1 circuit theory chapter 8 second-order circuits second-order circuits copyright © the mcgraw-hill...

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Circuit TheoryCircuit TheoryChapter 8Chapter 8

Second-Order CircuitsSecond-Order Circuits

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Second-Order CircuitsSecond-Order CircuitsCChapter 8hapter 8

8.1 Examples of 2nd order RLC circuit8.3 The source-free parallel RLC circuit8.5 Step response of a parallel RLC8.2 The source-free series RLC circuit8.4 Step response of a series RLC circuit

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8.1 Examples of Second 8.1 Examples of Second Order RLC circuits (1)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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8.3 Source-Free Parallel 8.3 Source-Free Parallel RLC Circuits (1)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

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

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

02

2,1 swhere

2. If = o, critical damped casetetAAtv )( )( 12

2,1swhere

3. If < o, 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.

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

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

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8.5 Step-Response Parallel 8.5 Step-Response Parallel RLC Circuits (1)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 s 12

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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8.2 Source-Free Series 8.2 Source-Free Series RLC Circuits (1)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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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 and

=> 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

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

1. If > o, over-damped casetsts eAeAti 21

21)( 20

22,1 swhere

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

3. If < o, 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

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8.4 Step-Response Series 8.4 Step-Response Series RLC Circuits (1)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

1)determine the qualitative form of the response 1)determine the qualitative form of the response vvcc(t)(t) as being as being either overdamped, underdamped, or critically damped.either overdamped, underdamped, or critically damped.

Overdamped

Underdamped

Underdamped

Critically damped

2) Determine the initial value of Vr(t), and dVr(t)/dt, 2) Determine the initial value of Vr(t), and dVr(t)/dt, and the final value of Vr(t)and the final value of Vr(t)

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Vr(0) = 6V;dVr(0)/dt = 3V/s;Vr(infinity) = 0V

http://www.rose-hulman.edu/CLEO/video/player.php?id=34&embed

3) For both the inductor current and voltage, determine their 3) For both the inductor current and voltage, determine their initial values, the initial values of their derivatives, and their initial values, the initial values of their derivatives, and their

final values. The capacitor has 9 J of stored energy before the final values. The capacitor has 9 J of stored energy before the switch closes.switch closes.

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4)Plot the inductor current 4)Plot the inductor current iiLL(t)(t) and inductor voltage and inductor voltage vvLL(t)(t) for time for time tt = = -0.1 seconds to -0.1 seconds to tt = 3 seconds. Confirm your results using a circuit = 3 seconds. Confirm your results using a circuit

simulator.simulator.

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iL(t) = 2 + 14.5 sin(4.15t) e -1.67t , t>=0

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http://www.rose-hulman.edu/CLEO/browse/?path=1/2/76/89/224/253

1 circuit theory chapter 8 second-order circuits second-order circuits copyright © the mcgraw-hill...

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