 and A2

24 V

6

i0.25 F

+

-

v

1H

21)(429.3)0( AAvv 2211

)0( AsAsdt

dv

21 236.5764.0714.13 AA 2124429.3 AA

448.02 A019.211 A

1 2

tt eetv 236.5764.0 448.00198.2124)(

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7. Find other circuit quantities for t > 0.

24V

6

i0.25 F

+

-

v

1H

tt eedtd

dttdvCti 236.5764.0 448.00198.212425.0)()(

tt eeti 236.5764.0 )448.0(236.5)764.0(019.2125.0)( tt ee 236.5764.0 586.0015.4

tt eetv 236.5764.0 448.00198.2124)(

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Verification Using PSpice

24V

6

i0.25 F

+

-

v

1H

0

1 2 3

-+

1h 2h

E1h G2h

1 1

Hand Calculation Verification: Example 1**Overdamped response: Case 1.Param R=6 L=1 C=0.25**************************************************************************************E1h 1h 0 value={24-21.019*EXP(-0.764*TIME)+0.448*EXP(-5.236*TIME)}R1h 1h 0 1 ; V(1h) = output voltage****************************************************************************************************************************************************************************G2h 0 2h value={4.015*EXP(-0.764*TIME)-0.586*EXP(-5.236*TIME)}R2h 2h 0 1 ; V(2h) = output current***************************************************************************************V1 1 0 DC 24VR1 1 2 {R}L1 2 3 {L} IC=3.429AC1 3 0 {C} IC=3.429V.Tran 5m 5s 0 5m UIC.Probe V(1h) V(3) .Probe I(R2h) I(L1).End

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(i) Find i(t) for t > 0.(ii) Find v(t) for t >0.

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Example 2 : Solution

vi

24 V

5

1 H

0.25 F

+

-

t = 0

t = 0

The switches have been opened for a long time and they are closed at t = 0.

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1. Draw the circuit for t = 0-

This statement means that the circuit is in steady state at t = 0-. Therefore, C is opened and L is shorted.

“The switch have been opened for a long time and it is closed at t = 0”

24 V

5

1 v(0-) = 24 V

+

-

Find i(0) and v(0)

Steps to solve this problem

i(0-) = 0 A

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2. Draw the circuit for t = 0+

This is a starting point for the circuit to experience transient. Therefore, C is not open and L is not shorted.

We know that i(0-) = i(0+) and v(0-) = v(0+)

24 Vi(0+) = 0

0.25 F

+

-v(0+) = 24 V

Then we can find dv(0+)/dt or/and di(0+)/dt

1

1H

20

i(0+) = 00.25 F

+

-v(0+) = 24 V 1

1H

iR(0+) = 24/1 = 24 AiC(0+)

iC(0+) = -iR(0+) = -24 A

sVC

idt

dv C /9625.024)0()0(

Lv

dtdi L )0()0(

and from the circuit 0)0( Lv

0

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3. Draw the circuit for t = ∞At t = ∞ the circuit reaches steady state again. Therefore, C is open and L is shorted.

Find v(∞) or/and i(∞)

24 Vi(∞) = 24 A +

-

v(∞) = 24 V 1

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4. Draw the source free circuit for t >0Voltage source is shorted and current source is open.

i(t)0.25 F

+

-

v(t) 1

1H

Generally for this step, we have to find the differential equation for the source free circuit, then its characteristic equation. Since the circuit is a parallel RLC, we can directly write down its characteristic equation, and determine the type of the response.

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For second order circuit:

02 22 oss

0112 LC

sRC

s

For second order circuit, its characteristics equation is

General form

Parallel RLC circuit

2)25.0)(1(2

12

1

RC 2

25.0111

LC

o rad/srad/s

o the response is critically damped

21 s

22 sThe roots are real and equal

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5. Write the general solution for the circuit for t > 0.

24 Vi(t)

0.25 F

+

-

v(t) 1

1H

tetAAiti )()()( 21

tetAAti 221 )(24)(

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6. Find A1 and A2

24 Vi(t)

0.25 F

+

-

v(t) 1

1H

1240)0( Ai 241 A

21)0( AA

dtdi

02)0(21 AA

dtdi 48)24)(2(2 12 AA

tt etetti 22 )4824(24)4824(24)(

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7. Find other circuit quantities for t > 0.

24 Vi(t)

0.25 F

+

-

v(t) 1

1H

tL et

dtditv 2)4824(241)(

tetti 2)4824(24)(

)48()4824(21)( 22 ttL eettv

tL tetv 248)(

tL tetvtv 24824)(24)(

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(i) Find i(t) for t > 0.(ii) Find iR(t) for t >0.

vi

24 V

R = 1 1 H

0.25 F

+

-

1

t = 0

Example 3: Solution

iR

The switch has been opened for a long time and it is closed at t = 0.

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1. Draw the circuit for t = 0-

This statement means that the circuit is in steady state at t = 0-. Therefore, C is opened and L is shorted.

“The switch has been opened for a long time and it is closed at t = 0”

24 V

1 i(0-) = 0

v(0-) = 24 V

+

-

Find i(0) and v(0)

Steps to solve this problem

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2. Draw the circuit for t = 0+

This is a starting point for the circuit to experience transient. Therefore, C is not opened and L is not shorted.

We know that i(0-) = i(0+) and v(0-) = v(0+)

24 Vi(0+) = 0

0.25 F

+

-

v(0+) = 24

Then we can find dv(0+)/dt or/and di(0+)/dt

1

1H

1 iR(0+) = 24A

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24 Vi(0+) = 0

0.25 F+

-v(0+) = 24 1

1H

1 iR(0+) = 24A

sVC

idt

dv C /9625.024)0()0(

iC(0+)

iC(0+) = -iR(0+)

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3. Draw the circuit for t = ∞At t = ∞ the circuit reaches steady state again. Therefore, C is open and L is shorted.

Find v(∞) or/and i(∞)

24 Vi(∞) = 12 A +

-

v(∞) = 12V 1

1

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4. Draw the source free circuit for t >0.Voltage source is shorted and current source is opened.

Since this circuit is neither a parallel nor a series RLC. This circuit must be a general second order. So, we have to find the differential equation in terms of i(t) or v(t).

Is this a series RLC circuit? Is this a parallel RLC circuit?

i(t)0.25 F

+

-

v(t) 1

1H

1

33

i(t)0.25 F

+

-

v(t) 1

1H

1

Applying KCL at node a:

a

dtdvvi

41

1

Applying KVL to the left mesh:

0411

41

2

2

vdt

vddtdv

dtdvv

011 vdtdii

0852

2

vdtdv

dtvd

34

For second order circuit:

02 22 oss

0852 ss

For second order circuit, its characteristics equation is

General form

The circuit under-study

5.2 828.28 o rad/srad/s

o the response is underdamped

djs 1

djs 2

The roots are complex conjugate

d o2 2 d 1.323

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5. Write the general solution for the circuit for t > 0.

24 Vi(t)

0.25 F

+

-

v(t) 1

1H

1

tdd etAtAvtv )sin()cos()()( 21

tetAtAtv 5.221 )323.1sin()323.1cos(12)(

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6. Find A1 and A2

24 Vi(t)

0.25 F

+

-

v(t) 1

1H1

2412)0( 1 Av

tetAtAtv 5.221 )323.1sin()323.1cos(12)(

121 A

12)0( AA

dtdv

d

12 5.2323.196 AA 89.492 A

tetttv 5.2)323.1sin(89.49)323.1cos(1212)(

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7. Find other circuit quantities for t > 0.

24 Vi(t)

0.25 F+

-v(t) 1

1 H1

tetttv 5.2)323.1sin(89.49)323.1cos(1212)(

iC(t)

tC ett

dtdti 5.2)323.1sin(89.49)323.1cos(121225.0)(

iR(t)

1)()( tvtiR )()()( tititi RC

38

Example 4 t = 0

5 H

3

8 A 0.05 F

i(t)

-

+

v(t)1

Find v(t) for t >0

39

Example 5

6

4u(t)

0.1 F1 H v(t)

+

-

i(t)

Find i(t) for t >0

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Example 1a

A series RLC circuit has R = 10 k, L = 0.1 mH, and C = 10 F. What type of damping is exhibited by the circuit.

Example 2a

A parallel RLC circuit has R = 10 k, L = 0.1 mH, and C = 10 F. What type of damping is exhibited by the circuit.

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Example 4aThe responses of a series RLC circuit are

ttL

ttc

eeti

eetv1020

1020

3040)(

301030)(

Determine the values of R, L, and C.

V

mA

(a) Overdamped (b) Critically damped(c) Underdamped

If R = 20 , L = 0.6 H, what value of C will make an RLC series circuit:

Example 3a

42

Example 5a

Find i(t) in the circuit of Fig. 8.10. Assume that the circuit has reached steady state at t = 0-.Fig 8.10

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Example 8.9

• Find v(t) for t > 0 in Fig. 8.29.

44

Example 8.9

• Find v(t) for t > 0 in Fig. 8.29.

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Problem 8.56

+ v -

iR

i

0.75

0.25 H

4

0.04 F

20 V

Find iR(t), v(t), and i(t) for t > 0.

Given v(0) = 0 and i(0) = 0