ECE 8443 – Pattern RecognitionEE 3512 – Signals: Continuous and Discrete

• Objectives:Stability and the s-PlaneStability of an RC Circuit1st and 2nd Order Systems

• Resources:MIT 6.003: Lecture 18GW: Underdamped 2nd-Order SystemsCRB: System ResponseRVJ: RLC CircuitsMH: Control Theory and StabilityEC: Step ResponseWiki: Routh-Hurwitz Stability TestTBCO: Routh-Hurwitz Tutorial

LECTURE 28: 1st and 2nd Order Systems

URL:

EE 3512: Lecture 28, Slide 2

Stability of CT Systems in the s-Plane

• Recall our stability condition for the Laplace transform of the impulse response of a CT linear time-invariant system:

• This implies the poles are in the left-half plane.This also implies:

• A system is said to be marginally stableif its impulse response is bounded:

In this case, at least one pole of the systemlies on the jω-axis.

• Recall periodic signals also have poles on the jω-axis because they are marginally stable.

• Also recall that the left-half plane maps to the inside of the unit circle in the z-plane for discrete-time (sampled) signals.

• We can show that circuits built from passive components (RLC) are always stable if there is some resistance in the circuit.

Nipssas

bsbsbsH iN

NN

MM

MM ...,,2,1for0Re

...

...)(

01

1

01

1

dtthtasth )(and0)(

tcth )(

LHP

EE 3512: Lecture 28, Slide 3

Stability of CT Systems in the s-Plane

• Example: Series RLC Circuit

Using the quadratic formula:

LCsLRs

LCsH

/1)/(

/1)(

2

LCL

R

L

Rpp

1

22,

2

21

stablealways2

Re

poles,complextwo01

2

1

2

L

Rp

LCL

R

:2Case

stablealwaysL

R

LCL

R

L

R

stablealways

LCL

R

L

R

LCL

R

2bemusttermquadratic

01

22

LHPinpole01

22

polesrealtwo01

2

1

2

2

2

:Case

The RLC circuit is always stable.

Why?

EE 3512: Lecture 28, Slide 4

Analysis of the Step Response For A 1st-Order System

• Recall the transfer function for a1st-order differential equation:

)(1)(

//)(

)()()()(

1)()(

)(

tuep

kty

ps

pk

s

pksY

pss

ksXsHsY

stusX

ps

ksH

pt

L

• Define a time constant as the time it takes for the response to reach 1/e (37%) of its value.

• The time constant in this case is equal to-1/p. Hence, the real part of the pole, which is the distance of the pole from the jω-axis, and is the bandwidth of the pole, is directly related to the time constant.

num = 1; den = [1 –p];

t = 0:0.05:10;

y = step(num, den, t);

EE 3512: Lecture 28, Slide 5

Second-Order Transfer Function

• Recall our expression for a simple, 2nd-order differential equation:

• Write this in terms of two parameters, ζ and ωn, related to the poles:

• From the quadratic equation:

• There are three types of interestingbehavior of this system:

)zeta""is(2

)(22

2

nn

n

sssH

012

0001

2

)()()(asas

bsHtxbtya

dt

dya

dt

yd

1, 221 nnpp

d)(overdampe

axis) real (negative poles two:1

damped)y (criticall

at pole double:1

ed)(underdamp

polescomplex:10

ns

Impulse Response

Step Response

EE 3512: Lecture 28, Slide 6

Step Response For Two Real Poles• When ζ > 1, both poles are

real and distinct:

0,1)(

1

))(()(

1)(

))((

2)(

2111

21

2

21

2

21

2

22

2

tekekpp

ty

spspssY

ssX

psps

sssH

tptpn

n

n

nn

n

• There are two componentsto this response:

)inputsteptodue()(

0,)(

21

2

1121

221

ppty

tekekpp

ty

nss

tptpntr

• When ζ = 1, both poles arereal (s = ωn) and repeated:

0,11)(

1

)()(

1)(

)(

2)(

21

2

21

2

22

2

ttety

spssH

ssX

ps

sssH

tn

n

n

nn

n

n

EE 3512: Lecture 28, Slide 7

Step Response For Two Real Poles (Cont.)

• ζ is referred to as the damping ratio because it controls the time constant of the impulse response (and the time to reach steady state);

• ωn is the natural frequency and controls the frequency of oscillation (which we will see next for the case of two complex poles).

• ζ > 1: The system is considered overdamped because it does not achieve oscillation and simply directly approaches its steady-state value.

• ζ = 1: The system is considered critically damped because it is on the verge of oscillation.

Two Real PolesBoth Real and

Repeated

EE 3512: Lecture 28, Slide 8

Step Response For Two Complex Poles• When 0 < ζ < 1, we have two complex conjugate poles:

• The transfer function can be rewritten as:

22

2

222222

2

22

2

2

2)(

dn

n

nnnn

n

nn

n

s

ss

sssH

• The step response, after some simplification, can be written as:

• Hence, the response of this system eventually settles to a steady-state value of 1. However, the response can overshoot the steady-state value and will oscillate around it, eventually settling in to its final value.

n

dd

t

n

n wherettety n

1tan0,sin1)(

dn

nd

nn

jpp

pp

21

2

221

,

1

1,

EE 3512: Lecture 28, Slide 9

Analysis of the Step Response For Two Complex Poles

• ζ > 1: the overdamped system experiences an exponential rise and decay. Its asymptotic behavior is a decaying exponential.

• ζ = 1: the critically damped system has a fast rise time, and converges to the steady-state value in an exponetial fashion.

• 0 < ζ > 1: the underdamped system oscillates about the steady-state behavior at a frequency of ωd.

• Note that you cannot control the rise time and the oscillation behavior independently!

• What can we conclude about the frequency response of this system?

Impulse Response

Step Response

EE 3512: Lecture 28, Slide 10

Implications in the s-Plane

Several important observations:

• The pole locations are:

• Since the frequencyresponse is computedalong the jω-axis, we can see that the pole islocated at ±ωd.

• The bandwidth of thepole is proportional to the distance from the jω-axis, and is given by ζωn.

• For a fixed ωn, the range 0 < < 1 describes a circle. We will make use of this concept in the next chapter when we discuss control systems.

• What happens if ζ is negative?

dn

nn

j

pp

1, 221

EE 3512: Lecture 28, Slide 11

RC Circuit

• Example: Find the response to asinewave:.

• Solution:

• Again we see the solution is the superposition of a transient and steady-state response.

• The steady-state response could have been found by simply evaluating the Fourier transform at ω0 and applying the magnitude scaling and phase shift to the input signal. Why?

• The Fourier transform is given by:

20

20 )()(cos)(

s

CssXtutCtx

0,tancos)(

)(

/1

/1)(

01022

0

220

20

2

tp

tp

Cke

p

kCpty

sps

kCssY

ps

k

RCs

RCsH

pt

RCj

RCsHeH jes

j

/1

/1)()(

EE 3512: Lecture 28, Slide 12

Summary

• Reviewed stability of CT systems in terms of the location of the poles in the s-plane.

• Demonstrated that an RLC circuit is unconditionally stable.

• Analyzed the properties of the impulse response of a first-order differential equation.

• Analyzed the behavior of stable 2nd-order systems.

• Characterized these systems in terms of three possible behaviors: overdamped, critically-damped, or overdamped.

• Discussed the implications of this in the time and frequency domains.

• Analyzed the response of an RC circuit to a sinewave.

• Next: Frequency response, Bode plots and filters.

EE 3512: Lecture 28, Slide 13

The Routh-Hurwitz Stability Test

• The procedures for determining stability do not require finding the roots of the denominator polynomial, which can be a daunting task for a high-order system (e.g., 32 poles).

• The Routh-Hurwitz stability test is a method of determining stability using simple algebraic operations on the polynomial coefficients. It is best demonstrated through an example.

• Consider:

• Construct the Routh array:

011

1 ...)( asasasasA NN

NN

1

54

1

5414

1

32

1

3212

00

11

022

7533

6422

5311

42

columns2/)1(:odd

columns1)2/(:even

00

00

0

N

NNN

N

NNNNN

N

NNN

N

NNNNN

NNNN

NNNN

NNNN

NNNN

a

aaa

a

aaaab

a

aaa

a

aaaab

NN

NN

fs

es

dds

cccs

bbbs

aaas

aaas

Number of sign changes in 1st column = number of poles in the RHPRLC circuit is always stable

EE 3512: Lecture 28, Slide 14

Routh-Hurwitz Examples

• Example: 012)( asassA

0)0)(1(

0

1

01

010

11

02

aa

aas

as

as

unstablepolesRHP2changessigntwo0and0if

unstablepoleRHP1changesignone0and0if

unstablepoleRHP1changesignone0and0if

01

01

01

aa

aa

aa

• Example: 012

23)( asasassA

0

0)1(

1

00

2

01

2

0121

022

13

asa

aa

a

aaas

aas

as

unstablepolesRHP2changessigntwo0and0if 02 aa