Lect16 EEE 202 1

System Responses

Dr. Holbert

March 24, 2008

Lect16 EEE 202 2

Introduction

• Today, we explore in greater depth the three cases for second-order systems– Real and unequal poles– Real and equal poles– Complex conjugate pair

• This material is ripe with new terminology

Lect16 EEE 202 3

Second-Order ODE

• Recall: the second-order ODE has a form of

• For zero-initial conditions, the transfer function would be

200

2

200

2

2

1

)(

)()(

)(2)(

sss

ssH

ssss

F

X

FX

)()()(

2)( 2

002

2

tftxdt

tdx

dt

txd

Lect16 EEE 202 4

Second-Order ODE

• The denominator of the transfer function is known as the characteristic equation

• To find the poles, we solve :

which has two roots: s1 and s2

02 200

2 ss

12

4)2(2, 2

00

20

200

21

ss

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Damping Ratio () andNatural Frequency (0)

• The damping ratio is ζ• The damping ratio determines what type of

solution we will obtain:– Exponentially decreasing ( >1)– Exponentially decreasing sinusoid ( < 1)

• The undamped natural frequency is 0

– Determines how fast sinusoids wiggle– Approximately equal to resonance frequency

Lect16 EEE 202 6

Characteristic Equation Roots

The roots of the characteristic equation determine whether the complementary (natural) solution wiggles

1

1

2002

2001

s

s

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1. Real and Unequal Roots

• If > 1, s1 and s2 are real and not equal

• This solution is overdamped

tt

c eKeKtx

1

2

1

1

200

200

)(

Lect16 EEE 202 8

Overdamped

0

0.2

0.4

0.6

0.8

1

0.0E+00 5.0E-06 1.0E-05

Time

i(t)

-0.2

0

0.2

0.4

0.6

0.8

0.0E+00 5.0E-06 1.0E-05

Time

i(t)

Both of these graphs have a response of the formi(t) = K1 exp(–t/τ1) + K2 exp(–t/τ2)

Lect16 EEE 202 9

2. Complex Roots

• If < 1, s1 and s2 are complex

• Define the following constants:

• This solution is underdamped

tAtAetx

jss

ddt

c

d

d

sincos)(

,

1

21

21

20

0

Frequency) Natural (Damped

t)Coefficien (Damping

Lect16 EEE 202 10

Underdamped

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1.00E-05 1.00E-05 3.00E-05

Time

i(t)

A curve having a response of the formi(t) = e–t/τ [K1 cos(ωt) + K2 sin(ωt)]

Lect16 EEE 202 11

3. Real and Equal Roots

• If = 1, then s1 and s2 are real and equal

• This solution is critically damped

ttc etKeKtx 00

21)(

Lect16 EEE 202 12

IF Amplifier Example

This is one possible implementation of the filter portion of an intermediate frequency (IF) amplifier

10

769pFvs(t)

i(t)

159H

+–

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IF Amplifier Example (cont’d.)

• The ODE describing the loop current is

• For this example, what are ζ and ω0?

)()()(

2)(

)(1)(

1)()(

2002

2

2

2

tftidt

tdi

dt

tid

dt

tdv

Lti

LCdt

tdi

L

R

dt

tid s

Lect16 EEE 202 14

IF Amplifier Example (cont’d.)

• Note that 0 = 2f = 2455,000 Hz)

• Is this system overdamped, underdamped, or critically damped?

• What will the current look like?

011.0μH159

102

rad/sec1086.2)pF769)(μH159(

11

0

60

20

L

R

LC

Lect16 EEE 202 15

IF Amplifier Example (cont’d.)

• The shape of the current depends on the initial capacitor voltage and inductor current

-1

-0.8

-0.6

-0.4

-0.2

00.2

0.4

0.6

0.8

1

-1.00E-05 1.00E-05 3.00E-05

Time

i(t)

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Slightly Different Example

• Increase the resistor to 1k• Exercise: what are and 0?

1k

769pFvs(t)

i(t)

159H

+–

Lect16 EEE 202 17

Different Example (cont’d.)

• The natural (resonance) frequency does not change: 0 = 2455,000 Hz)

• But the damping ratio becomes = 2.2

• Is this system overdamped, underdamped, or critically damped?

• What will the current look like?

Lect16 EEE 202 18

Different Example (cont’d.)

• The shape of the current depends on the initial capacitor voltage and inductor current

0

0.2

0.4

0.6

0.8

1

0.0E+00 5.0E-06 1.0E-05

Time

i(t)

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Damping Summary

Damping Ratio

Poles (s1, s2) Damping

ζ > 1 Real and unequal Overdamped

ζ = 1 Real and equal Critically damped

0 < ζ < 1 Complex conjugate pair set

Underdamped

ζ = 0 Purely imaginary pair Undamped

Lect16 EEE 202 20

Transient and Steady-State Responses

• The steady-state response of a circuit is the waveform after a long time has passed, and depends on the source(s) in the circuit– Constant sources give DC steady-state responses

• DC steady-state if response approaches a constant

– Sinusoidal sources give AC steady-state responses• AC steady-state if response approaches a sinusoid

• The transient response is the circuit response minus the steady-state response

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Transient and Steady-State Responses

• Consider a time-domain response from an earlier example this semester

tt eetf 32

3

10

2

5

6

5)(

SteadyState

Response

TransientResponse

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

TransientResponse

Steady-StateResponse

Lect16 EEE 202 22

Class Examples

• Drill Problems P7-6, P7-7, P7-8