Electrical Circuits (2) Shoubra... · 2016. 4. 11. · Electrical Circuits (2) - Basem ElHalawany 7 Second-Order RLC Transient (Step Response) The Switch “S” is closed at t=0

Electrical Circuits (2)

Lecture 8

Transient Analysis Part(2)

Dr.Eng. Basem ElHalawany

2

The switch “S” is closed at t = 0 Apply KVL to the circuit in figure:

First-Order RL TransientStep-Response

Rearranging and using “D” operator notation :

This Equation is a first order, linear differential equation

1. Complementary (Transient) Solution

2. Particular (Steady-State) Solution

The auxiliary equation is : 0 L

Rm

tL

R

mt AeAei

LR

Time constant

The steady-state value of the current for DC source is : R

VI ss

3

The total solution is:

First-Order RL TransientStep-Response

R

VAei

tL

R

Since The initial current is zero:

R

VA0

The voltage across the resistor is:

The voltage across the inductor is:

4First-Order RL Transient (Discharge)

The RL circuit shown in Figure contains an initial current of (V/R)

The Switch “S” is moved to position”2” at t=0

The solution is the transient (Complementary) part only.

Using the initial condition of the current, we get:

The corresponding voltages across the resistance and inductance are

Electrical Circuits (2) - Basem ElHalawany 5

Examples


Examples

(b) For the two voltage to be equal:each must be 50 volts since the applied voltage is 100,


Second-Order RLC Transient (Step Response)

The Switch “S” is closed at t=0 Applying KVL will produce the following

Integro-Differential equation:

Differentiating, we obtain

This second order, linear differential equation is of the homogeneous type with a particular solution of zero.

The complementary function can be one of three different types according to the roots of the auxiliary equation which depends upon the relative magnitudes of R, L and C.

012

LC

mL

Rm



012

LC

mL

Rm

02 2002 mm

We can Rewrite the auxiliary equation as:

2

0

0

1

2

LC

L

R

frequency natural undamped :

ratio dampling expontial :

0

The roots of the equation (or natural frequencies):

1

1

2

002

2

001

m

m

LCL

R

L

Rm

LCL

R

L

Rm

1

422

1

42

2

2

2

2

1

0

2

2

0

2

1

m

m

9

Case 1: Overdamped,

o

LC22

1

2L

R

1

Natural response is the sum of two decaying exponentials:

tmtm

tr eKeKi21

21

1

1

2

002

2

001

m

m


unequal and real are , 21 mm

Case 2: Critically damped, equal. and real are , 21 mm

o

LC22

1

2L

R

1

021 mm

)()( 211 tBBetxtm

c Use the initial conditions to get the constants

Usually it is reduced to:tm

c etBtx1..)(

10Second-Order RLC Transient (Step Response)

Case 3: Underdamped,

conjugate. andcomplex are , 21 mm

Natural response is an exponentially damped oscillatory response:

)}sin()cos({ 21 tAtAei ddt

tr

)1()(

)1()(

2

002

2

001

jjm

jjm

d

d

o

LC22

1

2L

R

1


tri

t t

Critically

damped

OverdampedUnderdamped

(ringing)

Envelope

tri

The current in all cases contains the exponential decaying factor (damping factor) assuring that the final value is zero

In other words, assuring that the complementary function decays in a relatively short time.


Alternating Current Transients

RL Sinusoidal Transient

1. Complementary (Transient) Solution is the solution of the homogeneous 1st order DE

The same as before, The auxiliary equation is : 0L

Rm

2. Particular (Steady-State) Solution

The steady-state value of the current for ac source is :

))/(( tan1

22

max RLwtSin

L

I

RX

Vss




Use the initial condition to find the value of c

Substituting by the constant values, we get:



RC Sinusoidal Transient



RLC Sinusoidal Transient

Particular (Steady-State) Solution

Complementary(Transient) Solution

The complementary function is identical to that of the DC series RLC circuit examined previously where the result was overdamped, critically

damped or oscillatory, depending upon R, L and C.

For the complete analysis Check Chapter 16 Schaum Series (Old version)


Transient Analysis using Laplace Transform

Solving differential equations Circuit analysis (Transient and general circuit

analysis) Digital Signal processing in Communications and Digital Control

Laplace transform is considered one of the most important tools in Electrical Engineering

It can be used for:


Transient Analysis using Laplace Transform


The switch “S” is closed at t = 0 to allow the step voltage to excite the circuit Apply KVL to the circuit in figure:

First-Order RL Transient (Step-Response)

s

VisIsLsIR )]0()(.[)(.

Apply Laplace Transform on both sides

i(0) = 0 >> initial value of the current at t = 0

s

VsLRsI ]).[(

][][)(

LRss

LV

sLRs

VsI

Apply the inverse Laplace Transform technique to get the expression of the current i(t)

22

First-Order RL Transient (Step-Response)

Use the partial fraction technique

LRs

A

s

A

LRss

LV

sI

21

][)(

Multiply both sides by ).( LRss

L

RAsAA

sAL

RsAL

V

.).(......

.).(

11

1

2

2

both sides Compare the coefficients

RVA

1 RVA 2

)11

()(

LRssR

VsI

So, the current in s-domain is given by:

Apply the inverse Laplace transform :0);1()(

teR

Vti

tL

R

R

VsILRsA

R

VsIsA

LRs

s

/2

01

|)}(*)/{(

|)}(*{

OR

The same as last lecture

23First-Order RL Transient (Discharge)

The RL circuit shown in Figure contains an initial current of (V/R)

The Switch “S” is moved to position”2” at t=0

The same as before

0)]0()(.[)(. isIsLsIR


i(0) = V/R >> initial value of the current at t = 0

][)(

LRs

RV

sI

0])(.[)(. RVsIsLsIR

Apply the inverse Laplace transform :

0;..)(

teIeR

Vti

tL

R

o

tL

R


First-Order RC Transient (Step-Response)

o Assume the switch S is closed at t = 0o Apply KVL to the series RC circuit shown:

s

VsIR

s

v

cs

sI c )(.])0()(

[


Vc(0) = 0 >> initial value of the voltage at t = 0

VtiRvdttic

c )(.)]0().(1

[

s

V

csRsI ]

1).[(

]1[]1[)(

cRs

RV

csR

sV

sI

Apply the inverse Laplace Transform technique to get the expression of the current i(t)

0;)(

1

teR

Vti

tRC

The same as last lecture

25


The Switch “S” is closed at t=0 Applying KVL will produce the following

Integro-Differential equation:

s

VsIRisIsL

s

v

cs

sI c )(.)]0()(..[])0()(

[


VtiRdt

tdiLvdtti

cc )(.

)()]0().(

1[

Assume:vc(0) = 0 & i(0) = 0

]1).([]1[)(

2

LcLRss

LV

cssLR

sV

sI

To convert this to time-domain, it will depend on the roots of the denominator which could be expressed as:

(s-S1).(s-S2) >>>>> similar to last lecture (m-m1).(m-m2)

012

LC

mL

Rm


LCL

R

L

RS

1)

2(

2

2

2,1

12

002,1 S

2

0

2

2,1 Sfrequency natural undamped :

ratio dampling expontial :

0


212 ]1).([

)(SS

B

SS

A

LcLRss

LV

sI

Apply Partial Fraction:

AL

VsISSB

L

VsISSA

o

ss

o

ss

1.2|)().(

1.2|)().(

22

21

2

1

]11

.[1.2

)(21

2 SSSSL

VsI

o

27


Apply Partial According to the values of the roots , we have 3 scenarios:

].[1.2

)( 212

tStS

o

eeL

Vti

1. Over-damped Case i.e. Two real distinct roots

ttS oetL

Vet

L

Vti

....)( 1

2. Critically-damped Case i.e. Two real equal roots

222

12 )()()(]1).([

)(oS

LV

S

LV

SS

LV

LcLRss

LV

sI

L

RSS

221

o

Convert by inverse L.T:at

n

ne

n

t

aS

.

)!1()(

1 1

28


)sin(......

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

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

)]sin()cos([)(

.

.

21.

21

.

tAe

ttAe

tA

At

A

AAe

tAtAeti

d

t

dd

t

dd

t

dd

t

3. Under-damped Case i.e. Two Complex-conjugate roots

djS 2

0

2

2,1

1A

2A

A

29


R = 5 ohms, L = 0.01 H, Vm = 100 volts, ф = 0, ω = 500

22100)]0()(.[)(.

sisIsLsIR


i(0) = 0 >> initial value of the current at t = 0

22100)]().[(

sL

RsLsI

)500).((

105

)).(.(

.100)(

22

6

22

ss

x

L

RssL

sI

jS

jS

S

3

2

1 500

30

RL Sinusoidal Transient Use Partial Fraction:

500)500).((

105)( 321

22

6

s

A

js

A

js

A

ss

xsI

OR500

3

22

21

s

A

s

BsB

tettti 500.10)500cos(.10)500sin(.10)(

Compare to find the constants:

Use Inverse L.T.

500

10

500.10

500

500.10

500

10]500[10)(

222222

ss

s

sss

ssI

tettti 500.10)]500cos()500.[sin(10)(

1

1

2)1(1 22 A

o45)1

1(tan 1

trss

to IIetti .500.10)45500sin(.2.10)(

Check EXAMPLE 11–12 ,13 Miller 31

Examples

The capacitor of Figure 11–24(a) is uncharged. The switch is moved to position 1 for 10 ms, then to position 2, where it remains.

Note that discharge is more rapid than charge since

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Electrical Circuits (2) Shoubra... · 2016. 4. 11. · Electrical Circuits (2) - Basem ElHalawany 7 Second-Order RLC Transient (Step Response) The Switch “S” is closed at t=0