LECTURE 11 T RESPONSE OF 2ND-ORDER CIRCUITS (NATURAL …vivekv/ELL100/L11_VV.pdf · 2020. 1. 23. · INTRODUCTION 12 • This lecture deals with the RLC circuits containing both an

ELL 100 - Introduction to Electrical Engineering

LECTURE 11:

TRANSIENT RESPONSE OF 2ND-ORDER CIRCUITS

(NATURAL RESPONSE)

INSIGHT AND REAL LIFE APPLICATIONS

Application of RC/RL/RLC

Antennas Can be modeled as RLC circuit


In-rush current limiters Electrical circuit breakers




Electronic oscillators/clock circuits



Lightning arresters



Surge arrester Line trap circuit



Electric Fan Ceiling fan wiring diagram



Tube light RL choke coil



RLC projector lamp Electric light dimmer



LCR meters


Application of RC/RL/RLC: Electronic Filters

High-pass RC High-pass RL Series RLC Band-pass

Low-Pass RL Filters Wide-Band Band-Pass Filters

INTRODUCTION

12

• This lecture deals with the RLC circuits containing both an inductor

and a capacitor, which are 2nd-order circuits

• 2nd-order circuit responses are described by 2nd-order differential

equations (containing a double derivative w.r.t time).

• The response of RLC circuits with DC sources and switches consist of

a natural response and a forced response:

v(t) = vf (t)+vn (t)

• The complete response must satisfy both the initial conditions and the

final conditions of the forced response.

BASIC CONCEPTS

13

Finding initial and final values

• Objective: Find v(0), i(0), dv(0)/dt, di(0)/dt, i(∞), v(∞)

• Two key points: (a) The direction of the current i(t)

(b) the polarity of voltage v(t).

• v and i are defined according to the passive sign convention.

BASIC CONCEPTS

14

• The capacitor voltage is always continuous:

v(0+) = v(0-)

• The inductor current is always continuous:

i(0+) = i(0-)

Learning objectives:

• Determine the natural responses of parallel and series RLC circuits.

• To understand the initial conditions in an RLC circuit and use them

to determine the expansion coefficients of the complete solution.

• What do the response curves of over, under, and critically-damped

circuits look like? How to choose R, L, C values to achieve

fast switching or to prevent overshooting damage.

15

Problem 1: The switch was closed for a long time and opened at t = 0.

Find: (a) i(0+), v(0+), (b) di(0+) ∕dt, dv(0+) ∕dt, (c) i(∞), v(∞).

16

i(0-) = (12)/(4 + 2) = 2 A, v(0-) = 2i(0-) = 4 V

(a) t = 0- (b) t = 0+ (c) t → ∞

As the inductor current and the capacitor voltage cannot change abruptly,

i(0+) = i(0-) = 2 A, v(0+) = v(0-) = 4 V

At t = 0+, the switch is open, same current flows through both the inductor

and capacitor. Hence, iC(0+) = i(0+) = 2 A

17

Since,

Also,

vL is obtained by applying KVL in the loop,

−12 + 4i(0+) + vL(0+) + v(0+) = 0

=> vL(0+) = 12 − 8 − 4 = 0

=>

;

(0 )(0 ) 220V/s

0.1

cc

c

idv dvi C

dt dt C

idv

dt C

; LLvdi di

v Ldt dt L

(0 )(0 ) 00A/s

0.25

Lvdi

dt L

t = 0+

18

For t > 0,

• The circuit first undergoes through some transients.

• As t → ∞, the circuit reaches steady state again.

The inductor acts like a short circuit and the

capacitor like an open circuit.

i(∞) = 0 A, v(∞) = 12 V

t → ∞

THE SOURCE-FREE SERIES RLC CIRCUIT

19

Consider the given series RLC circuit,

• The circuit is being excited by the

energy initially stored in the

capacitor and inductor.

• The energy is represented by the

initial capacitor voltage V0 and

initial inductor current I0.

• Thus, at t = 0,0

0

0

1(0)

(0)

v i dt VC

i I

20

On applying KVL,

This is a second-order differential equation for the current i in the circuit.

The initial values and the first derivative are related as,


2

2

1( ) 0

on differentiating,

0

tdi

Ri L i ddt C

d i R di i

dt L dt LC

0 0 0

(0) (0) 1(0) 0; ( )

di diRi L V RI V

dt dt L

21

Look for solutions of the form i = Aest where, A and s are constants.

Substitute this into differential equation,

Thus,

This quadratic equation is known as the characteristic equation


2

2

0

10

stst st

st

AR AseAs e se

L LC

RAe s s

L LC

2 1 0R

s sL LC

22

The roots of the equation dictate the character of i and they are given as,

A more compact way of expressing the roots is,

where,

2

1

2

2

1

2 2

1

2 2

R Rs

L L LC

R Rs

L L LC

2 2

1 0

2 2

2 0

s

s

0

1;

2

R

L LC SERIES RLC CIRCUIT

23

• The roots s1 and s2 are called natural frequencies, measured in

nepers per second (Np/s), because they are associated with the

natural response of the circuit

• ω0 is known as the resonant frequency or strictly as the undamped

natural frequency, expressed in radians per second (rad/s);

• α is the neper frequency (or damping constant) expressed in

nepers per second.

The expression given is modified in terms of α and ω0,


2

2 2

0

10

2 0

Rs s

L LC

s s

24

The two values of s indicate that there are two possible solutions for i,

• A complete or total solution would therefore require a linear

combination of i1 and i2.

• Thus, the natural response of the series RLC circuit is

where the constants A1 and A2 are determined from the initial values

i(0) and di(0) ∕dt.

Three types of solutions are inferred:

1. If α > ω0, we have the over-damped case.

2. If α = ω0, we have the critically-damped case.

3. If α < ω0, we have the under-damped case.


1 2

1 1 2 2;s t s ti A e i A e

1 2

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

• Overdamped Case (α > ω0)

α > ω0 implies R2 > 4L ∕ C

When this happens, both roots s1 and s2 are negative and real.

The response is given as,

Overdamped response

25

1 2

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

2 2

1 0

2 2

2 0

s

s

0

1;

2

R

L LC

26

• Critically Damped Case (α = ω0)

α = ω0 implies R2 = 4L ∕ C Thus

For this case,

where A3 = A1 + A2 . But this cannot be the solution, because the two

initial conditions cannot be satisfied with the single constant A3.

When α = ω0 = R ∕ 2L, then

1 22

Rs s

L

1 2 3( )t t ti t A e A e A e

22

22 0

0

d i dii

dt dt

d di dii i

dt dt dt

let, , then,

0

dif i

dt

dff

dt

this is a first-order differential

equation with solution f = A1e−αt,

where A1 is a constant.

27

The original equation for current i becomes,

1

1 1

1 2 1 2

;

on integration,

; ( )

t

t t t

t t

dii A e

dt

di de e i A e i A

dt dt

e i A t A i A t A e

Hence, the natural response of the critically damped circuit is a sum of

two terms: a negative exponential and a negative exponential multiplied

by a linear term.

Critically-damped response

28

• Underdamped Case (α < ω0)

α < ω0 implies R2 < 4L ∕ C. The roots may be written as,

where,

Both ω0 and ωd are natural frequencies because they help determine the

natural response; while ω0 is called the undamped natural frequency,

ωd is called the damped natural frequency. The natural response is

2 2

1 0

2 2

2 0

( )

( )

d

d

s j

s j

2 2

01; dj

( ) ( )

1 2

1 2

( )

(

d d

d d

j t j t

j t j tt

i t A e A e

e A e A e

)

29

By using Euler’s identities,

Note:

It is clear that the natural response for this case is exponentially damped

but also oscillatory in nature. The response has a time constant of 1/α and

a period of T = 2π/ωd.

1 2

1 2 1 2

1 2

( ) [ (cos sin ) (cos sin )]

[( ) cos ( )sin ]

( ) [ cos sin ]

t

d d d d

t

d d

t

d d

i t e A t j t A t j t

e A A t j A A t

i t e B t B t

Under-damped response

30

Conclusions:

(i) The damping effect is due to the presence of resistance R.

• The damping factor α determines the rate at which the response is

damped.

• If R = 0, then α = 0 and we have an LC circuit with as the

undamped natural frequency. The response in such a case is

undamped and purely oscillatory.

• The circuit is said to be lossless because the dissipating or damping

element (R) is absent.

• By adjusting the value of R, the response may be made undamped,

overdamped, critically damped or underdamped.

1/ LC

31

Conclusions:

(ii) Oscillatory response is possible due to the presence L and C.

• The damped oscillation exhibited by the underdamped response is

known as ringing. It stems from the ability of the storage elements

L and C to transfer energy back and forth between them.

(iii) The overdamped has the longest settling time because it takes the

longest time to dissipate the initial stored energy.

• If we desire the fastest response without oscillation or ringing, the

critically damped circuit is the right choice.

32

Problem 2: For the given circuit, R= 40 Ω, L = 4 H, and C = 1/4 F.

Calculate the characteristic roots of the circuit. Is the natural response

overdamped, underdamped, or critically damped?

2

1

2

2

5 5 1

5 5 1

s

s

33

The roots are,

s1 = -0.101; s2 = -9.899

Since α > ω0, we conclude that the

response is overdamped.

This is also evident from the fact that

the roots are real and negative.

2 2

1 0

2 2

2 0

s

s

0

15; 1

2

R

L LC R= 40 Ω, L = 4 H, C = 1/4 F =>

34

Problem 3: Find i(t) for t > 0. Assume that the circuit has reached

steady state before the switch is opened.

35

Solution:

(a) for t < 0 (b) for t > 0.

Under-damped response

10(0) 1A; (0) 6 (0) 6V

4 6i v i

0

19; 10

2

R

L LC

2

1

2

2

1,2

9 9 100

9 9 100

9 4.359

s

s

s j

36

Hence,

A1 and A2 are found using the initial conditions.

At t = 0, i(0) = 1 = A1

9

1 2( ) [ (cos 4.359 ) (sin 4.359 )]ti t e A t A t

0

9

1 2

9

1 2

1[ (0) (0)] 6 /

Taking the derivative of ( )

9 [ (cos 4.359 ) (sin 4.359 )]

(4.359)[ (sin 4.359 ) (cos 4.359 )]

t

t

t

diRi v A s

dt L

i t

die A t A t

dt

e A t A t

t > 0

37

Substituting the values of A1 and A2 yields the complete solution as,

for t > 0

1 2

1

2

2

6 9( 0) 4.359( 0 )

substituting 1,

6 9 4.359

0.6882

Α A

Α

A

A

9( ) [(cos 4.359 ) 0.6882(sin 4.359 )]Ati t e t t

38

• Assume initial inductor current I0 and

initial capacitor voltage V0,

• Three elements are in parallel, they

have the same voltage v across them.

• Applying KCL at the top node gives,

THE SOURCE-FREE PARALLEL RLC CIRCUIT

0

0

0

1(0) ( )

(0)

i I v t dtL

v V

01

( ) 0v dv

v d CR L dt

39

Taking another derivative w.r.t ‘t’,

The characteristic equation is given as,

THE SOURCE-FREE PARALLEL RLC CIRCUIT

2

2

1 10

d v dvv

dt RC dt LC

2

2

1,2

2 2

1,2 0 0

1 10

Roots of the characteristic equation are,

1 1 1

2 2

1 1 , ,

2

s sRC LC

sRC RC LC

sRC LC

40

1 2

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

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

2 2

1,2 0

1 2

;

v( ) ( cos sin )

d d

t

d d

s j

t e A t A t

• Overdamped Case (α > ω0)

α > ω0 => L/C > 4R2. The roots of the characteristic equation are real

and negative. The response is,

• Critically Damped Case (α = ω0)

α = ω0 => L/C = 4R2. The roots are real and equal so that the

response is,

• Underdamped Case (α < ω0)

α < ω0 => L/C < 4R2. In this case the roots are complex conjugates

expressed as

41

• The constants A1 and A2 in each case can be determined from the

initial conditions i.e. v(0) and dv(0) ∕dt

00

0 0

(0)0

(V )(0)

V dvI C

R dt

RIdv

dt RC

42

Problem 4: In the given parallel circuit, find v(t) for t > 0, assuming

v(0) = 5 V, i(0) = 0, L = 1 H, and C = 10 mF.

Consider three cases: 1) R = 1.923 Ω, 2) R = 5 Ω, and 3) R = 6.25 Ω.

Solution: Case 1: R = 1.923 Ω, L = 1 H, C = 10 mF

2 2

1 0

2 2

2 0

2 50

1 2

1 2

2 50

1 2

( )

Applying intial conditions,

v(

2

50

(0) (0) (0)260

on differentiating

0)=5=

,

2 50

t t

t t

s

s

dv v Ri

v t A e A

dt RC

d

e

A A

A e A ev

dt

43

3

03

1 126;

2 2 1.923 10 10

1 110

1 10 10

RC

LC

Since α > ω0, the response is overdamped.

The roots of the characteristic equation are

44

2 2

1 0

2 2

2 0

2 50

1 2

1 2

2 50

1 2

( )


v(

2

50

(0) (0) (0)260

on differentiating

0)=5=

,

2 50

t t

t t

s

s

dv v Ri

v t A e A

dt RC

d

e

A A

A e A ev

dt

1 2

1 2

2 50

at 0, -260=-2 50

The obtained values are,

0.2083, 5.208

( ) 0.2083 5.208t t

t A A

A A

v t e e

Case 1:

R = 1.923 Ω, L = 1 H, C = 10 mF

45

CASE 2: R = 5 Ω, L = 1 H, C = 10 mF

The response is critically-damped.

3

03

1 110;

2 2 5 10 10

1 110

1 10 10

RC

LC

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


1

10

1 2 2

1 2

10

v(0)=5=

( 10 1

(0) (0)

0 )

at

(0)100

o

0, -100=-10

The obtained values are,

( ) (5 50

n differentiating,

)

t

t

A

A A t A e

t A A

v t t e

dv v Ri

dt RC

V

dv

dt

46

CASE 3: R = 6.25 Ω, L = 1 H, C = 10 mF

The response is underdamped.

Obtain A1 and A2 from initial conditions:

3

03

1 18;

2 2 6.25 10 10

1 110

1 10 10

RC

LC

1,2

8

1 2

8 6

v( ) ( cos 6 sin 6 )

d

t

s j j

t e A t A t

1

(0)

v(0)

(0) (0)

=

80

=5

dv v Ri

dt C

A

R

1 2

1 2

8

at 0, -80=-8 6

5; 6.667

( ) (5cos 6 6.667sin 6 ) t

t A A

A A

v t t t e V

47

Problem 5: Find v(t) for t > 0 in the RLC circuit. Assume

that the switch has been open for a long time before closing.

48

Solution:

• For t < 0, the switch

is open.

• Inductor acts like a

short circuit,

capacitor behaves

like an open circuit.

• The initial voltage

across the capacitor

is the same as the

voltage across the

50-Ω resistor.

50(0) (40) 25;

30 50

40(0) 0.5

30 50

v

i A

(0) (0) (0)0

dv v Ri

dt RC

49

For t > 0, the switch is

closed. The voltage source

along with the 30-Ω resistor

is separated from the rest of

the circuit.

Since α > ω0, we have the

overdamped response.

6

0

1 1500;

2 2 50 20 10

1354

RC

LC

2 2

1,2 0

1 2

500 354

854; 146

s

s s

854 146

1 2( )t tv t A e A e

50

1 2

854 146

1 2

on differentiating,


v(0)=25=

854 146t td

A A

A e A ev

dt

1 2

1 2

(0)0 854 146

5.156; 30.16

dvA A

dt

A A

854 146( ) 5.156 30.16t tv t e e V

Thus, the complete solution is given as,

51

Problem 6:For the circuit, find:

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

(b) di(0+) ∕ dt and dv(0+) ∕ dt,

(c) i(∞) and v(∞)

Ans: (a) i(0+)= 2A, v(0+)=12V

(b) di(0+) ∕ dt = -4A/s, dv(0+) ∕ dt= -5V/s

(c) i(∞)=0A, v(∞)=0V

EXERCISE AND NUMERICAL EXAMPLES

Finding Initial and Final Values

52

Problem 7: Refer to the circuit. Calculate:

(a) iL(0+), vC(0

+), and vR(0+)

(b) diL(0+) ∕dt, dvC(0

+) ∕dt, and dvR(0+) ∕dt

(c) iL(∞), vC(∞), and vR(∞).

Ans: (a) iL(0+)=0A, vC(0

+)=-10V, vR(0+)=0V

(b) diL(0+) ∕dt= 0A/s ,dvC(0

+) ∕dt=8V/s, dvR(0+) ∕dt=8V/s

(c) iL(∞)=400mA, vC(∞)=6V, = vR(∞)=16V


53

Problem 8:Refer to the circuit. Determine:

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

(b) di ∕(0+)dt and dv(0+) ∕dt

(c) i(∞) and v(∞).

Ans: (a) i(0+) = 0A ,v(0+) = 0V

(b) di ∕(0+)dt = 4A/s, dv(0+) ∕dt =0V/s

(c) i(∞) = 2.4A, v(∞) = 9.6V


54

Problem 9: In the circuit, find:

(a) vR(0+) and vL(0

+)

(b) dvR(0+) ∕dt and dvL(0

+) ∕dt

(c) vR(∞) and vL(∞)

Ans: (a) vR(0+) =0V , vL(0

+) =0V

(b) dvR(0+) ∕dt =0V/s , dvL(0

+) ∕dt = Vs/(CRs)

(c) vR(∞) = [R/(R + Rs)]Vs, vL(∞) =0V


55

Source-Free Series RLC Circuit

Problem 10: The current in an RLC circuit is described by

If i(0) = 10 A and di(0) ∕dt = 0, find i(t) for t > 0.

Problem 11: The natural response of an RLC circuit is described by the

differential equation,

for which the initial conditions are v(0) = 10 V and dv(0) ∕dt = 0. Solve for

v(t).


2

210 25 0

d i dii

dt dt

2

22 0

d v dvv

dt dt

Ans: i(t) = [(10 + 50t)e-5t] A

Ans: v(t) = [(10 + 10t)e-t] V

56

Problem 12: The switch moves from position A to position B at t = 0

(please note that the switch must connect to point B before it breaks the

connection at A, a make-before-break switch). Let v(0) = 0, find v(t) for t

> 0.


Ans: v(t) =5.333e–2t–5.333e–0.5t V

57

Problem 13: In the circuit, the switch instantaneously moves from

position A to B at t = 0. Find v(t) for all t ≥ 0.


Ans: v(t) = [21.55e-2.679t – 1.55e-37.32t] V

58

Problem 14: The switch in the circuit has been closed for a long time but

is opened at t = 0. Determine i(t) for t > 0.


Ans: i(t) = (15cos(2t) + 15sin(2t))e-2t A

59

Source-Free Parallel RLC Circuit

Problem 15: For the network, what value of C is needed to make the

response underdamped with unity neper frequency (α = 1)?


Ans: C = 40 mF

60

Problem 16: The switch moves from position A to position B at t = 0

(please note that the switch must connect to point B before it breaks the

connection at A, a make-before-break switch). Determine i(t) for t > 0.


Ans: i(t) = e–5t[4cos(19.365t) + 1.0328sin(19.365t)] A

61

Problem 17: A source free RLC has R= 1Ω, C=1nF and L= 1pF.

Calculate (a) calculate α and ω0(b) s1and s2(c)What is the form of inductor current response for t>0.

Ans: (a) α =5x108 s-1 , ω0=3.16x1013 rad/s

(b) s1and s2=

(c) The circuit is underdamped since α

62

Problem 18:Assuming R=2kΩ, design a parallel RLC circuit that has the

characteristic equation

Ans: L=20H, C=50nF

Problem 19: Calculate io(t) and vo(t) for t > 0.


2 6100 10 0s s

Ans: vo(t) =(24cos1.9843t + 3.024sin1.9843t)e-t/4 V

io(t) =[– 12.095sin1.9843t]e–t/4 A.

REFERENCES

63

[1] Charles K. Alexander and Matthew N. O. Sadiku,

“Fundamentals of Electric Circuits”, 6th Ed., McGraw Hill,

Indian Edition, 2013.

[2] William H. Hayt, Jr., Jack E. Kemmerly, Steven M.

Durbin, “Engineering Circuit Analysis” 8th Ed., McGraw-

Hill, New York, 2012.

THANK YOU

Documents

LECTURE 11 T RESPONSE OF 2ND-ORDER CIRCUITS (NATURAL …vivekv/ELL100/L11_VV.pdf · 2020. 1. 23. · INTRODUCTION 12 • This lecture deals with the RLC circuits containing both an