1

Chapter 7.3

Natural and Step Responses of RLC Circuits

2

Initial conditions: v 0 !( ) = v 0 +( ) , dv 0 +( )dt

=i 0 +( )C

3

Initial conditions: i 0 !( ) = i 0 +( ) , di 0 +( )dt

=v 0 +( )L

4

5

■ In dc steady state, a capacitor looks like an open circuit and an inductor looks like a short circuit.

■ The voltage across a capacitor must be a continuous function of time

■ The current flowing through an inductor must be a continuous function of time

6

I. Parallel RLC Circuit: Step Response and Natural Response

II Series RLC Circuit: Step Response and Natural Response

7

I. Parallel RLC Circuit: Step Response and Natural Response

iL + iR + iC = I

iL 0 +( )+ 1L

v z( )dz0

t

!"#$

%&'+vR+C

dvdt

= IiL +vR+C

dvdt

= I

iL +LRdiLdt

+ LCd 2iLdt2

= I

•

•

d 2iLdt2

+1RC

diLdt

+1LC

iL =ILC

•

•

• vL+1Rdvdt

+Cd2vdt2

= 0

d 2vdt2

+1RC

dvdt

+1LC

v = 0•

8

Total Response =

{ zero-input response = natural response} + { zero-state response = forced response }

! Steady- State Response (Limit Response)

9

I-1 The Natural Response of a Parallel RLC Circuit

d2vdt 2

+ 1RC

dvdt

+ 1LC

v = 0

Second-Order, Constant - Coefficient, Homogenous, Linear DE

10

A Parametrization and General Solution

d2vdt 2

+ 1RC

dvdt

+ 1LC

v = 0

! = 12RC

- Neper frequency (rad/sec)

! 0 =1LC

- Resonant radian frequency (rad/sec)

! = "# 0 ! ! - damping ratio

d 2vdt2

+ 2!dvdt

+" 02v = 0

d2vdt 2

+ 2!" 0dvdt

+" 02v = 0

11

Let us seek the solution in the form: v t( ) = Aest , s can be complex

Substitute into the ODE, we got an algebraic (characteristic) equation

d2vdt 2

+ 2! dvdt

+" 02v = 0 ! s2 + 2!s +" 0

2 = 0

s2 + 2! s+" 02 = 0 Characteristic Equation

s1 = !! + ! 2 !" 0

2

s2 = !! ! ! 2 !" 02

"#$

%$

12

Since the ODE is linear ! linear combination of solutions remains a solution to the equation. The general solution for v t( )must be of the form

v t( ) = A1es1t + A2es2t ,

where A1 and A2 will be determined by the two initial conditions

v 0 +( ), dv 0 +( )dt

• v 0 !( ) = v 0 +( ) , dv 0 +( )dt

=i 0 +( )C

• ! circuit theory

13

• •

•

•

•

s2 = !4s1 = !2

overdamped underdampeds1 = !3+ j4

s1 = !3! j4

s1 = s2 = !1critically damped

14

B The Forms of the Natural Response ! = 1

2RC, ! 0 =

1LC

B1 The Overdamped Voltage Response: ! 2 >" 02

v t( ) = A1e! !! ! 2!"0

2{ }t + A2e! !+ ! 2!"0

2{ }t ! A1es1t + A2e

s2t

A1 , A2 - obtained from the initial conditions ! v 0 +( ), dv 0 +( )dt

!circuit th.

v 0 +( ) = A1 + A2dv 0 +( )dt

= s1A1 + s2A2

!"#

$#

15

The two initial conditions: need v 0 +( ) and dv 0 +( )dt

v 0 +( ) - the initial voltage on the capacitor ! V0

Cdv 0 +( )dt

= iC 0 +( ) ! dv 0 +( )dt

= 1CiC 0 +( )

KVL at 0 + : iC 0 +( ) = !V0R! I0

dv 0 +( )dt

= ! V0CR

! I0C

- the initial voltage change on the capacitor

16

B2 The Underdamped Voltage Response: ! 02 >" 2 ! = 1

2RC, ! 0 =

1LC

s1 = !! + ! 2 !" 02 = !! + j " 0

2 !! 2

s2 = !! ! ! 2 !" 02 = !! ! j " 0

2 !! 2 ! s2 = s1"

! d = ! 0

2 !" 2 - damped radian frequency (rad/sec)

v = B1e!"t cos # dt( )+ B2e!"t sin # dt( )

• The response has the oscillatory damped character if ! > 0 • If ! = 0 (R!" ) ! max. oscillations with ! d !! 0

17

B1 , B2 - obtained from the initial conditions ! v 0 +( ), dv 0 +( )dt

v 0 +( ) = B1dv 0 +( )dt

= !!B1 +" dB2

"#$

%$

! v t( ) = B1e!!t cos " dt( )+ B2e!!t sin " dt( )

• v 0 !( ) = v 0 +( ) , dv 0 +( )dt

=i 0 +( )C

• ! circuit theory

18

Euler’s Formula Euler, L. Introductio in Analysin Infinitorum, Vol. 1. Bosquet, Lucerne, Switzerland, 1748

The Master of Us All

e jx = cos x( )+ j sin x( )

e! jx = cos x( )! j sin x( )

cos x( ) = ejx + e! jx

2 , sin x( ) = e

jx ! e! jx

2 j

e j! +1= 0

How many mathematicians does it take to change a light bulb?

19

B3 The Critically Damped Voltage Response: ! 02 =" 2 ! 0 =

1LC

! = 12RC

s1 = !! + ! 2 !" 02 = !! !

s2 = !! ! ! 2 !" 02 = !! R = 1

2LC

v t( ) = D1te!"t + D2e!"t

D1 , D2 - obtained from the initial conditions ! v 0 +( ), dv 0 +( )dt

! circuit th.

v 0 +( ) = D2

dv 0 +( )dt

= D1 !!D2

"#$

%$

20

21

• •

•

•

•

s2 = !4s1 = !2

overdamped underdampeds1 = !3+ j4

s1 = !3! j4

s1 = s2 = !1critically damped

22

23

24

Examples of Second-Order Linear Homogenous Differential Equations with Constant Coefficients

1. y 2( ) +10y 1( ) + 21y = 0

Characteristic Equation: s2 +10s + 21= 0

! s2 +10s + 21= s + 3( ) s + 7( ) = 0 ! s1 = !3 , s2 = !7

Solution: y t( ) = C1e!3t +C2e!7t

2. y 2( ) + 7y 1( ) = 0 ! s2 + 7s = 0 ! s1 = 0 , s2 = !7 ,

Solution: y t( ) = C1 +C2e!7t

25

3. y 2( ) + 4y 1( ) + 5y = 0 ! s2 + 4s + 5 = 0

! s1,2 =!4 ± 42 ! 4 " 5

2 ! s1 = !2 + j , s2 = !2 ! j

Solution: y t( ) = C1e!2t cos t( )+C2e!2t sin t( )

4. y 2( ) + 4y = 0 ! s2 + 4 = 0 ! s1,2 = ±2 j

Solution: y t( ) = C1 cos 2t( )+C2 sin 2t( )

26

5. y 2( ) + 8y 1( ) +16y = 0 ! s2 + 8s +16 = 0 ! s + 4( )2 = 0

! s1 = s2 = !4

Solution: y t( ) = C1e!4 t +C2te!4 t

6. y 2( ) = 0 ! s2 = 0 ! s1 = s2 = 0

Solution: y t( ) = C1e0!t +C2te0!t = C1 +C2t

27

I-2 The Step Response of a Parallel RLC Circuit

d2iLdt 2

+ 1RC

diLdt

+ 1LC

iL =ILC

iL = iLnatural + I

!

d2iLdt 2

+ 2! diLdt

+" 02iL = 0

28

• iL = I + A1es1t + A2e

s2t - overdamped

• iL = I + B1e!!t cos " dt( )+ B2e!!t sin " dt( ) - underdamped

• iL = I + D1te

!!t + D2e!!t - critically damped

The required constants can be found from iL 0 +( ) , diL 0 +( )dt

• iL 0 !( ) = iL 0 +( ) , diL 0 +( )dt

=v 0 +( )L

• ! circuit theory

! v 0 +( ) = v 0 !( )

29

Example

I = 24mA

1. Finding the Overdamped Step (DC) Response of a Parallel RLC Circuit R = 400!

the initial energy stored in the circuit is zero ! iL 0 +( ) = 0 , vC 0 +( ) = 0

! = 12RC

, ! 0 =1LC

?

! ! 2 >" 02

30

(a) iL 0 +( ) , diL 0 +( )dt

I = 24mA

the initial energy stored = 0 ! iL 0 +( ) = 0

! vC 0 +( ) = 0 and vC 0 +( ) = v 0 +( ) = L diL 0 +( )dt

! diL 0 +( )dt

= 0

31

(b) Characteristic Equation ! = 12RC

, ! 0 =1LC

I = 24mA

• ! 02 = 1

LC = 1012

25 ! 25 = 16 !108

• ! = 12RC

= 109

2 ! 400 ! 25= 5 !104 ! ! 2 = 25 !108

Since ! 2 >" 02 ! overdamped

32

s1 = !! + ! 2 !" 02 = -5 !104+ 3!104 = -20,000

s2 = !! ! ! 2 !" 02 = 5 !104 - 3!104 = -80,000

33

(c) iL t( ) iL t( ) = I + A1es1t + A2es2t

iL 0 +( ) = I + A1 + A2 = 0 diL 0 +( )

dt= s1A1 + s2A2 = 0

! A1 = !32mA , A2 = 8mA

iL t( ) = 24 ! 32e!20,000t + 8e!80,000t mA, t ! 0

34

2. Finding the Underdamped Step Response of a Parallel RLC Circuit ! 0 =

1LC

R = 625! iL 0 +( ) = 0 , vC 0 +( ) = 0 ! = 12RC

! 0

2 = 16 !108 , ! 2 = 10.24 !108 ! ! 2 <" 02 ! underdamped

s1 = !! + j " 02 !! 2 = !3.2 "104 + j2.4 "104

s2 = !! ! j " 02 !! 2 = !3.2 "104 ! j2.4 "104

iL = I + B1e!!t cos " dt( )+ B2e!!t sin " dt( )

! d = ! 02 !" 2 = 24,000 rad/s

35

iL t( ) = I + B1e!!t cos " dt( )+ B2e!!t sin " dt( ) iL 0 +( ) = I + B1 = 0

diL 0( )dt

=! dB2 !"B1 = 0

! B1 = !24mA , B2 = !32mA

iL t( ) = 24 ! 24e!32,000t cos 24,000t( )! 32e!32,000t sin 24,000t( ) mA, t ! 0

36

3. Finding the Critically Damped Step Response of a Parallel RLC Circuit ! = 1

2RC , ! 0 =

1LC

R = 500! iL 0 +( ) = 0 , vC 0 +( ) = 0

! 02 = 16 !108 , ! 2 = 16 !108 ! ! 2 =" 0

2 critically damped

iL t( ) = I + D1te!!t + D2e!!t !

iL 0 +( ) = I + D2 = 0

diL 0( )dt

= D1 !!D2 = 0 ! D1 = !960,000mA / s , D2 = !24mA

iL t( ) = 24 ! 960,000te!40,000t ! 24e!40,000t mA, t ! 0

37

4. Comparing the Three-Step Response Forms

iL !( ) = 24

•

tod90% =130µs

tcd90% = 97µs

tud90% = 74µs

•

•

•

••

tud90% < tcd

90% < tod90%

38

On the basis of the results obtained, which response would you specify in a design that puts a premium on reaching 90% of the final value of the output in the shortest time ?

The underdamped response reaches 90% of the final value in the fastest time, so it is the desired response type when speed is the most important design specification.

39

Which response would you specify in a design that must ensure that the final value of the current is never exceeded? From the plot, you can see that the underdamped response overshoots the final value of current, whereas neither the critically damped nor the overdamped response produces currents in excess of 24 mA. Although specifying either of the latter two responses would meet the design specification, it is best to use the overdamped response. It would be impractical to require a design to achieve the exact component values that ensure a critically damped response.

! 2 =" 02 !"! R = 1

2LC

40

5. Finding Step Response of a Parallel RLC Circuit with Initial Stored Energy

R = 500! iL 0 +( ) = 29mA , vC 0 +( ) = 50V

I = 24mA

! 0

2 = 16 !108 , ! 2 = 16 !108 ! ! 2 =" 02 ! critically damped

iL t( ) = I + D1te!!t + D2e!!t

41

iL t( ) = I + D1te!!t + D2e

!!t

iL 0 +( ) = I + D2 = 29mA

diL 0( )dt

= D1 !!D2 ! LdiL 0 +( )dt

= vC 0 +( ) = 50V

! diL 0 +( )dt

= 2000A / s

iL 0 +( ) = I + D2 = 29mA ! D2 = 5mA

diL 0( )dt

= D1 !!D2 = 2000 ! D1 = 2.2 !106mA / s

42

iL t( ) = 24 + 2.2 !106 te"40,000t + 5e"40,000t mA, t ! 0

Also

v t( ) = L diL t( )dt

= !2.2 "106 te!40,000t + 50e!40,000t V, t ! 0

! vC 0 +( ) = 50V

43

iL t( ) , v t( )

44

II Series RLC Circuit: Step Response and Natural Response

KVL: V = Ri + L didt

+ vC and i = C dvCdt

d2vCdt 2

+ RLdvCdt

+ 1LC

vC = VLC

45

• ! parallel =12RC

• ! = R2L

- Neper frequency ! rad/sec

! 0 =1LC

- Resonant radian frequency!rad/sec

! d = ! 02 !" 2 - damped radian frequency

d 2vCdt 2

+ 2! dvCdt

+" 02vC = V

LC

s2 + 2!s +" 02 = 0 ! s1,s2

46

• vC =V + A1es1t + A2e

s2t - overdamped

• vC =V + B1e!!t cos " dt( )+ B2e!!t sin " dt( ) - underdamped

• vC =V + D1te

!!t + D2e!!t - critically damped

The required constants can be found from

vC 0 +( ) , dvC 0 +( )dt

! circuit theory

47

Example E7. 17 Find i0 t( ) and v0 t( )

v0 t( ) =18i0 t( ) +12

48

1. t > 0

KVL: !4 + 11/ 36

i0 z( )dz + vC 0 ±( )0

t

"#$%

&'(+ 2 di0 t( )

dt+18i0 t( )+12 = 0

! d 2i0dt 2

+ 9 di0dt

+18i0 = 0

49

d 2i0dt 2

+ 9 di0dt

+18i0 = 0

Ch. Eq: s2 + 9s +18 = 0 ! s1 = !3 , s2 = !6 ! overdamped

! i0 t( ) = A1e!3t + A2e!6t

Need i0 0 +( ) and di0 0 +( )dt

to determine A1,A2

We must go back to t = 0 !

50

2. t < 0

vC 0 !( ) = 0

iL 0 !( ) = i0 0 !( ) = 24 !126 +18

= 0.5A

51

3. t = 0 +

iL 0 +( ) = iL 0 !( ) = i0 0 +( ) = 0.5A ; vC 0 !( ) = vC 0 +( ) = 0

KVL: !4 ! vC 0 +( )=0

!"# $# + vL 0 +( )+18 i0 0 +( )=0.5!"# +12 = 0 ! vL 0 +( ) = !17 V

Also vL 0 +( ) = L di0 0 +( )dt

! di0 0 +( )dt

= vL 0 +( ) / L = !172

!

52

4. A1 , A2 from i0 0 +( ) = 0.5A and di0 0 +( )dt

= !172

i0 t( ) = A1e!3t + A2e!6t

i0 0 +( ) = A1 + A2 , di0 0 +( )dt

= !3A1 ! 6A2

A1 + A2 = 0.5

!3A1 ! 6A2 = !172

"#$

%$ ! A1 = !11/ 6 , A2 = 14 / 6

i0 t( ) = !116e!3t + 14

6e!6t , t > 0

53

i0 t( ) = !116e!3t + 14

6e!6t , t > 0

iL 0 !( ) = 0.5A i0 !( ) = 0

54

Examples: Tutorial Chapter 7