Differential Equations

EE 313 Linear Systems and Signals Fall 2010

Initial conversion of content to PowerPointby Dr. Wade C. Schwartzkopf

Prof. Brian L. Evans

Dept. of Electrical and Computer Engineering

The University of Texas at Austin

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)(on dependent

zero are conditions initial all

)( zero-non toresponse

)( oft independen

only conditions system internal from results

0)(when

response state-zero responseinput -zeroresponse Total

tx

tx

tx

tx

Time-Domain Analysis• For a system governed by a

linear constant coefficientdifferential equation,

Each component can be computed independently of other

System satisfies linearity property if zero-input response is zero (i.e. all initial conditions are zero)

Zero-state response is convolution of impulse response and input signal

x(t) y(t)

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Time-Domain Analysis• Zero-input response

Response when x(t) = 0

Results from internal system conditions only

Independent of x(t)

For most filtering applications (e.g. your stereo system), we want a zero-valued zero-input response

• Zero-state responseResponse to non-zero x(t)

when system is relaxed

A system in zero state cannot generate any response for zero input

Zero state corresponds to initial conditions being zero

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Zero-Input Response• Simplest case (first-order equation)

• Solution:• For arbitrary constant C

How is C determined?

Could C be complex-valued?

• How about the following Nth-order equation?

0010 tyatydt

d tyatydt

d010

taeCty 0

1

)(11

1

1 txtyatydt

daty

dt

daty

dt

dNNN

N

N

N

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tkk

k

t

t

eCdt

yd

eCdt

yd

eCdt

dy

0

22

02

0

Zero-Input Response• For the Nth-order equation

Guess solution has form y0(t) = C e t

Substitute form into differential equation

Factor common terms to obtain

y0(t) = C et is a solution provided that Q() = 0

Factor Q() to obtain N characteristic roots

011

1

zeronon

t

Q

NNNN

zeronon

eaaaC

0)( 21 NQ

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Zero-Input Response• Assuming that no two i terms are equal

• For repeated roots, solution changesSimplest case of root repeated twice:

With r repeated roots

• Characteristic modes e t

Determine zero-input responseInfluence zero-state response

tetCCty 210

trr etCtCCty 1

210

tN

tt NeCeCeCty 21210

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Zero-Input Response• Could i be complex?

If complex, we can write it in Cartesian form

Exponential solution et becomes product of two terms

• For conjugate symmetric roots, and conjugate symmetric constants,

iii j

termgoscillatin termdampening

sincos tjteeeee iittjttjt iiiiii

termgoscillatin

1

termdampening

111 cos2 CteCeCeC ittt iii

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Zero-Input Solution Example• Component values

L = 1 H, R = 4 , C = 1/40 FRealistic breadboard components?

• Loop equations(D2 + 4 D + 40) [y0(t)] = 0

• Characteristic polynomial 2 + 4 + 40 =

( + 2 - j 6)( + 2 + j 6)

• Initial conditionsy(0) = 2 Aý(0) = 16.78 A/s

L R

C

y(t)

x(t)

Envelope

y0(t) = 4 e-2t cos(6t - /3) A

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Impulse Response• Response to unit impulse

Set x(t) = (t) and solve for y(t)

• Linear constant coefficient differential equation

• With zero initial conditions, impulse response is

b0 is coefficient of dNx(t)/dtN term and could be 0

)( )()( )( 0 tutvtbth

txbtxdt

dbtx

dt

dbtx

dt

db

tyatydt

daty

dt

daty

dt

d

NNM

M

MNM

M

MN

NNN

N

N

N

11

1

1

11

1

1

tybtydt

dbty

dt

dbty

dt

dbtv NNM

M

MNM

M

MN 00101

1

10)(

x(t) y(t)

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Impulse Response• Where did b0 come from?

• In solving these differential equations for t 0,

• Funny things happen to y’(t) and y”(t)

• In differential equations class, solved for m(t)Likely ignored (t) and ’(t) terms

Solution for m(t) is really valid for t 0+

)( )()( tutgtx )( )()( tutmty

)( )()( )(')(' ttmtutmty

)(' )()( )(' 2)( )(")(" ttmttmtutmty