NATURAL AND STEP RESPONSES OF NATURAL AND STEP RESPONSES OF RLC CIRCUITSRLC CIRCUITS

Yaser_rahmati@hotmail.com

NATURAL RESPONSE OF A PARALLEL RALC CIRCUIT

V

R LVd I C

dV

dt

R

dV

dt

V

LC

d V

dt

d V

dt RC

dV

dt

V

LC

t

1

0

10

10

00

2

2

2

2

Ordinary, second-order differential equation with constant coefficients. Therefore, this circuit is called a second-order circuit.

CL R V

++V0

I0iciL iR

GENERAL SOLUTION OF SECOND-ORDER DIFFERENTIAL EQUATIONS

Assume that the solution of the differential equation is of exponential form V=Aest where A and s are unknown constants.

As eAs

RCe

Ae

LC

Ae ss

RC LC

st stst

st

2

2

0

10

( )

This equation can be satisfied for all values of t only if A=0 or the term in parentheses is zero.

A=0 cannot be used as a general solution because to do so implies that the voltage is zero for all time-a physical impossibility if energy is stored in either inductor or capacitor.

THE CHARACTERISTIC EQUATION

ss

RC LC2 1

0 is called the characteristic equation

The two roots of the characteristic equation are

sRC RC LC

sRC RC LC

1

2

2

2

1

2

1

2

1

1

2

1

2

1

If either s1 or s2 is substituted into Aest, the assumed solution satisfies the differential equation, regardless of the value of A

V Ae V A es t s t1 1 2 2

1 2 ,

These two solutions as well as their summation V=V1+V2 satisfy the differential equation

V Ae A es t s t 1 21 2

FREQUENCIES

The behavior of V(t) depends on the values of s1 and s2. Writing the roots in a notation as widely used in the literature

s s

LC

12

02

22

02

02 1

,

=1

2RC

The exponent of e must be dimensionless, so both s1 and s2 (and hence and 0) must have the dimension of the reciprocal of time (frequency). s1 and s2 are referred to as complex frequencies, is called the neper frequency, and 0 is called the resonant radian frequency. They have the unit radians per second (rad/s)

The nature of the roots s1 and s2 depends on the values of and 0. There are three possible cases:

1) If 02<2, both roots will be real and distinct. The voltage

response is said to be overdamped.

2) If 02>2, both roots will be complex and, in addition, will be

conjugates of each other. The voltage response is said to be underdamped.

3) If 02=2, roots will be real and equal. The voltage response

is said to be critically damped.

EXAMPLE

CL R V

+ Find the roots of the characteristic equation if R=200 , L=50 mH, and C=0.2 F.

1

2

10

400 0 2125 10

1 10 10

50 0 210

125 10 15625 10 10 5000

125 10 15625 10 10 20000

64

3 68 2

14 8 8

24 8 8

RC

LCs

s

s

( )( . ).

( )

( )( . )/

. .

. .

rad / s

rad

rad / s

rad / s

02 2

Overdamped

+V0

I0

Repeat the problem for R=312.5

1

2

10

625)(0 28000

1 10 10

50 0 210

8000 0 64 10 10 8000 6000

8000 0 64 10 10 8000 6000

6

3 68 2

18 8

28 8

RC

LCs

s j

s j

( . )

( )

( )( . )/

.

.

rad / s

rad

rad / s

rad / s

02 2

Underdamped

Find the value of R for a critically damped circuit.

For critical damping, 2=02

1

2

110

110

10

2 10 0 2250

28

46

4

RC LC

RCR

( )( . )

THE OVERDAMPED RESPONSE

When the roots of the characteristic equation are real and distinct, the response is said to be overdamped in the form

VtAeAestst

() 1212

The constant A1 and A2 are to be determined by the initial conditions V(0+) and dV(0+)/dt.

V A AdV

dts A s A

V VdV

dt

i

C

iV

RI

c

c

( ) ,( )

( )( ) ( )

( )

00

00 0

0

1 2 1 1 2 2

0

00

EXAMPLE

CL R V

+ Fin V(t), if R=200 , L=50 mH, and C=0.2 F. V0=12V, I0=30 mA.

+V0

I0iciL iR

i I mA

iV

R

V

RmA

i i i mA

dV

dt

i

CkV s

L

Rc

c L R

c

( )

( )( )

( ) ( ) ( )

( ) ( )

./

0 30

00 12

20060

0 0 0 90

0 0 90 10

0 2 10450

0

0

3

6

From the previous example, we determined s1=-5000 rad/s and s2=-20000 rad/s. Then

V t Ae A e

V A A

dV t

dtAe A e

dV

dtA A

A V A V

V t e e V t

t t

t t

t t

( )

( )

( )

( )

( ) ( )

15000

220000

1 2

15000

220000

1 23

1 2

5000 20000

0 12

5000 20000

05000 20000 450 10

14 26

14 26 0

0 0.5 1 1.5 2 2.5

x 10-4

-6

-4

-2

0

2

4

6

8

10

12

i tV t

e e mA t

i t CdV t

dte e

e e mA t

i t i t i t

e e mA t

Rt t

ct t

t t

L R c

t t

( )( )

( )

( )( )

. ( )

( )

( ) ( ) ( )

( )

20070 130 0

0 2 10 70000 520000

14 104 0

56 26 0

5000 20000

6 5000 20000

5000 20000

5000 20000

THE UNDERDAMPED RESPONSE

When 02>2, the roots of the characteristic equation are complex,

and the response is underdamped.

s j

s j

d

d d

1 02 2

2 02 2

( )

d is called the damped radian frequency.

V t A e A e

A e e A e e

e A t jA t A t jA t

e A A t j A A t

j t j t

t j t t j t

td d d d

td d

d d

d d

( )

( cos sin cos sin )

[( )cos ( )sin ]

( ) ( )

1 2

1 2

1 1 2 2

1 2 1 2

A1 and A2 are complex conjugates. Therefore, their sum is a real number and their difference is imaginary. Then j(A1-A2) is also a real number. Denoting B1=A1+A2, and B2=j(A1-A2)

V t e B t B ttd d( ) ( cos sin ) 1 2

DAMPING FACTOR

The trigonometric functions indicate that the response is oscillatory; that is, the voltage alternates between positive and negative values. The rate at which the voltage oscillates is fixed with d. The rate at which the amplitude decreases is determined by . Because determines how quickly the amplitude decreases, it is called as the damping factor. If there is no damping, =0 and the frequency of oscillations is 0. When there is a dissipative element, R, in the circuit, is not zero and the frequency of oscillations is, d, less than 0. Thus, when is not zero, the frequency of oscillation is said to be damped.

EXAMPLE

CL R V

++V0

I0iciL iR Find V(t) if R=20 k, L=8H, C=0.125F,

V0=0, and I0=-12.25 mA

1

2

10

2 20 10 0125)200

1 10

8 0125)10

979 8

200 979 8

200 979 8

6

3

63

202 2

RC

LC

j j

j j

d

d

d

( ) ( .

( .

.

.

.

rad / s

rad / s

rad / s

s rad / s

s rad / s

0

02

1

2

V(0+)=V0=0, then iR(0+)=V(0+)/R=0.

ic(0+)=-iL(0+)=12.25 mA

dV

dt

B B V

V t e t V t

d

t

( ) .

.

,

( ) sin .

0 12 25 10

0125 1098000

098000

100

100 979 8 0

3

6

1 2

200

V / s

THE CRITICALLY DAMPED RESPONSE

The second-order circuit is critically damped when 2=02. The

two roots of the characteristic equation are equal

s sRC1 2

1

2

For a critically damped circuit, the solution takes the following form

V t D te D et t( ) 1 2

D1 and D2 are constant which must be determined using the initial conditions V(0+) and dV(0+)/dt

EXAMPLE

CL R V

++V0

I0iciL iR Determine the value of R for a critically

damped response when, L=8H, C=0.125F, V0=0, and I0=-12.25 mA

From the previous example 02=106. Then

101

240003

RCR

Again, from the previous example V(0+)=0 and dV(0+)/dt=98000 V/s Then D1=0 and D2=98000 V/s.

V t te V tt( ) 98000 01000

THE STEP RESPONSE OF A PARALLEL RLC CIRCUIT

IC

L R V+t=0 ic iL iR

i i i I

iV

RC

dV

dtI

L R c

L

V Ldi

dt

dV

dtL

d i

dt

iL

R

di

dtLC

d i

dtI

d i

dt RC

di

dt

i

LC

I

LC

L L

LL L

L L L

2

2

2

2

2

2

1

The equation describing the step response of a second-order circuit is a second-order differential equation with constant coefficients and with a constant forcing function. The solution of this differential equation equals the forced response which is in the same form of the forcing function (constant for a step input) plus a response function identical in form to the natural response. Thus, the solution for the inductor current is in the form

i IL f

function of the same form

as the natural response

EXAMPLE

IC

L R V+t=0 ic iL iR C=25nF, L=25mH, R=400

The initial energy in the circuit is zero. I=24 mA. Find iL(t)

Since there is no initial energy in the circuit, iL(0+)=0 and V(0+)=0V(0+)=L[diL(0+)/dt]=0, then diL(0+)/dt=0

02

128

94

802

1 10

25)(25)16 10

1

2

10

2 400 25)5 10

25 10

LC

RC

(

( )( rad / s

2

Overdamped circuit

s

s

14 4

24 4

5 10 3 10 20000

5 10 3 10 80000

rad / s

rad / s

As t , circuit reaches dc steady state where inductor is short and capacitor is open. All of the input current flows through the inductor. Then If=24 mA.

i t Ae A e A

i A A

di

dtAe A e

di

dtA A

Lt t

L

L t t

L

( )

( )

( )

24 10

0 24 10

20000 80000

020000 80000 0

31

200002

80000

31 2

120000

280000

1 2

A mA A mA

i t e e mA tLt t

1 2

20000 80000

32 8

24 32 8 0

( ) ( )

EXAMPLE: If the resistor in the circuit is increased to 625, find iL(t) in the circuit.

Since L and C remain fixed, resonant frequency has the same value. But neper frequency decreases to 3.2x104 rad/s. With these values circuit is underdamped with complex conjugate roots.

s j

s j

i t I B e t B e t

I A

i B

di

dte B t B t

e B t B t

di

L ft

dt

d

f

L

L td d

td d

L

14 4

24 4

1 2

3

31

1 2

1 2

32 10 2 4 10

32 10 2 4 10

24 10

0 24 10 0

0

. .

. .

( ) cos sin

( )

( cos sin )

( sin cos )

(

rad / s

rad / s

+ d

)

dtB Bd 1 2 0

B mA B mA

i t e t t mA tLt

1 2

32000

24 32

24 24 24000 32 24000 0

( ) ( ( cos sin ))

EXAMPLE: Find iL(t) if R=500

The resonant frequency remains the same, but neper frequency becomes 4x104 rad/s. These values correspond to critical damping. Roots of the characteristic equation are real and equal at s=-40000

i t I D te D e AL ft t( )

140000

240000

I A

i D

di

dte D t D D e

di

dtD D

D mA s D mA

i t te e mA t

f

L

L t t

L

Lt t

24 10

0 24 10 0

00

960000 24

24 960000 24 0

3

32

1 2 1

2 1

1 2

40000 40000

( )

( )

( )

/

( ) ( )

0 0.5 1 1.5 2 2.5

x 10-4

0

0.005

0.01

0.015

0.02

0.025

Underdamped Critically damped

Overdamped

EXAMPLE

IC

L R V+t=0 ic iL iR

C=25nF, L=25mH, R=500 The initial current in the inductor is 29 mA, the initial voltage across the capacitor is 50 V. I=24 mA. Find iL(t) and V(t)

From the previous example, we know that this circuit is critically damped with s1=s2=-40000 rad/s

i i mA

V V V

L L

c c

( ) ( )

( ) ( )

0 0 29

0 0 50

+

Vc

V V Ldi

dt

di

dtA s

i t I D te D e A

I A

i D D A

di

dtD D D A s

i t te

L cL L

L ft t

f

L

L

Lt

( ) ( )( ) ( )

/

( )

( )

( )/

( ) ( .

0 00 0 50

25 102000

24 10

0 24 10 29 10 5 10

02000 2200

24 2 2 10 5

3

140000

240000

3

32

32

3

1 2 1

6 40000

e mA tt 40000 0)

V t Ldi

dt

te

e e

te e V

L

t

t t

t t

( )

( )[( . )( )

. ( ) ]

.

25 10 2 2 10 40000

2 2 10 5 40000 10

2 2 10 50

3 6 40000

6 40000 40000 3

6 40000 40000

THE NATURAL AND STEP RESPONSE OF A SERIES RLC CIRCUIT

R L

C +

V0

I0

i

Ri Ldi

dt Cid V

Rdi

dtL

d i

dt

i

C

t

1

0

0

0 0

2

2

d i

dt

R

L

di

dt

i

LC

sR

Ls

LC

2

2

2

0

10

sR

L

R

L LC

R

L LC

1 2

22

02

0

2 2

1

2

1

,

rad / s rad / s

These equations are in the same form that of the equations for the parallel RLC circuit. Therefore, the solution will be overdamped, critically damped, or underdamped depending on relative magnitudes of the resonant frequency and neper frequency.

i t Ae A e

i t B e t B e t

i t D te D e

s t s t

td

td

t t

( )

( ) cos sin

( )

1 2

1 2

1 2

1 2 (overdamped)

(underdamped)

(critically damped)

R L

C +

Vci

+ VR + VL

+V

t=0V Ri L

di

dtV

i CdV

dt

di

dtC

d V

dt

c

c c

2

2

d V

dt

R

L

dV

dt

V

LC

V

LC

V t V Ae A e

V t V B e t B e t

V t V D te D e

c c c

c fs t s t

c ft

dt

d

c ft t

2

2

1 2

1 2

1 2

1 2

( )

( ) cos sin

( )

(overdamped)

(underdamped)

(critically damped)

t=0

+

100V

100mH

+

Vc 560i

The capacitor is initially charged to 100V. At t=0, switch closes. Find i(t) and Vc(t) for t0

02

3 68

3

602

02 2

1 10 10

100 0110

2

560

2 10010 2800

7 84 10

9600

LC

R

L

( )( )

( . )

( )

.

rad / s

Underdamped

rad / s

2

d

i t B e t B e t

i B

V Ri Ldi

dtdi

dt

V

LA s

di

dtB e t B e t

di

dtB B A

td

td

c

c

t t

( ) cos sin

( )

( ) ( )( )

( ) ( )/

sin cos

( )

1 2

1

3

22800

22800

2 2

0 0

0 00

0

0 0 100

10010 1000

2800 9600 9600 9600

09600 1000

1000

9600

i t e t t

V t iR Ldi

dt

V t t t e V t

t

c

ct

( ).

sin

( )

( ) ( cos . sin )

1

9 69600 0

100 9600 2917 9600 0

2800

2800

A

EXAMPLE

0.4F

+

Vc

+48V

t=0

280 0.1H There is no initial stored energy in the circuit at t=0. Find Vc(t) for t0.

s

j

s j

V t V B e t B e t tc cft t

1

2 6

2

11400

21400

280

2 01

280

0 2

10

01 0 4

1400 4800

1400 4800

4800 4800 0

( . ) . ( . )( . )

( ) cos sin ,

rad / s

rad / s

As t , circuit reaches dc steady state (inductor short, capacitor open). Thus, Vcf=48V

VdV

dt

i

C

i

C

V B B V

dV

dte B t B t

e B t B t

cc c L

c

c t

t

( ) ,( ) ( ) ( )

( )

( cos sin )

( cos )

0 00 0 0

0

0 48 0 48

1400 4800 4800

4800 4800 4800

1 1

14001 2

14002 1

dV

dtB B

B V

V t e t e t V t

c

ct t

( )

( ) ( cos sin ) ,

04800 1400 0

14

48 48 4800 14 4800 0

2 1

2

1400 1400

A CIRCUIT WITH TWO INTEGRATING AMPLIFIERS

++Vo1

R1

++Vo

R2

C2

C1

+Vg

Assuming ideal opamps, find the relation between Vo and Vg

00 0

1

11 1

1

1 1

V

RC

d

dtV

dV

dt R CV

g

oo

g( )

00 0

1

1

1 1

1

22

2 21

20

22 2

1

20

21 1 2 2

V

RC

d

dtV

dV

dt R CV

d V

dt R C

dV

dt

d V

dt R C R CV

oo

oo

o

g

( )

EXAMPLE

++Vo1

250k

++Vo

0.1F

+Vg

500k

1F

-5V-9V

9V5V

No energy is stored in the circuit when the input voltage Vg jumps instantaneously from 0 to 25 mV.

Derive the expression for Vo(t) for 0 ttsat.

How long is it before the circuit saturates?

1 1000

250 0140

1 1000

500 12

40 2 25 10 2

2 2

2

1 1

2 2

2

23

0

2

0

R C

R C

d V

dtdV

dtdx t

V xdx t

o

ot

o

t

( )( . )

( )( )

( )( )

0 t tsat

The second integrating amplifier saturates when Vo reaches 9V or t=3s. But it is possible that the first opamp saturates before t=3s. To explore this possibility use the following equation

dV

dt R CV

V t

og

o

1

1 1

3

1

140 25 10 1

( )

Thus, at t=3s, Vo1=-3V. The first opamp does not reach saturation at t=3s. The circuit reaches saturation when the second amplifier saturates.

TWO INTEGRATING AMPLIFIERS WITH FEEDBACK RESISTORS

++Vo1

Ra

++Vo

Rb

C2

C1

+Vg

R1 R2

The reason that the op amp saturates in the integrating amplifier is the feedback capacitor’s accumulation of charge. To overcome this problem, a resistor is placed in parallel with each feedback capacitor.

0 00 0

1

0 00 0

1

11 1

1

1 11

1

1 1 11 1

1 1

1

22

2

1

22 2 2

V

R

V

RC

d

dtV

dV

dt R CV

V

R C

R CdV

dt

V V

R C

V

R

V

RC

d

dtV

dV

dt

V V

R CR C

g

a

oo

oo

g

a

o o g

a

o

b

oo

o o o

b

( )

( )

Let

,

d V

dt

dV

dt R C

dV

dt

dV

dt

V V

R C

V R CdV

dt

R CV

d V

dt

dV

dtV

V

R C R C

o

b

o

o o g

a

o bo b

o

o oo

g

a b

20

22 2

1

1 1

1 1

1 22

2

2

21 2 1 2 1 2

1 1

1 1 1

The characteristic equation is

s s2

1 2 1 2

1 1 10

The roots of the characteristic equation are real

s s11

22

1 1

EXAMPLE

The parameters of the circuit are Ra=100 k, Rb=25 k, R1=500 k, R2=100 k, C1=0.1F, and C2=1F. The power supply of each op amp is 6V. The input voltage jumps from 0 to 250 mV at t=0. No energy is stored in the feedback capacitors at the instant the input is applied. Find Vo(t) for t0.

1 1 1 21 2 2

1 2

2

2

0 05 01

30 200 1000

R C s R C s

V

R C R C

d V

dt

dV

dtV

g

a b

o oo

. .

= 1000 V / s

+

2

The characteristic equation is s2+30s+200=0. Thens1=-20 rad/s and s2=-10 rad/s. The final values of the output is the input voltage times the gain of each stage, because capacitors behave as open circuits as t

V V

V t Ae A e

VdV

dtA V A V

V t e e V t

o

ot t

oo

ot t

( ) ( )

( )

( )( )

( ) ( )

250 10500

100

100

255

5

0 00

0

10 5

5 10 5 0

3

110

220

1 2

10 20

and