06 Yaser Rahmati

8/10/2019 06 Yaser Rahmati

1/54

NATURAL AND STEP RESPONSES OF

RLC CIRCUI TS

[email protected]


2/54

NATURAL RESPONSE OF A

PARALLEL RALC CIRCUIT

V

R LVd I C

dV

dt

R

dV

dt

V

L Cd 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


3/54

GENERAL SOLUTION OF SECOND-

ORDER DIFFERENTIAL EQUATIONS

Assume that the solution of the differential equation is of

exponential form V=Aestwhere A and s are unknown constants.

As e

As

RCe

Ae

LC

Ae ss

RC LC

st st

st

st

2

2

0

10

( )

This equation can besatisfied 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.


4/54

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


5/54

If either s1or s2is 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 21 2 ,

These two solutions as well as their summation V=V1+V2satisfy

the differential equation

V A e A es t s t 1 21 2


6/54

FREQUENCIES

The behavior of V(t) depends on the values of s1and s2.

Writing the roots in a notation as widely used in the literature

s s

LC

1

2

0

2

2

2

0

2

0

2 1

,

=1

2RC

The exponent of e must be dimensionless, so both s1and s2(and

hence and 0) must have the dimension of the reciprocal oftime (frequency). s1and s2are 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)


7/54

The nature of the roots s1and s2depends on the values of and

0. There are three possible cases:

1) If 022, 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.


8/54

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 2

10

125 10 15625 10 10 5000

125 10 15625 10 10 20000

64

3 6

8 2

1

4 8 8

2

4 8 8

RC

LC

s

s

s

( )( . ).

( )

( )( . )

/

. .

. .

rad / s

rad

rad / s

rad / s

0

2 2

Overdamped

+

V0

I0


9/54

Repeat the problem for R=312.5

1

2

10

625)(02

8000

1 10 10

50 0 210

8000 0 64 10 10 8000 6000

8000 0 64 10 10 8000 6000

6

3 6

8 2

1

8 8

2

8 8

RC

LCs

s j

s j

( . )( )

( )( . )/

.

.

rad / s

rad

rad / s

rad / s

0

2 2

Underdamped


10/54

Find the value of R for a critically damped circuit.

For critical damping, 2=02

1

2

110

1 10 102 10 0 2

250

2

8

4

6

4

RC LC

RCR

( )( . )


11/54

THE OVERDAMPED RESPONSE

When the roots of the characteristic equation are real and

distinct, the response is said to be overdamped in the form

V t A e A es t s t

( ) 1 21 2

The constant A1and A2are 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

R

I

c

c

( ) ,( )

( )( ) ( )

( )

00

00 0

0

1 2 1 1 2 2

0

0

0


12/54

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

R

c

c L R

c

( )

( )( )

( ) ( ) ( )

( ) ( )

./

0 30

00 12

20060

0 0 0 90

0 0 90 10

0 2 10450

0

0

3

6


13/54

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

dVdt

A A

A V A V

V t e e V t

t t

t t

t t

( )

( )

( )

( )

( ) ( )

1

5000

2

20000

1 2

1

5000

2

20000

1 23

1 2

5000 20000

0 12

5000 20000

0 5000 20000 450 10

14 26

14 26 0


14/54

0 0.5 1 1.5 2 2.5

x 10-4

-6

-4

-2

0

2

4

6

8

10

12


15/54

i tV t

e e mA t

i t CdV tdt

e e

e e mA t

i t i t i t e e mA t

R

t t

c

t 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


16/54

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 0

2 2

2 0

2 2

( )

dis called the damped radian frequency.


17/54

V t Ae A e

Ae 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

t d d d d

t

d d

d d

d d

( )

( cos sin cos sin )

[( )cos ( )sin ]

( ) ( )

1 2

1 2

1 1 2 2

1 2 1 2

A1and A2are 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 t t d d( ) ( cos sin )

1 2


18/54

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 determinedby . 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 isnot zero, the frequency of oscillation is said to be damped.


19/54

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

12

102 20 10 0125)

200

1 10

8 0125)10

9798

200 979 8

200 979 8

6

3

6

3

2

0

2 2

RC

LC

j j

j j

d

d

d

( ) ( .

( .

.

.

.

rad / s

rad / s

rad / s

s rad / s

s rad / s

0

0

2

1

2


20/54

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


21/54

THE CRITICALLY DAMPED

RESPONSE

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

two roots of the characteristic equation are equal

s s RC1 21

2

For a critically damped circuit, the solution takes the following

formV t D te D et t( ) 1 2

D1and D2are constant which must be determined using the

initial conditions V(0+) and dV(0+)/dt


22/54

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

RC R

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

V t te V t t( ) 98000 01000


23/54

THE STEP RESPONSE OF A

PARALLEL RLC CIRCUIT

IC

L R V

+t=0ic iL iR

i i i I

iV

RC

dV

dtI

L R c

L

V Ldi

dt

dV

dtL

d i

dt

i

L

R

di

dt LC

d i

dt I

d i

dt RC

di

dt

i

LC

I

LC

L L

L

L L

L L L

2

2

2

2

2

2

1


24/54

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 inthe 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


25/54

EXAMPLE

IC

L R V

+t=0ic 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+)=0

V(0+)=L[diL(0+)/dt]=0, then diL(0

+)/dt=0

0

2

12

8

9

4

8

0

2

1 10

25)(25)16 10

1

2

10

2 400 25)5 10

25 10

LC

RC

(

( )(rad / s

2

Overdamped

circuit


26/54

s

s

1

4 4

2

4 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 theinductor. Then If=24 mA.

i t A e A e A

i A A

di

dtA e A e

di

dtA A

L

t t

L

L t t

L

( )

( )

( )

24 10

0 24 10

20000 80000

020000 80000 0

3

1

20000

2

80000

3

1 2

1

20000

2

80000

1 2


27/54

A mA A mA

i t e e mA t Lt 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 3.2x104rad/s. Withthese values circuit is underdamped with complex conjugate

roots.


28/54

s j

s j

i t I B e t B e t

I A

i B

di

dt

e B t B t

e B t B t

di

L f

t

d

t

d

f

L

L t

d d

t

d d

L

14 4

2

4 4

1 2

3

3

1

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


29/54

B mA B mA

i t e t t mA t Lt

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 4x104rad/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( ) 1

40000

2

40000


30/54

I A

i D

didt

e 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

L

t t

24 10

0 24 10 0

00

960000 24

24 960000 24 0

3

3

2

1 2 1

2 1

1 2

40000 40000

( )

( )

( )

/

( ) ( )


31/54

0 0.5 1 1.5 2 2.5

-4

0

0.005

0.01

0.015

0.02

0.025

Underdamped

Critically dampedOverdamped


32/54


33/54

V V Ldi

dt

di

dtA s

i t I D te D e AI A

i D D A

didt

D D D A s

i t te

L c

L L

L f

t t

f

L

L

L

t

( ) ( )( ) ( )

/

( )

( )

( ) /

( ) ( .

0 00 0 50

25 102000

24 10

0 24 10 29 10 5 10

0 2000 2200

24 2 2 10 5

3

1

40000

2

40000

3

3

2

3

2

3

1 2 1

6 40000

e mA t t 40000 0)


34/54

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


35/54

THE NATURAL AND STEP RESPONSE

OF A SERIES RLC CIRCUIT

R L

C+

V0

I0

i

Ri Ldi

dt Cid V

Rdi

dt

Ld i

dt

i

C

t

1

0

0

00

2

2

d i

dt

R

L

di

dt

i

LC

s RL

sLC

2

2

2

0

1 0

sR

L

R

L LC

R

L LC

1 2

2

2

0

2

0

2 2

1

2

1

,

rad / s rad / s


36/54

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 neperfrequency.

i t Ae A e

i t B e t B e t i t D te D e

s t s t

t

d

t

dt t

( )

( ) cos sin( )

1 2

1 2

1 2

1 2 (overdamped)

(underdamped) (critically damped)


37/54

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 eV t V B e t B e t

V t V D te D e

c c c

c f

s t s t

c f

t

d

t

d

c f

t t

2

2

1 2

1 2

1 2

1 2

( )( ) cos sin

( )

(overdamped) (underdamped)

(critically damped)


38/54

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

0

2

3 6

8

3

6

0

2

0

2 2

1 10 10

100 0110

2

560

2 100

10 2800

784 10

9600

LC

R

L

( )( )

( . )

( ).

rad / s

Underdamped

rad / s

2

d


39/54

i t B e t B e t

i B

V Ri L didt

di

dt

V

LA s

didt

B e t B e t

di

dtB B A

t

d

t

d

c

c

t t

( ) cos sin

( )

( ) ( ) ( )

( ) ( )/

sin cos

( )

1 2

1

3

22800

22800

2 2

0 0

0 0 0 0

0 0 100

10010 1000

2800 9600 9600 9600

09600 1000

1000

9600


40/54

i t e t t

V t iR Ldi

dt

V t t t e V t

t

c

c

t

( )

.

sin

( )

( ) ( cos . sin )

1

9 6

9600 0

100 9600 2917 9600 0

2800

2800

A


41/54

EXAMPLE

0.4

F

+

Vc+48V

t=0

280 0.1HThere is no initial stored

energy in the circuit at t=0.

Find Vc(t) for t0.

s

js j

V t V B e t B e t t c cft t

1

2 6

2

1

1400

2

1400

280

2 01

280

0 2

10

01 0 4

1400 48001400 4800

4800 4800 0

( . ) . ( . )( . )

( ) cos sin ,

rad / srad / s


42/54


43/54

dV

dt

B B

B V

V t e t e t V t

c

c

t t

( )

( ) ( cos sin ) ,

04800 1400 0

14

48 48 4800 14 4800 0

2 1

2

1400 1400

A CIRCUIT WITH TWO


44/54

A CIRCUIT WITH TWO

INTEGRATING AMPLIFIERS

++

Vo1

R1

++

Vo

R2

C2C1

+

Vg

Assuming ideal opamps, find the relation between Voand Vg

00 0

1

1

1 1

1

1 1

V

RC

d

dtV

dV

dt R C V

g

o

o

g( )


45/54

00 0

1

1

1 1

1

2

2

2 2

1

2

0

2

2 2

1

2

0

21 1 2 2

V

RC

d

dtV

dV

dt R C V

d V

dt R C

dV

dt

d V

dt R C R C V

o

o

o

o

o

g

( )


46/54

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?


47/54

1 1000

250 0140

1 1000

500 1

2

40 2 25 10 2

2 2

2

1 1

2 2

2

2

3

0

2

0

R C

R Cd V

dt

dV

dt dx t

V xdx t

o

o t

o

t

( )( . )

( )( )

( )( )

0 t tsat


48/54

The second integrating amplifier saturates when Voreaches 9V

or t=3s. But it is possible that the first opamp saturates before

t=3s. To explore this possibility use the following equation

dVdt R C

V

V t

o

g

o

1

1 1

3

1

1 40 25 10 1

( )

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

second amplifier saturates.


49/54

TWO INTEGRATING AMPLIFIERS

WITH FEEDBACK RESISTORS

++

Vo1

Ra

++

Vo

Rb

C2C1

+Vg

R1 R2

The reason that the op amp saturates in the integratingamplifier is the feedback capacitors accumulation of charge.

To overcome this problem, a resistor is placed in parallel

with each feedback capacitor.


50/54

0 00 0

1

0 0 0 0

1

1

1 1

1

1 11

1

1 1 1

1 1

1 1

1

2

2

2

1

2

2 2 2

V

R

V

RC

d

dtV

dV

dt R C V

V

R C

R CdV

dt

V V

R C

VR

VR

C ddt

V

dV

dt

V V

R CR C

g

a

o

o

o

o

g

a

o o g

a

o

b

oo

o o o

b

( )

( )

Let

,


51/54

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 b

o b

o

o o

o

g

a b

2

0

2

2 2

1

1 1

1 1

1 2

2

2

2

2

1 2 1 2 1 2

1 1

1 1 1


52/54

The characteristic equation is

s s2

1 2 1 2

1 1 10

The roots of the characteristic equation are real

s s11

2

2

1 1


53/54

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 capacitorsat 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 o

o

. .

= 1000 V / s

+

2


54/54

The characteristic equation is s2+30s+200=0. Then

s1=-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

dt

A V A V

V t e e V t

o

o

t t

oo

o

t t

( ) ( )

( )

( )( )

( ) ( )

250 10500

100

100

255

5

0 00

0

10 5

5 10 5 0

3

1

10

2

20

1 2

10 20

and

06 Yaser Rahmati

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