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

•Objectives:First-OrderSecond-OrderNth-OrderComputation of the Output SignalTransfer Functions

• Resources:PD: Differential EquationsWiki: Applications to DEsGS: Laplace Transforms and DEsIntMath: Solving DEs Using Laplace

LECTURE 23: DIFFERENTIAL EQUATIONS

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EE 3512: Lecture 23, Slide 2

First-Order Differential Equations• Consider a linear time-invariant system defined by:

• Apply the one-sided Laplace transform:

• We can now use simple algebraic manipulations to find the solution:

• If the initial condition is zero, we can find the transfer function:

• Why is this transfer function, which ignores the initial condition, of interest?(Hints: stability, steady-state response)

• Note we can also find the frequency response of the system:

• How does this relate to the frequency response found using the Fourier transform? Under what assumptions is this expression valid?

)()()( tbxtaydttdy

)()()0()( sbXsaYyssY

assbX

asysY

sbXysYas

)()0()(

)()0()()(

asb

sXsYsH

)()()(

ajbsHeH

jsj

)()(

EE 3512: Lecture 23, Slide 3

RC Circuit

)(/1

/1/1

)0()(

)(1)(1)(

sXRCsRC

RCsysY

txRC

tyRCdt

tdy

RCssRCsy

sRCsRC

RCsysY

ssXtutx

/111

/1)0(1

/1/1

/1)0()(

1)()()(

• The input/output differential equation:

• Assume the input is a unit step function:

• We can take the inverse Laplace transform to recover the output signal:

• For a zero initial condition:

• Observations: How can we find the impulse response? Implications of stability on the transient response? What conclusions can we draw about the complete response to a sinusoid?

0,1)0()( )/1()/1( teeyty tRCtRC

0,1)( )/1( tety tRC

EE 3512: Lecture 23, Slide 4

Second-Order Differential Equation• Consider a linear time-invariant system defined by:

• Apply the Laplace transform:

• If the initial conditions are zero:

• Example:

)()0()0()0(

)(

)()()()]0()([)()0()(

012

01

012

1

01010

2

sXasas

bsbasasyaysy

sY

sXbssXbsYayssYadttdysysYs

t

0)0(assume)()()()()(01012

2

xtxbdttdxbtya

dttdya

dttyd

012

01

)()()(

asasbsb

sXsYsH

0,25.05.025.0)(4

25.02

5.025.0186

2)(

/1)()()(86

2)()(2)(8)(6)(

42

2

22

2

teetyssssss

sY

ssXtutxss

sHtxtydttdy

dttyd

tt

• What is the nature of the impulse response of this system?

• How do the coefficients a0 and a1 influence the impulse response?

EE 3512: Lecture 23, Slide 5

Nth-Order Case• Consider a linear time-invariant system defined by:

• Example:

Could we have predicted the final value of the signal?

• Note that all circuits involving discrete lumped components (e.g., RLC) can be solved in terms of rational transfer functions. Further, since typical inputs are impulse functions, step functions, and periodic signals, the computations for the output signal always follows the approach described above.

• Transfer functions can be easily created in MATLAB using tf(num,den).

N

MM

M

ii

i

i

N

ii

i

iN

N

ssasaasbsbsbb

sH

NMdttxdb

dttyda

dttyd

......

)(

)()()()(

2210

2210

0

1

0

0,212sin212cos)(

221

4)2(1

284162)()()(

21)(

84162)(

22

223

2

23

2

tettety

ssss

sssssssXsHsY

ssX

ssssssH

tt

EE 3512: Lecture 23, Slide 6

Circuit Analysis• Voltage/Current Relationships:

• Series Connections (Voltage Divider):

)0()()()()(

)0(1)(1)()(1)(

)()()()(

:TransformLaplace:Eq.Diff.

LisLsIsVdttdiLtv

vs

sICs

sVtiCdt

tdv

sRIsVtRitv

)()()(

)()(

)()()(

)()(

21

22

21

11

sVsZsZ

sZsV

sVsZsZ

sZsV

EE 3512: Lecture 23, Slide 7

Circuit Analysis (Cont.)• Parallel Connections (Current Divider):

• Example:

Note the denominator of the transfer function did not change. Why?

)()()(

)()(

)()()(

)()(

21

12

21

21

sIsZsZ

sZsI

sIsZsZ

sZsI

)/1()/(/1

)()(

)(

)()/1(

/1)(

2 LCsLRsLC

sXsV

sH

sXCsRLs

CssV

c

c

)/1()/()/(

)()(

)(

)()/1(

)(

2 LCsLRssLR

sXsV

sH

sXCsRLs

RsV

c

R

EE 3512: Lecture 23, Slide 8

RLC Circuit• Consider computation of the transfer

function relating the current in the capacitor to the input voltage.

• Strategy: convert the circuit to its Laplace transform representation, and use normal circuit analysis tools.

• Compute the voltage across the capacitor using a voltage divider, and then compute the current through the capacitor.

• Alternately, can use KVL, KVC, mesh analysis, etc.

• The Laplace transform allows us to reduce circuit analysis to algebraic manipulations.

• Note, however, that we can solve for both the steady state and transient responses simultaneously.

• See the textbook for the details of this example.

EE 3512: Lecture 23, Slide 9

Interconnections of Other Components• There are several useful building blocks in signal processing: integrator,

differentiator, adder, subtractor and scalar multiplication.

• Graphs that describe interconnections of these components are often referred to as signal flow graphs.

• MATLAB includes a very nice tool, SIMULINK, to deal with such systems.

EE 3512: Lecture 23, Slide 10

Example

EE 3512: Lecture 23, Slide 11

Example (Cont.)

)()()()()(3)()(

)()(4)(

2

212

11

sXsQsYsXsQsQssQ

sXsQssQ

• Write equations at each node:

• Solve for the first for Q1(s):

)(4

1)(1 sXs

sQ

• Subst. this into the second:

)()4)(3(

5)]()([3

1)( 12 sXss

ssXsQs

sQ

• Subst. into the third and solve for Y(s)/X(s):

)4)(3(178)(

2

sssssH

EE 3512: Lecture 23, Slide 12

Interconnections• Blocks can be thought of as subsystems that make up a system described by

a signal flow graph.• We can reduce such graphs to a transfer function.• Consider a parallel connection:

• Consider a series connection:

)()()()()(

)()()()()()()()()()()(

)()()(

21

21

22

11

21

sHsHsXsYsH

sXsHsXsHsYsXsHsYsXsHsYsYsYsY

)()()()()(

)()()()()()()(

)()()()()()(

21

21

12

122

11

sHsHsXsYsH

sXsHsHsYsXsHsH

sYsHsYsXsHsY

EE 3512: Lecture 23, Slide 13

Feedback• Feedback plays a major role in

signals and systems. For example,it is one way to stabilize an unstablesystem.

• Assuming interconnection does not loadthe other systems:

• How does the addition of feedback influence the stability of the system?• What if connection of the feedback system changes the properties of the

systems? How can we mitigate this?

)()(1)(

)()()(

)()()(1

)()(

)()()()()(1)()()()()(

)()()()()()()()()(

21

1

21

1

121

21

2221

111

sHsHsH

sXsYsH

sXsHsH

sHsY

sXsHsYsHsHsYsHsXsHsY

sYsHsXsYsXsXsXsHsY

EE 3512: Lecture 23, Slide 14

Summary• Demonstrated how to solve 1st and 2nd-order differential equations using

Laplace transforms.

• Generalized this to Nth-order differential equations.

• Demonstrated how the Laplace transform can be used in circuit analysis.

• Generalize this approach to other useful building blocks (e.g., integrator).

• Next: Generalize this approach to other block diagrams. Work another circuit example demonstrating transient and steady-state

response.

EE 3512: Lecture 23, Slide 15

Circuit Analysis Example• Assume R = 1, C = 2, and:

• Also assume • Can we predict the form of the output

signal? Or solve using Laplace:

• Class assignment: find y(t) using:• Analytic/PPT – Laplace transform (last name begins with A-M)• Numerical – MATLAB code + plot (last name begins with N-P)• Email me the results by 8 AM Monday for 1 point extra credit (maximum)

)(]7)5sin(3[)( tuttx

0)( tvc

)shifted/phase(amplitude sinewavelexponentiadecayingtermDC)(52/1)5))(2/1((

)5)(2/7()2/3(

))2/1((()2/7(

))2/1()(5()2/3(1)7(

51)3(

2/12/1)()()(

2/12/1

/1/1)(

1)7(5

5)3(1)7()3()(

2222

22

2222

2222

tys

DCssB

sA

sssss

ssssssssXsHsY

sRCsRCsH

sssssX