ECE 8443 – Pattern RecognitionECE 3163 – Signals and Systems

•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

• URL: .../publications/courses/ece_3163/lectures/current/lecture_24.ppt• MP3: .../publications/courses/ece_3163/lectures/current/lecture_24.mp3

LECTURE 24: DIFFERENTIAL EQUATIONS

ECE 3163: Lecture 24, 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?

)()()(

tbxtaydt

tdy

)()()0()( sbXsaYyssY

as

sbX

as

ysY

sbXysYas

)()0()(

)()0()()(

as

b

sX

sYsH

)(

)()(

aj

bsHeH

js

j

)()(

ECE 3163: Lecture 24, Slide 3

RC Circuit

)(/1

/1

/1

)0()(

)(1

)(1)(

sXRCs

RC

RCs

ysY

txRC

tyRCdt

tdy

RCssRCs

y

sRCs

RC

RCs

ysY

ssXtutx

/1

11

/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

ECE 3163: Lecture 24, 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

bsb

asas

yaysysY

sXbssXbsYayssYadt

tdysysYs

t

0)0(assume)()(

)()()(

01012

2

xtxbdt

tdxbtya

dt

tdya

dt

tyd

012

01

)(

)()(

asas

bsb

sX

sYsH

0,25.05.025.0)(

4

25.0

2

5.025.01

86

2)(

/1)()()(86

2)()(2)(8

)(6

)(

42

2

22

2

teety

sssssssY

ssXtutxss

sHtxtydt

tdy

dt

tyd

tt

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

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

ECE 3163: Lecture 24, 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

ssasaa

sbsbsbbsH

NMdt

txdb

dt

tyda

dt

tyd

...

...)(

)()()()(

2210

2210

0

1

0

0,212sin2

12cos)(

2

21

4)2(

1

284

162)()()(

2

1)(

84

162)(

22

223

2

23

2

tettety

sss

s

ssss

sssXsHsY

ssX

sss

sssH

tt

ECE 3163: Lecture 24, Slide 6

Circuit Analysis

• Voltage/Current Relationships:

• Series Connections (Voltage Divider):

)0()()()()(

)0(1

)(1

)()(1)(

)()()()(

:TransformLaplace:Eq.Diff.

LisLsIsV

dt

tdiLtv

vs

sICs

sVtiCdt

tdv

sRIsVtRitv

)()()(

)()(

)()()(

)()(

21

22

21

11

sVsZsZ

sZsV

sVsZsZ

sZsV

ECE 3163: Lecture 24, 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 LCsLRs

LC

sX

sVsH

sXCsRLs

CssV

c

c

)/1()/(

)/(

)(

)()(

)()/1(

)(

2 LCsLRs

sLR

sX

sVsH

sXCsRLs

RsV

c

R

ECE 3163: Lecture 24, 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.

ECE 3163: Lecture 24, 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.

ECE 3163: Lecture 24, Slide 10

Example

ECE 3163: Lecture 24, Slide 11

Example (Cont.)

)()()(

)()(3)()(

)()(4)(

2

212

11

sXsQsY

sXsQsQssQ

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 sX

ss

ssXsQ

ssQ

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

)4)(3(

178)(

2

ss

sssH

ECE 3163: Lecture 24, Slide 12

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.

Review for exam no. 2.