1 EKT101 Electric Circuit Theory Chapter 5 First-Order and Second Circuits

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EKT101Electric Circuit

TheoryChapter 5

First-Order and Second Circuits

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First-Order and Second CircuitsChapter 5

5.1Natural response of RL and RC Circuit5.2Force response of RL and RC Circuit5.3Solution of natural response and force

response in RL and RC Circuit5.4 Natural and force response in series

RLC Circuit5.5Natural and force response in parallel

RLC Circuit

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5.1 Natural response of RL and RC circuit (1)

• A first-order circuit is characterized by a first-order differential equation.

• 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 iiBy KCL

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• 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.

• 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 t and slower for large t.

CRTime constantDecays more slowly

Decays faster

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

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

2. The time constant = RC.

/0)( teVtv CRwhere

DC source disconnected

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Example 1

Refer to the circuit below, determine vC, vx, and io for t ≥ 0.

Assume that vC(0) = 30 V.

• Please refer to lecture or textbook for more detail elaboration.Answer: vC = 30e–0.25t V ; vx = 10e–0.25t ; io = –2.5e–0.25t A

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Solution 1

vC(0) = 30 V.

= vC = 30e–0.25t V

vx = 10e–0.25t

io = –2.5e–0.25t A

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Example 2

The switch in circuit below is opened at t = 0, find v(t) for t ≥ 0.

• Please refer to lecture or textbook for more detail elaboration.Answer: V(t) = 8e–2t V

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Solution 2

1/6 F

V(t) = 8e–2t V

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• A first-order RL circuit consists of a inductor L (or its equivalent) and a resistor (or its equivalent)

0 RL vvBy KVL

0 iRdt

diL

Inductors law Ohms law

dtL

R

i

di LtReIti /

0 )(

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• 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)( teIti

R

L

A general form representing a RL

where

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5.1 Natural response of RL and RC circuit(8)

/0)( teIti

R

L

A RL source-free circuit

where /0)( teVtv RC

A RC source-free circuit

where

Comparison between a RL and RC circuit

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

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

2. The time constant = L/R.

/0)( teIti

R

Lwhere

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Example 3

Find i and vx in the circuit.

Assume that i(0) = 5 A.

Answer: i(t) = 5e–53t A

Solution 3find rth

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1

5

3

?

? ?

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Example 4

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

• Please refer to lecture or textbook for more detail elaboration.

Answer: i(t) = 2e–2t A

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t=0, SC

i(0) = current division ??

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Unit-Step Function (1)• The unit step function u(t) is 0 for negative

values of t and 1 for positive values of t.

0,1

0,0)(

t

ttu

o

oo tt

ttttu

,1

,0)(

o

oo tt

ttttu

,1

,0)(

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Unit-Step Function (2)

1. voltage source.

2. for current source:

Represent an abrupt change for:

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• Initial condition: v(0-) = v(0+) = V0

• Applying KCL,

or

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

5.2 Force response of RL and 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.

0)(

R

tuVv

dt

dvc s

)(tuRC

Vv

dt

dv s

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5.3 Solution of natural and force response in RL and RC Circuit

• Integrating both sides and considering the initial conditions, the solution of the equation is:

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)

= V0e–t/τ + Vs(1–e–t/τ)

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5.2 Force response of RC Circuit(3)

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

1. The initial capacitor voltage v(0).2. The final capacitor voltage v() — DC voltage

across C.3. The 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.

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Example 5

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.

5.2 Force response of RC Circuit(4)


Answer: and v(0.5) = 0.5182V515)( 2 tetv

Solution Ex 5• 1 -The initial capacitor voltage v(0).

• 2- final capacitor voltage v() – Use KCL at nod to get v(∞)

• 3- time constant . – Use Rth =RC

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5.2 Force response of 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.

• Initial currenti(0-) = i(0+) = Io

• Final inductor current i(∞) = Vs/R

• Time constant t = L/R

)()()( tueR

VI

R

Vti

ts

os

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

1. The initial inductor current i(0) at t = 0+.2. The final inductor current i().

3. The 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

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Example 6

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.



Answer: teti 102)(

Example 6 (solution)

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30V

Apply source transformation

Ai 3)0(

30

2A

teti 102)(

5.4 NATURAL AND FORCE RESPONSE IN SERIES RLC CIRCUIT

Second order circuit

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Examples of Second Order RLC circuits (1)

What is a 2nd order circuit?

A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.

RLC Series RLC Parallel RL T-config RC Pi-config

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Source-Free Series RLC Circuits (1)

• The solution of the source-free series RLC circuit is called as the natural response of the circuit.

• The circuit is excited by the energy initially stored in the capacitor and inductor.

02

2

LC

i

dt

di

L

R

dt

idThe 2nd order of expression

How to derive and how to solve?

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• At t=0,

• So,

• Eliminate integral, differentiate to t, rearrange

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There are three possible solutions for the following 2nd order differential equation:

02

2

LC

i

dt

di

L

R

dt

id

The types of solutions for i(t) depend on the relative values of a and .w

=> 02 202

2

idt

di

dt

id LC

andL

R 1

2 0

General 2nd order Form

where

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02 202

2

idt

di

dt

id


1. If a > wo, over-damped casetsts eAeAti 21

21)( 20

22,1 swhere

2. If a = wo, critical damped casetetAAti )()( 12 2,1swhere

3. If a < wo, under-damped case

)sincos()( 21 tBtBeti ddt

where 220 d


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Example 1

If R = 10 Ω, L = 5 H, and C = 2 mF in 8.8, find α, ω0, s1 and s2.

What type of natural response will the circuit have?


Answer: underdamped

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Example 2

The circuit shown below has reached steady state at t = 0-.

If the make-before-break switch moves to position b at t = 0, calculate i(t) for t > 0.


Answer: i(t) = e–2.5t[5cos1.6583t – 7.538sin1.6583t] A

t>0

under-damped case

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5.4 NATURAL AND FORCE RESPONSE IN PARALLEL RLC CIRCUIT

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Source-Free Parallel RLC Circuits (1)

The 2nd order of expression

011

2

2

vLCdt

dv

RCdt

vd

0

0 )(1

)0( dttvL

IiLet

v(0) = V0

Apply KCL to the top node:

t

dt

dvCvdt

LR

v0

1

Taking the derivative with respect to t and dividing by C

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LCRCv

dt

dv

dt

vd 1and

2

1 where0 2 0

202

2


1. If a > wo, over-damped casetsts eAeAtv 21 )( 21 2

02

2,1 swhere

2. If a = wo, critical damped casetetAAtv )( )( 12

2,1swhere

3. If a < wo, under-damped case

)sincos()( 21 tBtBetv ddt

where 220 d


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Example 3

Refer to the circuit shown below. Find v(t) for t > 0.


Answer: v(t) = 66.67(e–10t – e–2.5t) V

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: v(t) = 66.67(e–10t – e–2.5t) V

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Step-Response Series RLC Circuits (1)

• The step response is obtained by the sudden application of a dc source.

The 2nd order of expression LC

v

LC

v

dt

dv

L

R

dt

vd s2

2

The above equation has the same form as the equation for source-free series RLC circuit. • The same coefficients (important in determining the

frequency parameters). • Different circuit variable in the equation.

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The solution of the equation should have two components:the transient response vt(t) & the steady-state response vss(t):

)()()( tvtvtv sst

The transient response vt is the same as that for source-free case

The steady-state response is the final value of v(t). vss(t) = v(∞)

The values of A1 and A2 are obtained from the initial conditions: v(0) and dv(0)/dt.

tstst eAeAtv 21

21)( (over-damped)t

t etAAtv )()( 21 (critically damped)

)sincos()( 21 tAtAetv ddt

t (under-damped)

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Example 4

Having been in position for a long time, the switch in the circuit below is moved to position b at t = 0. Find v(t) and vR(t) for t > 0.


Answer: v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e–2t]} V vR(t)= [2.31sin3.464t]e–2t V

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Step-Response Parallel RLC Circuits (1)

• The step response is obtained by the sudden application of a dc source.

The 2nd order of expression

It has the same form as the equation for source-free parallel RLC circuit.

• The same coefficients (important in determining the frequency parameters).

• Different circuit variable in the equation.

LC

I

LC

i

dt

di

RCdt

id s1

2

2

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The solution of the equation should have two components:the transient response vt(t) & the steady-state response vss(t):

)()()( tititi sst

The transient response it is the same as that for source-free case

The steady-state response is the final value of i(t). iss(t) = i(∞) = Is

The values of A1 and A2 are obtained from the initial conditions: i(0) and di(0)/dt.

tstst eAeAti 21

21)( (over-damped)t

t etAAti )()( 21 (critical damped)

)sincos()( 21 tAtAeti ddt

t (under-damped)

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Example 5

Find i(t) and v(t) for t > 0 in the circuit shown in circuit shown below:


Answer: v(t) = Ldi/dt = 5x20sint = 100sint V

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v(t) = Ldi/dt = 5x20sint = 100sint V

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1 EKT101 Electric Circuit Theory Chapter 5 First-Order and Second Circuits