Apply KCL to the top node ,we have

(i.e, ( ) )v t

Apply KCL to the top node ,we have ( ) ( ) ( ) 0CLR ii t ti t

We normalize the highest derivative by dividing by C , we get

Since the highest derivative in the equation is 2 , then this equation is

We can not solve this equation by separating variables and integrating as we did inThe first order equation in Chapter 7

The classical approach is to assume a solution

Question : what is a solution we might assume that satisfies the above equation

what is a solution that when differentiated twice and added to its first derivative multiplied by a constant and then added to it the solution itself divided by a constant will give zero

The only candidate ( مرشح ) that satisfies the above equation will be an exponential function similar to chapter 7

Substituting the proposed or assumed solution into the differential equation we get

A ≠ 0 else the whole proposed solution will be zero

That leave

The equation is called the characteristic equation of the differential equation because the roots of this quadratic equation determine the mathematical character of v(t)

characteristic equation of the differential equation

The two roots of characteristic equation are

The solutions either one satisfies the differential equation

Denoting these two solutions as v1(t) and v2(t)

characteristic equation

are solutions regardless of A1 and A2

0 0

are solutions

Also a combinations of the two solution is also a solution

is a solutions as can be shown:

Substituting the above in the differential equation , we have

0 0

is a solutions

are solutions

Is the most general solution because it containall possible solutions

however

( Recall how the time constant on RL and RC circuits depend on R,L and C)

1,2

21 1 1

2s

RC RC LC

1,2

21 1 1

2s

RC RC LC

Rewriting the roots s1,2 as follows:

characteristic equation of the differential equation

Roots of the characteristic equation

where

characteristic equation of the differential equation

where 1time Frequency

1V.s A.sA V

21

s 2 Frequency

0S o both and are freque e nci s

So to distinguish between the two frequencies we name them as

0 Neper frequency and Resonant radian frequ cy en

characteristic equation of the differential equation

where Neper frequency

Resonant radian frequency

s1,2 are also frequencies since they are summation of frequencies

To distinguish them from the Neper and Resonant frequencies and because s1,2 can be complex we call them complex frequencies

All these frequencies have the dimension of angular frequency per timer (rad/s)

complex frequencies

1 2 0There are three nature of the roots and depends on the values of an s s d

2 20 1 2

< and are real and distincts overdamped s sIf 2 2

0 1 2 < and are complex conjugate and distincts unders s dampedIf

1 22 2

0 and are real and identical criticall y dampeds sIf

We will discuss each case separately

20000 rad/s

did not change

Therefore, the first step in finding that natural response is to calculate these values and determine whether the response is overdamped, underdamped or critically damped

So far we have seen that the behavior of a second-order RLC circuit depends on the values of s1 and s2, which in turn depend on the circuit parameters R,L, and C.

Completing the description of the natural response requires finding two unknown coefficients, such as A1 and A2 .The method used to do this is based on matching the solution for the natural response to the initial conditions imposed by the circuit

In this section, we analyze the natural response form for each of the three types of damping, beginning with the overdamped response as will be shown next

2 20 1 2

< and are real and distincts overdamped s sIf

where

1 2 0(0 ) + ------v A A V (1)

But what is-------(2)

A1 , A2 are determined from Initial conditions as follows:

1 2 0(0 ) + ------v A A V (1)

C

d (t)Since (t) = C

dtv

iC

++ (0 )d (0 )dt C

iv

C L R (0 ) (0 ) (0 )i i i KCL

00

VI

R

00

1 1 2 2 + ------C

VI

Rs A s A

(2)

1 1 2 2d (0 )

+ ------dt

vs A s A

(2)

(1) From the circuit elements R,L,C we can find

But what is

(2) A1 , A2 are determined from Initial conditions as follows:

Summarizing

KCL

C

++ (0 )d (0 )dt C

iv

Because the roots are real and distinct, we know that the response is overdamped

+d (0 )450 kV/s

dtv

+d (0 )450 kV/s

dtv

1 2 0(0 ) + v A A V

00

1 1 2 2 + C

VI

Rs A s A

2 20 1 2

< and are real and distincts overdamped s sIf

1 2 and are determined by solving the following equations:A A

where

21,2

20

s =

2 20

<

2 20

j dj

2 20d

were

Damped Radian Frequency

Neper frequency

Resonant radian frequency

1 2

1 2(t) s t s tv A e A e ( ) ( )1 2

d dt tj jA e A e 1 2 +d dtj j tte A e A e

Using Euler Identities cos( ) sin( )je j

1 2

1 2(t) s t s tv A e A e ( ) ( )1 2

d dt tj jA e A e 1 2 +d dtj j tte A e A e

Using Euler Identities cos( ) sin( )je j

1 2cos( ) sin( ) + cos( ) sin( )td d d de A t t A jtj t

1 2 1 2cos( ) + sin( )td de A A t A Aj t

1 2 1 21 2Let B BA A A A

1 2( ) cos( ) sin( )td dv t e B t B t

1 2( ) cos( ) sin( )td dv t e B t B t

It is clear that the natural response for this case is exponentially damped and oscillatory in nature

1 2( ) cos( ) sin( )td dv t e B t B t

The constants B1 and B2 are real because v(t) real (the left hand side)

Therefore the right hand side is also real

The constants A1 and A2 are complex conjugate ( can be shown )

Therefore

*1 2 1 1 1

*1 2 1 1 1

2

2

Re

Im

A A A A A

A jA A A A

1

2

1 2 1

1 2 1

2

Re

2Im

B

B

A A A

A A Aj j

1 2cos( ) sin( )td de B t B t

Summary 2 20

<

21,2

20

s = 2 20

j dj

2 20d were

1 2

1 2(t) s t s tv A e A e 1 2 1 2cos( ) + sin( )td de A A t A Aj t

The constants B1 and B2 can be determined from the initial conditions +

0 (0 ) v V +0 (0 ) i I

+ 00 1 2(0 ) cos(0) sin(0)v V e B B

1 -----B (1)

0

(0 ) ( )

t

dv dv tdt dt

1 1

2 2

( ) cos( ) sin( )

sin( ) cos( )

t td d d

t td d d

dv tB e t B e t

dt

B e t B e t

1 2

(0 )d

dvB B

dt

This was shown previously using KCL and +

+ dv(0 ) (0 ) C

dtCi 0

0

VI

RC

-----(2)

Solving (1) and (2) , w obtain B1 and B2

0 00 V 12.25 mA V I

(a) Calculate the roots of the characteristic equation

underdamped

For the underdamped case, we do not ordinarily solve for s1 and s2 because we do not use them explicitly.

0 00 V 12.25 mA V I

(0 )(b) calculte (0 ) and

dvv

dt

+0(0 ) 0v V

0+0(0 )

VIdv R

dt C

6

0 ( 12.25 mA) 20000

0.125 X 10 98,000 V/s

(c) Calculate the voltage response for t ≥ 0

0 00 V 12.25 mA V I

+(0 ) 0v +(0 )

98,000 V/sdv

dt

1 2( ) cos( ) sin( )td dv t e B t B t

1 0 0B V 1 2

(0 )d

dvB B

dt

1

2

(0 )

d

dvB

dtB

98,000 (200)(0)100

979.8

200( ) 100 sin(979.80 ) V t 0tv t e t

(d) Plot v(t) versus t for the time interval 0 ≤ t ≤ 11 ms

0 00 V 12.25 mA V I

+(0 ) 0v +(0 )

98,000 V/sdv

dt

200( ) 100 sin(979.80 ) V t 0tv t e t

The Critically Damped Voltage Response

The second-order circuit is critically damped when 2 20

0

or

21,2

20

s = 12RC

1 2

1 2(t) s t s tv A e A e 1 2t tA e A e 1 2

tA A e 0tA e

where A0 is an arbitrary constant

The solution above can not satisfies the two initial conditions (V0, I0) with only on constant A0

It seems that there is a problem ?

We can trace this dilemma back to the assumption that the solution takes the form of

1 2

1 2(t) s t s tv A e A e

When the roots of the characteristic equation are equal, the solution for the differential equation takes a different form, namely

1 2(t) t tv D te D e

The Critically Damped Voltage Response

0

1,2

1 s2RC

1 2

1 2(t) s t s tv A e A e

Another solution for the differential equation takes a different form, namely

1 2(t) t tv D te D e

Is not possible assumption because it leads to 0( ) tv t A e which can not satisfies (V0, I0 )

The justification of is left for an introductory course in differential equations.

Finding the solution involves obtaining D1 and D2 by following the same pattern set in the overdamped and underdamped cases:

We use the initial values of the voltage v(0+)and the derivative of the voltage with respect to time dv(0+)/dt to write two equations containing D1 and D2

0 00 V 12.25 mA V I

(a) For the circuit above ,find the value of R that results in a critically damped voltage response.

31 1 10 rad/s2RC LC

31

102

RC

3

6

110 4000

2(0.125 10 )X

0 critically damped

0 00 V 12.25 mA V I

0+0(0 )

VIdv R

dt C

6

0 ( 12.25 mA) 20000

0.125 X 10 98,000 V/s

(b) Calculate v(t) for t > 0

1 2(t) t tv D te D e critically damped

0 0 VV

310 rad/s

2 0 0 VD V +

1

(0 )98000

dvD

dt

1000(t) 98000 V t 0tv te

(c) Plot v(t) versus t for 0 ≤ t ≤ 7 ms

1000(t) 98000 V t 0tv te

iov

7.5 V

0t

ov

0t Initial voltage on the Capacitor

Initial current through the Indictor

ov

i

7.5 V ov

0t 0t

ov

ov

0

ov

ov

2 20 1 2

< and are real and distincts overdamped s sIf

where

1 2 0(0 ) + ------v A A V (1)

-------(2)

A1 , A2 are determined from Initial conditions as follows:

C

++ (0 )d (0 )dt C

iv

2 20 1 2

< and are complex conjugate and distincts unders s dampedIf

2 201

s = = dj j 2 202

s = = dj j

where

1 2 and are determined by solving the following equations:B B

1 22 2

0 and are real and identical criticall y dampeds sIf

1 2 and are determined by solving the following equations:D D

1 2

1 s2

sRC

2 20 <

1 2 and are real and distis s ncts

overd amped

1 2 and are determined byA A

1 2 and are complex conjugate

and dist

s s

incts

2 20

<

underdamped

2 201

s = = dj j

2 202

s = = dj j

1 2 and are determined byB B

critically d d ampe

1 2 and are real and idents s ical

2 20

1 2

1 s2

sRC

1 2 and are determined byD D

Summary

follows

We will show first the indirect approach solution , then the direct approach

Differentiating once we obtain

t

0

1i(t)= v dτ + i(0)

L

KCL

(Natural Response Solutions)

the equations into

L Land solve for as v dvi i I C

R dt

' ' ' ' ' '1 2 1 2 1 2where , , , , and

are arbitray constants

A A B B D D

' '1 2

1 2

The primed constants can be found

interms of

cumbersome

time-consumi

( or

ng)

A D

A D

' '1 2 1 2

dv(0)The primed constants can be found interms of using v(0) and

dtA D A Dindirectly

L' '1 2 L

d (0)or we can find using (0) and

dti

A D idirectly

L

f

f

(step respoThe solution (t) for the 2ed order differential equation with constant forcing function

(

nse)

natural r) + esponfunction of the same form as the

( ) + function of the same

s

f

e

o

i

i t I

v t V

naturalrm as t reshe ponse

f fwhere , represent the the final value of the response function which can be zero as

in the example above

I V v

The initial energy stored in the RLC circuit above is zero

At t = 0 , a dc current source of 24 mA is applied to the circuit.

The value of R is 400

24 mA 400

24 mA 400

24 mA 400

2 20 1 2

Because < and are real and distincts overda es p ds m

24 mA 400

24 mA 400

1 2 20,000 80,s s 000 20,000t 80,000t

1 2v(t) A Ae e

1 2v(0) 0 A A ------(1)

20,000t 80,000t1 2

dv(t)20,000A 80,000A

dte e

+

1 2

dv(0 )20,000A 80,000A

dt

00 24

C

VI mA

R

3

9

0 0 24X10960000

25X10R

1 220,000A 80,000A 960000 1 2A 4A 48------(2)

1 2Solving (1) and (2) we get A 16 and A 16 20,000t 80,000tv(t) 16 16e e

24 mA 400

20,000t 80,000tv(t) 16 16e e

+

0

1(t) ( ) (0 )t

L Li v d iL

Now

20,000 80,000t3

0

1 16 16 025X10

t

e e d

20,000t 80,000t 24 32 8 mA t 0e e

(t) (t) (t)L R Ci I i i OR v(t) v(t)24

R dtd

mA C

20,000t 80,000t24 32 8 mA t 0e e =

24 mA 400 20,000t 80,000tv(t) 16 16e e

20,000t 80,000t24 32 8 mA t 0e e =

380 8000 rad/sec

2 2(5X10 )RL

0 3 6

1 1 (5X10 )(2X10 )LC

410 rad/s

0 > uderd amped

1 2 and are complex and conjug ss s ate

1,2s dj 2 24 310 8 X 10 6000 rad/sd

18000 6000 r /ss adj 2

8000 6000 r /ss adj

18000 6000 r /ss adj

28000 6000 r /ss adj

1 2( ) cos( ) sin( )td di t e B t B t

80001 2cos(6000 ) sin(6000 )te B t B t

1(0 ) (0 ) 0 i i B 80002( ) sin(6000 )ti t e B t

80002 2

( )8000 sin(6000 ) 6000 cos(6000 )tdi t

B e t B tdt

42

(0 )10 6000

diB

dt

4

210 1.676000

B

8000( ) 1.67 sin(6000 )ti t e t

18000 6000 r /ss adj

28000 6000 r /ss adj

8000( ) 1.67 sin(6000 )ti t e t

L

( )v ( )

di tt L

dt

3 8000 80005X10 (1.67) ( 8000) sin(6000 ) (6000) cos(6000 )t te t e t

8000 800066.8 sin(6000 ) 50.1 cos(6000 )t te t e t

Rv ( ) 80 ( )t i t 8000(80) 1.67 sin(6000 )te t 8000133.6 sin(6000 )te t

L R CKVL 100 + v ( ) + v ( )+ v ( ) 0t t t

C R Cv ( ) 100 (v ( ) v ( ))t t t 100 50.1cos(6000 ) 66.8sin(6000 )t t

18000 6000 r /ss adj

28000 6000 r /ss adj

8000( ) 1.67 sin(6000 )ti t e t

C C

0

1v ( ) ( ) v (0)C

t

t i d OR

8000-6

0

1 1.67 sin(6000 ) 502X10

t

e d

8000 100.1 50.1cos(6000 ) 66.8sin(6000 )te t t

Differentiating once we obtain

KVL

The 2ed order differential equation has the same form as the parallel RLC

2ed order differential equation

The characteristic equation for the series RLC circuit is

where

The step response for a series RLC circuit

KVL

Substitute in the KVL equation

The differential equation has the same form as the step parallel RLC

Therefore the procedure of finding in the series RLC similar the one used to find for the parallel RLC

CvLi

Underdammped

28001 2( ) cos(9600 ) sin(9600 )ti t e B t B t

From the circuit , we note the followings:

(0 ) 0 (0 )i i

(0 ) 0Rv

(0 ) 100Lv 3 (0 )(100X10 )

didt

3X(0 ) 100 1000 A/s

100 10di

dt

28001 2( ) cos(9600 ) sin(9600 )ti t e B t B t

28002( ) sin(9600 )ti t B e t

28002( ) sin(9600 )ti t B e t

(0 )1000 A/s

didt

2Finding continueB

2800 28002 2

( )( 2800) sin(9600 ) + (9600)cos(9600 )t tdi t

B e t B e tdt

28002400 24cos(9600 ) 7sin(9600 )tB e t t

(0 )1000

didt

2400 (24)B 29600B

21000 0.10429600

B

2800( ) 0.1042 sin(9600 ) t 0ti t e t

2Solving for B

2800( ) 0.1042 sin(9600 ) t 0ti t e t

C C

0 100C1( ) ( ) + (0)

t

v t i d v

C ( ) div t Ri Ldt

OR

Whichever expression is used ( the second is recommended ) , we get

C

2800( ) 100cos(9600 ) + 29.17sin(9600 ) V t 0tv t e t t