electrical engineering 1 12026105 lecture 7 first-order ... · 7.1 the source-free rc circuit (1) a...

Electrical Engineering 1

12026105

Lecture 7

First-Order Circuits

1

First-Order Circuits

7.1 The Source-Free RC Circuit (วงจร RC ทีไ่มม่แีหลง่จา่ย)

7.2 The Source-Free RL Circuit (วงจร RL ทีไ่มม่แีหลง่จา่ย)

7.3 Unit-step Function

7.4 Step Response of an RC Circuit (วงจร RC ทีม่แีหลง่จา่ย)

7.5 Step Response of an RL Circuit (วงจร RC ทีม่แีหลง่จา่ย)

2

7.1 The Source-Free RC Circuit (1)

A first-order circuit is characterized by a first-order

differential equation.

3

• Apply Kirchhoff’s laws to purely resistive circuit results in algebraic equations.

• Apply the laws to RC and RL circuits produces differential equations.

Ohms law Capacitor law

0 dt

dvC

R

v0 CR ii

By KCL


• The natural response of a circuit refers to the

behavior (in terms of voltages and currents) of the

circuit itself, with no external sources of excitation.

4

• The time constant of a circuit is the time required for

the response to decay by a factor of 1/e or 36.8% of its

initial value.

• v decays faster for small and slower for large .

CRTime constant

Decays more slowly

Decays faster


The key to working with a source-free RC circuit is

finding:

5

1. The initial voltage v(0) = V0 across the capacitor.

2. = RC.

/0)( teVtv CRwhere


Example 1

Refer to the circuit below, determine vC , vx , and io for t ≥ 0. Assume that vC(0) = 60 V.

6 Answer: vC = 60e–0.25t V ; vx = 20e–0.25t V; io = -5e–0.25t A

4//0 60)( tt eeVtv

4/20)(84

4)( t

Cx etvtv

4/5)( tCo e

dt

dvCti


Example 2 The switch is opened at t = 0, find v(t) for t ≥ 0.

7 Answer: v(t) = 8e–2t V

VvVv 824)0(?)0(63

30

สวทิซปิ์ด

สวทิซเ์ปิด


Example 3 Find for vo(t) for t>0. Determine the time for the

capacitor voltage to decay to one-third of its value at t=0

8 Answer: v(t) = 9e–t / 0.06 V; t = 0.066 s

VvVv 936)0(?)0(93

30



ste t 066.0)9/3ln(06.0;933/9 06.0/

7.2 The Source-Free RL Circuit (1)

• A first-order RL circuit consists of L and R (or

their equivalent)

9

0 RL vvBy KVL

0 iRdt

diL

Inductors law Ohms law

dtL

R

i

di

t

L

R

eIti 0 )(แก้สมการ 1st order dif eq


10

• The time constant of a circuit is the time required for the

response to decay by a factor of 1/e or 36.8% of its initial

value.

• i(t) decays faster for small t and slower for large t.

• The general form is very similar to a RC source-free circuit.

/

0)( teItiR

L

A general form representing RL

where


11

/

0)( teItiR

L

RL source-free circuit

where /0)( teVtv RC

RC source-free circuit

where

Comparison between RL and RC circuit


The key to working with a source-free RL circuit is

finding:

12

1. The initial current i(0) = I0 through the inductor.

2. The time constant = L/R.

/0)( teIti

R

Lwhere


Example 4 Assume that i (0) = 10 A. Find i (t) and ix(t).

13

Answer:

A6667.1)( A;10)( 3/23/2 tx

t etieti

วธิทีี1่ ใช้ KVL


14

วธิทีี2่ ใช้ Req และ KVL


Example 5 Find i and vx. Assume that i (0) = 12 A.

15 Answer:

+ VL -

03)(22,1 loop 2 xviidt

di ivx

ivivivii xxx 5.05.0,048,0228,2 loop 222

0204203)5.0(22 idt

dii

dt

diiii

dt

di

V12)()( A;12)( 22 tx

t etitveti

toeIitidt

i

di 22)ln(2


Example 5 Find i and vx. Assume that i (0) = 12 A

16 Answer:

+ VL -

125223,1 loop 21121 iiiii

21211 2822,2 loop iiiii

4

1,

8

112 ii

V12)()( A;12)( 22 tx

t etitveti

2

1

4

2,4

eqo

oeq

R

L

i

vR


Example 6 For the circuit, find i (t) for t > 0.

17 Answer: i(t) = 5e–2t A , t >0

45//)812( eqR

A515)0(?)0(12)8//24(

8//240

iIi



7.3 Unit-Step Function (1)

• The unit step function u(t) is 0 for negative

values of t and 1 for positive values of t.

18

0,1

0,0)(

t

ttu

o

o

ott

ttttu

,1

,0)(

o

o

ott

ttttu

,1

,0)(

7.3 Unit-Step Function (2)

1. voltage source:

2. current source:

19

Represent an abrupt change for:

7.4 The Step-Response of a RC Circuit (1)

• The step response of a circuit is its behavior

when the excitation is the step function,

which may be a voltage or a current source.

20

• Initial condition:

• Applying KCL,

or

• Where u(t) is the unit-step function

0)(

R

tuVv

dt

dvC s

RC

tuVv

dt

dv s )(

0)0()0( Vvv

RC

dt

Vv

dv

s

ttv

VsRC

tVv

0

)(

0)ln(


21

t

tv

VsRC

tVv

0

)(

0)ln(

RC

tVVVtv ss )ln())(ln( 0

RC

t

VV

Vtv

s

s

0

)(ln

RC

t

s

s eVV

Vtv

0

)(

RC

t

ss eVVVtv

0)(

t

evvvtv

)()0()()(


• Integrating both sides and considering the initial

conditions, the solution of the equation is:

22

0)(

0)(

/0

0

teVVV

tVtv

tss

Final value at t ∞ Initial value at t = 0 Source-free Response

Complete Response = Natural response + Forced Response

(stored energy) (independent source) (เกดิจาก no source) (เกดิจาก source)

Transient Response

(temporary part)

Steady-state Response

(permanent part)

)1( //0

ts

t eVeV


3 steps to find out the step response of an RC circuit:

23

1. Initial capacitor voltage v(0).

2. Final capacitor voltage v() — DC voltage across C.

3. Time constant .

/ )]( )0( [ )( )( tevvvtv

Note: The above method is a short-cut method. You may

also determine the solution by setting up the circuit formula

directly using KCL, KVL , ohms law, capacitor and inductor

VI laws.


Example 7

Find v(t) for t>0 in the circuit in below. Assume the switch has

been open for a long time and is closed at t=0. Calculate v(t) at

t=0.5.

24 Answer: เม่ือ t>0, v (0.5) = 11.44 V Vetv t )375.9625.5()( 2

7.5 The Step-response of a RL Circuit (1)

• The step response of a circuit is its behavior when the

excitation is the step function, which may be a voltage or a

current source.

25

• Initial current

• Final inductor current

i(∞) = Vs /R

• Time constant = L/ R

t

so

s eR

VI

R

Vti

)()(

0)0()0( Iii

t

eiiiti

)( )0( )( )(

7.5 The Step-Response of a RL Circuit (2)

3 steps to find out the step response of an RL circuit:

26

1. Initial inductor current i(0) at t = 0+.

2. Final inductor current i().

3. Time constant .

Note: The above method is a short-cut method. You may

also determine the solution by setting up the circuit formula

directly using KCL, KVL , ohms law, capacitor and inductor

VI laws.

/ )]( )0( [ )( )( teiiiti


Example 8

The switch in the circuit shown below has been closed for

a long time. It opens at t = 0. Find i(t) for t > 0.

27 Answer: teti 1024)(

A 66 010

10 )0(

i

A 46 105

10 )(

i

10

1

15

5.1 ; 15105

eqeq

R

LR

ttt eeeiiiti 1010/ 24)46(4 )]( )0( [ )( )(


Example 9 At t=0 switch 1 is closed, and switch 2 is

closed 4 s later. Find i(t) for t>0. Calculate i for t=2 s

and t=5 s.

28

Answer: AeiAei 02.3273.1727.2)5(,93.314)2( 4667.14

0)0()0( :0 iit

three time intervals

10 ,464

40)( :40 eqRit

4 ,40 ,0 ttt

sR

L

eq 2

1

Aeiit 414)4()4( :4 8

11

180

62

10

4

40 :P nodeat

v

vvvKCL

ttt eeeiiiti 22/ 14)40(4 )]( )0( [ )( )(

3

2262//4 ,727.2

11

30

6 )( eqRA

vi s

R

L

eq 22

15

44667.1/4/ 273.1727.2)727.24(727.2 )]( )4( [ )( )( ttt eeeiiiti

electrical engineering 1 12026105 lecture 7 first-order ... · 7.1 the source-free rc circuit (1) a...

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